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Unformatted text preview: Problem 4.1 Given: Data on mass and spring Find: [1] Maximum spring compression Solution:
The given data is M = 3⋅ kg h = 5⋅ m k = 400⋅ N
m Apply the First Law of Thermodynamics: for the system consisting of the mass and the spring (the spring has gravitional potential
energy and the spring elastic potential energy)
E1 = M⋅ g⋅ h Total mechanical energy at initial state Total mechanical energy at instant of maximum compression x E2 = M⋅ g⋅ ( −x) + 1
2
⋅k ⋅x
2 Note: The datum for zero potential is the top of the uncompressed spring
But E1 = E2 so M⋅ g⋅ h = M⋅ g⋅ ( −x) + Solving for x x− 2⋅ M⋅ g
2⋅ M⋅ g⋅ h
⋅x −
=0
k
k x= M⋅ g
+
k 2 1
2
⋅k ⋅x
2 2 ⎛ M⋅ g ⎞ + 2⋅ M⋅ g⋅ h
⎜
⎟
k
⎝k⎠
m m
x = 3⋅ kg × 9.81⋅ ×
+
2 400⋅ N
s
x = 0.934 m Note that ignoring the loss of potential of the mass due to spring compression x gives
x= 2⋅ M⋅ g⋅ h
k x = 0.858 m Note that the deflection if the mass is dropped from immediately above the spring is
x= 2⋅ M⋅ g
k 2 ⎛ 3⋅ kg × 9.81⋅ m × m ⎞ + 2 × 3⋅ kg × 9.81⋅ m × 5⋅ m × m
⎜
2 400⋅ N ⎟
2
400⋅ N
s
s
⎝
⎠ x = 0.147 m Problem 4.2 [1] Problem 4.3 Given: Data on Boeing 777200 jet Find: [2] Minimum runway length for takeoff Solution:
dV
dV
= M⋅ V⋅
= Ft = constant
dt
dx Basic equation ΣFx = M⋅ Separating variables M⋅ V⋅ dV = Ft⋅ dx Integrating x= M⋅ V
2⋅ Ft x= 1⋅ km
1
km
1⋅ hr ⎞
1
1 N⋅ s
3
× 325 × 10 kg × ⎛ 225
×
×
⋅×
⎜
⎟×
3 N kg⋅ m
2
hr 1000⋅ m 3600⋅ s ⎠
⎝
2 × 425 × 10 Note that the "weight" is already in mass units! 2 2 For time calculation M⋅ Integrating t= dV
= Ft
dt dV = Ft
M 2 x = 747 m ⋅ dt M⋅ V
Ft
3 t = 325 × 10 kg × 225 2 1⋅ km
km
1⋅ hr
1
1 N⋅ s
×
×
×
⋅×
hr 1000⋅ m 3600⋅ s 2 × 425 × 103 N kg⋅ m Aerodynamic and rolling resistances would significantly increase both these results t = 23.9 s Problem 4.5 Problem 4.4 [2] Problem 4.4 Problem 4.5 [2] Problem 4.6 Given: Data on air compression process Find: [2] Internal energy change Solution:
Basic equation δQ − δW = dE Assumptions: 1) Adiabatic so δQ = 0 2) Stationary system dE =dU 3) Frictionless process δW = pdV = Mpdv
Then dU = −δW = −M⋅ p⋅ dv Before integrating we need to relate p and v. An adiabatic frictionless (reversible) process is isentropic, which for an ideal gas gives
cp
k
where
p⋅ v = C
k=
cv Hence Substituting 1
1
−
k
k v = C ⋅p du = ⎛ C
⎜
Δu =
⋅ p2
k−1 ⎝ But Hence From Table A.6 C ⋅p k−1
k k−1
k 1
1
−
k
k = C ⋅p − p1 k−1 ⎞
k⎟ ⎠ 1
k 1
− −1
k dU
= −p⋅ dv = −p⋅ C ⋅ ⋅ p
M
k 1
k 1
− −1
k dv = C ⋅ ⋅ p
k and 1
k1 Integrating between states 1 and 2 1
k 1
k1 ⋅ dp = k−1
k 1
k = C ⋅ p1
k−1 ⋅ dp − −C
⋅p
k 1
k ⋅ dp k−1
⎤
⎡
⎢
⎥
k
⎢⎛ p2 ⎞
⎥
⋅ ⎢⎜ ⎟
− 1⎥
⎣⎝ p1 ⎠
⎦ ⋅ p = p⋅ v = Rair⋅ T k−1
⎤
⎡
⎢
⎥
k
Rair⋅ T1 ⎢⎛ p2 ⎞
⎥
Δu =
⋅ ⎢⎜ ⎟
− 1⎥
k−1
⎣⎝ p1 ⎠
⎦ Rair = 53.33⋅ ft⋅ lbf
lbm⋅ R and k = 1.4 1.4−1
⎡
⎤
⎢
⎥
1.4
1
ft⋅ lbf
⎞
⎢⎛ 3 ⎟
Δu =
× 53.33⋅
× ( 68 + 460) R × ⎜
− 1⎥
0.4
lbm⋅ R
⎣⎝ 1 ⎠
⎦ Δu = 33.4⋅ Btu
lbm Δu = 1073⋅ Btu
slug 4 ft⋅ lbf Δu = 2.6 × 10 ⋅ lbm (Using conversions from Table G.2) Problem 4.7 [2] Given: Data on cooling of a can of soda in a refrigerator Find: How long it takes to warm up in a room Solution:
The First Law of Thermodynamics for the can (either warming or cooling) is
M⋅ c⋅ ( ) dT
= −k⋅ T − Tamb
dt or ( ) dT
= −A⋅ T − Tamb
dt where A= k
M⋅ c where M is the can mass, c is the average specific heat of the can and its contents, T is the temperature, and Tamb is the
ambient temperature
Separating variables dT
= −A⋅ dt
T − Tamb Integrating T ( t) = Tamb + Tinit − Tamb ⋅ e ( −At ) where Tinit is the initial temperature. The available data from the coolling can now be used to obtain a value for constant A Hence Tinit = ( 25 + 273) ⋅ K Tinit = 298 K Tamb = ( 5 + 273) ⋅ K Tamb = 278 K T = ( 10 + 273) ⋅ K Given data for cooling T = 283 K when t = τ = 10⋅ hr A= 1⋅ hr
1 ⎛ Tinit − Tamb ⎞
1
298 − 278 ⎞
⋅ ln ⎜
×
× ln ⎛
⎟=
⎜
⎟
τ
T − Tamb
3⋅ hr 3600⋅ s
⎝ 283 − 278 ⎠ ⎝ A = 1.284 × 10 ⎠ Then, for the warming up process
Tinit = ( 10 + 273) ⋅ K Tinit = 283 K Tend = ( 15 + 273) ⋅ K Tend = 288 K ( ) Tamb = ( 20 + 273) ⋅ K Tamb = 293 K −Aτ with Tend = Tamb + Tinit − Tamb ⋅ e Hence the time τ is τ= 1
A ⎛ Tinit − Tamb ⎞
s
283 − 293 ⎞
⋅ ln ⎛
⎟=
⎜
⎟
− 4 ⎝ 288 − 293 ⎠
Tend − Tamb
⎝
⎠ 1.284⋅ 10 ⋅ ln ⎜ 3 τ = 5.40 × 10 s τ = 1.50 hr −4 −1 s Problem 4.8 [2] Given: Data on heat loss from persons, and peoplefilled auditorium Find: Internal energy change of air and of system; air temperature rise Solution:
Basic equation Q − W = ΔE Assumptions: 1) Stationary system dE =dU 2) No work W = 0
W
60⋅ s
× 6000⋅ people × 15⋅ min ×
person
min Then for the air ΔU = Q = 85⋅ For the air and people ΔU = 459 MJ ΔU = Qsurroundings = 0 The increase in air energy is equal and opposite to the loss in people energy
For the air
Hence
From Table A.6 ΔU = Q but for air (an ideal gas) Rair⋅ Q⋅ T
Q
=
M⋅ cv
cv⋅ p⋅ V
J
and
Rair = 286.9⋅
kg⋅ K ΔU = M⋅ cv⋅ ΔT with M = ρ⋅ V = p⋅ V
Rair⋅ T ΔT = ΔT = 286.9
717.4 6 × 459 × 10 ⋅ J × ( 20 + 273) K × This is the temperature change in 15 min. The rate of change is then J
cv = 717.4⋅
kg⋅ K
1 2 m
1
1
×
⋅
3N
53
101 × 10
3.5 × 10 m
⋅ ΔT
K
= 6.09
15⋅ min
hr ΔT = 1.521 K Problem 4.9 [3] Part 1/2 Problem 4.9 [3] Part 2/2 Problem 4.10 [3] Given: Data on velocity field and control volume geometry Find: Several surface integrals Solution:
r
ˆ
dA1 = − wdzˆ + wdyk
j r
ˆ
dA1 = − dzˆ + dyk
j r
dA2 = wdzˆ
j r
dA2 = dzˆ
j ( r
ˆ
V = azˆ + bk
j (a) (b) ) ( r
ˆ
V = 10 zˆ + 5k
j ( )( ) ) r
ˆ
ˆ
V ⋅ dA1 = 10 zˆ + 5k ⋅ − dzˆ + dyk = −10 zdz + 5dy
j
j ∫ A1 1
1
r
1
1
V ⋅ dA1 = − 10 zdz + 5dy = − 5 z 2 + 5 y 0 = 0 ∫ ∫ 0 0 0 ( )( ) )( ) (c) r
ˆ
V ⋅ dA2 = 10 zˆ + 5k ⋅ dzˆ = 10 zdz
j
j (d) rr
ˆ
V V ⋅ dA2 = 10 zˆ + 5k 10 zdz
j (e) ∫( ( A2 ) ∫ (10 zˆj + 5kˆ )10 zdz = 100 z
3 rr
V V ⋅ dA2 = 1 0 1
3 1 ˆ
ˆ
ˆ + 25 z 2 k = 33.3 ˆ + 25k
j
j
0 0 Problem 4.11 [3] Given: Geometry of 3D surface Find: Volume flow rate and momentum flux through area
r
ˆ
dA = dxdzˆ + dxdyk
j Solution: r
ˆ
V = axi − byˆ
j r
ˆj
V = xi − yˆ We will need the equation of the surface: z = 3 − a) 1
y or y = 6 − 2 z
2 Volume flow rate )( ( r
ˆ
ˆ
Q = ∫ V ⋅ dA = ∫ xi − yˆ ⋅ dxdzˆ + dxdyk
j
j
A A 10 3 3 3 00 = 0 ) 0 2
∫ ∫ − ydzdx = ∫ − 10 ydz = ∫ − 10(6 − 2 z )dz = − 60 z + 10 z Q = (− 180 + 90 ) Q = −90 3
0 ft 3
s ft 3
s b) Momentum flux ( rr r ) ( ) ρ ∫ V V ⋅ dA = ρ ∫ xiˆ − yˆ (− ydxdz )
j
A A 10 3 3 = ρ ∫ ∫ (− xy )dzdxiˆ + ρ ∫ 10 y 2dz ˆ
j
00 0 10 3 3 0 0 = − ρ ∫ xdx ∫ (6 − 2 z )dziˆ + ρ ∫ 10(6 − 2 z )2 dzˆ
j
0 ⎛x
= ρ⎜−
⎜2
⎝ 3
⎞
⎛
⎞
⎟⎛ 6 z − z 2 3 ⎞i + ρ ⎜10⎛ 36 z − 12 z 2 + 4 z 3 ⎞ ⎟ ˆ
ˆ
⎜
⎟
⎜
⎟j
0⎠
⎜⎝
⎟⎝
3 ⎠0⎟
0⎠
⎝
⎠
ˆ + ρ (10(108 − 108 + 36)) ˆ
= ρ (− 50 )(18 − 9 )i
j
2 10 ˆ
= −450 ρi + 360 ρˆ
j ⎛ slug ⋅ ft
⎞
⎜
s if ρ is in slug ⎟
⎜
s
ft 3 ⎟
⎝
⎠ Problem 4.12 Problem 4.12 [2] Problem 4.13 Given: Geometry of 3D surface Find: [3] Surface integrals r
ˆ
dA = dydzi − dxdzˆ
j Solution: r
ˆ
ˆ
V = − axi + byˆ + ck
j r
ˆ
ˆ
V = − 2 x i + 2 y ˆ + 2. 5k
j We will need the equation of the surface: y = 3
2
x or x = y
2
3 ∫V ⋅ dA = ∫ (− axiˆ + byjˆ + ckˆ )⋅ (dydziˆ − dxdzˆj )
r A A 23 22 2 00 00 0 3 2 Q = (− 6a − 6b )
Q = −24 3 2 m3
s We will again need the equation of the surface: y = 3
2
3
x or x = y , and also dy = dx and a = b
2
3
2 j
∫ V (V ⋅ dA) = ∫ (− axiˆ + byˆ + ckˆ )(− axiˆ + byˆj + ckˆ)⋅ (dydziˆ − dxdzˆj )
ˆ
j
= ∫ (− axiˆ + byˆ + ck )(− axdydz − bydxdz )
rr A r A A ⎛
ˆ
= ∫ ⎜ − axi +
A
⎝
⎛
ˆ
= ∫ ⎜ − axi +
A
⎝ 3ˆ
3
3
⎞
ˆ ⎞⎛
axj + ck ⎟⎜ − ax dxdz − a xdxdz ⎟
2
2
2
⎠⎝
⎠
3ˆ
ˆ⎞
axj + ck ⎟(− 3axdxdz )
2
⎠ 22 22 22 ˆ
ˆ9
j
= 3∫ ∫ a 2 x 2 dxdzi − ∫ ∫ a 2 x 2 dxdzˆ − 3∫ ∫ acxdxdzk
200
00
00
3
⎛ 2 x3 2 ⎞
⎛
⎟iˆ − (9 )⎜ a 2 x
= (6)⎜ a
⎜
⎜3
3 0⎟
⎝
⎠
⎝
2ˆ
2ˆ
ˆ
= 16a i − 24a j − 12ack ˆ
ˆ
= 64iˆ − 96 j − 60k m4
s2 2 2
3
1
3
ydy − b∫ dz ∫ xdx = − 2a y 2 − 2b x 2
3
2
30
40
0
0
0 = ∫ ∫ − axdydz − ∫∫ bydxdz = −a ∫ dz ∫ 2
⎞
⎛
⎟ j − (6 )⎜ ac x
ˆ
⎟
⎜2
0⎠
⎝
2 ⎞
⎟
⎟
0⎠
2 Problem 4.14
Problem 4.12 [2] Problem 4.15 [2] Problem 4.16 [2] Problem 4.17 Given: Data on flow through nozzles Find: [1] Average velocity in head feeder; flow rate Solution:
Basic equation (→ → ) = 0
∑ V⋅ A
CS Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
Then for the nozzle flow (→ → ) = −Vfeeder⋅ Afeeder + 10⋅ Vnozzle⋅ Anozzle = 0
∑ V⋅ A
CS Hence 10⋅ Anozzle Vfeeder = Vnozzle⋅
Afeeder ⎛1⎞
⎜8⎟
ft
Vfeeder = 10⋅ × 10 × ⎜ ⎟
s
⎝1⎠ ⎛ Dnozzle ⎞
= Vnozzle⋅ 10⋅ ⎜
⎟
⎝ Dfeeder ⎠ 2 2 Vfeeder = 1.56⋅
2 The flow rate is Q = Vfeeder⋅ Afeeder = Vfeeder⋅ π⋅ Dfeeder
4
2 Q = 1.56⋅ ft π ⎛
1⋅ ft ⎞
7.48⋅ gal
60⋅ s
× × ⎜ 1⋅ in ×
×
⎟×
3
s4⎝
12⋅ in ⎠
1⋅ min
1⋅ ft Q = 3.82⋅ gpm ft
s Problem 4.18 [3] Given: Data on flow into and out of tank Find: Time at which exit pump is switched on; time at which drain is opened; flow rate into drain Solution:
Basic equation ∂
∂t MCV + ( ρ→ → ) = 0
∑ ⋅ V⋅ A
CS Assumptions: 1) Uniform flow 2) Incompressible flow
∂ After inlet pump is on ∂t MCV + ( ρ→ → ) = ∂ Mtank − ρ⋅ Vin⋅ Ain = 0
∑ ⋅ V⋅ A
∂t CS Ain
⎛ Din ⎞
dh
= Vin⋅
= Vin⋅ ⎜
⎟
dt
Atank
⎝ Dtank ⎠
Hence the time to reach hexit = 0.7 m is
∂ After exit pump is on ∂t MCV + ∂ texit = hexit
dh
dt ∂t dh
= ρ⋅ Vin⋅ Ain
Mtank = ρ⋅ Atank⋅
dt 2 hexit ⎛ Dtank ⎞
=
⋅⎜
⎟
Vin
⎝ Din ⎠ 2 1s
3⋅ m ⎞
texit = 0.7⋅ m × ⋅ × ⎛
⎜
⎟
5 m ⎝ 0.1⋅ m ⎠ ∑ (ρ⋅ V⋅ A) = ∂t Mtank − ρ⋅ Vin⋅ Ain + ρ⋅ Vexit⋅ Aexit = 0
→→ ∂ Atank⋅ where h is the
level of water
in the tank 2 texit = 126 s dh
= Vin⋅ Ain − Vexit⋅ Aexit
dt CS 2 Ain
Aexit
⎛ Din ⎞
⎛ Dexit ⎞
dh
= Vin⋅
− Vexit⋅
= Vin⋅ ⎜
⎟ − Vexit⋅ ⎜
⎟
dt
Atank
Atank
⎝ Dtank ⎠
⎝ Dtank ⎠ Hence the time to reach hdrain = 2 m is tdrain = texit + (hdrain − hexit) (hdrain − hexit) = dh
dt 2 2 ⎛ Din ⎞
⎛ Dexit ⎞
Vin⋅ ⎜
⎟ − Vexit⋅ ⎜
⎟
⎝ Dtank ⎠
⎝ Dtank ⎠ 2 1 tdrain = 126⋅ s + ( 2 − 0.7) ⋅ m × 2 5⋅ m ⎛ 0.08⋅ m ⎞
m ⎛ 0.1⋅ m ⎞
×⎜
⎟ − 3⋅ × ⎜
⎟
s ⎝ 3⋅ m ⎠
s ⎝ 3⋅ m ⎠ tdrain = 506 s 2 The flow rate into the drain is equal to the net inflow (the level in the tank is now constant)
2 Qdrain = Vin⋅ π⋅ Din
4 − Vexit⋅ π⋅ Dexit
4 2 Qdrain = 5⋅ m
s × π
4 2 × ( 0.1⋅ m) − 3⋅ m
s × π
4 × ( 0.08⋅ m) 2 3 m
Qdrain = 0.0242
s Problem 4.19 [4] Moist air CS Warm water Given: Data on flow into and out of cooling tower Find: Cool water Volume and mass flow rate of cool water; mass flow rate of moist and dry air Solution:
Basic equation ( ρ→ → ) = 0
∑ ⋅ V⋅ A Q = V⋅ A and at each inlet/exit CS Assumptions: 1) Uniform flow 2) Incompressible flow
At the cool water exit Qcool = V⋅ A Qcool = 5.55⋅ ft π
2
× × ( 0.5⋅ ft)
s4 The mass flow rate is mcool = ρ⋅ Qcool mcool = 1.94⋅ slug
ft 3 3 × 1.09⋅ ft
s 3 Qcool = 1.09 ft
s slug
mcool = 2.11
s Qcool = 489 gpm
5 lb
mcool = 2.45 × 10
hr NOTE: Software does not allow dots over terms, so m represents mass flow rate, not mass!
