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Problem 4.56
Problem 4.64
Problem 4.65
Problem 4.68 2
is deflected by a hinged plate of length 2 m supported by a spring with spring constant k = 1 N/m and uncompressed length x0 = 1 m. Find and plot the deflection angle θ as a function of jet speed V . What jet speed has a deflection of 10°?
Given: Data on flow and system geometry Find: Deflection angle as a function of speed; jet speed for 10 o deflection
Solution The given data are
999
kg 3
A
2
0.005 m
L
2m
m
Governing equation:
Momentum
Fspring
V sin
But
Fspring
kx
Hence
k x0
L sin
VA
k x0
L sin
2
V A sin
k
1
N m
x0
1m
Solving for θ
k x0
asin kL
AV
2
For the speed at which θ = 10o, solve
V
k x0
L sin
A sin
1 V 999
N
(1
m
kg 3
2 si sin (10) ) m
kg m 2
2 0.00 .005 m sin ( 10) N s
m
V
0.867
m s
The deflection is plotted in the corresponding Excel workbook, where the above velocity is obtained using Goal Seek
Problem 4.68 (In Excel) 2
A free jet of water with constant cross-section area 0.005 m is deflected by a hinged plate of length 2 m supported by a spring with spring constant k = 1 N/m and uncompressed length x 0 = 1 m. Find and plot the deflection angle θ as a function of jet speed V . What jet speed has a deflection of 10°? Given: Geometry of system o
Find: Speed for angle to be 10 ; plot angle versus speed
Problem 4.81 The horizontal velocity in the wake behind an object in an air stream of velocity U is given by u ( r)
U 1
u ( r)
U
cos
r
2
2
r
1
r
1
where r is the non-dimensional radial coordinate, measured perpendicular to the flow. Find an expression for the drag on the object.
Given: Data on wake behind object
Find: An expression for the drag
Solution Governing equation:
Momentum
Applying this to the horizontal motion
F
1
2
U
1 U
u ( r) 0
F
U
2
1
2
2
r u ( r) dr 0
2
r u (r) dr
1
F
U
2
1
2
r 1
cos
r
2
2
dr
2
0
1
F
U
2
1
2
r
2 r co cos
0
Integrating and using the limits
F
F
U
5 8
2
1
2
3
2 2
8
U
2
r 2
2
r cos
r 2
4
dr
Problem 4.82
Problem 4.83
Problem 4.86
Problem *4.91
Problem 4.107
Problem 4.108
Problem 4.109 A jet boat takes in water at a constant volumetric rate Q through side vents and ejects it at a high jet speed V j at the rear. A variable-area variable-are a exit orifice controls the jet speed. The drag on the 2 boat is given by F drag drag = kV , where V speed V . If a jet speed V j = 25 m/s produces a boat speed of 10 m/s, what jet speed will be required to double the boat speed?
Given: Data on jet boat Find: Formula for boat speed; jet speed to double boat speed
Solution
CV in boat coordinates
Governing equation:
Momentum
Applying the horizontal component of momentum
Fdrag
Hence
kV
kV
Solving for V
V
2
2
V
Q V j
QV
Q
V j
Q
QV
Q V j
Q
Q
2 k
2 k
2
0
Q V j k
Q
Let
2 k
2
V
2
V j
We can use given data at V = 10 m/s to find α
10
m
2
s
2
50
Hence
V
100
For V = 20 m/s
20
100 9
V j
s
10
100
20
3
9
3
10
100
20
3
9
3
20
V 3 j
80
m s
70 3
m
V j
s
s
10 m 3
10
m
2 25
2
10
V
V j
V j
20
2
25
m s
Problem 4.110
Problem 4.112
Problem 4.113
Problem 4.114
Problem *4.168
