Lab 4: Linear Momentum Experiments
Purpose:
The purpose of this lab is to demonstrate the momentum theorem, with the scope of this lab
being restricted to linear momentum concepts. The lab will consist of two parts, one with water
and the other with air a
2517821f77b76b175b1a1c3242a9c8274a8fc304.xls
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NOZZLE FLOW METER
mdot = C E A2 Y (2gc rho1 (p1 - p2)^.5
INPUT
patm =
in. hg
Temp =
deg. F
E
ME335
F
G
NOZFM335.XLS
14.06
P7=
Beta =
Y=
mu =
C=
E
Pre Lab Linear Momentum Equation
Fig. 1: Apparatus for Linear Momentum Experiments with Water as the Fluid.
For this lab, a pump will be used to push water from the intake, through the flow control valve,
and to form a jet of water that rises until it hit
Problem 2.86
[Difficulty: 3]
Given:
Geometry of and flow rate through tapered nozzle
Find:
At which point becomes turbulent
Solution:
Basic equation
Re =
For pipe flow (Section 2-6)
V D
= 2300
for transition to turbulence
2
D
Q=
Also flow rate Q is give
Problem 2.88
[Difficulty: 3]
Given:
Data on supersonic aircraft
Find:
Mach number; Point at which boundary layer becomes turbulent
Solution:
Basic equation
V = M c
Hence
M=
V
c
c=
and
k R T
For air at STP, k = 1.40 and R = 286.9J/kg.K (53.33 ft.lbf/lbmoR)
Problem 2.6
[Difficulty: 1]
Given:
Velocity field
Find:
Whether field is 1D, 2D or 3D; Velocity components at (2,1/2); Equation for streamlines; Plot
Solution:
The velocity field is a function of x and y. It is therefore 2D.
u = a x y = 2
At point (2,1/2
Problem 2.87
[Difficulty: 2]
Given:
Data on water tube
Find:
Reynolds number of flow; Temperature at which flow becomes turbulent
Solution:
Basic equation
At 20oC, from Fig. A.3 = 9 10
For the heated pipe
Hence
Re =
For pipe flow (Section 2-6)
Re =
=
V D
Problem 2.5
[Difficulty: 2]
Given:
Velocity field
Find:
Equation for streamlines; Plot several in the first quadrant, including one that passes through point (0,0)
Solution:
Governing equation: For streamlines
v
u
=
dy
=
dy
dx
Assumption: 2D flow
Hence
v
Problem 2.87
[Difficulty: 2]
Given:
Data on water tube
Find:
Reynolds number of flow; Temperature at which flow becomes turbulent
Solution:
Basic equation
At 20oC, from Fig. A.3 = 9 10
For the heated pipe
Hence
Re =
For pipe flow (Section 2-6)
Re =
=
V D
Problem 2.2
Given:
Velocity fields
Find:
[Difficulty: 1]
Whether flows are 1, 2 or 3D, steady or unsteady.
Solution:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
V = V ( y)
V = V ( x)
V = V ( x , y)
V = V ( x , y)
V = V ( x)
V = V ( x , y , z)
V = V ( x , y)
V = V ( x
Problem 2.4
Given:
Velocity field
Find:
[Difficulty: 1]
Equation for streamlines
Streamline Plots
Solution:
v
u
So, separating variables
dy
=
dy
y
dx
=
=
B x y
2
2
A x y
=
C=1
C=2
C=3
C=4
B y
4
A x
B dx
Ax
y (m)
For streamlines
5
3
2
Integrating
The solut
Problem 2.86
[Difficulty: 3]
Given:
Geometry of and flow rate through tapered nozzle
Find:
At which point becomes turbulent
Solution:
Basic equation
Re =
For pipe flow (Section 2-6)
V D
= 2300
for transition to turbulence
2
D
Q=
Also flow rate Q is give
Problem 2.3
Given:
Viscous liquid sheared between parallel disks.
Upper disk rotates, lower fixed.
Velocity field is:
r
r z
V = e
h
Find:
a.
Dimensions of velocity field.
b.