→→
For the air flow we need to use
to balance the water flow
ρ⋅ V ⋅ A = 0 ∑( ) CS We have −mwarm + mcool + mv = 0 mv = mwarm − mcool This is the mass flow rate of water vapor. We need to use this to obtain air flow rates. From psychrometrics ρmoist = 0.066⋅ x= mv
mair ρmoist where x is the relative humidity. It is also known (try Googling "density of moist air") that We are given lb
mv = 5073
hr ρdry 1+x = 1 + x⋅ RH2O
Rair lb
ft 3 p
For dry air we could use the ideal gas equation ρdry =
R⋅ T
ρdry = 0.002377⋅ slug
ft 3 Note that moist air is less dense than dry air! but here we use atmospheric air density (Table A.3) ρdry = 0.002377⋅ slug
ft 3 × 32.2⋅ lb
slug ρdry = 0.0765 lb
ft 3 Hence 0.066
=
0.0765 x= Hence 1+x
85.78 1 + x⋅ 53.33 0.0765 − 0.066
85.78
− .0765
0.066⋅
53.33 mv
=x
mair using data from Table A.6 leads to Finally, the mass flow rate of moist air is x = 0.354 mv
mair =
x lb
1
mair = 5073⋅ ×
hr 0.354 mmoist = mv + mair lb
mmoist = 19404
hr lb
mair = 14331
hr Problem 4.20 Given: Data on wind tunnel geometry Find: [1] Average speeds in wind tunnel Solution:
Basic equation Q = V⋅ A Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
2 Between sections 1 and 2 Q = V1⋅ A1 = V1⋅ Hence ⎛ D1 ⎞
V2 = V1⋅ ⎜
⎟
⎝ D2 ⎠ 2 2 Similarly ⎛ D1 ⎞
V3 = V1⋅ ⎜
⎟
⎝ D3 ⎠ π⋅ D 1
4 2 = V2⋅ A2 = V2⋅ π⋅ D 2
4 5
V2 = 20⋅ mph⋅ ⎛ ⎞
⎜⎟
⎝ 3⎠ 2 5
V3 = 20⋅ mph⋅ ⎛ ⎞
⎜⎟
⎝ 2⎠ 2 V2 = 55.6 mph V3 = 125 mph Problem 4.21 Given: Data on flow through box Find: [1] Velocity at station 3 Solution:
Basic equation (→ → ) = 0
∑ V⋅ A
CS Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
→→
Then for the box
V⋅ A = −V1⋅ A1 + V2⋅ A2 + V3⋅ A3 = 0 ∑( )
CS Note that the vectors indicate that flow is in at location 1 and out at location 2; we assume outflow at location 3
Hence A1
A2
V3 = V1⋅
− V2⋅
A3
A3 V3 = 10⋅ Based on geometry Vx = V3⋅ sin( 60⋅ deg) Vx = 4.33 ft
s Vy = −V3⋅ cos ( 60⋅ deg) Vy = −2.5 ft
s ⎯
→
ft
ft
V3 = ⎛ 4.33⋅ , −2.5⋅ ⎞
⎜
⎟
s
s ⎝ ⎠ ft 0.5
ft 0.1
×
− 20⋅ ×
s 0.6
s 0.6 V3 = 5 ft
s Problem 4.22 Given: Data on flow through device Find: [1] Volume flow rate at port 3 Solution:
Basic equation (→ → ) = 0
∑ V⋅ A
CS Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
→→
Then for the box
V⋅ A = −V1⋅ A1 + V2⋅ A2 + V3⋅ A3 = −V1⋅ A1 + V2⋅ A2 + Q3 ∑( )
CS Note we assume outflow at port 3
Hence Q3 = V1⋅ A1 − V2⋅ A2 The negative sign indicates the flow at port 3 is inwards. Q3 = 3⋅ m
m
2
2
× 0.1⋅ m − 10⋅ × 0.05⋅ m
s
s Flow rate at port 3 is 0.2 m3/s inwards 3 Q3 = −0.2⋅ m
s Problem 4.23 Given: Water needs of farmer Find: [1] Number of 6 in. pipes needed Solution:
Basic equation Q = V⋅ A Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
Then Q = n⋅ V ⋅ π⋅ D
4 2 where n is the number of pipes, V is the average velocity in the pipes, and D is the pipe diameter
2 Q= The flow rate is given by 5⋅ acre⋅ 0.25⋅ ft
5⋅ acre⋅ 0.25⋅ ft 43560⋅ ft
1⋅ hr
=
×
×
1⋅ acre
1⋅ hr
1⋅ hr
3600⋅ s Data on acres from Googling! 3 Q = 15.1⋅
Hence n= 2 4⋅ Q
π⋅ V ⋅ D ft
s 2 Hence we need at least eight pipes n= 3 s
4
1⎞
ft
×
×⎛
⎜
⎟ × 15.1⋅
s
π 10⋅ ft ⎝ 0.5⋅ ft ⎠ n = 7.69 Problem 4.24 Given: Data on filling of gas tank Find: [1] Crosssection area of tank CS Solution: Rising level We can treat this as a steady state problem if we choose a CS as the original volume of
gas in the tank, so that additional gas "leaves" the gas as the gas level in the tank rises, OR
as an unsteady problem if we choose the CS as the entire gas tank. We choose the latter
Basic equation ∂
∂t MCV + Inflow ( ρ→ → ) = 0
∑ ⋅ V⋅ A
CS Assumptions: 1) Incompressible flow 2) Uniform flow
Hence ∂
∂t ( ρ→ → ) = ρ⋅ Q
∑ ⋅ V⋅ A dh
=−
MCV = ρ⋅ A⋅
dt CS where Q is the gas fill rate, A is the tank crosssection area, and h is the rate of rise in the gas tank
Hence A= 3 Q A = 5.3⋅ dh
dt A = 1.98 ft 2 gal
1⋅ ft
1 min 12⋅ in
×
×
⋅
×
min 7.48⋅ gal 4.3 in
1⋅ ft
2 A = 285 in Data on gals from Table G.2 This seems like a reasonable area e.g., 1 ft x 2 ft Problem 4.25 Given: Data on filling of a sink Find: [1] Accumulation rate under various circumstances Solution:
This is an unsteady problem if we choose the CS as the entire sink
Basic equation ∂
∂t MCV + ( ρ→ → ) = 0
∑ ⋅ V⋅ A
CS Assumptions: 1) Incompressible flow
Hence ∂
∂t ( ρ→ → ) = Inflow− Outflow
∑ ⋅ V⋅ A MCV = Accumulationrate= − CS Accumulationrate= Inflow− Outflow
For the first case Accumulationrate= 5000⋅ units
units 60⋅ min
− 60⋅
×
hr
min
hr Accumulationrate= 1400⋅ units
hr For the second case Accumulationrate= 5000⋅ units
units 60⋅ min
− 13⋅
×
hr
min
hr Accumulationrate= 4220⋅ units
hr For the third case Outflow = Inflow − Accumulationrate
Outflow = 5⋅ units
units
− ( − 4) ⋅
s
s Outflow = 9⋅ units
s Problem 4.26 Given: Data on filling of a basement during a storm Find: [1] Flow rate of storm into basement Solution:
This is an unsteady problem if we choose the CS as the entire basement
Basic equation ∂
∂t MCV + ( ρ→ → ) = 0
∑ ⋅ V⋅ A
CS Assumptions: 1) Incompressible flow
Hence or (ρ→ → ) = ρ⋅ Qstorm − ρ⋅ Qpump
∑ ⋅ V⋅ A ∂ dh
MCV = ρ⋅ A⋅
=−
dt
∂t CS dh
Qstorm = Qpump − A⋅
dt
Qstorm = 10⋅ gal
1 ft ⎞ 7.48⋅ gal
1⋅ hr
− 25⋅ ft × 20⋅ ft × ⎛ − ⋅ ⎟ ×
×
⎜
3
min
60⋅ min
⎝ 12 hr ⎠
ft Qstorm = 15.2 gpm where A is the basement area and dh/dt is
the rate at which the height of water in the
basement changes. Data on gals from Table G.2 Problem 4.27 Given: Data on flow through device Find: [1] Volume flow rate at port 3 Solution:
Basic equation ( ρ→ → ) = 0
∑ ⋅ V⋅ A
CS Assumptions: 1) Steady flow 2) Uniform flow
→→
Then for the box
ρ⋅ V⋅ A = −ρu⋅ Vu⋅ Au + ρd ⋅ Vd⋅ Ad = 0 ∑( ) CS Hence Vd⋅ Ad
ρu = ρd ⋅
Vu⋅ Au ρ u = 4⋅ lb
ft 3 × 10
1
×
15 0.25 ρu = 10.7 lb
ft 3 Problem 4.28 [2] Given: Data on flow through device Find: Velocity V3; plot V3 against time; find when V3 is zero; total mean flow Solution:
Governing equation: Applying to the device (assuming V3 is out) V3 = V1⋅ A1 + V2⋅ A2
A3
− V3 = 6.67⋅ e t
2 →→
V⋅ A = 0 ∑ −V1⋅ A1 − V2⋅ A2 + V3⋅ A3 = 0
− The velocity at A3 is ⌠→→
⎮
⎮ V dA =
⌡ For incompressible flow (Eq. 4.13) and uniform flow 10⋅ e t
2m = ⋅ s 2 × 0.1⋅ m + 2⋅ cos ( 2⋅ π⋅ t) ⋅ m
2
× 0.2⋅ m
s 2 0.15⋅ m + 2.67⋅ cos ( 2⋅ π⋅ t) The total mean volumetric flow at A3 is
∞ ∞ ⌠
Q=⎮
⌡0 ⌠
t
⎞
⎮⎛
−
⎜
⎟
m 2⎞
2
⎮ ⎝ 6.67⋅ e
V3⋅ A3 dt =
+ 2.67⋅ cos ( 2⋅ π⋅ t)⎠ ⋅ 0.15 dt⋅ ⎛ ⋅ m ⎟
⎜
⌡0
⎝s
⎠ t
⎛
⎞
−
⎜
⎟
1
3
2
Q = lim ⎜ −2⋅ e
+
⋅ sin ( 2⋅ π⋅ t)⎟ − ( −2) = 2⋅ m
5⋅ π
⎠
t→∞⎝ The time at which V3 first is zero, and the plot of V3 is shown in the corresponding Excel workbook 3 Q = 2⋅ m t = 2.39⋅ s t (s) V 3 (m/s)
9.33
8.50
6.86
4.91
3.30
2.53
2.78
3.87
5.29
6.41
6.71
6.00
4.48
2.66
1.15
0.48
0.84
2.03
3.53
4.74
5.12 2.10
2.20
2.30
2.40
2.50 4.49
3.04
1.29
0.15
0.76 Exit Velocity vs Time
10
8 V 3 (m/s) 0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00 6
4
2
0
0.0 0.5 1.0 1.5 2.0 2 t (s) The time at which V 3 first becomes zero can be found using Goal Seek
t (s ) V 3 (m/s) 2.39 0.00 2.5 Problem 4.29 [2] Problem 4.30 [2] y 2h x CS Given: Data on flow at inlet and outlet of channel Find: Find umax Solution:
Basic equation r r ∫ ρV ⋅ dA = 0
CS Assumptions: 1) Steady flow 2) Incompressible flow
h h Evaluating at 1 and 2 ⌠
−ρ⋅ U⋅ 2⋅ h ⋅ w + ⎮ ρ⋅ u ( y) dy = 0
⌡− h ⎡ ⎡ h 3 ⎛ h 3 ⎞⎤⎤
⎟⎥⎥ = 2⋅ h ⋅ U
− ⎜−
⎢ 3⋅ h 2 ⎜ 3⋅ h 2 ⎟⎥⎥
⎣
⎝
⎠⎦⎦ umax⋅ ⎢[ h − ( −h ) ] − ⎢ ⎢
⎣ Hence umax = 3
3
m
⋅ U = × 2.5⋅
2
2
s ⌠
2
⎡
y⎞ ⎤
⎮
umax⋅ ⎢1 − ⎛ ⎟ ⎥ dy = 2⋅ h ⋅ U
⎜
⎮
⎣ ⎝h⎠ ⎦
⌡− h
4
umax⋅ ⋅ h = 2⋅ h ⋅ U
3
umax = 3.75⋅ m
s Problem 4.31 Given: Data on flow at inlet and outlet of pipe Find: [2] Find U Solution:
Basic equation r r ∫ ρV ⋅ dA = 0
CS Assumptions: 1) Steady flow 2) Incompressible flow
R 2⌠
Evaluating at inlet and exit −ρ⋅ U⋅ π⋅ R + ⎮ ρ⋅ u ( r ) ⋅ 2⋅ π⋅ r dr = 0
⌡0 umax⋅ ⎛ R −
⎜
2 ⎝ Hence 1 2⎞
2
⋅R ⎟ = R ⋅U
2
⎠ 1
m
U = × 3⋅
2
s R ⌠
2
⎡
r⎤
⎮
2
umax⋅ ⎢1 − ⎛ ⎞ ⎥ ⋅ 2⋅ r dr = R ⋅ U
⎜⎟
⎮
R⎠ ⎦
⎣⎝
⌡0
U= 1
⋅u
2 max U = 1.5⋅ m
s Problem 4.32 [2] Problem 4.33 [3] Given: Velocity distribution in annulus Find: Volume flow rate; average velocity; maximum velocity; plot velocity distribution Solution:
Governing equation The given data is Ro = 5⋅ mm ⌠→→
⎮
Q = ⎮ V dA
⌡ For the flow rate (Eq. 4.14a) and average velocity (Eq. 4.14b)
Δp
kPa
= −10⋅
L
m Ri = 1⋅ mm μ = 0.1⋅ ⎝ 2 ⎝ ⎠ N⋅ s
2 Q
A (From Fig. A.2) m ⎛
⎛ Ro ⎞ ⎞
−Δp ⎜ 2 2 Ro − Ri
u ( r) =
⋅ R −r +
⋅ ln ⎜ ⎟ ⎟
4⋅ μ⋅ L ⎜ o
⎛ Ri ⎞ ⎝ r ⎠ ⎟
ln ⎜ ⎟
⎜
⎟
Ro
2 Vav = ⎠ Ro The flow rate is ⌠
Q = ⎮ u ( r) ⋅ 2⋅ π⋅ r dr
⌡R
i 2⎞
⎤
⎡⎛ 2
Δp⋅ π ⎛ 2
2⎞ ⎢ ⎝ Ro − Ri ⎠ ⎛ 2
2⎞⎥
Q=
⋅ R − Ri ⋅
⎠ ⎢ ⎛ R ⎞ − ⎝ Ri + Ro ⎠⎥
8⋅ μ⋅ L ⎝ o
o
⎢ ln⎜ ⎟
⎥
Ri
⎣⎝⎠
⎦ Considerable mathematical manipulation leads to Substituting values ( ) 2 ( ) π
N
m
m⎞
3
2
2
Q = ⋅ −10⋅ 10 ⋅
⋅
⋅ 5 − 1 ⋅⎛
⎜
⎟
2
0.1⋅ N⋅ s
8
⎝ 1000 ⎠
m ⋅m
Q = 1.045 × 10 3
−5m Q = 10.45⋅ s Q
Q
The average velocity is Vav =
=
2
2
A
π⋅ ⎛ R o − R i ⎞ ⎝ 2
⎡ 52 − 12
⎤
⎢
⎞
(52 + 12)⎥ ⋅ ⎛ m ⎟
⋅
−
⎢ ⎛ 5⎞
⎥ ⎜ 1000 ⎠
⎝
ln ⎜ ⎟
⎢1
⎥
⎣ ⎝⎠
⎦ 2 mL
s Vav = ⎠ 1
π × 1.045 × 10 3
−5 m ⋅ s × 1 ⋅⎛
⎜ 1000 ⎞ ⎟
2
2
5 −1 ⎝ m ⎠ ⎡
⎛
⎡
− Ri
⎛ Ro ⎞ ⎞⎤
Δp ⎢
The maximum velocity occurs when du
d ⎢ −Δp ⎜ 2 2 Ro
=0=
⋅ Ro − r +
⋅ ln ⎜ ⎟ ⎟⎥ = −
⋅ − 2⋅ r −
dr
4⋅ μ⋅ L ⎢
dx ⎢ 4⋅ μ⋅ L ⎜
⎛ Ri ⎞ ⎝ r ⎠ ⎟⎥
ln ⎜ ⎟
⎢
⎜
⎟⎥
⎢
Ro
2 ⎣ 2 r= Ri − Ro ⎝ ⎝ 2 ⎠ ⎠⎦ ⎣ 2 Vav = 0.139 m
s ⎛ R 2 − R 2⎞ ⎤
i ⎠⎥
⎝o
⎛ Ri ⎞ ⎥
ln ⎜ ⎟ ⋅ r ⎥
⎝ Ro ⎠ ⎦ 2 ⎛ Ri ⎞ r = 2.73⋅ mm Substituting in u(r) umax = u ( 2.73⋅ mm) = 0.213⋅ 2⋅ ln ⎜ ⎟
⎝ Ro ⎠ The maximum velocity using Solver instead, and the plot, are also shown in the corresponding Excel workbook m
s Ro = 5 Ri =
Δp /L = 1
10 μ= 0.1 mm
mm
kPa/m
N.s/m2 r (mm) u (m/s)
0.000
0.069
0.120
0.157
0.183
0.201
0.210
0.213
0.210
0.200
0.186
0.166
0.142
0.113
0.079
0.042
0.000 Annular Velocity Distribution
6
5
r (mm) 1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
5.00 4
3
2
1
0
0.00 The maximum velocity can be found using Solver
r (mm) u (m/s)
2.73 0.213 0.05 0.10 0.15
u (m/s) 0.20 0.25 Problem 4.25 Problem 4.34 [2] Problem 4.26 Problem 4.35 [2] Problem 4.27 Problem 4.36 [2] Problem 4.28 Problem 4.37 [2] Problem 4.38 [2] CS
Outflow Given: Data on airflow out of tank Find: Find rate of change of density of air in tank Solution:
Basic equation rr
∂
ρdV + ∫ ρV ⋅ dA = 0
∫
∂t CV
CS Assumptions: 1) Density in tank is uniform 2) Uniform flow 3) Air is an ideal gas
Hence Vtank⋅ dρtank dρtank
dt dt + ρexit⋅ V⋅ A = 0 dρtank
dt =− ρexit⋅ V⋅ A
Vtank =− pexit⋅ V⋅ A
Rair⋅ Texit⋅ Vtank 2
m
1⋅ m ⎞
1 kg⋅ K
1
3N
2
× 250⋅ × 100⋅ mm × ⎛
⋅
×
×
⎜
⎟×
2
s
⎝ 1000⋅ mm ⎠ 286.9 N ⋅ m ( −20 + 273) ⋅ K = −300 × 10 ⋅ m kg dρtank
Hence dt = −0.258⋅ m 3 s The mass in the tank is decreasing, as expected 1
0.4⋅ m 3 Problem 4.30 Problem 4.39 [2] Problem 4.32 Problem 4.40 [2] Problem 4.31 Problem 4.41 [2] Problem 4.33 Problem 4.42 [2] Problem 4.35 Problem 4.43 [2] Problem 4.44 [3] Part 1/2 Problem 4.44 [3] Part 2/2 Problem 4.45 [3] Part 1/2 Problem 4.45 [3] Part 2/2 Problem 4.38 Problem 4.46 [3] Problem 4.39 Problem 4.47 [3] Problem 4.40 Problem 4.48 [3] Problem 4.41 Problem 4.49
P4.48. [3] Problem 4.42 Problem 4.50 [4] Problem 4.51 [4] Part 1/2 Problem 4.51 [4] Part 2/2 Problem 4.52 [4] Part 1/2 Problem 4.52 [4] Part 2/2 Problem 4.53 Given: Data on flow through a control surface Find: [3] Net rate of momentum flux Solution:
Basic equation: We need to evaluate ∫ CS rr
VρV ⋅ dA Assumptions: 1) Uniform flow at each section
From Problem 4.21 V1 = 10⋅ Then for the control surface ft
s A1 = 0.5⋅ ft 2 V2 = 20⋅ ft
s A2 = 0.1⋅ ft 2 A3 = 0.6⋅ ft 2 V 3 = 5⋅ ft
s It is an outlet rr
rrr rrr
rrr
VρV ⋅ dA = V1ρV1 ⋅ A1 + V2 ρV2 ⋅ A2 + V3 ρV3 ⋅ A3
∫CS
rr
rr
rr
= V1iˆρ V1 ⋅ A1 + V2 ˆρ V2 ⋅ A2 + V3 sin(60 )iˆ − V3 cos(60 ) ˆ ρ V3 ⋅ A3
j
j
ˆ
= −V1i ρV1 A1 + V2 ˆρV2 A2 + V3 sin(60 )iˆ − V3 cos(60 ) ˆ ρV3 A3
j
j ( ) ( [ [ )[ [
( ) = ρ − V12 A1 + V32 A3 sin(60) iˆ + ρ V22 A2 − V32 A3 cos(60) ˆ
j Hence the x component is ρ [− V12 A1 + V32 A3 sin(60)] = 65⋅ lbm
ft and the y component is ρ [V22 A2 − V32 A3 cos(60)] = 65⋅ 3 lbm
ft 3 ( 2 2 ) ft2 × lbf⋅ s
4 × −10 × 0.5 + 5 × 0.6 × sin( 60⋅ deg) ⋅ s ( 2 2 2 lbm⋅ ft ) ft2 × lbf ⋅ s
4 × 20 × 0.1 − 5 × 0.6 × cos ( 60⋅ deg) ⋅ s = −2406 lbf 2 lbm⋅ ft = 2113 lbf Problem 4.54 [3] y 2h x CS Given: Data on flow at inlet and outlet of channel Find: Ratio of outlet to inlet momentum flux Solution:
Basic equation: Momentum flux in x direction at a section r
mf x = ∫ uρV ⋅ dA
A Assumptions: 1) Steady flow 2) Incompressible flow
2 Evaluating at 1 and 2 mfx1 = U⋅ ρ⋅ ( −U⋅ 2⋅ h) ⋅ w mfx1 = 2⋅ ρ⋅ w⋅ U ⋅ h Hence h
⌠
2
⌠⎡
⎮⎡
2⎤
2
4
⌠
y⎞
y⎞
y⎞ ⎤
2
2
2⎮
mfx2 = ⎮ ρ⋅ u ⋅ w dy = ρ⋅ w⋅ umax ⋅ ⎮ ⎢1 − ⎛ ⎟ ⎥ dy = ρ⋅ w⋅ umax ⋅ ⎮ ⎢1 − 2⋅ ⎛ ⎟ + ⎛ ⎟ ⎥ dy
⎜
⎜
⎜
⎮ ⎣ ⎝ h⎠ ⎦
⌡− h
⎝ h⎠ ⎝ h⎠ ⎦
⌡− h
⌡− h ⎣ h h 4
2⎞
2
2 16
mfx2 = ρ⋅ w⋅ umax ⋅ ⎛ 2⋅ h − ⋅ h + ⋅ h⎟ = ρ⋅ w⋅ umax ⋅ ⋅ h
⎜
3
5⎠
15
⎝
Then the ratio of momentum fluxes is
mfx2
mfx1 = 16
2
⋅ ρ⋅ w⋅ umax ⋅ h
15
2 2⋅ ρ⋅ w⋅ U ⋅ h 8 ⎛ umax ⎞
=
⋅⎜
⎟
15 ⎝ U ⎠ 2 2 But, from Problem 4.30 umax = 3
⋅U
2 ⎛ 3 ⋅U ⎞
mfx2
8 ⎜2 ⎟
6
=
⋅⎜
⎟ = = 1.2
mfx1
15 ⎝ U ⎠
5 Hence the momentum increases as it flows in the entrance region of the channel. This appears to contradict common sense, as
friction should reduce flow momentum. What happens is the pressure drops significantly along the channel so the net force on the
CV is to the right. Problem 4.55 Given: Data on flow at inlet and outlet of pipe Find: [3] Ratio of outlet to inlet momentum flux Solution:
Basic equation: Momentum flux in x direction at a section r
mf x = ∫ uρV ⋅ dA
A Assumptions: 1) Steady flow 2) Incompressible flow ( 2 ) 2 mfx1 = ρ⋅ π⋅ U ⋅ R 2 Evaluating at 1 and 2 mfx1 = U⋅ ρ⋅ −U⋅ π⋅ R Hence R
⌠
2
⌠
⎮
⎡ ⎛ r ⎞ 2⎤
⎮
⌠
2
2
2
mfx2 = ⎮ ρ⋅ u ⋅ 2⋅ π⋅ r dr = 2⋅ ρ⋅ π⋅ umax ⋅ ⎮ r ⋅ ⎢1 − ⎜ ⎟ ⎥ dr = 2⋅ ρ⋅ π⋅ umax ⋅ ⎮
⎮
⌡0
R⎠ ⎦
⎝
⎮
⌡0 ⎣
⌡ R R 2
⎛ R2 R2 R2 ⎞
R
⎟ = ρ⋅ π⋅ umax2⋅
mfx2 = 2⋅ ρ⋅ π⋅ umax ⋅ ⎜
−
+
2
6⎠
3
⎝2 0 3
5⎞
⎛
⎜ r − 2⋅ r + r ⎟ dy
2
4
⎜
R
R⎟
⎝
⎠ 2 Then the ratio of momentum fluxes is
mfx2
mfx1 But, from Problem 4.31 = 1
22
⋅ ρ⋅ π⋅ umax ⋅ R
3 umax = 2⋅ U 2 ρ⋅ π ⋅ U ⋅ R 2 1 ⎛ umax⎞
= ⋅⎜
⎟
3⎝ U ⎠
mfx2
mfx1 2 2 = 1 ⎛ 2⋅ U ⎞
4
⋅⎜
⎟ = = 1.33
3⎝U ⎠
3 Hence the momentum increases as it flows in the entrance region of the pipe This appears to contradict common sense, as
friction should reduce flow momentum. What happens is the pressure drops significantly along the pipe so the net force on the
CV is to the right. Problem 4.48 Problem 4.56 [2] Problem 4.49 Problem 4.57 [2] Problem 4.58 Given: Water jet hitting wall Find: [2] Force generated on wall CS
y x Solution:
U Basic equation: Momentum flux in x direction Rx Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow 5) Water leaves vertically
Hence ( ) 2 2 π⋅ D Rx = u1⋅ ρ⋅ −u1⋅ A1 = −ρ⋅ U ⋅ A = −ρ⋅ U ⋅ 2 4 2 ⎛1
π⋅ ⎜ ⋅ ft⎞
⎟
2
slug ⎛ ft ⎞
6⎠
lbf ⋅ s
Rx = −1.94⋅
× ⎜ 20⋅ ⎟ × ⎝
×
3
slug⋅ ft
4
⎝ s⎠
ft
2 Rx = −16.9⋅ lbf Problem 4.59 Given: Fully developed flow in pipe Find: Why pressure drops if momentum is constant Solution:
Basic equation: Momentum flux in x direction Assumptions: 1) Steady flow 2) Fully developed flow
Hence Fx = Δp
− τw ⋅ A s = 0
L Δp = L⋅ τw⋅ As where Δp is the pressure drop over length L, τw is the wall friction and As is the pipe surface area
The sum of forces in the x direction is zero. The friction force on the fluid is in the negative x direction, so the net pressure
force must be in the positive direction. Hence pressure drops in the x direction so that pressure and friction forces balance [1] Problem 4.60 Given: Data on flow and system geometry Find: [2] Force required to hold plug Solution:
3 The given data is D1 = 0.25⋅ m D2 = 0.2⋅ m Q = 1.5⋅ m
s p1 = 3500⋅ kPa ρ = 999⋅ Then A1 = A1 = 0.0491 m 4 3 m 2 π⋅ D 1 kg 2
2 A2 = π⎛ 2
2
⋅ D − D2 ⎞
⎠
4⎝ 1 A2 = 0.0177 m V1 = Q
A1 V1 = 30.6 m
s V2 = Q
A2 V2 = 84.9 m
s Governing equation:
Momentum (4.18a) Applying this to the current system ( ) ( ) −F + p1⋅ A2 − p2⋅ A2 = 0 + V1⋅ −ρ⋅ V1⋅ A1 + V2⋅ ρ⋅ V2⋅ A2
Hence p2 = 0 and (gage) F = p1⋅ A1 + ρ⋅ ⎛ V1 ⋅ A1 − V2 ⋅ A2⎞
⎝
⎠
2 F = 3500 × kN
2 m 2 2 ⋅ 0.0491⋅ m + 999⋅ kg
3 m ⎡ × ⎢⎛ 30.6⋅
⎜ ⎣⎝ 2 ⎛
⎟ ⋅ 0.0491⋅ m − ⎜ 84.9⋅ m⎞ s⎠ 2 ⎝ 2⎤
⎟ ⋅ 0.0177⋅ m ⎥
s⎠
⎦ m⎞ 2 F = 90.4 kN Problem 4.61 Given: Large tank with nozzle and wire Find: [2] Tension in wire; plot for range of water depths Solution:
Basic equation: Momentum flux in x direction for the tank
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow
2 2 Hence Rx = T = V⋅ ρ⋅ ( V⋅ A) = ρ⋅ V ⋅ A = ρ⋅ ( 2⋅ g⋅ y) ⋅ When y = 0.9 m T= π⋅ d
4 T= 1
2
⋅ ρ ⋅ g⋅ y⋅ π ⋅ d
2 (1) 2 π
kg
m
2 N⋅ s
× 1000⋅
× 9.81⋅ × 0.9⋅ m × ( 0.015⋅ m) ×
3
2
kg⋅ m
2
m
s From Eq 1 T = 3.12 N 4 T (N) 3
2
1
0 0.3 0.6 y (m)
This graph can be plotted in Excel 0.9 Problem 4.62 [2] CS y V Rx Given: Nozzle hitting stationary cart Find: Value of M to hold stationary; plot M versu θ Solution:
Basic equation: Momentum flux in x direction for the tank
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow 5) Exit velocity is V
Hence 2 2 Rx = −M⋅ g = V⋅ ρ⋅ ( −V⋅ A) + V⋅ cos ( θ) ⋅ ( V⋅ A) = ρ⋅ V ⋅ A⋅ ( cos ( θ) − 1)
2 When θ = 40o M= From Eq 1 M= ρ⋅ V ⋅ A
⋅ ( 1 − cos ( θ) )
g 2 s
kg ⎛
m⎞
2
× 1000⋅
× ⎜ 10⋅ ⎟ × 0.1⋅ m × ( 1 − cos ( 40⋅ deg) )
3⎝
9.81⋅ m
s⎠
m M = 238 kg M (kg) 3000
2000
1000 0 45 90 Angle (deg)
This graph can be plotted in Excel 135 180 (1) Problem 4.63 [3] Given: Water jet hitting plate with opening Find: Force generated on plate; plot force versus diameter d CS
y Solution: x V V Basic equation: Momentum flux in x direction
Rx Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow ( ) ( ) 2 π⋅ D 2 2 π⋅ d Hence Rx = u1⋅ ρ⋅ −u1⋅ A1 + u2⋅ ρ⋅ u2⋅ A2 = −ρ⋅ V ⋅ For given data 2
2⎡
2
2
π
slug ⎛ ft ⎞
1
1 ⎞ ⎤ lbf ⋅ s
Rx = − ⋅ 1.94⋅
× ⎜ 15⋅ ⎟ × ⎛ ⋅ ft⎞ × ⎢1 − ⎛ ⎟ ⎥ ×
⎜⎟
⎜
3
4
⎝ s ⎠ ⎝ 3 ⎠ ⎣ ⎝ 4 ⎠ ⎦ slug⋅ ft
ft 4 + ρ⋅ V ⋅ 4 2 Rx = − 22
2
π ⋅ ρ⋅ V ⋅ D ⎡ ⎛ d ⎞ ⎤
⋅ ⎢1 − ⎜ ⎟ ⎥
4
⎣ ⎝D⎠ ⎦ Rx = −35.7⋅ lbf From Eq 1 (using the absolute value of Rx) Force (lbf) 40
30
20
10
0 0.2 0.4 0.6 Diameter Ratio (d/D)
This graph can be plotted in Excel 0.8 1 (1) Problem 4.64 Given: Water flowing past cylinder Find: [3] Horizontal force on cylinder V y
x Solution: CS Rx Basic equation: Momentum flux in x direction V Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow ( ) ( ) Hence For given data Rx = −1000⋅ 2 Rx = u1⋅ ρ⋅ −u1⋅ A1 + u2⋅ ρ⋅ u2⋅ A2 = 0 + ρ⋅ ( −V⋅ sin( θ) ) ⋅ ( V⋅ a ⋅ b )
kg
m 3 × ⎛ 3⋅
⎜ ⎝ 2 Rx = −ρ⋅ V ⋅ a ⋅ b ⋅ sin( θ)
2 m⎞
N ⋅s
⎟ × 0.0125⋅ m × 0.0025⋅ m × sin( 20⋅ deg) ×
kg⋅ m
s⎠ This is the force on the fluid (it is to the left). Hence the force on the cylinder is θ R x = −R x Rx = −0.0962 N
Rx = 0.0962 N Problem 4.65 [5] y V
x
CS W Rx Given: Water flowing into tank Find: Mass flow rates estimated by students. Explain discrepancy Solution:
Basic equation: Momentum flux in y direction
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow
For the first student m1 = ρ⋅ V
t where m1 represents mass flow rate (software cannot render a dot above it!) m1 = 1.94⋅ slug
ft For the second student 3 × 15⋅ ft × 3 M
m2 =
t 1
30⋅ s slug
m1 = 0.97⋅
s lbm
m1 = 31.2⋅
s where m2 represents mass flow rate 1
m2 = 960⋅ lb ×
30⋅ s slug
m2 = 0.995⋅
s lbm
m2 = 32⋅
s There is a discrepancy because the second student is measuring instantaneous weight PLUS the force generated as the pipe
flow momentum is "killed".
To analyse this we first need to find the speed at which the water stream enters the tank, 5 ft below the pipe exit. This would be a good
place to use the Bernoulli equation, but this problem is in the set before Bernoulli is covered. Instead we use the simple concept that the
fluid is falling under gravity (a conclusion supported by the Bernoulli equation). From the equations for falling under gravity:
2 2 Vtank = Vpipe + 2⋅ g⋅ h
where Vtank is the speed entering the tank, Vpipe is the speed at the pipe, and h = 5 ft is the distance traveled. Vpipe is obtained from
m1 Vpipe = 2 ρ⋅
Vpipe = Then Vtank = 4
π 4⋅ m1 = π⋅ dpipe 2 π⋅ ρ⋅ dpipe 4
× 31.2⋅ 2 lbm
s 3 × Vpipe + 2⋅ g⋅ h ft
1⋅ slug
1⎞
×
×⎛
1.94⋅ slug 32.2⋅ lbm ⎜ 1 ⎟
⎜ ⋅ ft ⎟
⎝6 ⎠ 2 Vpipe = 22.9 ft Vtank = 29.1 ft 2 Vtank = ⎛ 22.9⋅ ft ⎞ + 2 × 32.2⋅ ft × 5ft
⎜
⎟
2
s⎠
⎝
s s s We can now use the y momentum equation for the CS shown above ( ) Ry − W = −Vtank⋅ ρ⋅ −Vtank⋅ Atank
where Atank is the area of the water flow as it enters the tank. But for the water flow Vtank⋅ Atank = Vpipe⋅ Apipe 2 Hence ΔW = Ry − W = ρ⋅ Vtank⋅ Vpipe⋅ π⋅ dpipe
4 This equation indicate the instantaneous difference ΔW between the scale reading (Ry ) and the actual weight of water (W) in the tank
ΔW = 1.94⋅ slug
ft 3 2 × 29.1⋅ 2 ft
ft π ⎛ 1
lbf ⋅ s
× 22.9⋅ × × ⎜ ⋅ ft⎞ ×
⎟
slug⋅ ft
s
s 4 ⎝6 ⎠ ΔW = 28.2 lbf Hence the scale overestimates the weight of water by 28.2 lbf, or a mass of 28.2 lbm
For the second student M = 960⋅ lbm − 28.2⋅ lbm = 932⋅ lbm Hence M
m2 =
t where m2 represents mass flow rate 1
m2 = 932⋅ lb ×
30⋅ s slug
m2 = 0.966⋅
s lbm
m2 = 31.1⋅
s Comparing with the answer obtained from student 1, we see the students now agree! The discrepancy was entirely caused by the fact that t
second student was measuring the weight of tank water PLUS the momentum lost by the water as it entered the tank! Problem 4.66 [3] CS V
y
x
Rx Given: Water tank attached to mass Find: Whether tank starts moving Solution:
Basic equation: Momentum flux in x direction for the tank
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure at exit 4) Uniform flow
Hence 2
2 π⋅ D Rx = V⋅ cos ( θ) ⋅ ρ⋅ ( V⋅ A) = ρ⋅ V ⋅ 4 ⋅ cos ( θ) We need to find V. We could use the Bernoulli equation, but here it is known that
V= 2 × 9.81⋅ m
2 × 4⋅ m V = 8.86 s
Hence Rx = 1000⋅ kg
m 3 × ⎛ 8.86⋅
⎜ ⎝ 2⋅ g⋅ h where h = 4 m is the
height of fluid in the tank m
s 2 m⎞
π
2
⎟ × × ( 0.04⋅ m) × cos ( 60⋅ deg)
s⎠
4 This force is equal to the tension T in the wire V= T = Rx T = 49.3 N For the block, the maximum friction force a mass of M = 9 kg can generate is
Fmax = 9⋅ kg × 9.81⋅ m
2 s Rx = 49.3 N Fmax = M⋅ g⋅ μ 2 × 0.5 × N⋅ s
kg⋅ m Fmax = 44.1 N Hence the tension T created by the water jet is larger than the maximum friction Fmax; the tank starts to move where μ is static friction Problem 4.67 [4]
CS y’
y FR
x Given: Gate held in place by water jet Find: Required jet speed for various water depths Solution:
Basic equation: Momentum flux in x direction for the wall
Note: We use this equation ONLY for the jet impacting the wall. For the hydrostatic force and location we use computing equations
Ixx
FR = pc⋅ A
y' = yc +
A⋅ yc
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
Hence ( 2 π⋅ D ) Rx = V⋅ ρ⋅ −V⋅ Ajet = −ρ⋅ V ⋅ 2 4 This force is the force generated by the wall on the jet; the force of the jet hitting the wall is then
2 π⋅ D Fjet = −Rx = ρ⋅ V ⋅
For the hydrostatic force 2 where D is the jet diameter 4 3 h
1
2
FR = pc⋅ A = ρ⋅ g⋅ ⋅ h⋅ w = ⋅ ρ⋅ g⋅ w⋅ h
2
2 where h is the water depth and w is the gate width w⋅ h
12 Ixx
h
2
y' = yc +
=+
= ⋅h
h
A⋅ yc
2
3
w ⋅ h⋅
2
h
−Fjet⋅ hjet + FR⋅ ( h − y') = −Fjet⋅ hjet + FR⋅ = 0
3 For the gate, we can take moments about the hinge to obtain
where hjet is the height of the jet from the ground Hence For the first case (h = 0.5 m) 2 π⋅ D Fjet = ρ⋅ V ⋅ V= For the second case (h = 0.25 m) V= For the first case (h = 0.6 m) V= 2
3⋅ π 4 2 3 h
1
2h
⋅ hjet = FR⋅ = ⋅ ρ⋅ g⋅ w⋅ h ⋅
3
2
3 × 9.81⋅ m
2 × 0.5⋅ m × ( 0.5⋅ m) × ⎛
⎜ s 1 2 2
3⋅ π m × 9.81⋅ m
2 s 3 × 0.5⋅ m × ( 0.6⋅ m) × ⎛
⎜
3 V = 51 m V = 18 m 2 1⎞
1
× 9.81⋅ × 0.5⋅ m × ( 0.25⋅ m) × ⎛
⎜
⎟×
2
3⋅ π
⎝ 0.01⋅ m ⎠ 0.5⋅ m
s
2 s s 2 ⎞× 1
⎟
0.01⋅ m ⎠
0.5⋅ m
⎝
1 2 3⋅ π⋅ D ⋅ hj ⎞× 1
⎟
0.01⋅ m ⎠
0.5⋅ m
⎝ 3 2⋅ g⋅ w⋅ h V= V = 67.1 m
s Problem 4.55 Problem 4.68 [2] Problem 4.56 Problem 4.69 [2] Problem 4.70 [3] Given: Flow into and out of CV Find: Expressions for rate of change of mass, and force Solution:
Basic equations: Mass and momentum flux Assumptions: 1) Incompressible flow 2) Uniform flow
dMCV
+
dt For the mass equation (ρ→ → ) = dMCV + ρ⋅ (−V1⋅ A1 − V2⋅ A2 + V3⋅ A3 + V4⋅ A4) = 0
∑ ⋅ V⋅ A
dt CS dMCV
= ρ⋅ V1⋅ A1 + V2⋅ A2 − V3⋅ A3 − V4⋅ A4
dt ( Fx +
For the x momentum p1⋅ A1 + 2 ) V1
5
4
5
5
⋅ p2⋅ A2 − ⋅ p3⋅ A3 −
⋅ p4⋅ A4 = 0 +
⋅ − ρ ⋅ V 1⋅ A 1 +
⋅ V ⋅ −ρ⋅ V2⋅ A2 ...
13
5
13
13 2
2
4
5
+ ⋅ V3⋅ ρ⋅ V3⋅ A3 +
⋅ V ⋅ ρ ⋅ V 3⋅ A 3
5
13 3 (
( Fx = − p1⋅ A1 − 2 Fy + p1⋅ A1
2 − 12
13 ⋅ p2⋅ A2 − 3
5 ⋅ p3⋅ A3 + 12
13 ⋅ p4⋅ A4 = 0 + V1 2 + ) ( (
2 ) ⋅ − ρ ⋅ V 1⋅ A 1 − ) ) ( 12 ) ⋅ V ⋅ −ρ⋅ V2⋅ A2 ...