Problem *4.171 Water flows in a uniform flow out of the 5 mm slots of the rotating spray system as shown. Th flow rate is 15 kg/s. Find the torque required to hold the system stationary, and the steady-state speed of rotation after it is released. Given: Data on rotating spray system
Solution The given data is
999
kg 3
m
D
0.015 m
mflow
ro
15
0.25 m
kg s
ri
0.05 m
Governing equation: Rotating CV
For no rotation ( ω = 0) this equation reduces to a single scalar equation
Tshaft
r
V xyz
Vxyz dA
0.0 0.005 m
ro
or
Tshaft
2
rV
V dr
2
V
ro
2
r dr
ri
V
2
ri
where V is the exit velocity with respect to the CV mflow
V
2
ro
ri
2
mflow
Hence
2
Tshaft
2
ro
2
Tshaft
Tshaft
Tshaft
mflow 4
1 4
15
ri
ro
ri
ro
ri
kg s
ro
2
3
2
ri
m
1
( 0.25
0.05)
999 kg
0.00 .005 m
( 0.25
0.05)
V xyz
Vxyz dA
16.9N m
For the steady rotation speed the equation becomes
r
2
Vxyz
dV
r
2
ro
2
ri
The volume integral term
r
2
Vxyz
dV must must be be evalu evaluate ated d for for the the CV. 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 cross-section A at radial position r , if the flow in is Q, the flow out is Q + dQ, and the loss through the slot is V δdr . Hence mass conservation leads to (Q
dQ)
dQ
V
Q ( r)
At the inlet ( r = r i)
Hence
Q
Q
Q
V
Qi
r
0
const
mflow 2
V
mflow 2
Q
dr
V
Qi
dr
ri
1
mflow
r
2
ri
r
ro
ri
and along each rotor the water speed is v ( r)
mflow 2
mflow 2
ro
ro
r
ro
ri
ri
Q
mflow
ro
r
A
2
ro
ri
A
ri
r
Hence the term -
r
2
V xyz
dV becomes
ro
ro
r
2
Vxyz
dV
4
A
r v ( r) dr
4
mflow
r
ri
2
ro
r
ro
ri
dr
ri
or
ro
r
2
Vxyz
dV
2 mflow
r
ro
r
ro
ri
dr
mflow
ro
3
2
ri
3 ro
ri
Recall that
Hence equation
r
Vxyz
r
2
mflow
Solving for ω
V xyz dA
ro
3 ro mflow
Vxyz
3
2
ri
ri ro
2
ro
r
3 ro
2
2
ri
2
ro 2 ri
2
ri
Vxyz dA becomes
V xyz
V
ri
V 3
2
dV
2 ri
3 ro
V
2
2
ro
2
ri
2
ri
3 ro
2 ri
461rpm
ri
3 ro
Problem *4.172 If the same flow rate in the rotating spray system of Problem 4.171 is not uniform but instead varies linearly from a maximum at the outer radius to zero at a point 50 mm from the axis, find the torque required to hold it stationary, and the steady-state speed of rotation. Given: Data on rotating spray system
Solution The given data is
999
kg
mflow
3
m
D
0.015 m
ro
15
0.25 m
kg s
ri
0.05 m
Governing equation: Rotating CV
For no rotation ( ω = 0) this equation reduces to a single scalar equation
Tshaft
r
V xyz
Vxyz dA
ro
or
Tshaft
2
rV ri
V dr
0.0 0.005 m
where V is the exit velocity with respect to the CV. We need to find V (r ). ). To do this we use ma conservation, and the fact that the distribution is linear
V ( r)
and
so
2
1 2
Vmax
r
ri
ro
ri
V max ro
V ( r)
mflow
ri
mflow
r
ri
ro
ri
ro
Hence
Tshaft
2
2
2
2
r V dr
2
ro
mflow
r r ro
ri
2
ri ri
2
dr
ri
2
Tshaft
Tshaft
Tshaft
mflow 6
1 6
15
30 N m
ri
3 ro
ro
ri
kg s
2
3
m
1
999 kg
0.00 .005 m
( 0.05 ( 0.25
3 0.25) 0.05)
For the steady rotation speed the equation becomes
r
2
Vxyz
The volume integral term
r
2
dV
r
Vxyz
V xyz
Vxyz dA
dV must must be be evalu evaluate ated d for for the the CV. 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 cross-section A at radial position r , if the flow in is Q, the flow out is Q + dQ, and the loss through the slot is V δdr . Hence mass conservation leads to (Q