Satisfy physical boundary conditions.
r
r
To find dimensions, compare to V = V (
Problem 2.88
[Difficulty: 3]
Given:
Data on supersonic aircraft
Find:
Mach number; Point at which boundary layer becomes turbulent
Solution:
Basic equation
V = M c
Hence
M=
V
c
c=
and
k R T
For air at STP, k = 1.40 and R = 286.9J/kg.K (53.33 ft.lbf/lbmoR)
Problem 2.14
[Difficulty: 2]
Given:
Velocity field
Find:
Proof that the parametric equations for particle motion are x p = c1 e
A t
and y p = c2 e
A t
; pathline that was at
(2,2) at t = 0; compare to streamline through same point, and explain why they a
Problem 2.13
[Difficulty: 3]
Given:
Flow field
Find:
Plot of velocity magnitude along axes, and y = x; Equations of streamlines
Solution:
q x
u=
On the x axis, y = 0, so
(2
2 x + y
)
2
=
Plotting
q
2 x
v=
q y
(2
2 x + y
)
2
=0
100
u (m/s)
50
1
0.5
0
0.5
[Difficulty: 3]
Problem 2.12
Given:
Flow field
Find:
Plot of velocity magnitude along axes, and y = x; Equation of streamlines
Solution:
K y
u=
On the x axis, y = 0, so
(2
2 x + y
)
2
Plotting
=0
K x
v=
(2
2 x + y
)
2
=
K
2 x
160
v( m/s)
80
1
0.5
0
0.5
1
Problem 2.90
Given:
Data on seaplane
Find:
[Difficulty: 2]
Transition point of boundary layer
Solution:
For boundary layer transition, from Section 2-6
Retrans = 500000
Then
Retrans =
At 45oF = 7.2 oC (Fig A.3)
V x trans
2
5 m
= 0.8 10
s
V x trans
=
10.
Problem 2.11
[Difficulty: 3]
Given:
Flow field
Find:
Plot of velocity magnitude along axes, and y = x; Equation for streamlines
Solution:
u=
On the x axis, y = 0, so
Plotting
M y
2
=0
v=
M x
2
200
v (m/s)
150
100
50
0
0.2
0.4
0.6
0.8
1
x (km)
The veloci
Problem 2.90
Given:
Data on seaplane
Find:
[Difficulty: 2]
Transition point of boundary layer
Solution:
For boundary layer transition, from Section 2-6
Retrans = 500000
Then
Retrans =
At 45oF = 7.2 oC (Fig A.3)
V x trans
2
5 m
= 0.8 10
s
V x trans
=
10.
Problem 2.10
[Difficulty: 2]
Given:
Velocity field
Find:
Equation for streamline through (1,3)
Solution:
For streamlines
v
u
So, separating variables
y
A
dy
y
=
=
dy
dx
x
=
2
A
=
y
x
x
dx
x
Integrating
ln( y ) = ln( x ) + c
The solution is
y = C x
which i
Problem 2.89
Given:
Type of oil, flow rate, and tube geometry
Find:
[Difficulty: 2]
Whether flow is laminar or turbulent
Solution:
=
Data on SAE 30 oil SG or density is limited in the Appendix. We can Google it or use the following
At 100 oC, from Figs. A
Problem 2.89
Given:
Type of oil, flow rate, and tube geometry
Find:
[Difficulty: 2]
Whether flow is laminar or turbulent
Solution:
=
Data on SAE 30 oil SG or density is limited in the Appendix. We can Google it or use the following
At 100 oC, from Figs. A
Problem 2.85
Given:
Local temperature
Find:
[Difficulty: 1]
Minimum speed for compressibility effects
Solution:
Basic equation
V = M c
c=
Hence
and
k R T
M = 0.3
for compressibility effects
For air at STP, k = 1.40 and R = 286.9J/kg.K (53.33 ft.lbf/lbmoR)
Problem 2.1
Given:
Velocity fields
Find:
[Difficulty: 1]
Whether flows are 1, 2 or 3D, steady or unsteady.
Solution:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
V = V ( x , y)
V = V ( x , y)
V = V ( x)
V = V ( x)
V = V ( x)
V = V ( x , y)
V = V ( x , y)
V = V ( x , y
Problem 1.47
[Difficulty: 3]
Given:
Soda can with estimated dimensions D = 66.0 0.5 mm, H = 110 0.5 mm. Soda has SG = 1.055
Find:
Volume of soda in the can (based on measured mass of full and empty can); Estimate average
depth to which the can is filled a
Homework 7
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