13 2
3
12
+ ⋅ V3⋅ ρ⋅ V3⋅ A3 −
⋅ V ⋅ ρ ⋅ V 3⋅ A 3
5
13 3 ( p1⋅ A1 ( 5
4
5
1
5
4
5
2
2
2
2
⋅p ⋅A + ⋅p ⋅A +
⋅ p ⋅ A + ρ⋅ ⎛ −
⋅ V1 ⋅ A1 −
⋅ V ⋅ A2 + ⋅ V3 ⋅ A3 +
⋅ V ⋅ A3⎞
⎜
⎟
13 2 2 5 3 3 13 4 4
13 2
5
13 3
⎝2
⎠ For the y momentum Fy = − ) ) ( ) 3
12
1
12
3
12
2
2
2
2
⋅p ⋅A + ⋅p ⋅A −
⋅ p ⋅ A + ρ⋅ ⎛ −
⋅ V1 ⋅ A1 −
⋅ V ⋅ A2 + ⋅ V3 ⋅ A3 −
⋅ V ⋅ A3⎞
⎜
⎟
13 2 2 5 3 3 13 4 4
13 2
5
13 3
⎝2
⎠
12 Problem 4.71 [2] Problem 4.72 [2] y
CS x Rx Given: Water flow through elbow Find: Force to hold elbow Solution:
Basic equation: Momentum flux in x direction for the elbow
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure at exit 4) Uniform flow
Hence ( ) From continuity V2⋅ A2 = V1⋅ A1
Hence ( Rx + p1g⋅ A1 = V1⋅ −ρ⋅ V1⋅ A1 − V2⋅ ρ⋅ V2⋅ A2 Rx = −15⋅ lbf
2 in so
2 × 4⋅ in − 1.94⋅ ) Rx = −p1g⋅ A1 − ρ⋅ ⎛ V1 ⋅ A1 + V2 ⋅ A2 ⎞
⎝
⎠
2 A1
V2 = V1⋅
A2 2 ft 4
V2 = 10⋅ ⋅
s1 2
2
2
2
slug ⎡⎛
ft ⎞
ft ⎞
1⋅ ft ⎞
lbf⋅ s
2
2⎤
× ⎢⎜ 10⋅ ⎟ ⋅ 4⋅ in + ⎛ 40⋅ ⎟ ⋅ 1⋅ in ⎥ × ⎛
⎜
⎜
⎟×
3 ⎣⎝
⋅
s⎠
⎝ s⎠
⎦ ⎝ 12⋅ in ⎠ slugft
ft V2 = 40 ft
s Rx = −86.9⋅ lbf The force is to the left: It is needed to hold the elbow on against the high pressure, plus it generates the large change in x momentum Problem 4.73 [2] y
x CS Rx Given: Water flow through elbow Find: Force to hold elbow Solution:
Basic equation: Momentum flux in x direction for the elbow
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
Hence Rx + p1g⋅ A1 + p2g⋅ A2 = V1⋅ −ρ⋅ V1⋅ A1 − V2⋅ ρ⋅ V2⋅ A2 ( ) ⎛ D1 ⎞
V2 = V1⋅
= V1⋅ ⎜
⎟
A2
⎝ D2 ⎠ Hence so ( Rx = −p1g⋅ A1 − p2g⋅ A2 − ρ⋅ ⎛ V1 ⋅ A1 + V2 ⋅ A2 ⎞
⎝
⎠ 2 From continuity V2⋅ A2 = V1⋅ A1 ) A1 2 V2 = 0.8⋅ m ⎛ 0.2 ⎞
⋅⎜
⎟
s ⎝ 0.04 ⎠ 2
2
π⋅ ( 0.2⋅ m )
π⋅ ( 0.04⋅ m )
3N
3N
...
×
− 75 × 10 ⋅
×
2
2
4
4 Rx = −350 × 10 ⋅ m + −1000⋅ kg
m 3 ⎡ × ⎢⎛ 0.8⋅
⎜ ⎣⎝ 2 V2 = 20 2 m
s Rx = −11.6⋅ kN m 2
2
2
2
2
π⋅ ( 0.2⋅ m )
π⋅ ( .04⋅ m ) ⎤ N ⋅ s
m⎞
m⎞
⎥×
+ ⎛ 20⋅ ⎟ ×
⎟×
⎜
4
4
s⎠
⎝ s⎠
⎦ kg⋅ m The force is to the left: It is needed to hold the elbow on against the high pressures, plus it generates the large change in x momentum Problem 4.74 [2] y
x Rx Given: Water flow through nozzle Find: CS Force to hold nozzle Solution:
Basic equation: Momentum flux in x direction for the elbow
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
Hence ( From continuity V2⋅ A2 = V1⋅ A1 Hence ) ( ) Rx + p1g⋅ A1 + p2g⋅ A2 = V1⋅ −ρ⋅ V1⋅ A1 + V2⋅ cos ( θ) ⋅ ρ⋅ V2⋅ A2 so ⎛ D1 ⎞
V 2 = V 1⋅
= V1⋅ ⎜
⎟
A2
⎝ D2 ⎠
A1 Rx = −p1g⋅ A1 + ρ⋅ ⎛ V2 ⋅ A2⋅ cos ( θ) − V1 ⋅ A1⎞
⎝
⎠
2 2 V2 = 1.5⋅ m ⎛ 30 ⎞
⋅⎜ ⎟
s ⎝ 15 ⎠ 2 V 2 = 6⋅ m
s 2
2
2
2
2
2
π⋅ ( 0.15⋅ m)
π⋅ ( .3⋅ m) ⎤ N⋅ s
π⋅ ( 0.3⋅ m)
kg ⎡⎛ m ⎞
m
3N
⎥×
×
+ 1000⋅
× ⎢⎜ 6⋅ ⎟ ×
⋅ cos ( 30⋅ deg) − ⎛ 1.5⋅ ⎞ ×
⎜
⎟
2
3 ⎣⎝ s ⎠
4
4
4
s⎠
⎝
⎦ kg⋅ m Rx = −15 × 10 ⋅ m Rx = −668⋅ N 2 m The joint is in tension: It is needed to hold the elbow on against the high pressure, plus it
generates the large change in x momentum Problem 4.61 Problem 4.75 [2] Problem 4.76 [2] CS y
x Given: Water flow through orifice plate Find: Rx Force to hold plate Solution:
Basic equation: Momentum flux in x direction for the elbow
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
Hence ( ) ( ft
s and Rx + p1g⋅ A1 − p2g⋅ A2 = V1⋅ −ρ⋅ V1⋅ A1 + V2⋅ ρ⋅ V2⋅ A2 ) Rx = −p1g⋅ A1 + ρ⋅ ⎛ V2 ⋅ A2 − V1 ⋅ A1 ⎞
⎝
⎠
2 2 From continuity Q = V1⋅ A1 = V2⋅ A2
3 so V1 = Q
ft
= 20⋅
×
s
A1 4 ⎛1 ⎞
π⋅ ⎜ ⋅ ft ⎟
⎝3 ⎠ 2 = 229⋅ 2
2
A1
D
ft
4⎞
ft
V2 = V1⋅
= V1⋅ ⎛ ⎞ = 229⋅ × ⎛
⎜⎟
⎜ ⎟ = 1628⋅
A2
s ⎝ 1.5 ⎠
s
⎝d⎠ NOTE: problem has an error: Flow rate should be 2 ft3/s not 20 ft3/s! We will provide answers to both Hence 2
2
2
2
2⎤
2
2
π⋅ ( 1.5⋅ in)
π⋅ ( 4⋅ in)
slug ⎡⎛
ft ⎞
⎛ 229⋅ ft ⎞ × π⋅ ( 4⋅ in) ⎥ × ⎛ 1⋅ ft ⎞ × lbf⋅ s
⎢⎜ 1628⋅ ⎟ ×
Rx = −200⋅
×
+ 1.94⋅
×
−⎜
⎟
⎜
⎟
2
3 ⎣⎝
4
4
4
⋅
s⎠
s⎠
⎝
⎦ ⎝ 12⋅ in ⎠ slugft
in
ft lbf Rx = 51707⋅ lbf
With more realistic velocities
Hence Rx = −200⋅ lbf
2 in
Rx = −1970⋅ lbf × 2
2
2
2
2
2
2
π⋅ ( 1.5⋅ in)
π⋅ ( 4⋅ in) ⎤ ⎛ 1⋅ ft ⎞
π⋅ ( 4⋅ in)
slug ⎡⎛
ft
ft ⎞
lbf ⋅ s
⎥×⎜
+ 1.94⋅
× ⎢⎜ 163⋅ ⎞ ×
− ⎛ 22.9⋅ ⎟ ×
×
⎟
⎜
⎟
3
4
4
4
s⎠
s⎠
⎣⎝
⎝
⎦ ⎝ 12⋅ in ⎠ slug⋅ ft
ft Problem 4.63 Problem 4.77 [2] Problem 4.64 Problem 4.78 [2] Problem 4.79 [2] CS
Ve y
x
Rx
Given: Data on rocket motor Find: Thrust produced Solution:
Basic equation: Momentum flux in x direction for the elbow
Assumptions: 1) Steady flow 2) Neglect change of momentum within CV 3) Uniform flow
Hence ( ) Rx − peg⋅ Ae = Ve⋅ ρe⋅ Ve⋅ Ae = me⋅ Ve Rx = peg⋅ Ae + me⋅ Ve where peg is the exit pressure (gage), me is the mass flow rate at the exit (software cannot render dot over m!) and Ve is the xit velocity
For the mass flow rate kg
kg
me = mnitricacid + maniline = 80⋅
+ 32⋅
s
s Hence Rx = ( 110 − 101) × 10 ⋅ 2
2
π⋅ ( 0.6⋅ m)
kg
m N⋅ s
3N
×
+ 112⋅
× 180⋅ ×
2
4
s
s kg⋅ m m kg
me = 112⋅
s
Rx = 22.7 kN Problem 4.65 Problem 4.80 [2] Problem 4.81 [3] Problem 4.82 [2] Given: Data on flow and system geometry Find: Deflection angle as a function of speed; jet speed for 10o deflection Solution:
The given data is ρ = 999⋅ kg
3 A = 0.005⋅ m 2 L = 2⋅ m k = 1⋅ m N
m x0 = 1⋅ m Governing equation:
y momentum (4.18b) Applying this to the current system in the vertical direction
Fspring = V⋅ sin( θ) ⋅ ( ρ⋅ V⋅ A) ( ) But Fspring = k ⋅ x = k ⋅ x0 − L⋅ sin( θ) Hence k ⋅ x0 − L⋅ sin( θ) = ρ⋅ V ⋅ A⋅ sin( θ) Solving for θ θ = asin⎜ ( ) 2 k ⋅ x0
⎞
⎛
⎟
⎜ k ⋅ L + ρ⋅ A⋅ V2 ⎟
⎝
⎠ For the speed at which θ = 10o, solve V= ( k⋅ x0 − L⋅ sin ( θ)
ρ⋅ A⋅ sin ( θ) ) 1⋅
V= N
⋅ ( 1 − 2⋅ sin ( 10⋅ deg) ) ⋅ m
m 999⋅ kg
3 ⋅ kg⋅ m
2 2
⋅ 0.005⋅ m ⋅ sin ( 10⋅ deg) N⋅ s m The deflection is plotted in the corresponding Excel workbook, where the above velocity is obtained using Goal Seek V = 0.867 m
s ρ=
xo =
L=
k= 999
1
2 A= 0.005 1 30.0
29.2
27.0
24.1
20.9
17.9
15.3
13.0
11.1
9.52
8.22
7.14
6.25
5.50
4.87
4.33 N/m V (m/s) θ (o) 2 0.867 10 m o
θ( ) 0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5 To find when θ = 10o, use Goal Seek m
m Deflection Angle vs Jet Speed
35
30
θ (deg) V (m/s) kg/m3 25
20
15
10
5
0
0.00 0.25 0.50 0.75
V (m/s) 1.00 1.25 1.50 Problem 4.69 Problem 4.83 [3] Problem 4.84 [2] y
x
Ry
Rx
CS Given: Data on nozzle assembly Find: Reaction force Solution:
Basic equation: Momentum flux in x and y directions Assumptions: 1) Steady flow 2) Incompressible flow CV 3) Uniform flow
For x momentum ( ) Rx = V2⋅ cos ( θ) ⋅ ρ⋅ V2⋅ A2 = ρ⋅ V2 ⋅ From continuity Hence Rx = 1000⋅ kg
3 ⋅ cos ( θ) ⎛ D1 ⎞
V2 = V1⋅
= V1⋅ ⎜
⎟
A2
⎝ D2 ⎠ × ⎛ 18⋅
⎜ m 4 A1 A1⋅ V1 = A2⋅ V2 For y momentum 2
2 π⋅ D 2 ⎝ 2 V2 = 2⋅ 2 m ⎛ 7.5 ⎞
×⎜
⎟
s ⎝ 2.5 ⎠ 2 V2 = 18 m
s 2 m⎞
π
N⋅ s
2
⎟ × × ( 0.025⋅ m) × cos ( 30⋅ deg) ×
kg⋅ m
s⎠
4 ( ) ( Rx = 138 N ) Ry − p1⋅ A1 − W − ρ⋅ Vol⋅ g = −V1⋅ −ρ⋅ V1⋅ A1 − V2⋅ sin ( θ) ⋅ ρ⋅ V2⋅ A2
Ry = p1⋅ π⋅ D 1 2 + W + ρ⋅ Vol⋅ g + 4 ρ⋅ π ⎛ 2
2
2
2
⋅ V ⋅ D1 − V2 ⋅ D2 ⋅ sin ( θ)⎞
⎠
4⎝1 2 where N⋅ s
W = 4.5⋅ kg × 9.81⋅ ×
2 kg⋅ m
s Hence Ry = 125 × 10 ⋅ m 3 W = 44.1 N Vol = 0.002⋅ m 2
2
π⋅ ( 0.075⋅ m)
kg
m N⋅ s
3N
3
...
×
+ 44.1⋅ N + 1000⋅
× 0.002⋅ m × 9.81⋅ ×
2
3
2 kg⋅ m
4 m m s 2
2
⎤ N⋅ s2
m⎞
kg π ⎡⎛ m ⎞
⎢⎜ 2⋅ ⎟ × ( 0.075⋅ m) 2 − ⎛ 18⋅ ⎟ × ( 0.025⋅ m) 2 × sin ( 30⋅ deg)⎥ ×
+ 1000⋅
××
⎜
3 4 ⎣⎝ s ⎠
⎝ s⎠
⎦ kg⋅ m
m Ry = 554 N Problem 4.71 Problem 4.85 [3] Problem 4.86 Given: Data on water jet pump Find: [3] Speed at pump exit; pressure rise Solution:
Basic equation: Continuity, and momentum flux in x direction Assumptions: 1) Steady flow 2) Incompressible flow CV 3) Uniform flow
From continuity V2 = 10⋅
For x momentum As
Aj
Aj
⎛ A2 − Aj ⎞
V2 = Vs⋅
+ Vj⋅
= Vs⋅ ⎜
⎟ + Vj⋅
A2
A2
A2
⎝ A2 ⎠
ft
V2 = 22
s −ρ⋅ Vs⋅ As − ρ⋅ Vj⋅ Aj + ρ⋅ V2⋅ A2 = 0
ft ⎛ 0.75 − 0.1 ⎞
ft
0.1
×⎜
⎟ + 100⋅ ×
s ⎝ 0.75 ⎠
s 0.75 ( ) ( ) ( ⎛ 2 Aj
2 As
2⎞
Δp = p2 − p1 = ρ⋅ ⎜ Vj ⋅
+ Vs ⋅
− V2 ⎟
A2
A2
⎝
⎠ 2
2
2
2
⎡
0.1 ⎛ ft ⎞
( 0.75 − 0.1) ⎛ ft ⎞ ⎤ lbf ⋅ s
ft
× ⎢⎛ 100⋅ ⎞ ×
+ ⎜ 10⋅ ⎟ ×
− ⎜ 22⋅ ⎟ ⎥ ×
⎜
⎟
3
0.75 ⎝
0.75
s⎠
s⎠
⎣⎝
⎝ s ⎠ ⎦ slug⋅ ft
ft Δp = 1.94⋅ Hence ) p1⋅ A2 − p2⋅ A2 = Vj⋅ −ρ⋅ Vj⋅ Aj + Vs⋅ −ρ⋅ Vs⋅ As + V2⋅ ρ⋅ V2⋅ A2 slug Δp = 1816 lbf
ft 2 Δp = 12.6 psi Problem 4.73 Problem 4.87 [3] Problem 4.74 Problem 4.88 [3] Problem 4.89 [3] V1 V2
CS p1 p2
Rx y
x Given: Data on adiabatic flow of air Find: Force of air on pipe Solution:
Basic equation: Continuity, and momentum flux in x direction, plus ideal gas equation
p = ρ⋅ R ⋅ T Assumptions: 1) Steady flow 2) Ideal gas CV 3) Uniform flow
From continuity −ρ1⋅ V1⋅ A1 + ρ2⋅ V2⋅ A2 = 0 ρ1⋅ V1⋅ A = ρ2⋅ V2⋅ A For x momentum Rx + p1 ⋅ A − p2 ⋅ A = V1⋅ −ρ1⋅ V1⋅ A + V2⋅ ρ2⋅ V2⋅ A = ρ1⋅ V1⋅ A⋅ V2 − V1 ( ( ) ) ( Rx = p2 − p1 ⋅ A + ρ1⋅ V1⋅ A⋅ V2 − V1
For the air ρ1 = P1
Rair⋅ T1 ( ) ( ρ1⋅ V1 = ρ2⋅ V2 ) )
kg⋅ K
1
3N
×
×
2 286.9⋅ N ⋅ m ( 60 + 273) ⋅ K ρ1 = ( 200 + 101) × 10 ⋅ ρ1 = 3.15 m kg
m 3 2
kg
m
m N ⋅s
3N
2
2
Rx = ( 80 − 200) × 10 ⋅
× 0.05⋅ m + 3.15⋅
× 150⋅ × 0.05⋅ m × ( 300 − 150) ⋅ ×
2
3
s
s kg⋅ m m Hence m Rx = −2456 N This is the force of the pipe on the air; the pipe is opposing flow. Hence the force of the air on the pipe is
Fpipe = 2456 N The air is dragging the pipe to the right Fpipe = −Rx Problem 4.90 [3] V1 V2
CS p1 p2 ρ1 Rx y V3 ρ2 x Given: Data on heated flow of gas Find: Force of gas on pipe Solution:
Basic equation: Continuity, and momentum flux in x direction
p = ρ⋅ R ⋅ T Assumptions: 1) Steady flow 2) Uniform flow
From continuity ρ1
m3
V 2 = V 1⋅
−
ρ2 ρ2⋅ A −ρ1⋅ V1⋅ A1 + ρ2⋅ V2⋅ A2 + m3 = 0 where m3 = 20 kg/s is the mass leaving through
the walls (the software does not allow a dot) 3 V2 = 170⋅
For x momentum m
6
kg
m
1
×
− 20⋅
×
×
2
2.75⋅ kg
s 2.75
s
0.15⋅ m ( ) ( V2 = 322 Rx + p1⋅ A − p2⋅ A = V1⋅ −ρ1⋅ V1⋅ A + V2⋅ ρ2⋅ V2⋅ A m
s ) 2
2
Rx = ⎡ p2 − p1 + ρ2⋅ V2 − ρ1⋅ V1 ⎤ ⋅ A
⎣
⎦ ( ) ⎡
⎡
kg ⎛
m⎞
kg ⎛
m⎞
3N
Rx = ⎢( 300 − 400) × 10 ⋅
+ ⎢2.75⋅
× ⎜ 322⋅ ⎟ − 6⋅
× ⎜ 170⋅ ⎟
2⎢
3⎝
3⎝
⎢
s⎠
s⎠
m
m
m
⎣
⎣
2 Hence Rx = 1760 N 2⎤ 2⎤
× N⋅ s ⎥ × 0.15⋅ m2
⎥ kg⋅ m⎥
⎦
⎦ Problem 4.77 Problem 4.91 [3] Problem 4.78 Problem 4.92 [3] Problem 4.79 Problem 4.93 [3] Problem 4.94 [4] Given: Data on flow in wind tunnel Find: Mass flow rate in tunnel; Maximum velocity at section 2; Drag on object Solution:
Basic equations: Continuity, and momentum flux in x direction; ideal gas equation
p = ρ⋅ R ⋅ T
Assumptions: 1) Steady flow 2) Uniform density at each section
From continuity mflow = ρ1⋅ V1⋅ A1 = ρ1⋅ V1⋅ π⋅ D 1 where mflow is the mass flow rate 4 ρair = We take ambient conditions for the air density 2 patm
Rair⋅ Tatm kg
m π⋅ ( 0.75⋅ m)
mflow = 1.2⋅
× 12.5⋅ ×
3
4
s
m 2 ρair = 101000⋅ N
2 m × kg⋅ K
1
×
