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Unformatted text preview: 2.4: PROBLEM DEFINITION Situation: An engineer needs to know the local density for an experiment with a glider.
z = 2500 ft. Find: Calculate density using local conditions. Compare calculated density with the value from Table A2, and make a recommen
dation. Properties:
From Table A2, Rm, 2 287 J p = 1.22 kg/ m3. k K’
Local temperature : 74.3013 =g 296.7 K. Local pressure : 27.3 in.—Hg : 92.45 kPa. Apply the ideal gas law for local conditions. Ideal gas law L
RT 92,450 N/1112
(287kg/ m3) (296.7K) 1.086 kg/m3 p = 1.09 kg/m3 (local conditions) Table value. From Table A2 H P = 122 kg/m3 (table value) The density differenCe (local conditions versus table value) is about 12%. Most
of this difference is due to the effect of elevation on atmospheric pressure. Recommendationeuse the local value of density because the effects of elevation are signiﬁcant . Note: Always use absolute pressure when working with the ideal gas law. 2.5: PROBLEM DEFINITION Carbon dioxide.
Find:
Density and speciﬁc weight of C02. Properties:
From Table A2, RCOQ = 189 J/kgK.
p = 300 kPa, T —— 60 °C. 1. First, apply the ideal gas law to ﬁnd density.
2. Then, calculate speciﬁc weight using 7 2 pg SOLUTION 1. Ideal gas law P
pCOQ = i 300, 000 kPa
I (189 J/lch) (60+273)K
p002 : 4.767 kg/m?’
2. Speciﬁc weight
7 = P9
Thus
’Ycog = F7002 X 9
= 4.767kg/ m3 x 9.81 m/ s?
7602 = N/m3
REVIEW Always use absolute pressure when working with the ideal gas law. 2.9: PROBLEM DEFINITION Situation;
Oxygen ﬁlls la tank.
mG = 10 ft3, Wt...k = 150155 Find:
Weight (tank plus oxygen). Properties:
From Table A2, R02 2 1555 ftlbf/(slug ~° R).
p = 500 psia, T 2 70°F. Apply the ideal gas law to ﬁnd density of oxygen.
Find the weight of the oxygen using speciﬁc weight (7) and add this to the weight of
the tank. 1. Ideal gas law pabs. = 500 psia x 144 psf/psi = 72, 000 psf
T = 460 4 70 = 530°R
p _ 1
RT
i 72, 000 psf
5 (1555 ftlbf/slugOR) (530m)
,0 = 0.087 slugs/ft3 2. Speciﬁc weight 5
 P9 0087 slug ft fts X 32.2? 7 : 2.80 lbf/ft3 3. Weight of ﬁlled tank Waxygen : 2.80 Raf/£133 x 1053
= 281bf Wmml = l/Voxygon + WW1}; 28.0lbf+1501bf Wm1 : 1781bf H REVIEW 1. For compressed gas in a tank, pressures are often very high and the ideal gas
assumption is invalid. For this problem the pressure is about 34 atmospheres—it is
a good idea to check a thermodynamics reference to analyze whether or not real gas
effects are signiﬁcant. 2. Always use absolute pressure when working with the ideal gas law. 10 2.10: PROBLEM DEFINITION Situation: Oxygen is released from a tank through a valve.
V = 10 1113. Find:
Mass of oxygen that has been released. Properties:
R02 : 260$.
p1 = 800 kPa, TJ = 15 00‘
p2 = 600 kPa, T2 = 20°C. 1. Use ideal gas lawJ expressed in terms of density and the gas—speciﬁc (not universal)
gas constant. 2. Find the density for the case before the gas is released; and then mass from
density, given the tank volume. 3. Find the density for the case after the gas is released, and the corresponding mass.