286.9⋅ N⋅ m 293⋅ K ρair = 1.2 kg
mflow = 6.63
s R Vmax = For x momentum 3⋅ mflow
2⋅ π⋅ ρair⋅ R 3 3
kg
m
1
⎞
Vmax =
× 6.63⋅
×
×⎛
⎜
⎟
1.2⋅ kg ⎝ 0.375⋅ m ⎠
2⋅ π
s 2 3 m 2⋅ π⋅ ρair⋅ Vmax⋅ R
⌠
2⋅ π⋅ ρair⋅ Vmax ⌠ R 2
⌠
r
⋅ ⎮ r dr =
mflow = ⎮ ρ2⋅ u2 dA2 = ρair⋅ ⎮ Vmax⋅ ⋅ 2⋅ π⋅ r dr =
⎮
⎮
⌡0
3
R
R
⌡
⌡0 Also kg 2 2 Vmax = 18.8 m
s ⌠
Rx + p1⋅ A − p2⋅ A = V1⋅ −ρ1⋅ V1⋅ A + ⎮ ρ2⋅ u2⋅ u2 dA2
⎮
⌡ ( ) R 2
⌠
2
2⋅ π⋅ ρair⋅ Vmax ⌠ R 3
⎮
⎛ V ⋅ r ⎞ ⋅ 2⋅ π⋅ r dr = p − p ⋅ A − V ⋅ m
⋅ ⎮ r dr
Rx = p2 − p1 ⋅ A − V1⋅ mflow + ⎮ ρair⋅ ⎜ max ⎟
2
1
1 flow +
2
⌡0
R⎠
⎝
R
⌡0 ( ) ( ( ) ) π
22
Rx = p2 − p1 ⋅ A − V1⋅ mflow + ⋅ ρair⋅ Vmax ⋅ R
2
We also have p1 = ρ⋅ g⋅ h1 p1 = 1000⋅ kg
3 × 9.81⋅ m
Hence m
2 × 0.03⋅ m p1 = 294 Pa p2 = ρ⋅ g⋅ h2 p2 = 147⋅ Pa s ⎡
π⋅ ( 0.75⋅ m)
kg
mπ
kg ⎛
m
2⎤ N⋅ s
Rx = ( 147 − 294) ⋅
×
+ ⎢−6.63⋅
× 12.5⋅ + × 1.2⋅
× ⎜ 18.8⋅ ⎞ × ( 0.375⋅ m) ⎥ ×
⎟
2
3⎝
4
⎢
⎥ kg⋅ m
s
s
2
s⎠
m
m
⎣
⎦
N Rx = −54 N 2 The drag on the object is equal and opposite 2 Fdrag = −Rx 2 Fdrag = 54.1 N Problem 4.95 Given: Data on wake behind object Find: [2] An expression for the drag Solution:
Governing equation:
Momentum (4.18a) Applying this to the horizontal motion
⌠
2
−F = U⋅ ( −ρ⋅ π⋅ 1 ⋅ U) + ⎮
⌡ 1 u ( r ) ⋅ ρ⋅ 2⋅ π⋅ r ⋅ u ( r ) dr 0 1
⎡
⎤
⌠
2⎥
⎢
⎮
2⎞
⎛
π⋅ r ⎞
2
F = π ρ⋅ U ⋅ ⎢1 − 2⋅ ⎮ r ⋅ ⎜ 1 − cos ⎛
⎜ ⎟ ⎟ dr⎥
⎮
⎢
⎝2⎠⎠ ⎥
⌡0 ⎝
⎣
⎦
1
⎛
⎞
⌠
2
4⎟
⎜
2
⎮
⎛ π⋅ r ⎞ + r⋅ cos ⎛ π⋅ r ⎞ dr⎟
F = π ρ⋅ U ⋅ ⎜ 1 − 2⋅ ⎮ r − 2⋅ r⋅ cos ⎜
⎟
⎜⎟
⎝2⎠
⎝2⎠ ⎟
⎜
⌡0
⎝
⎠ 2 ⎞⎤
3
2
Integrating and using the limits F = π ρ⋅ U ⋅ ⎡1 − ⎛ +
⎢ ⎜8
2 ⎟⎥
π ⎠⎦
⎣⎝ 1
⎛2
⎞
⎜ U − 2⋅ ⌠ r ⋅ u ( r ) 2 dr ⎟
⎮
F = π ρ⋅
⎜
⎟
⌡0
⎝
⎠ ⎛ 5⋅ π − 2 ⎞ ⋅ ρ⋅ U2
⎟
π⎠
⎝8 F=⎜ Problem 4.96 Given: Data on flow in 2D channel Find: [4] Maximum velocity; Pressure drop y 2h x Solution:
CS Basic equations: Continuity, and momentum flux in x direction; ideal gas equation Assumptions: 1) Steady flow 2) Neglect frition
From continuity ⌠
−ρ⋅ U1⋅ A1 + ⎮ ρ⋅ u2 dA = 0
⎮
⌡
h ⌠
2⎞
⎛
⎮
h
h ⎤⎤
y⎟
4
dy = w⋅ umax⋅ ⎡[ h − ( −h ) ] − ⎡ − ⎛ − ⎞⎥⎥ = w⋅ umax⋅ ⋅ h
U1⋅ 2⋅ h ⋅ w = w⋅ ⎮ umax⋅ ⎜ 1 −
⎢
⎢
⎜⎟
2⎟
⎜
3
⎣
⎣ 3 ⎝ 3 ⎠⎦⎦
⎮
h⎠
⎝
⌡− h
3
⋅U
21 umax = 3
m
× 7.5⋅
2
s Hence umax = For x momentum ⌠
p1⋅ A − p2⋅ A = V1⋅ −ρ1⋅ V1⋅ A + ⎮ ρ2⋅ u2⋅ u2 dA2
⎮
⌡ ( umax = 11.3 ) h ⌠
⎮
2w
2⎛
p1 − p2 = −ρ⋅ U1 + ⋅ ⎮ ρ⋅ umax ⋅ ⎜ 1 −
⎜
A⎮
⎝
⌡− h 2
y⎞
⎟
2⎟ h⎠ m
s Note that there is no Rx (no friction) 2 2 2 dy = −ρ⋅ U1 + ρ⋅ umax
h 2
1⎤
⋅ ⎡2⋅ h − 2⋅ ⎛ ⋅ h⎞ + 2⋅ ⎛ ⋅ h⎞⎥
⎢
⎜⎟
⎜⎟
3⎠
⎣
⎝
⎝ 5 ⎠⎦ ⎡ 8 3⎞
⎤1
8
2
2
Δp = p1 − p2 = −ρ⋅ U1 +
⋅ ρ⋅ umax = ρ⋅ U1⋅ ⎢ ⋅ ⎛ ⎟ − 1⎥ = ⋅ ρ⋅ U1
⎜
15
⎣ 15 ⎝ 2 ⎠
⎦5
2 Hence Δp = 1
5 × 1.24⋅ kg
3 m × ⎛ 7.5⋅
⎜ ⎝ m⎞ 2 ⎟× s⎠ 2 N⋅ s
kg⋅ m Δp = 14 Pa Problem 4.83 Problem 4.97 [3] Problem 4.84 Problem 4.98 [3] Problem 4.86 Problem 4.99 [3] Problem 4.100 [4] CS b c
y
x a d
Ff Given: Data on flow of boundary layer Find: Force on plate per unit width Solution:
Basic equations: Continuity, and momentum flux in x direction Assumptions: 1) Steady flow 2) Incompressible 3) No net pressure force
δ From continuity ⌠
−ρ⋅ U0⋅ w⋅ δ + mbc + ⎮ ρ⋅ u⋅ w dy = 0
⌡0 Hence ⌠
mbc = ⎮ ρ⋅ U0 − u ⋅ w dy
⌡0 For x momentum ⌠
⌠
2
2
−Ff = U0⋅ −ρ⋅ U0⋅ w⋅ δ + U0⋅ mbc + ⎮ u⋅ ρ⋅ u⋅ w dy = ⎮ ⎡−U0 + u + U0⋅ U0 − u ⎤ ⋅ w dy
⎣
⎦
⌡0
⌡0 δ ( where mbc is the mass flow rate across bc (Note:
sotware cannot render a dot!) ) ( δ δ ) ( ) δ δ
⌠
⌠
u⎞
2u⎛
Then the drag force is Ff = ⎮ ρ⋅ u⋅ U0 − u ⋅ w dy = ⎮ ρ⋅ U0 ⋅
⋅ ⎜1 −
⎟ dy
U0
⎮
⌡0
⎝ U0 ⎠
⌡0 ( But we have u
3
13
= ⋅η − ⋅η
U0
2
2
η=1 ⌠
=⎮
w
⎮
⌡0 Ff Ff
w
Hence ) Ff
w
Ff
w y = δ⋅ η where we have used substitution
1 u⎛
u⎞
3
9 2 1 3 3 4 1 6⎞
2⌠
dη = ρ⋅ U0 ⋅ δ⋅ ⎮ ⎛ ⋅ η − ⋅ η − ⋅ η + ⋅ η − ⋅ η ⎟ dη
ρ⋅ U 0 ⋅ δ ⋅
⋅ ⎜1 −
⎟
⎮ ⎜2
U0
4
2
2
4⎠
⎝ U0 ⎠
⌡⎝
2 0 = ρ⋅ U 0 ⋅ δ ⋅ ⎛
⎜
2 3 − ⎝4 31
3
1
2
−+
− ⎞ = 0.139⋅ ρ⋅ U0 ⋅ δ
⎟
4 8 10 28 ⎠ = 0.139 × 0.002377⋅ = 2.48 × 10 − 3 lbf ⋅ ft 2 2 ft
0.1
lbf ⋅ s
× ⎛ 30⋅ ⎞ ×
⋅ ft ×
⎜
⎟
3
slug⋅ ft
⎝ s ⎠ 12
ft slug (using standard atmosphere density) Problem 4.101 [4] CS b c
y
x a d
Ff Given: Data on flow of boundary layer Find: Force on plate per unit width Solution:
Basic equations: Continuity, and momentum flux in x direction Assumptions: 1) Steady flow 2) Incompressible 3) No net pressure force
δ From continuity ⌠
−ρ⋅ U0⋅ w⋅ δ + mbc + ⎮ ρ⋅ u⋅ w dy = 0
⌡0 Hence ⌠
mbc = ⎮ ρ⋅ U0 − u ⋅ w dy
⌡0 For x momentum ⌠
⌠
2
2
−Ff = U0⋅ −ρ⋅ U0⋅ w⋅ δ + U0⋅ mbc + ⎮ u⋅ ρ⋅ u⋅ w dy = ⎮ ⎡−U0 + u + U0⋅ U0 − u ⎤ ⋅ w dy
⎣
⎦
⌡0
⌡0 δ ( where mbc is the mass flow rate across bc (Note:
sotware cannot render a dot!) ) ( δ δ ) ( ) δ δ
⌠
⌠
u⎞
2u⎛
Then the drag force is Ff = ⎮ ρ⋅ u⋅ U0 − u ⋅ w dy = ⎮ ρ⋅ U0 ⋅
dy
⋅ 1−
U0 ⎜
U0 ⎟
⎮
⌡0
⎝
⎠
⌡0 ( But we have u
y
=
U0
δ where we have used substitution η=1 ⌠
=⎮
w
⎮
⌡0 Ff Ff
w
Hence ) Ff
w
Ff
w 2 1 ρ⋅ U 0 ⋅ δ ⋅ = ρ⋅ U 0 ⋅ δ ⋅ ⎛
⎜
2 1 ⎝2 u⎛
u⎞
2⌠
⋅ ⎜1 −
⎟ dη = ρ⋅ U0 ⋅ δ⋅ ⎮ η⋅ ( 1 − η) dη
⌡0
U0
⎝ U0 ⎠ 1
1
2
− ⎞ = ⋅ ρ⋅ U 0 ⋅ δ
⎟
3⎠ 6
2 = 2 1
kg ⎛ m ⎞
2
N⋅ s
× 1.225⋅
× ⎜ 20⋅ ⎟ ×
⋅m ×
3⎝
6
kg⋅ m
s⎠
1000
m = 0.163⋅ N
m y = δ⋅ η (using standard atmosphere density) Problem 4.102 [4] Part 1/2 Problem 4.102 [4] Part 2/2 Problem 4.103 [4] Problem 4.104 [4] Problem *4.91 Problem *4.105 [4] Problem *4.106 [4] CS Given: Air jet striking disk Find: Manometer deflection; Force to hold disk Solution:
Basic equations: Hydrostatic pressure, Bernoulli, and momentum flux in x direction
2 pV
+
+ g⋅ z = constant
ρ
2
Assumptions: 1) Steady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (gx = 0)
Applying Bernoulli between jet exit and stagnation point
2
p0
p
V
+
=
+0
ρair
2
ρair But from hydrostatics p0 − p = SG⋅ ρ⋅ g⋅ Δh p0 − p = Δh = so
2 Δh = 0.002377⋅ For x momentum ( 1
2
⋅ ρair⋅ V
2 1
2
⋅ρ ⋅V
2 air
SG⋅ ρ⋅ g 3 2 = 2⋅ SG⋅ ρ⋅ g 2 slug ⎛
ft ⎞
1
ft
s
× ⎜ 225⋅ ⎟ ×
×
×
3⎝
s⎠
2⋅ 1.75 1.94⋅ slug 32.2⋅ ft
ft ) ρair⋅ V Δh = 0.55⋅ ft 2
2 π⋅ D Rx = V⋅ −ρair⋅ A⋅ V = −ρair⋅ V ⋅ 4
2 ⎛ 0.5 ⋅ ft ⎞
2 π⋅ ⎜
⎟
2
slug ⎛
ft ⎞
12 ⎠
lbf⋅ s
Rx = −0.002377⋅
× ⎜ 225⋅ ⎟ × ⎝
×
3⎝
slugft
⋅
s⎠
4
ft
The force of the jet on the plate is then F = −Rx Rx = −0.164⋅ lbf
F = 0.164⋅ lbf Δh = 6.6⋅ in Problem *4.107 [2] CS
y x V, A Given: Water jet striking surface Find: Rx Force on surface Solution:
Basic equations: Momentum flux in x direction Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow
Hence 2 Q⎞
ρ⋅ Q
2
Rx = u1⋅ −ρ⋅ u1⋅ A1 = −ρ⋅ V ⋅ A = −ρ⋅ ⎛ ⎟ ⋅ A = −
⎜
A
A⎠
⎝ ( ) The force of the jet on the surface is then F = −R x = For a fixed flow rate Q, the force of a jet varies as 4⋅ ρ⋅ Q
π⋅ D 2 =− 4⋅ ρ⋅ Q 2 2 where Q is the flow rate π⋅ D 2 2 1 : A smaller diameter leads to a larger force. This is because as
2
D
the diameter decreases the speed increases, and the impact force varies as the square of the speed, but linearly with area
For a force of F = 650 N
2 Q= π⋅ D ⋅ F
4⋅ ρ 2 Q= 3 1⋅ L
π⎛6
m
kg⋅ m
60⋅ s
⎞
×⎜
⋅ m ⎟ × 650⋅ N ×
×
×
×
2
− 3 3 1⋅ min
1000⋅ kg
4 ⎝ 1000 ⎠
s ⋅ N 10 ⋅ m Q = 257⋅ L
min Problem *4.108 [3] Problem *4.109 [3] CS Given: Water jet striking disk Find: Expression for speed of jet as function of height; Height for stationary disk Solution:
Basic equations: Bernoulli; Momentum flux in z direction
2 pV
+
+ g⋅ z = constant
ρ
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow
2 The Bernoulli equation becomes V0
2 2 V
+ g⋅ h
2 + g⋅ 0 = ( 2 2 V = V0 − 2⋅ g⋅ h ) 2 V= V0 − 2⋅ g⋅ h 2 Hence −M⋅ g = w1⋅ −ρ⋅ w1⋅ A1 = −ρ⋅ V ⋅ A But from continuity ρ ⋅ V 0⋅ A 0 = ρ ⋅ V ⋅ A Hence we get M⋅ g = ρ⋅ V⋅ V⋅ A = ρ⋅ V0⋅ A0⋅ V0 − 2⋅ g⋅ h Solving for h 2⎤
1 ⎡ 2 ⎛ M⋅ g ⎞ ⎥
⎢V −
h=
⋅
⎜ ρ⋅ V ⋅ A ⎟ ⎥
2⋅ g ⎢ 0
00 so V⋅ A = V0⋅ A0 2 ⎣ ⎝ ⎠⎦ ⎡ ⎡
s
m⎞
9.81⋅ m
m
s
h= ×
× ⎢⎛ 10⋅ ⎟ − ⎢2⋅ kg ×
×
×
×
⎜
2
1000⋅ kg 10⋅ m
2 9.81⋅ m ⎢⎝
s⎠
⎢
s
1 2 ⎢
⎣ h = 4.28 m 2 3 ⎢
⎣ ⎤
⎥
2⎥
25
⎛
⎞
⋅ m⎟ ⎥
π⋅ ⎜
⎝ 1000 ⎠ ⎦
4 2⎤ ⎥
⎥
⎥
⎦ Problem *4.96 Problem *4.110 [4] Part 1/2 Problem *4.96 cont'd Problem *4.110 [4] Part 2/2 Problem *4.95 Problem *4.111 [3] Problem *4.112 Given: Data on flow and venturi geometry Find: [2] Force on convergent section Solution:
The given data is ρ = 999⋅ kg D = 0.1⋅ m 3 d = 0.04⋅ m p1 = 600⋅ kPa V 1 = 5⋅ m
A1 = Then π⋅ D
4 2 A2 = π2
⋅d
4 A2 = 0.00126 m V2 = 2 A1 = 0.00785 m Q
A2 V2 = 31.3 3 Q = V1⋅ A1 Q = 0.0393 m
s m
s 2 m
s Governing equations:
2 Bernoulli equation pV
+
+ g⋅ z = const
ρ
2 (4.24) Momentum (4.18a)
p1 Applying Bernoulli between inlet and throat ρ
Solving for p2 2 + ρ
2
2
p2 = p1 + ⋅ ⎛ V1 − V2 ⎞
⎠
2⎝ V1
2 p2 = ρ 2 + V2
2 p2 = 600⋅ kPa + 999⋅ kg
3 (2 ) 2
2
N⋅ s
kN
2m
×
×
2
kg⋅ m 1000⋅ N × 5 − 31.3 ⋅ p2 = 125⋅ kPa s m Applying the horizontal component of momentum ( ) ( ) −F + p1⋅ A2 − p2⋅ A2 = V1⋅ −ρ⋅ V1⋅ A1 + V2⋅ ρ⋅ V2⋅ A2 F = 600⋅ kN
2 m 2 × 0.00785⋅ m − 125⋅ kN
2 m 2 × 0.00126⋅ m + 999⋅ F = p1⋅ A1 − p2⋅ A2 + ρ⋅ ⎛ V1 ⋅ A1 − V2 ⋅ A2⎞
⎝
⎠
2 or kg
3 m ⎡ × ⎢⎛ 5⋅
⎜ 2 ⎛
⎟ ⋅ 0.00785⋅ m − ⎜ 31.3⋅ m⎞ ⎣⎝ s ⎠ 2 ⎝ m⎞ 2 2⎤ 2 N⋅ s
kg × m
⎦ ⎟ ⋅ 0.00126⋅ m ⎥ ⋅ s⎠ 2 F = 3.52⋅ kN Problem *4.98 Problem *4.113 [4] Problem *4.99 Problem *4.114 [4] Problem *4.101 Problem *4.115 [4] Problem *4.100 Problem *4.116 [4] Problem *4.102 Problem *4.117 [4] Problem *4.118 [4] Part 1/2 Problem *4.118 [4] Part 2/2 Problem *4.105 Problem *4.119 [5] Problem *4.104 Problem *4.120 [5] Part 1/2 Problem *4.104 cont'd Problem *4.120 [5] Part 2/2 Problem *4.121 [4] Part 1/2 Problem *4.121 [4] Part 2/2 Problem *4.122 [3] CS (moves
at speed U) y
x Rx
Ry Given: Water jet striking moving vane Find: Force needed to hold vane to speed U = 5 m/s Solution:
Basic equations: Momentum flux in x and y directions Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Jet relative velocity is constant
Then ( ) ( ) Rx = u1⋅ −ρ⋅ V1⋅ A1 + u2⋅ ρ⋅ V2⋅ A2 = −( V − U) ⋅ [ ρ⋅ ( V − U) ⋅ A] + ( V − U) ⋅ cos ( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A]
2 Rx = ρ ( V − U) ⋅ A⋅ ( cos ( θ) − 1) A= π ⎛ 40
⎞
⋅⎜
⋅ m⎟
4 ⎝ 1000 ⎠ 2 A = 1.26 × 10 −3 2 m Using given data
Rx = 1000⋅
Then 2 2 m
N⋅ s
−3 2
× ⎡( 25 − 5) ⋅ ⎤ × 1.26 × 10 ⋅ m × ( cos ( 150⋅ deg) − 1) ×
⎢
⎥
3⎣
kg⋅ m
s⎦
m
kg ( ) ( Rx = −940 N ) Ry = v1⋅ −ρ⋅ V1⋅ A1 + v2⋅ ρ⋅ V2⋅ A2 = −0 + ( V − U) ⋅ sin ( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A]
2 2 m
N⋅ s
−3 2
Ry = ρ ( V − U) ⋅ A⋅ sin ( θ) Ry = 1000⋅
× ⎡( 25 − 5) ⋅ ⎤ × 1.26 × 10 ⋅ m × sin ( 150⋅ deg) ×
⎢
⎥
3⎣
kg⋅ m
s⎦
m
2 kg Hence the force required is 940 N to the left and 252 N upwards to maintain motion at 5 m/s Ry = 252 N Problem 4.123 [3] CS (moves
at speed U) y
Ry Given: x Water jet striking moving vane Find: Rx Force needed to hold vane to speed U = 10 m/s Solution:
Basic equations: Momentum flux in x and y directions Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Jet relative velocity is constant
Then ( ) ( ) Rx = u1⋅ −ρ⋅ V1⋅ A1 + u2⋅ ρ⋅ V2⋅ A2 = −( V − U) ⋅ [ ρ⋅ ( V − U) ⋅ A] + ( V − U) ⋅ cos ( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A]
2 Rx = ρ ( V − U) ⋅ A⋅ ( cos ( θ) − 1)
Using given data
Rx = 1000⋅
Then 2 2 m⎤
N⋅ s
2
× ⎡( 30 − 10) ⋅ ⎥ × 0.004⋅ m × ( cos ( 120⋅ deg) − 1) ×
⎢
3⎣
kg⋅ m
s⎦
m
kg ( ) ( Rx = −2400 N ) Ry = v1⋅ −ρ⋅ V1⋅ A1 + v2⋅ ρ⋅ V2⋅ A2 = −0 + ( V − U) ⋅ sin ( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A]
2 2 m⎤
N⋅ s
2
Ry = ρ ( V − U) ⋅ A⋅ sin ( θ) Ry = 1000⋅
× ⎡( 30 − 10) ⋅ ⎥ × 0.004⋅ m × sin ( 120⋅ deg) ×
⎢
3⎣
kg⋅ m
s⎦
m
2 kg Hence the force required is 2400 N to the left and 1390 N upwards to maintain motion at 10 m/s Ry = 1386 N Problem 4.124 [2] Problem 4.125 [2] Given: Data on jet boat Find: Formula for boat speed; jet speed to double boat speed Solution:
CV in boat coordinates Governing equation:
Momentum (4.26) Applying the horizontal component of momentum
Fdrag = V⋅ ( −ρ⋅ Q) + Vj⋅ ( ρ⋅ Q) Fdrag = k⋅ V or, with 2 2 k⋅ V = ρ⋅ Q⋅ Vj − ρ⋅ Q⋅ V 2 k⋅ V + ρ⋅ Q⋅ V − ρ⋅ Q⋅ Vj = 0 Solving for V V=− Let α= 2
⎛ ρ⋅ Q ⎞ + ρ⋅ Q ⋅ V j
⎜
⎟
k
⎝ 2⋅ k ⎠ ρ⋅ Q
+
2⋅ k ρ⋅ Q
2⋅ k
2 V = −α + α + 2⋅ α⋅ Vj
V = 10⋅ We can use given data at V = 10 m/s to find α
10⋅ m
s = −α + 10 Hence V=− For V = 20 m/s 20 = − 3
10
3 2 α + 2⋅ 25⋅ + 100 + 100 9 9 m
s + 20
⋅V
3j Vj = 25⋅ 2 2 m
s
2 10 m
⋅
3s α + 50⋅ α = ( 10 + α) = 100 + 20⋅ α + α α= 100 20
70
+
⋅V =
9
3j
3 Vj = 80⋅ 20
⋅V
3j + ⋅α m
s m
s Problem 4.110 Problem 4.126 [2] Problem 4.112 Problem 4.127 [2] Problem 4.128 [3] CS (moves
at speed U) y
Rx Ry Given: Water jet striking moving vane Find: x Expressions for force and power; Show that maximum power is when U = V/3 Solution:
Basic equation: Momentum flux in x direction
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Jet relative velocity is constant
Then ( ) ( ) Rx = u1⋅ −ρ⋅ V1⋅ A1 + u2⋅ ρ⋅ V2⋅ A2 = −( V − U) ⋅ [ ρ⋅ ( V − U) ⋅ A] + ( V − U) ⋅ cos ( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A]
2 Rx = ρ ( V − U) ⋅ A⋅ ( cos ( θ) − 1)
2 This is force on vane; Force exerted by vane is equal and opposite Fx = ρ⋅ ( V − U) ⋅ A⋅ ( 1 − cos ( θ) ) The power produced is then P = U⋅ Fx = ρ⋅ U⋅ ( V − U) ⋅ A⋅ ( 1 − cos ( θ) ) 2 To maximize power wrt to U dP
2
= ρ⋅ ( V − U) ⋅ A⋅ ( 1 − cos ( θ) ) + ρ⋅ ( 2) ⋅ ( −1) ⋅ ( V − U) ⋅ U⋅ A⋅ ( 1 − cos ( θ) ) = 0
dU Hence V − U − 2⋅ U = V − 3⋅ U = 0 Note that there is a vertical force, but it generates no power U= V
3 for maximum power Problem 4.114 Problem 4.129 [3] Problem 4.130 [3] CS (moves to
left at speed Vc) V j + Vc V j + Vc
R
y Rx
x
t Given: Water jet striking moving cone Find: Thickness of jet sheet; Force needed to move cone Solution:
Basic equations: Mass conservation; Momentum flux in x direction Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Jet relative velocity is constant
2 Then ( − ρ ⋅ V 1⋅ A 1 + ρ ⋅ V 2⋅ A 2 = 0 ) − ρ⋅ V j + V c ⋅ π⋅ D j
4 ( ) + ρ ⋅ V j + V c ⋅ 2⋅ π ⋅ R ⋅ t = 0 (Refer to sketch) 2 Hence t= Dj t= 8⋅ R 1
1
2
× ( 4⋅ in) ×
8
9⋅ in t = 0.222 in Using relative velocities, x momentum is ( ) ( ) ( )( ) ( ) ( ) Rx = u1⋅ −ρ⋅ V1⋅ A1 + u2⋅ ρ⋅ V2⋅ A2 = − Vj + Vc ⋅ ⎡ρ⋅ Vj + Vc ⋅ Aj⎤ + Vj + Vc ⋅ cos ( θ) ⋅ ⎡ρ⋅ Vj + Vc ⋅ Aj⎤
⎣
⎦
⎣
⎦ ( ) 2 Rx = ρ Vj + Vc ⋅ Aj⋅ ( cos ( θ) − 1)
Using given data
2 ⎛4 ⎞
2 π⋅ ⎜ ⋅ ft⎟
2
slug ⎡
ft
12 ⎠
lbf ⋅ s
Rx = 1.94⋅
× ⎢( 100 + 45) ⋅ ⎤ × ⎝
× ( cos ( 60⋅ deg) − 1) ×
⎥
3
slug⋅ ft
s⎦
4
⎣
ft
Hence the force is 1780 lbf to the left; the upwards equals the weight Rx = −1780⋅ lbf Problem 4.116 Problem 4.131 [3] Problem 4.117 Problem 4.132 [3] Problem 4.133 [2] Problem 4.119 Problem 4.134 [3] Problem 4.120 Problem 4.133 Problem 4.135 [2] Problem 4.136 [2] Problem 4.137 [2] Problem 4.138 [4] Given: Data on vane/slider Find: Formula for acceleration, speed, and position; plot Solution:
The given data is ρ = 999⋅ kg 2 M = 30⋅ kg 3 A = 0.005⋅ m V = 20⋅ m m
s 2 The equation of motion, from Problem 4.136, is dU
ρ⋅ ( V − U ) ⋅ A
=
− g⋅ μk
dt
M 2 The acceleration is thus a= ρ⋅ ( V − U ) ⋅ A
− g⋅ μk
M
dU Separating variables = dt 2 ρ⋅ ( V − U ) ⋅ A
− g⋅ μk
M
Substitute u = V−U dU = −du du
2 = −dt ρ⋅ A ⋅ u
− g⋅ μk
M
⌠
M
⎛ ρ⋅ A ⎞
1
⎮
du = −
⋅ atanh ⎜
⋅ u⎟
⎮⎛
2
g⋅ μk⋅ ρ⋅ A
⎞
ρ⋅ A ⋅ u
⎝ g⋅ μk⋅ M ⎠
⎮⎜
− g⋅ μk⎟
⎮⎝M
⎠
⌡ and u = V  U so − ⎤
M
⎛ ρ⋅ A ⎞
⎡ ρ⋅ A
M
⋅ atanh ⎜
⋅ u⎟ = −
⋅ atanh ⎢
⋅ ( V − U)⎥
g⋅ μk⋅ ρ⋅ A
g⋅ μk⋅ ρ⋅ A
⎝ g⋅ μk⋅ M ⎠
⎣ g⋅ μk⋅ M
⎦ Using initial conditions − ⎤
M
⎡ ρ⋅ A
⋅ atanh ⎢
⋅ ( V − U)⎥ +
g⋅ μk⋅ ρ⋅ A
⎣ g⋅ μk⋅ M
⎦ V−U = U = V− g⋅ μk⋅ M
ρ⋅ A g⋅ μk⋅ M
ρ⋅ A M
g⋅ μk⋅ ρ⋅ A ⎛ ⋅ atanh ⎜ ⎞ ⋅ V⎟ = − t ⎝ g⋅ μk⋅ M ⎠ ⎛ g⋅ μk⋅ ρ⋅ A
⎞⎞
⎛ ρ⋅ A
⋅ t + atanh ⎜
⋅ V⎟ ⎟
⎜
⎟
M
⎝
⎝ g⋅ μk⋅ M ⎠ ⎠ ⋅ tanh⎜ ⎛ g⋅ μk⋅ ρ⋅ A
⎞⎞
⎛ ρ⋅ A
⋅ t + atanh ⎜
⋅ V⎟ ⎟
⎜
⎟
M
⎝
⎝ g⋅ μk⋅ M ⎠ ⎠ ⋅ tanh⎜ ρ⋅ A μk = 0.3 Note that ⎛ atanh ⎜ ⎝ ⎞
ρ⋅ A
π
⋅ V⎟ = 0.213 − ⋅ i
g⋅ μk⋅ M
2
⎠ which is complex and difficult to handle in Excel, so we use the identity so and finally the identity tanh⎜ x − g⋅ μk⋅ M U = V− ⎛
⎝ ρ⋅ A ⎛ 1⎞ π
atanh ( x) = atanh ⎜ ⎟ − ⋅ i
⎝ x⎠ 2 for x > 1 ⎞
⎛ g⋅ μk⋅ ρ⋅ A
1
⎛
⎞ − π ⋅ i⎟
⋅ t + atanh ⎜
⎟
M
2⎟
⎜
ρ⋅ A
⎜
⋅V ⎟
⎜
⎟
⎜ g⋅ μk⋅ M ⎟
⎝
⎝
⎠
⎠ ⋅ tanh⎜ 1
π⎞
⋅ i⎟ =
tanh( x)
2⎠
g⋅ μk⋅ M to obtain ρ⋅ A U = V− ⎛ g⋅ μk⋅ ρ⋅ A
⎛ g⋅ μk⋅ M 1 ⎞ ⎞
⋅ t + atanh ⎜
⋅ ⎟⎟
M
⎝
⎝ ρ⋅ A V ⎠ ⎠ tanh⎜ g⋅ μk⋅ M
For the position x dx
= V−
dt ρ⋅ A ⎛ g⋅ μk⋅ ρ⋅ A
⎛ g⋅ μk⋅ M 1 ⎞ ⎞
⋅ t + atanh ⎜
⋅ ⎟⎟
M
⎝
⎝ ρ⋅ A V ⎠ ⎠ tanh⎜ This can be solved analytically, but is quite messy. Instead, in the corresponding Excel workbook, it is solved numerically
using a simple Euler method. The complete set of equations is
g⋅ μk⋅ M
U = V− ρ⋅ A ⎛ g⋅ μk⋅ ρ⋅ A
⎛ g⋅ μk⋅ M 1 ⎞ ⎞
⋅ t + atanh ⎜
⋅ ⎟⎟
M
⎝
⎝ ρ⋅ A V ⎠ ⎠ tanh⎜ 2 ρ⋅ ( V − U ) ⋅ A
a=
− g⋅ μk
M
g⋅ μk⋅ M
⎞
⎛
⎜
⎟
ρ⋅ A
⎜V −
⎟ ⋅ Δt
x ( n + 1) = x ( n) +
⎜
⎛ g⋅ μk⋅ ρ⋅ A
⎛ g⋅ μk⋅ M 1 ⎞ ⎞ ⎟
⎜
tanh⎜
⋅ t + atanh ⎜
⋅ ⎟⎟ ⎟
M
⎝
⎝
⎝ ρ⋅ A V ⎠ ⎠ ⎠
The plots are presented in the Excel workbook The equations are ρ = 999 kg/m3
μk = 0.3 0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0 0.0
0.0
0.5
1.2
2.2
3.3
4.4
5.7
7.0
8.4
9.7
11.2
12.6
14.1
15.5
17.0
18.5
20.1
21.6
23.1
24.7
26.2
27.8
29.3
30.9
32.4
34.0
35.6
37.1
38.7
40.3 0.0
4.8
7.6
9.5
10.8
11.8
12.5
13.1
13.5
13.9
14.2
14.4
14.6
14.8
14.9
15.1
15.2
15.3
15.3
15.4
15.4
15.5
15.5
15.6
15.6
15.6
15.6
15.7
15.7
15.7
15.7 63.7
35.7
22.6
15.5
11.2
8.4
6.4
5.1
4.0
3.3
2.7
2.2
1.9
1.6
1.3
1.1
0.9
0.8
0.7
0.6
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.2
0.2
0.1
0.1 x (m) 0.0 0.5 1.0 1.5 2.0 2.5 3.0 2.0 2.5 3.0 t (s) Velocity U vs Time
18
16
14
12
10
8
6
4
2
0 U (m/s) t (s) x (m) U (m/s) a (m/s2) 45
40
35
30
25
20
15
10
5
0 0.0 0.5 1.0 1.5
t (s) 70 Acceleration a vs Time 60
2 = 0.005 m
= 20 m/s
= 30 kg
= 0.1 s a (m/s ) A
V
M
Δt Position x vs Time 2 50
40
30
20
10
0
0.0 0.5 1.0 1.5 t (s) 2.0 2.5 3.0 Problem 1.24 Problem 4.133 Problem 4.139 [3] Problem 4.140 [4] CS (moves at
speed
instantaneous
speed U)
y
x Given: Water jet striking moving vane/cart assembly Find: Angle θ at t = 5 s; Plot θ(t) Solution:
Basic equation: Momentum flux in x direction for accelerating CV Assumptions: 1) cahnges in CV 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Constant jet relative velocity
Then ( ) ( ) −M⋅ arfx = u1⋅ −ρ⋅ V1⋅ A1 + u2⋅ ρ⋅ V2⋅ A2 = −( V − U) ⋅ [ ρ⋅ ( V − U) ⋅ A] + ( V − U) ⋅ cos ( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A]
2 −M⋅ arfx = ρ ( V − U) ⋅ A⋅ ( cos ( θ) − 1)
Since or arfx = constant U = arfx⋅ t ⎡ θ = acos ⎢1 − ⎢
⎣ then M⋅ arfx cos ( θ) = 1 − cos ( θ) = 1 − 2 ρ⋅ ( V − U ) ⋅ A
M⋅ arfx ( ) 2 ρ⋅ V − arfx⋅ t ⋅ A ⎤
⎥
2
ρ⋅ (V − arfx⋅ t) ⋅ A⎥
⎦
M⋅ arfx Using given data ⎡ θ = acos ⎢1 − 55⋅ kg × 1.5⋅ ⎢
⎢
⎣ m
2 s 3 × m
×
1000⋅ kg 1 ⎛ 15⋅ m − 1.5⋅ m × 5⋅ s⎞
⎜s
⎟
2
s
⎝ ⎠ 2 × ⎤
⎥
2
0.025⋅ m ⎥
⎥
⎦
1 θ = 19.7 deg at t = 5 s 20 135 15 90 10 45 5 0 0 2.5 5 7.5 Time t (s) The solution is only valid for θ up to 180o (when t = 9.14 s). This graph can be plotted in Excel 0
10 Speed U (m/s) Angle (deg) 180 Problem 4.126 Problem 4.141 [3] Problem 4.142 [3] Part 1/2 Problem 4.142 [3] Part 2/2 Problem 4.143 [3] Problem 4.144 [3] Part 1/2 Problem 4.144 [3] Part 2/2 Problem 4.130 Problem 4.145 [3] Problem 4.146 [4] Part 1/3 Problem 4.146 [4] Part 2/3 Problem 4.146 [4] Part 3/3 Problem 4.132 Problem 4.147 [3] Problem 4.148 [4] Given: Data on vane/slider Find: Formula for acceleration, speed, and position; plot Solution:
The given data is ρ = 999⋅ kg M = 30⋅ kg 3 A = 0.005⋅ m m 2 V = 20⋅ m
s k = 7.5⋅ 2 The equation of motion, from Problem 4.147, is dU
ρ⋅ ( V − U) ⋅ A k ⋅ U
=
−
dt
M
M 2 The acceleration is thus a = ρ⋅ ( V − U) ⋅ A k ⋅ U
−
M
M The differential equation for U can be solved analytically, but is quite messy. Instead we use a simple numerical method Euler's method ⎡ ρ⋅ ( V − U) 2⋅ A U( n + 1) = U( n ) + ⎢ ⎣ For the position x k ⋅ U⎤
⎥ ⋅ Δt
M⎦ dx
=U
dt so M − x ( n + 1) = x ( n) + U⋅ Δt The final set of equations is ⎡ ρ⋅ ( V − U) 2⋅ A k⋅ U⎤
⎥ ⋅ Δt
U ( n + 1) = U ( n) + ⎢
−
M
M⎦
⎣
2 a= ρ⋅ ( V − U) ⋅ A k⋅ U
−
M
M x ( n + 1) = x ( n) + U⋅ Δt
The results are plotted in the corresponding Excel workbook where Δt is the time step N⋅ s
m kg/m3
N.s/m
m2
m/s
kg
s Position x vs Time x (m) ρ = 999
k=
7.5
A = 0.005
V=
20
M=
30
Δt =
0.1 U (m/s) a (m/s2) 0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0 0.0
0.0
0.7
1.6
2.7
3.9
5.2
6.6
7.9
9.3
10.8
12.2
13.7
15.2
16.6
18.1
19.6
21.1
22.6
24.1
25.7
27.2
28.7
30.2
31.7
33.2
34.8
36.3
37.8
39.3
40.8 0.0
6.7
9.5
11.1
12.1
12.9
13.4
13.8
14.1
14.3
14.5
14.6
14.7
14.8
14.9
15.0
15.0
15.1
15.1
15.1
15.1
15.1
15.2
15.2
15.2
15.2
15.2
15.2
15.2
15.2
15.2 66.6
28.0
16.1
10.5
7.30
5.29
3.95
3.01
2.32
1.82
1.43
1.14
0.907
0.727
0.585
0.472
0.381
0.309
0.250
0.203
0.165
0.134
0.109
0.0889
0.0724
0.0590
0.0481
0.0392
0.0319
0.0260
0.0212 0.5 1.0 1.5 2.0 2.5 3.0 2.5 3.0 t (s) Velocity U vs Time
16
14
U (m/s) x (m) 12
10
8
6
4
2
0
0.0 0.5 1.0 1.5 2.0 t (s) 70 Acceleration a vs Time 60
2
a (m/s ) t (s) 45
40
35
30
25
20
15
10
5
0
5 0.0 50
40
30
20
10
0
0 1 1 2
t (s) 2 3 3 Problem 4.134 Problem 4.149 [3] Problem 4.136 Problem 4.150 [3] Problem 4.151 [3] Given: Data on system Find: Jet speed to stop cart after 1 s; plot speed & position; maximum x; time to return to origin Solution:
The given data is ρ = 999⋅ kg 2 M = 100⋅ kg 3 A = 0.01⋅ m U0 = 5⋅ m 2 dU