4. Calculate the mass difference, which is the mass released. SOLUTION 1. Ideal gas law p = L
RT
2. Density and mass for case 1
_ 800,000%
p1 “ (260 gyms K)
kg
k
= iota—g3 >< 10m3
In
M1 : 1068 kg
3. Density and mass for case 2
_ 600, 000%
p2 _ (200 gymss K)
k0" kcr 8.01—33 x 10 mg
m' M1 = 80.1 kg 4. Mass released from tank Ml—Mg 2 1068—801 12 2.19: PROBLEM DEF INITIO i Situation:
Air at certain temperatures.
T1=10°C,T2 = 70°C. Find:
Change in kinematic viscosity, Properties:
From Table A3, 1/70 : 1.99 x 10—5 KHZ/S, 1/10 —— 1.41 x 10‘5 m2/s. PLAN
Use properties found in Table A3. SOLUTION Aviiriioqm : (1.99 — 141) x 10*5 Avair710a70 : 5.8x10’6rn2/s REVIEW Sutherland’s equation could also be used to solve this problem. 2.31: PROBLEM DEFINITION Situation:
Oil (SAE 10W30) ﬁlls the space between two plates.
Ay = 1/8 = 0.125in, u = 25ft/s. Lower plate is at rest. Find:
Shear stress in oil. Properties:
Oi] (SAE 10W30 @ 150 OF) from Figure A2: p : 5.2 X 10“1 ibfs/ftg. Assumptions:
1.) Assume oil is a Newtonian ﬂuid.
2.) Assume Couette ﬂow (linear velocity proﬁle). Rate of strain d_u _ &
dy _ Ay
_ 25ft/s
_ (0.125/12)ft
du
— = 240 '1
dy U s Newton’s law of viscosity If
A
en
[0
X
H
O
I
ppEE
,0.
U}
V
X
A
[\D
,5;
o
O
wit4
\_/ II
E“
M
q;
00 .M 37 2.32: PROBLEM DEFINITION Situation: Properties of air and water.
T :2: 40°C, p = 170 kPa. Find:
Kinematic and dynamic viscosities of air and water. Properties:
Air data from Table A3, um : 1.91 X 10—5 N~s/m2
Water data from Table A6, umm = 6.53 x 10—4 Ns/1112> pwatm = 992 kg/m3. Apply the ideal gas law to find density. Find kinematic Viscosity as the ratio of
dynamic and absolute Viscosity. SOLUTION
A.) Air
Ideal gas law 1
RT pair 170, 000 kPa
(287J/ kg K) (3132 K)
1.89 kg/m3 I! ll N s
in2 am = 1.91 x 105 E p 1.91 X 10*5Ns/ml2
1.89 kg/m3 um 10.1 X 10’6 rn2/s #wam = 6.53 X 10*5 N.S/m2 B.) water B p 6.53 x 10—4Ns/ m2
992kg/ m3 H I/wate, : 6.58 X 10’7 1112/3 38 2.33: PROBLEM DEFINITION Situation: Sliding plate viscometer is used to measure ﬂuid viscosity.
A =—= 50 X 100mm, Ay = lrnrn.
u:10rn/s,F= 3N. Find:
Viscosity of the ﬂuid. Assumptions:
Linear velocity distribution. 1. The shear force 7' is a, force/ area.
2. Use equation for viscosity to relate shear force to the velocity distribution. SOLUTION 1. Calculate shear force T _ Force
— Area
7 _ 3N
_ 50mm x 100 mm
7' : SOON
2. Find viscosity
M ¥ 7'
du.
(a)
_ 600N
“ ‘ [lOrn/sl/[lnim]
ur6x10‘2121'25 39 2.36: PROBLEM DEFINITIO _ Situation:
A liquid ﬂows between parallel boundaries.
yo = 0.0 mm, V0 = 0.0m/s.
yl 21.0mm, V1 = 1.0 m/s.
yg = 2.0 mm, V2 = 1.99 m/s.
y3 = 3.0 mm, V3 = 298 m/s, F ind:
(a) Maximum shear stress.
(b) Location Where minimum shear stress occurs. (21) Maximum shear stress 7' = udV/dy
MAV/Ay) next to wall
(103N  5/113qu m/s)/0.001 m) Tmax : 1‘0 N/m2
(b)The minimum shear stress will OCCUI midway between the two walls . Its mag— nitude W1 e zero ecause t 6 ve ocity gra ient is zero at t e mipoint. 22 Tr“ ZLX ’ max 2.37: PROBLEM DEFINITION m:
Glycerin is ﬂowing in between two stationary plates. The velocity distribution is
1 03p
' 2 — —— B1 ~ 2
U 211 dm ( J y ) dp/dzc = —1.6kPa/m, B = 5cm. Find:
Velocity and shear stress at a distance of 12 mm from wall (Le. at y = 12 mm).
Velocity and shear stress at the wall (ie at y = 0 mm). Properties:
Glycerin (20°C), Table A4: it = 1.41 N A 8/1112. PLAN Find velocity by direct substitutiOn into the speciﬁed velocity distribution.
Find shear stress using the deﬁnition of viscosity: 7‘ = p. (du/dy), where the rateof
strain (i.e. the derivative du/dy) is found by differentiating the velocity distribution. SOLUTION a.) Velocity (at y :r— 12 mm) 1 dp
= ——_... B] i 2
u 2nd33( y y)
1 .
='—— 416ON3 .5 .1 _l 2
2(1_41N.S/m2)< 0 /m)((00 m)(002m) (0012110))
= 0.25873
S u (y 12mm) = 0259 m/s du d 1 Lip 2
_ 2 _ ____ B _
dy dy < 2/1 dCL' ( y y I!
/'I\
she
\_._/
b 31%
V
el e
A m <2 1 EM
V = es (a Rate of strain (at y : 12 HUD) = «a (e 1 N (~1600r—ng) (0.05m — 2 X 0.012 m)
14.75s‘L 43 Deﬁnition of viscosity du
I“ dy = (1.41 N' 5) (14755—1) m2
= 20. 798 Pa 7' (y 12 mm) : 20.8 Pa
b.) Velocity (at y = 0 mm) d
U = ﬁlm—$39412)
1 T i # m2
= f2(1_41N_3/m2) (—1600x/m3) ((0.05m)(0m) (0 )j
—— 0.00E u(y =0mm) =01n/s Rate of strain (at y = 0 mm) clu _ / 1\ Lip
E ‘ lsmllﬁlw‘zy)
1 \ _ N
= 28.37s'1 Shear stress (at y : 0mm) du
T = _..
Mdy
N s _ _1
= (1.41 mg > (28.378 )
= 4000 Pa
7’ (y : 0mm) 2 40.0 Pa
REVIEW
1. As expected, the velocity at the wall (Le. at y — 0) is zero due to the no slip
condition. 2. As expected, the shear stress at the wall is larger than the shear stress away
from the wall. This is because shear stress is maximum at the wall and zero
along the centerline (Le. at y = 8/2). 44 2.45: PROBLEM DEFINITION M:
Pressure is applied to a mass of water.
V: 2000cm3, p = 2 x 106 N/rr12. Find:
Volume after pressure applied (Cm3). Properties:
FrOm Table A5, E = 2.2 X 109 Pa 1. Use modulus of elasticity equation to calculate volume change resulting from
pressure change.
2. Calculate ﬁnal volume based on original volume and volume change. SOLUTION
1. Elasticity equation m
n
l 5‘
l AV = wav (2 x 106) Pa
(2.2 x 109) Pa
—1.82 (71113 2000 cm3 ll 2. Final volume mea; : (2000 — 1.82) cm3 me = 1998 cm3 2.47: PROBLEM DEFINITION Situation:
Open tank of water.
T20 : 2000, : 80°C.
V: 400], d = 3m.
Hint: Volume change is due to temperature. Find:
Percentage change in volume.