ρ⋅ ( V + U ) ⋅ A
=−
dt
M The equation of motion, from Problem 4.149, is which leads to d ( V + U)
( V + U) 2 m
s ρ⋅ A ⎞
= −⎛
⋅ dt⎟
⎜
⎝M ⎠
V + U0 U = −V + Integrating and using the IC U = U0 at t = 0 1+ ( ) ⋅t ρ⋅ A ⋅ V + U 0
M To find the jet speed V to stop the cart after 1 s, solve the above equation for V, with U = 0 and t = 1 s. (The
equation becomes a quadratic in V). Instead we use Excel's Goal Seek in the associated workbook
From Excel V = 5⋅ m
s
dx
= U = −V +
dt For the position x we need to integrate The result is x = −V ⋅ t + ⎡ M
ρ⋅ A ⋅ ln ⎢1 + ⎣ ( V + U0
1+ ( ) ⋅t ρ⋅ A ⋅ V + U 0
M ) ρ⋅ A ⋅ V + U 0 ⎤
⋅ t⎥
M
⎦ This equation (or the one for U with U = 0) can be easily used to find the maximum value of x by differentiating, as well as the
time for x to be zero again. Instead we use Excel's Goal Seek and Solver in the associated workbook
From Excel xmax = 1.93⋅ m t ( x = 0) = 2.51⋅ s The complete set of equations is
V + U0 U = −V +
1+ ( ) ⋅t ρ⋅ A ⋅ V + U 0 The plots are presented in the Excel workbook M x = −V ⋅ t + M
ρ⋅ A ⎡ ⋅ ln ⎢1 + ⎣ ( ) ρ⋅ A ⋅ V + U 0 ⎤
⋅ t⎥
M
⎦ M= 100 ρ= 999 kg
kg/m3 A=
Uo = 0.01
5 m2
m/s t (s ) x ( m) U (m/s) 0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0 0.00
0.82
1.36
1.70
1.88
1.93
1.88
1.75
1.56
1.30
0.99
0.63
0.24
0.19
0.65
1.14 5.00
3.33
2.14
1.25
0.56
0.00
0.45
0.83
1.15
1.43
1.67
1.88
2.06
2.22
2.37
2.50 To find V for U = 0 in 1 s, use Goal Seek
t (s ) U (m/s) V (m/s) 1.0 0.00 5.00 To find the maximum x , use Solver
t (s ) x (m ) 1.0 1.93 To find the time at which x = 0 use Goal Seek
t (s ) x (m ) 2.51 0.00 Cart Position x vs Time
2.5
2.0 x (m) 1.5
1.0
0.5
0.0
0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 2.5 3.0 1.0
1.5 t (s) Cart Speed U vs Time
6 U (m/s) 5
4
3
2
1
0
1 0.0 0.5 1.0 1.5 2
3 t (s) 2.0 Problem 4.137 Problem 4.152 [3] Problem *4.153 [3] CS moving
at speed U Given: Water jet striking moving disk Find: Acceleration of disk when at a height of 3 m Solution:
Basic equations: Bernoulli; Momentum flux in z direction (treated as upwards) for linear accelerating CV
2 pV
+
+ g⋅ z = constant
ρ
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow in jet)
(All
2 The Bernoulli equation becomes V0
2 2 + g⋅ 0 = V1 ( + g⋅ z − z0 2 ) V1 = 2 ⎛ 15⋅ m ⎞ + 2 × 9.81⋅ m ⋅ ( 0 − 3) ⋅ m
⎜
⎟
2
⎝ s⎠
s V1 = 2 ( ) V0 + 2⋅ g⋅ z0 − z V1 = 12.9 m
s The momentum equation becomes ( ) ( )( ) ( ) −W − M⋅ arfz = w1⋅ −ρ⋅ V1⋅ A1 + w2⋅ ρ⋅ V2⋅ A2 = V1 − U ⋅ ⎡−ρ⋅ V1 − U ⋅ A1⎤ + 0
⎣
⎦ Hence V0
2
ρ⋅ V1 − U ⋅ A0⋅
ρ⋅ V 1 − U ⋅ A 1 − W
ρ⋅ V 1 − U ⋅ A 1
V1
arfz =
=
−g =
−g
M
M
M ( ) 2 ( 2 ) 2 ( ) kg ⎡
m
15
1
m
2
arfz = 1000⋅
× ⎢( 12.9 − 5) ⋅ ⎤ × 0.005⋅ m ×
×
− 9.81⋅
⎥
3⎣
2
s⎦
12.9 30⋅ kg
m
s using m
arfz = 2.28
2
s V1⋅ A1 = V0⋅ A0 Problem *4.154 [4] M = 35 kg CS moving
at speed U
D = 75 mm Given: Water jet striking disk Find: Plot mass versus flow rate to find flow rate for a steady height of 3 m Solution:
Basic equations: Bernoulli; Momentum flux in z direction (treated as upwards)
2 pV
+
+ g⋅ z = constant
ρ
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow in jet)
(All
2 V0 The Bernoulli equation becomes 2 2 + g⋅ 0 = V1
2 + g⋅ h 2 V1 = V0 − 2⋅ g⋅ h The momentum equation becomes ( ) ( ) ( ) −M⋅ g = w1⋅ −ρ⋅ V1⋅ A1 + w2⋅ ρ⋅ V2⋅ A2 = V1⋅ −ρ⋅ V1⋅ A1 + 0
2 ρ⋅ V 1 ⋅ A 1
M=
g Hence M= ρ ⋅ V 1⋅ V 0⋅ A 0
g V1⋅ A1 = V0⋅ A0 but from continuity
2 π ρ⋅ V0⋅ D0
2
=⋅
⋅ V0 − 2⋅ g⋅ h
g
4 and also Q = V0⋅ A0 This equation is difficult to solve for V0 for a given M. Instead we plot first:
100 M (kg) 80
60
40
20
0.02 0.03 0.04 0.05 0.06 Q (cubic meter/s)
This graph can be parametrically plotted in Excel. The Goal Seek or Solver feature can be used to find Q when M = 35 kg
3 Q = 0.0469⋅ m
s Problem 4.155 [3] Problem 4.156 [3] Problem 4.142 Problem 4.157 [3] Part 1/2 Problem 4.142 cont'd Problem 4.157 [3] Part 2/2 Problem 4.158 [3] Part 1/2 Problem 4.158 [3] Part 2/2 Problem 4.159 [3] CS at speed U y
x Ve
Y
X
Given: Data on rocket sled Find: Minimum fuel to get to 265 m/s Solution:
Basic equation: Momentum flux in x direction
Assumptions: 1) No resistance 2) pe = patm 3) Uniform flow 4) Use relative velocities
From continuity dM
= mrate = constant
dt Hence from momentum −arfx⋅ M = − Separating variables dU = Integrating so ( M = M0 − mrate⋅ t ) ( (Note: Software cannot render a dot!) ) dU
⋅ M0 − mrate⋅ t = ue⋅ ρe⋅ Ve⋅ Ae = −Ve⋅ mrate
dt Ve⋅ mrate ⋅ dt
M0 − mrate⋅ t
M0
⎞
⎛
⎛ mrate⋅ t ⎞
U = Ve⋅ ln ⎜
⎟ = −Ve⋅ ln⎜ 1 −
⎟
M0
⎝ M0 − mrate⋅ t ⎠
⎝
⎠ The mass of fuel consumed is mf Hence mf U⎞
⎛
−
⎜
Ve ⎟
= mrate⋅ t = M0⋅ ⎝ 1 − e
⎠
265 ⎞
⎛
−
⎜
2750 ⎟
= 900⋅ kg × ⎝ 1 − e
⎠ ⎛ or − U ⎞ M0 ⎜
Ve ⎟
t=
⋅ ⎝1 − e
⎠
mrate mf = 82.7 kg Problem 4.160 [3] CS at speed U y
x Ve
Y
X
Given: Data on rocket weapon Find: Expression for speed of weapon; minimum fraction of mass that must be fuel Solution:
Basic equation: Momentum flux in x direction
Assumptions: 1) No resistance 2) pe = patm 3) Uniform flow 4) Use relative velocities 5) Constant mass flow rate
From continuity dM
= mrate = constant
dt M = M0 − mrate⋅ t so ( ) ( (Note: Software cannot render a dot!) ) dU
Hence from momentum −arfx⋅ M = − ⋅ M0 − mrate⋅ t = ue⋅ ρe⋅ Ve⋅ Ae = −Ve⋅ mrate
dt
Separating variables dU = Ve⋅ mrate
M0 − mrate⋅ t ⋅ dt Integrating from U = U0 at t = 0 to U = U at t = t (( ) ⎛ ( )) = −Ve⋅ ln⎜1 − U − U0 = −Ve⋅ ln M0 − mrate⋅ t − ln M0 ⎝ mrate⋅ t ⎞
⎟
M0 ⎠ ⎛ mrate⋅ t ⎞
U = U0 − Ve⋅ ln ⎜ 1
⎟
M0
⎝ Rearranging MassFractionConsumed = ⎠ mrate⋅ t
M0 − = 1−e ( U−U0)
Ve − = 1−e ( 3500−600)
6000 = 0.383 Hence 38.3% of the mass must be fuel to accomplish the task. In reality, a much higher percentage would be needed due to drag effects Problem 4.161 [3] Part 1/2 Problem 4.161 [3] Part 2/2 Problem 4.147 Problem 4.162 [3] Problem 4.163 [3] Part 1/2 Problem 4.163 [3] Part 2/2 Problem 4.148 Problem 4.164 [3] Problem 4.165 [3] CS at speed V y
x Y
Ve X Given: Data on rocket Find: Speed after 8 s; Plot of speed versus time Solution:
Basic equation: Momentum flux in y direction Assumptions: 1) No resistance 2) pe = patm 3) Uniform flow 4) Use relative velocities 5) Constant mass flow rate
From continuity dM
= mrate = constant
dt M = M0 − mrate⋅ t so ( (Note: Software cannot render a dot!) ) Hence from momentum −M⋅ g − arfy⋅ M = ue⋅ ρe⋅ Ve⋅ Ae = −Ve⋅ mrate Hence Separating variables Ve⋅ mrate
Ve⋅ mrate
dV
arfy =
=
−g =
−g
dt
M
M0 − mrate⋅ t (1) ⎞ ⎛ Ve⋅ mrate dV = ⎜ ⎝ M0 − mrate⋅ t − g⎟ ⋅ dt ⎠ Integrating from V = at t = 0 to V = V at t = t (( ) ⎛ mrate⋅ t ⎞ ⎝ ( )) M0
⎠ V = −Ve⋅ ln M0 − mrate⋅ t − ln M0 − g⋅ t = −Ve⋅ ln ⎜ 1 − ⎛ mrate⋅ t ⎞ ⎝ M0
⎠ V = −Ve⋅ ln ⎜ 1 − At t = 8 s V = −3000⋅ m
s ⎟ − g⋅ t ⋅ ln ⎛ 1 − 8⋅
⎜ ⎝ kg
s × ⎟ − g⋅ t (2) 1
300⋅ kg × 8⋅ s⎞ − 9.81⋅
⎟ ⎠ m
2 × 8⋅ s s The speed and acceleration as functions of time are plotted below. These are obtained
from Eqs 2 and 1, respectively, and can be plotted in Excel V = 641 m
s 5000 V (m/s) 4000
3000
2000
1000
0 10 20 30 20 30 Time (s) 400 a (m/s2) 300
200
100 0 10 Time (s) Problem 4.151 Problem 4.166 [3] Problem 4.167 [4] y
x
CS (moves
at speed U) Ry Ff Given: Water jet striking moving vane Find: Plot of terminal speed versus turning angle; angle to overcome static friction Solution:
Basic equations: Momentum flux in x and y directions Assumptions: 1) Incompressible flow 2) Atmospheric pressure in jet 3) Uniform flow 4) Jet relative velocity is constant
Then ( ) ( ) −Ff − M⋅ arfx = u1⋅ −ρ⋅ V1⋅ A1 + u2⋅ ρ⋅ V2⋅ A2 = −( V − U) ⋅ [ ρ⋅ ( V − U) ⋅ A] + ( V − U) ⋅ cos ( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A]
2 arfx =
Also ρ ( V − U) ⋅ A⋅ ( 1 − cos ( θ) ) − Ff (1) M ( ) Ry − M⋅ g = v1⋅ −ρ⋅ V1⋅ A1 + v2⋅ ρ⋅ V2⋅ A2 = 0 + ( V − U) ⋅ sin ( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A]
2 Ry = M⋅ g + ρ ( V − U) ⋅ A⋅ sin ( θ)
At terminal speed arfx = 0 and Ff = μkRy . Hence in Eq 1 0= or ρ⋅ V − Ut ⋅ A⋅ ( 1 − cos ( θ) ) − μk⋅ ⎡M⋅ g + ρ⋅ V − Ut ⋅ A⋅ sin ( θ)⎤ ( V − Ut = ) 2 ( ⎣ ) 2 M ( μk⋅ M⋅ g ρ⋅ A⋅ 1 − cos ( θ) − μk⋅ sin ( θ) ) Ut = V − ⎦ = ρ⋅ (V − Ut) ⋅ A⋅ (1 − cos ( θ) − μk⋅ sin ( θ) ) − μ ⋅ g
k
M
2 ( μk⋅ M⋅ g ρ⋅ A⋅ 1 − cos ( θ) − μk⋅ sin ( θ) The terminal speed as a function of angle is plotted below; it can be generated in Excel ) Terminal Speed (m/s) 20
15
10
5 0 10 20 30 40 50 60 70 80 Angle (deg)
For the static case Ff = μs⋅ Ry and Substituting in Eq 1, with U = 0 or ( (the cart is about to move, but hasn't) ) ρ⋅ V ⋅ A⋅ ⎡1 − cos ( θ) − μs⋅ ρ⋅ V ⋅ A⋅ sin ( θ) + M⋅ g
2 0= arfx = 0 ⎣ cos ( θ) + μs⋅ sin ( θ) = 1 − 2 M
μs⋅ M⋅ g
2 ρ⋅ V ⋅ A We need to solve this for θ! This can be done by hand or by using Excel's Goal Seek or Solver θ = 19 deg Note that we need θ = 19o, but once started we can throttle back to about θ = 12.5o and still keep moving! 90 Problem 4.168 [4] Problem 4.169 [4] Problem 4.170 [4] Problem 4.171 [5] CS at speed V y
x Y
Ve X Given: Data on rocket Find: Maximum speed and height; Plot of speed and distance versus time Solution:
Basic equation: Momentum flux in y direction Assumptions: 1) No resistance 2) pe = patm 3) Uniform flow 4) Use relative velocities 5) Constant mass flow rate
From continuity dM
= mrate = constant
dt M = M0 − mrate⋅ t so ( (Note: Software cannot render a dot!) ) Hence from momentum −M⋅ g − arfy⋅ M = ue⋅ ρe⋅ Ve⋅ Ae = −Ve⋅ mrate
Hence Ve⋅ mrate
Ve⋅ mrate
dV
arfy =
=
−g =
−g
dt
M
M0 − mrate⋅ t Separating variables dV = ⎜ ⎞ ⎛ Ve⋅ mrate
⎝ M0 − mrate⋅ t − g⎟ ⋅ dt ⎠ Integrating from V = at t = 0 to V = V at t = t (( ) ⎛ mrate⋅ t ⎞ ⎝ ( )) M0
⎠ V = −Ve⋅ ln M0 − mrate⋅ t − ln M0 − g⋅ t = −Ve⋅ ln ⎜ 1 − ⎛ mrate⋅ t ⎞ ⎝ M0
⎠ V = −Ve⋅ ln ⎜ 1 − ⎟ − g⋅ t t ≤ tb for mf
To evaluate at tb = 1.7 s, we need Ve and mrate mrate =
tb mrate = Also note that the thrust Ft is due to
momentum flux from the rocket Ve = Hence Ft = mrate⋅ Ve ⎟ − g⋅ t 12.5⋅ gm
1.7⋅ s
Ft mrate (burn time) (1)
− 3 kg mrate = 7.35 × 10
Ve = s 5.75⋅ N
7.35 × 10 − 3 kg ⋅ ⎛ mrate⋅ tb ⎞
Vmax = −Ve⋅ ln ⎜ 1 −
⎟ − g⋅ tb
M0
⎝
⎠
Vmax = −782⋅ m
s ⋅ ln ⎛ 1 − 7.35 × 10
⎜ ⎝ − 3 kg ⋅ s × 1
0.0696⋅ kg × 1.7⋅ s⎞ − 9.81⋅
⎟ ⎠ m
2 s × 1.7⋅ s × kg⋅ m
2 s ⋅N Ve = 782 s Vmax = 138 m
s m
s To obtain Y(t) we set V = dY/dt in Eq 1, and integrate to find
Y= Ve⋅ M0 ⎡⎛
mrate⋅ t ⎞ ⎛ ⎛
mrate⋅ t ⎞
⎞⎤12
⋅ ⎢⎜ 1 −
⎟ ⋅ ⎜ ln⎜ 1 −
⎟ − 1⎟ + 1⎥ − ⋅ g⋅ t
mrate
M0
M0
⎣⎝
⎠⎝ ⎝
⎠⎠⎦2 t ≤ tb tb = 1.7⋅ s (2) m
s
⎡⎛ 0.00735⋅ 1.7 ⎞ ⎛ ln⎛ 1 − .00735⋅ 1.7 ⎞ − 1⎞ + 1⎤ ...
× 0.0696⋅ kg ×
⋅ ⎢⎜ 1 −
⎟⎜ ⎜
⎟⎟⎥
−3
s
0.0696 ⎠ ⎝ ⎝
.0696 ⎠
⎠⎦
7.35 × 10 ⋅ kg ⎣⎝
1
m
2
+ − × 9.81⋅ × ( 1.7⋅ s)
2
2
s
Yb = 113 m
Yb = 782⋅ At t = tb After burnout the rocket is in free assent. Ignoring drag ( V ( t) = Vmax − g⋅ t − tb ( ) (3) ) ( ) 1
2
Y ( t) = Yb + Vmax⋅ t − tb − ⋅ g⋅ t − tb
2 t > tb (4) The speed and position as functions of time are plotted below. These are obtained from Eqs 1 through 4, and can be plotted in Excel 150 V (m/s) 100
50 0 5 10 15 20 − 50 Time (s) Y (m) 1500 1000 500 0 5 10 15 20 Time (s)
Using Solver, or by differentiating y(t) and setting to zero, or by setting V(t) = 0, we find for the maximumt y 15.8 s
= ymax = 1085 m Problem 4.172 [4] Problem *4.173 [5] Part 1/3 Problem *4.173 [5] Part 2/3 Problem *4.173 [5] Part 3/3 Problem *4.174 [5] Part 1/2 Problem *4.174 [5] Part 2/2 Problem *4.175 [5] CS moving
at speed U Given: Water jet striking moving disk Find: Motion of disk; steady state height Solution:
Basic equations: Bernoulli; Momentum flux in z direction (treated as upwards) for linear accelerating CV
2 pV
+
+ g⋅ z = constant
ρ
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure 4) Uniform flow 5) velocities wrt CV
2 V0 The Bernoulli equation becomes 2 (All in jet) 2 + g⋅ 0 = V1 + g⋅ h 2 V1 = 2 V0 − 2⋅ g⋅ h 2 V1 = ⎛ 15⋅ m ⎞ + 2 × 9.81⋅ m ⋅ ( 0 − 3) ⋅ m
⎜
⎟
2
⎝ s⎠
s V1 = 12.9 (1) m
s The momentum equation becomes ( ) ( )( ) ( ) −M⋅ g − M⋅ arfz = w1⋅ −ρ⋅ V1⋅ A1 + w2⋅ ρ⋅ V2⋅ A2 = V1 − U ⋅ ⎡−ρ⋅ V1 − U ⋅ A1⎤ + 0
⎣
⎦
2 With arfz = dh U= and 2 dt dh
dt 2 −M⋅ g − M⋅ we get 2 dh
= − ρ⋅ ⎛ V 1 − ⎞ ⋅ A 1
⎜
⎟
2
dt ⎠
⎝
dt dh Using Eq 1, and from continuity V1⋅ A1 = V0⋅ A0
2
ρ ⋅ A 0⋅ V 0
dh
2
= ⎛ V0 − 2⋅ g⋅ h − ⎞ ⋅
−g
⎜
⎟
2
dt ⎠
2
⎝
dt
M⋅ V0 − 2⋅ g⋅ h
2 dh (2) This must be solved numerically! One approach is to use Euler's method (see the Excel solution)
At equilibrium h = h0 dh
=0
dt 2 dh
2 =0 so dt Hence 2 V0 ⎡
h0 =
⋅ ⎢1 − ⎛
⎜
2⋅ g ⎢ ⎜ ⎞⎤
⎟⎥
2
ρ⋅ V 0 ⋅ A 0 ⎟ ⎥
⎣⎝
⎠⎦
2⎤
⎡⎡
2
2
3
2
1
m
s
m
m
s⎞
1 ⎤⎥
⎢
⎥
h0 = × ⎛ 15⋅ ⎟ ×
× 1 − ⎢30⋅ kg × 9.81⋅ ×
×⎛
h0 = 10.7 m
⎜
⎜
⎟×
2 1000⋅ kg ⎝ 15⋅ m ⎠
2⎥ ⎥
9.81⋅ m ⎢ ⎢
2⎝
s⎠
s
.005⋅ m ⎦ ⎦
⎣⎣
⎛ V 2 − 2⋅ g⋅ h ⎞ ⋅ ρ⋅ A ⋅ V − M⋅ g = 0
0⎠
00
⎝0 and M⋅ g 2 Problem *4.175 (In Excel) Δt = 0.05 s
2
A 0 = 0.005 m
g=
V=
M=
ρ= 9.81
15
30
1000 h i +1 = h i + Δ t ⋅ m/s2
m/s
kg
kg/m3 dh
dt i d 2h
⎛ dh ⎞
⎛ dh ⎞
⎜ ⎟ = ⎜ ⎟ + Δt ⋅ 2
dt
⎝ dt ⎠ i +1 ⎝ dt ⎠ i h (m) dh/dt (m/s) d 2h/dt 2 (m/s2)
2.000
2.000
2.061
2.167
2.310
2.481
2.673
2.883
3.107
3.340
3.582
3.829
4.080
4.333
4.587
4.840
5.092
5.341
5.588
5.830
6.069
6.302
6.530
6.753
6.969
7.179
7.383
7.579
7.769
7.952
8.127
8.296
8.457
8.611
8.757
8.896
9.029
9.154
9.272
9.384
9.488 0.000
1.213
2.137
2.852
3.412
3.853
4.199
4.468
4.675
4.830
4.942
5.016
5.059
5.074
5.066
5.038
4.991
4.930
4.854
4.767
4.669
4.563
4.449
4.328
4.201
4.069
3.934
3.795
3.654
3.510
3.366
3.221
3.076
2.931
2.787
2.645
2.504
2.365
2.230
2.097
1.967 24.263
18.468
14.311
11.206
8.811
6.917
5.391
4.140
3.100
2.227
1.486
0.854
0.309
0.161
0.570
0.926
1.236
1.507
1.744
1.951
2.130
2.286
2.420
2.535
2.631
2.711
2.776
2.826
2.864
2.889
2.902
2.904
2.896
2.878
2.850
2.814
2.769
2.716
2.655
2.588
2.514 12 6 10 5 8 4
Position
Speed 6 3 4 2 2 1 0 0
0 1 2 3
Time t (s) 4 5 Speed (m/s) 0.000
0.050
0.100
0.150
0.200
0.250
0.300
0.350
0.400
0.450
0.500
0.550
0.600
0.650
0.700
0.750
0.800
0.850
0.900
0.950
1.000
1.050
1.100
1.150
1.200
1.250
1.300
1.350
1.400
1.450
1.500
1.550
1.600
1.650
1.700
1.750
1.800
1.850
1.900
1.950
2.000 i Position (m) t (s) [3] Problem 4.176 [5] Part 1/2 Problem 4.176 [5] Part 2/2 Problem *4.177 [5] Part 1/3 Problem 4.133 Problem *4.177 [5] Part 2/3 Problem *4.177 [5] Part 3/3 Problem *4.178 [5] Part 1/2 *4.179
*4.179 *4.179 Problem *4.178 [5] Part 2/2 Problem *4.179 4.137 [5] Part 1/4 Problem *4.179 [5] Part 2/4 Problem *4.179 [5] Part 3/4 Problem *4.179 [5] Part 4/4 Problem *4.180 [3] Part 1/2 Problem *4.180 [3] Part 2/2 Problem *4.165 Problem *4.181 [2]
Example 4.6 Problem *4.182 [3] Problem *4.168 Problem *4.183 [3] Problem *4.169 Problem *4.184 [3] Problem *4.170 Problem *4.185 [3] Problem *4.186 [3] Given: Data on rotating spray system Find: Torque required to hold stationary; steadystate speed Solution:
The given data is ρ = 999⋅ kg kg
mflow = 15⋅
s 3 m D = 0.015⋅ m ro = 0.25⋅ m ri = 0.05⋅ m δ = 0.005⋅ m Governing equation: Rotating CV For no rotation (ω = 0) this equation reduces to a single scalar equation
ro ro i ⌠ → ⎯⎯ ⎯⎯ →
→
→
Tshaft = ⎮ r × Vxyz⋅ ρ⋅ Vxyz dA
⎮
⌡ i ⌠
2⌠
2
2
2
Tshaft = 2⋅ δ⋅ ⎮ r⋅ V⋅ ρ⋅ V dr = 2⋅ ρ⋅ V ⋅ δ⋅ ⎮ r dr = ρ⋅ V ⋅ δ⋅ ⎛ ro − ri ⎞
⎝
⎠
⌡r
⌡r or mflow V= where V is the exit velocity with respect to the CV ( ρ ) 2⋅ δ⋅ ro − ri 2 Hence ⎡ mflow ⎤
⎢
⎥
ρ
2
2
Tshaft = ρ⋅ ⎢
⎥ ⋅ δ⋅ ⎛ ro − ri ⎞
⎝
⎠
2⋅ δ⋅ (ro − ri)
⎣
⎦
Tshaft = 1
4 × ⎛ 15⋅
⎜ ⎝ kg ⎞ 2 3 m ( 0.25 + 0.05) 1 ×
×
⎟×
999⋅ kg 0.005⋅ m ( 0.25 − 0.05)
s⎠ For the steady rotation speed the equation becomes The volume integral term − 2 (
( )
) mflow ro + ri
Tshaft =
⋅
4⋅ ρ⋅ δ ro − ri Tshaft = 16.9 N⋅ m ⌠→
⌠ → ⎯⎯ ⎯⎯ →
→ ⎯⎯
→
→
→
−⎮ r × ⎛ 2⋅ ω × Vxyz⎞ ⋅ ρ dV = ⎮ r × Vxyz⋅ ρ⋅ Vxyz dA
⎮
⎮
⎝
⎠
⌡
⌡ ⌠→
→ ⎯⎯
→
⎮
r × ⎛ 2⋅ ω × Vxyz⎞ ⋅ ρ dV must be evaluated for the CV. The velocity in the CV
⎮
⎝
⎠
⌡ varies with r. This variation can be found from mass conservation
For an infinitesmal CV of length dr and crosssection A at radial position r, if the flow in is Q, the flow out is Q +
dQ, and the loss through the slot is Vδdr. Hence mass conservation leads to
( Q + dQ) + V⋅ δ⋅ drdQQ = V⋅ δ⋅ dr
− = −0 Q ( r) = −V⋅ δ⋅ r + const mflow
2⋅ ρ At the inlet (r = ri) Q = Qi = Hence Q = Qi + V⋅ δ⋅ ri − r = ( )( ( v ( r) = and along each rotor the water speed is Hence the term  mflow
mflow
+
⋅ δ⋅ ri − r
2⋅ ρ⋅ δ⋅ ro − ri
2⋅ ρ ) ) Q= mflow ⎛
⋅ ⎜1 +
2⋅ ρ ⎝ ri − r ⎞
mflow ⎛ ro − r ⎞
⋅⎜
⎟=
⎟
ro − ri
ro − ri
2⋅ ρ ⎠ ⎝ mflow ⎛ ro − r ⎞
Q
=
⋅⎜
⎟
A
2⋅ ρ⋅ A ro − ri ⎝ ⎠ ⌠→
→ ⎯⎯
→
⎮ r × ⎛ 2⋅ ω × V ⎞ ⋅ ρ dV
becomes
xyz⎠
⎮
⎝
⌡
ro r o
⌠→
⌠
→ ⎯⎯
→
mflow ⎛ ro − r ⎞
⌠
⎮
−⎮ r × ⎛ 2⋅ ω × Vxyz⎞ ⋅ ρ dV = 4⋅ ρ⋅ A⋅ ω⋅ ⎮ r⋅ v ( r) dr = 4⋅ ρ⋅ ω⋅ ⎮ r⋅
⋅⎜
⎟ dr
⎮
⎝
⎠
2⋅ ρ
⌡r
⌡
⎝ ro − ri ⎠
⎮
i
⌡r
i ro or 3 ( i Recall that ⌠ → ⎯⎯ ⎯⎯ →
→
→
2⎛2
2⎞
⎮ r × V ⋅ ρ⋅ V
xyz
xyz dA = ρ⋅ V ⋅ δ⋅ ⎝ ro − ri ⎠
⎮
⌡ Hence equation ⌠→
⌠ → ⎯⎯ ⎯⎯ →
→ ⎯⎯
→
→
→
−⎮ r × ⎛ 2⋅ ω × Vxyz⎞ ⋅ ρ dV = ⎮ r × Vxyz⋅ ρ⋅ Vxyz dA
⎮
⎮
⎝
⎠
⌡
⌡
3 2 ( becomes ) ro + ri ⋅ 2⋅ ri − 3⋅ ro
2
2
2
mflow⋅ ω⋅
= ρ⋅ V ⋅ δ⋅ ⎛ ro − ri ⎞
⎝
⎠
3⋅ ro − ri ( Solving for ω 2 ( ⌠→
⌠
ro + ri ⋅ 2⋅ ri − 3⋅ ro
→ ⎯⎯
→
⎛ ro − r ⎞
⎮
−⎮ r × ⎛ 2⋅ ω × Vxyz⎞ ⋅ ρ dV = 2⋅ mflow⋅ ω⋅ ⎮ r⋅ ⎜
⎟ dr = mflow⋅ ω⋅
⎮
⎝
⎠
3⋅ ro − ri
⌡
⎝ ro − ri ⎠
⎮
⌡r ) 3⋅ ro − ri ⋅ ρ⋅ V ⋅ δ⋅ ⎛ ro − ri ⎞
⎝
⎠
ω=
3
2
mflow⋅ ⎡ro + ri ⋅ 2⋅ ri − 3⋅ ro ⎤ ( ) ⎣ 2 2 ( 2 )⎦ ω = 461 rpm ) ) ⎠ Problem *4.187 [3] Given: Data on rotating spray system Find: Torque required to hold stationary; steadystate speed Solution:
The given data is ρ = 999⋅ kg kg
mflow = 15⋅
s 3 m D = 0.015⋅ m ro = 0.25⋅ m ri = 0.05⋅ m δ = 0.005⋅ m Governing equation: Rotating CV For no rotation (ω = 0) this equation reduces to a single scalar equation
⌠ → ⎯→ ⎯→ →
⎯
⎯
⎮
Tshaft = ⎮ r × Vxyz ρ⋅ VxyzdA
⋅
⌡ ro ⌠
Tshaft = 2⋅ δ⋅ ⎮
⌡r or r ⋅ V⋅ ρ⋅ V dr
i where V is the exit velocity with respect to the CV. We need to find V(r). To do this we use mass conservation, and the fact
that the distribution is linear (
( )
) r − ri
V ( r) = Vmax⋅
ro − ri
so V ( r) = ( ( ) )
) mflow r − ri
⋅
2
ρ⋅ δ
ro − ri ( 2⌠
⎮ ro Hence mflow
1
2⋅ ⋅ Vmax⋅ ro − ri ⋅ δ =
2
ρ and ro ( ⎤
)⎥
2
)⎥
⎦ mflow
⌠
⎡ r − ri
2
⋅ ⎮ r⋅ ⎢
Tshaft = 2⋅ ρ⋅ δ⋅ ⎮ r⋅ V dr = 2⋅
ρ⋅ δ ⎮
⌡r
⎢ r −r
i
⎣o i
⎮
⌡r ( 2 2 dr i Tshaft = 1
6 × ⎛ 15⋅
⎜ ⎝ kg ⎞ 2 ⎟× s⎠ 3 m
1
( 0.05 + 3⋅ 0.25)
×
×
999⋅ kg 0.005⋅ m
( 0.25 − 0.05) For the steady rotation speed the equation becomes
⌠→
⌠ → ⎯⎯ ⎯⎯ →
→ ⎯⎯
→
→
→
−⎮ r × ⎛ 2⋅ ω × Vxyz⎞ ⋅ ρ dV = ⎮ r × Vxyz⋅ ρ⋅ Vxyz dA
⎮
⎮
⎝
⎠
⌡
⌡ (
( mflow ⋅ ri + 3⋅ ro
Tshaft =
6⋅ ρ⋅ δ⋅ ro − ri Tshaft = 30⋅ N⋅ m ) ) ⌠→
→ ⎯⎯
→
⎮ r × ⎛ 2⋅ ω × V ⎞ ⋅ ρ dV
The volume integral term −
must be evaluated for the CV. The velocity in the CV
xyz⎠
⎮
⎝
⌡
varies with r. This variation can be found from mass conservation For an infinitesmal CV of length dr and crosssection A at radial position r, if the flow in is Q, the flow out is Q +
dQ, and the loss through the slot is Vδdr Hence mass conservation leads to
r ( Q + dQ) + V⋅ δ⋅ dr − Q = 0 ( r )
) dQ = −V⋅ δ⋅ dr ( Hence Q = Qi = i mflow
2⋅ ρ ⎤
)⎥
2
)⎥
⎦ mflow ⎡
r − ri
Q ( r) =
⋅ ⎢1 −
2⋅ ρ ⎢
ro − ri
⎣ (
( 2 ⎤
)⎥
2
)⎥
⎦ and along each rotor the water speed is mflow ⎡
r − ri
Q
v ( r) =
=
⋅ ⎢1 −
A
2⋅ ρ⋅ A ⎢
ro − ri
⎣ ⌠→
→ ⎯⎯
→
⎮ r × ⎛ 2⋅ ω × V ⎞ ⋅ ρ dV
Hence the term xyz⎠
⎮
⎝
⌡ ⌠
⎞
⎛ ⌠ ro
⎮
⎜ ⎮ r⋅ v ( r) dr⎟ = 4⋅ ρ⋅ ω⋅ ⎮
4⋅ ρ⋅ A⋅ ω⋅
⎜ ⌡r
⎟
⎮
⎝i
⎠
⎮ (
( 2 ro ⌡r becomes
⌠
⎡
⎮
⎮ r⋅ ⎢1⋅ −
2⋅ mflow⋅ ω⋅
⎢
⎮
⎣
⎮
⌡r ⎤
(ro − r) ⎥ dr = m ⋅ ω⋅ ⎛ 1 ⋅ r 2 + 1 ⋅ r ⋅ r − 1 ⋅ r 2⎞
⎟
flow ⎜ 6 o
2
3 i o 2 i⎠
⎝
ro − ri) ⎥
(
⎦
2 i 2 ( ) Recall that ⌠ → ⎯⎯ ⎯⎯ →
→
→
mflow ⋅ ri + 3⋅ ro
⎮ r × V ⋅ ρ⋅ V
xyz
xyz dA = 6⋅ r − r ⋅ ρ⋅ δ
⎮
oi
⌡ Hence equation ⌠→
⌠ → ⎯⎯ ⎯⎯ →
→ ⎯⎯
→
→
→
−⎮ r × ⎛ 2⋅ ω × Vxyz⎞ ⋅ ρ dV = ⎮ r × Vxyz⋅ ρ⋅ Vxyz dA
⎮
⎮
⎝
⎠
⌡
⌡ becomes 121
1 2⎞ mflow ⋅ (ri + 3⋅ ro)
mflow⋅ ω⋅ ⎛ ⋅ ro + ⋅ ri⋅ ro − ⋅ ri ⎟ =
⎜
6⋅ (ro − ri)⋅ ρ⋅ δ
3
2⎠
⎝6 Solving for ω ω= ( ) 2 ( mflow⋅ ri + 3⋅ ro ) ⎛ r 2 + 2⋅ r ⋅ r − 3⋅ r 2⎞ ⋅ r − r ⋅ ρ⋅ δ
io
i ⎠ ( o i)
⎝o ⎤
) ⎥ dr
2
)⎥
⎦ mflow ⎡
r − ri
⋅ r⋅ ⎢1 −
⎢
2⋅ ρ
ro − ri
⎣ i ro or )
) ( i At the inlet (r = ri) ( ⌠m
⌠m
flow r − ri
flow r − ri
⎮
⎮
dr = Qi −
dr
Q ( r) = Qi −δ⋅ ⎮
⋅
⋅
⎮
2
2
ρ⋅ δ
ρ
ro − ri
ro − ri
⎮
⎮
⌡r
⌡r ω = 1434⋅ rpm (
( 2 Problem *4.188 [3] Problem *4.189 [3] Problem *4.175 Problem *4.190 [3] Problem *4.176 Problem *4.191 [3] Problem *4.192 [4] Problem *4.178 Problem *4.193 [4] Problem *4.179 Problem *4.194 [4] Part 1/2 Problem *4.179 cont'd Problem *4.194 [4] Part 2/2 Problem *4.180 Problem *4.195 [4] Part 1/3 Problem *4.180 cont'd Problem *4.195 [4] Part 2/3 Problem *4.180 cont'd Problem *4.195 [4] Part 3/3 Problem *4.181 Problem *4.196 [5] Part 1/2 Problem *4.181 cont'd Problem *4.196 [5] Part 2/2 Problem *4.197 [5] Part 1/2 Problem *4.197 [5] Part 2/2 Problem 4.183 Problem 4.198 [2] Problem 4.199 Given: Compressed air bottle Find: [3] Rate of temperature change Solution:
Basic equations: Continuity; First Law of Thermodynamics for a CV Assumptions: 1) Adiabatic 2) No work 3) Neglect KE 4) Uniform properties at exit 5) Ideal gas
From continuity ∂
∂t
∂
∂t MCV + mexit = 0 where mexit is the mass flow rate at the exit (Note: Software does not allow a dot!) MCV = −mexit ⎛∂ ⎞
⎛∂ ⎞ ⎛
p⎞
p⎞
∂⌠
⎮
⎛
⎮ u dM + ⎜ u + ρ ⎟ ⋅ mexit = u⋅ ⎜ M ⎟ + M⋅ ⎜ u ⎟ + ⎜ u + ρ ⎟ ⋅ mexit
⎝
⎠
⎠
∂t ⌡
⎝ ∂t ⎠
⎝ ∂t ⎠ ⎝ From the 1st law 0= Hence dT
p
u⋅ −mexit + M⋅ cv⋅
+ u⋅ mexit + ⋅ mexit = 0
dt
ρ But M = ρ⋅ Vol For air ρ= ( ) p
R⋅ T so mexit⋅ p
dT
=−
dt
M⋅ cv⋅ ρ
mexit⋅ p
dT
=−
2
dt
Vol⋅ cv⋅ ρ kg⋅ K
1
6N
×
×
2 286.9⋅ N ⋅ m ( 60 + 273) ⋅ K ρ = 20 × 10 ⋅ m 3 Hence 2 ⎛m ⎞
dT
kg
1
kg⋅ K
K
6N
⎟ = −0.064⋅
= −0.05⋅
× 20 × 10 ⋅
×
×
×⎜
2
3 717.4⋅ N ⋅ m ⎝ 209⋅ kg ⎠
dt
s
s
m
0.5⋅ m ρ = 209 kg
3 m Problem 4.200 Given: Data on centrifugal water pump Find: [3] Pump efficiency Solution:
Basic equations:
(4.56)
Ws
Pin Δp = SGHg⋅ ρ⋅ g⋅ Δh η= D1 = 0.1⋅ m D2 = 0.1⋅ m Q = 0.02⋅ SGHg = 13.6 h1 = −0.2⋅ m 3 Available data: ρ = 1000 kg
3 m
s Pin = 6.75⋅ kW
p2 = 240⋅ kPa m Assumptions: 1) Adiabatic 2) Only shaft work 3) Steady 4) Neglect Δu 5) Δz = 0 6) Incompressible 7) Uniform flow Then 2⎞
2⎞
⎛
⎛
V1 ⎟
V2 ⎟
⎜
⎜
−Ws = ⎜ p1 ⋅ v1 +
⋅ −m
+ p ⋅v +
⋅m
2 ⎟ ( rate) ⎜ 2 2
2 ⎟ ( rate)
⎝
⎠
⎝
⎠ Since mrate = ρ⋅ Q V1 = V2 and ( ) ( −Ws = ρ⋅ Q⋅ p2⋅ v2 − p1⋅ v1 = Q⋅ p2 − p1
p1 = ρHg⋅ g⋅ h ( Ws = Q⋅ p1 − p2
η= Ws
Pin or ) (from continuity) ) p1 = SGHg⋅ ρ⋅ g⋅ h1 p1 = −26.7 kPa Ws = −5.33 kW The negative sign indicates work in η = 79.0 % Problem 4.187 Problem 4.201 [2] Problem 4.186 Problem 4.202 [2] Problem 4.188 Problem 4.203 [2] Problem 4.204 [3] zmax CV (b)
V2
CV (a) z
x Given: Data on fire boat hose system Find: Volume flow rate of nozzle; Maximum water height; Force on boat Solution:
Basic equation: First Law of Thermodynamics for a CV Assumptions: 1) Neglect losses 2) No work 3) Neglect KE at 1 4) Uniform properties at exit 5) Incompressible 6) patm at 1 and 2 Hence for CV (a) ⎛V 2
⎞
⎜2
⎟
−Ws = ⎜
+ g⋅ z2⎟ ⋅ mexit
⎝2
⎠ mexit = ρ⋅ V2⋅ A2 where mexit is mass flow rate (Note:
Software cannot render a dot!) ⎛ 1 ⋅ V 2 + g⋅ z ⎞ ⋅ ρ⋅ V ⋅ A = −W
⎜
2
2⎟
22
s
⎝2
⎠ Hence, for V2 (to get the flow rate) we need to solve which is a cubic for V2! To solve this we could ignore the gravity term, solve for velocity, and then check that the gravity term is in fact minor.
Alternatively we could manually iterate, or use a calculator or Excel, to solve. The answer is
Hence the flow rate is Q = V2⋅ A2 = V2⋅ π⋅ D 2 2 ft π
1
Q = 114⋅ × × ⎛ ⋅ ft⎞
⎜
⎟
s 4 ⎝ 12 ⎠
−Ws = g⋅ zmax⋅ mexit 4 To find zmax, use the first law again to (to CV (b)) to get
Ws Ws zmax = −
=−
g⋅ mexit
g⋅ ρ⋅ Q zmax = 15⋅ hp × 550⋅ ft⋅ lbf
s 1⋅ hp 2 × 3 2 s
ft
×
×
32.2⋅ ft 1.94⋅ slug V2 = 114 ft
s 3 ft
Q = 0.622
s s
0.622⋅ ft 3 × slug⋅ ft
2 s ⋅ lbf Q = 279 gpm zmax = 212 ft For the force in the x direction when jet is horizontal we need x momentum Then ( ) ( ) Rx = u1⋅ −ρ⋅ V1⋅ A1 + u2⋅ ρ⋅ V2⋅ A2 = 0 + V2⋅ ρ⋅ Q
Rx = 1.94⋅ slug
ft 3 3 × 0.622⋅ R x = ρ⋅ Q ⋅ V 2 2 ft
ft lbf ⋅ s
× 114⋅ ×
s
s slug⋅ ft Rx = 138 lbf Problem 4.189 Problem 4.205 [3] Problem *4.191 Problem *4.206 [4] Part 1/2 Problem *4.191 cont'd Problem *4.206 [4] Part 2/2 Problem 4.192 Problem 4.207 [4] Part 1/2 Problem 4.192 cont'd Problem 4.207 [4] Part 2/2 ...
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This note was uploaded on 07/10/2011 for the course CHE 144 taught by Professor Tuzla during the Spring '11 term at Lehigh University .
 Spring '11
 TUZLA

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