Water level rise for given diameter. Preperties:
From Table A5: p20 : 998%,and p80 : 972%. This problem is NOT solved using the elasticity equation, because the volume change
results from a change in temperature causing a density change, NOT a change in
pressure. The tank is open, so the pressure at the surface of the tank is always
atmospheric. a. Percentage Change in volume must be calculated for this mass of water at two
temperatures. For the first temperature, the volume is given as V20 : 4001 = 0.4 m3.ItS density is
p20 = 998%. Therefore, the mass for both cases is given by. la .
m = 9.98523 x 0.41113
H1“ : 399.2 kg For the second temperature, that mass takes up a larger volume: on 399.2kg
V50 : — : —k—
p 972855 : 0.4llkg‘ Therefore, the percentage change in volume is 0.411kg w 0.4 kg
0.4kg volume % change 2 b. If the tank has D = 3m, then V: 7r'r2/1 = 7.68/1.Therefore: : 0.0275 57 hgo : hgo Z .0541’11 And water level rise is 0054 a 0.52m = 0.002111 2 2mm.
water level rise is 2 0.002 m : 2 mm Density changes can result from temperature changes, as well as pressure changes. 58 Problem 4) Given = constant for an ideal gas that undergoes an isentropic process and
P
d E i v = , show that EV = kp for an iscntropic process.
dp/ ,0 Solution: 2‘49: PROBLEM DEFINITION m:
Very small spherical droplet of water. Find:
Pressure inside. Refer to Fig. 26(a). The surface tension force7 27717, will be resisted by the pressme
force acting on the cut section of the spherical droplet or p(7r7"2) : 27rr0
2C7
p : —
7“
J _40
1”— d 60 2.53: PROBLEM DEFINITION em:
Two vertical glass plates
y = 1 mm Find:
Capillary rise (h) between the plates. Properties:
From Table A4, surface tension of water is 7.3 x 10“2 N/m. PLAN I Apply equilibrium, then the surface tension force equation. Equilibrium ll
0 2a Force due to surface tension
(28) a H Weight of ﬂuid that has been pulled upward
(111%)? Solve for capillary rise (/1) 20€~MW : 0 20' t 2 X (7.3 x10’2N/m)
9810N/1‘n3 X 0.001m
0.0149 In h: h: h: 64 I 2.59: PROBLEM DEFINITIONI Situation:
A liquid reaches the vapor pressure Find:
What happens to the liquid If a liquid reaches its vapor pressure for a given temperature, it boils . 7O 2.60: PROBLEM DEFINITION Find: How does vapor pressure Change with increasing temperature? The vapor pressure increases with increasing temperature .
for this, note from the Appendix that the vapor pressure of water at 212 °F (100°C)
is 101 kPa (14.7 psia). To get water to boil at a lower temperature, you would have
to exert a vacuum on the water. To keep it from boiling until a higher temperature,
you would have to pressurize it. To get an everyday feel 71 2.61: PROBLEM DEFINITION Situation:
Water at 60 OF Find:
The pressure that must be imposed for water to boil Properties:
Water (60 °F), Table A.5; P0 = 0.363 psia The pressure to which the ﬂuid must be exposed is P =:: 0.363 psia. This is lower than atmospheric pressure. Therefore, assuming atmosp eric pressure is 14.7 sia
gage, or 14.7 pSig; the pressure needed could also be reported as P 2 —14.34 psig . 2.63: PROBLEM DEFINITION Situation:
Water in a, closed tank
T = 20 °C, p : 10400 Pa Find;
Whether water will bubble into the vapor phase (boil). Properties:
From Table A.5, at T : 20 OC, PU t: 2340 Pa abs SOLUTION ‘
The tank pressure is 103400 Pa abs. and PU = 2340 Pa abs. So the tank pressure is higher than the PU. Therefore the water will not boil .
REVIEW The water can be made to boil at this temperature only if the pressure is reduced
to 2340 Pa abs. Or, the water can be made to boil at this pressure only if the
temperature is raised to apprOXimater 50°C. 74 ...
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