Solution 3.95
For water take
= 998 kg/m3. The control volume surrounds the plate and yields
Problem 3.96
Extend Prob. 3.90 to the case of the liquid motion in a frictionless U-tube whose liquid column is
displaced a distance Z upward and then released, as in Fig. P3.96. Neglect the short horizontal
leg and combine control-volume analyses for the left and right legs to derive a single differential
equation for V(t) of the liquid column.
)05.0)(4/(
)4/(
m
h
Solution 3.96
As in Prob. 3.92, break it up into two moving CV’s, one for each leg, as shown. By mass
conservation, the velocity V(t) is the same in each leg. Let pI be the bottom pressure in the (very
short) cross-over leg. Neglect wall shear stress. Now apply vertical momentum to each leg:
Problem 3.97*
Extend Prob. 3.96 to include a linear (laminar) average wall shear stress resistance of the form
8
V/D, where
is the fluid viscosity. Find the differential equation for dV/dt and then solve
for V(t), assuming an initial displacement z = zo, V = 0 at t = 0. The result should be a damped
oscillation tending toward z = 0.
Solution 3.97
The derivation now includes wall shear stress on each leg (see Prob. 3.96):
Problem 3.98*
As an extension of Example 3.9, let the plate and its cart (see Fig. 3.9 a ) be unrestrained
horizontally, with frictionless wheels. Derive (a) the equation of motion for cart velocity Vc(t)
and (b) a formula for the time required for the cart to accelerate from rest to 90 percent of the jet
velocity (assuming the jet continues to strike the plate horizontally). (c) Compute numerical
values for part (b) using the conditions of Example 3.9 and a cart mass of 2 kg.
Solution 3.98
(a) Use Eq. (3.49) with arel equal to the cart acceleration and Fx = 0:
Problem 3.99
Let the rocket of Fig. E3.12 start at z = 0, with constant exit velocity and exit mass flow, and rise
vertically with zero drag. (a) Show that, as long as fuel burning continues, the vertical height S(t)
reached is given by
[ ln 1], 1
eo
o
VM mt
S where
mM
= − + = −
(b) Apply this to the case Ve = 1500 m/s and Mo = 1000 kg to find the height reached after a burn
of 30 seconds, when the final rocket mass is 400 kg.
Solution 3.99
(a) Ignoring gravity effects, integrate the equation of the projectile’s velocity (from E3.12):
0
( ) ( ) ln 1
t
e
o
mt
S t V t dt V dt
M
= = − −
Problem 3.100
Suppose that the solid-propellant rocket of Prob. 3.35 is built into a missile of diameter 70 cm
and length 4 m. The system weighs 1800 N, which includes 700 N of propellant. Neglect air
drag. If the missile is fired vertically from rest at sea level, estimate (a) its velocity and height at
fuel burnout and (b) the maximum height it will attain.
Solution 3.100
The theory of Example 3.12 holds until burnout. Now Mo = 1800/9.81 = 183.5 kg, and recall
from Prob. 3.35 that Ve = 1150 m/s and the exit mass flow is 11.8 kg/s. The fuel mass is
Problem 3.101
Water at 20C flows steadily through the tank in Fig. P3.101. Known conditions are D1 = 8 cm,
V1 = 6 m/s, and D2 = 4 cm. A rightward force F = 70 N is required to keep the tank fixed.
(a) What is the velocity leaving section 2? (b) If the tank cross-section is 1.2 m2, how fast is the
water surface h(t) rising or falling?
Solution 3.101
First, for water at 20C,
= 998 kg/m3. (a) For a control volume around the tank,
Problem 3.102
As can often be seen in a kitchen sink when the faucet is running, a high-speed channel flow
(V1, h1) may “jump” to a low-speed, low-energy condition (V2, h2) as in Fig. P3.102. The
pressure at sections 1 and 2 is approximately hydrostatic, and wall friction is negligible. Use the
continuity and momentum relations to find h2 and V2 in terms of (h1, V1).
Solution 3.102
The CV cuts through sections 1 and 2 and surrounds the jump, as shown. Wall shear is neglected.
There are no obstacles. The only forces are due to hydrostatic pressure:
Problem 3.103*
Suppose that the solid-propellant rocket of Prob. 3.35 is mounted on a 1000-kg car to propel it up
a long slope of 15. The rocket motor weighs 900 N, which includes 500 N of propellant. If the
car starts from rest when the rocket is fired, and if air drag and wheel friction are neglected,
estimate the maximum distance that the car will travel up the hill.
Solution 3.103
This is a variation of Prob. 3.100, except that “g” is now replaced by “g sin
.” Recall from
Prob. 3.35 that the rocket mass flow is 11.8 kg/s and its exit velocity is 1150 m/s. The rocket fires
Problem 3.104
A rocket is attached to a rigid horizontal rod hinged at the origin as in Fig. P3.104. Its initial
mass is Mo, and its exit properties are
m
and Ve relative to the rocket. Set up the differential
equation for rocket motion, and solve for the angular velocity
(t) of the rod. Neglect gravity, air
drag, and the rod mass.
Solution 3.104
The CV encloses the rocket and moves at (accelerating) rocket speed (t). The rocket arm is
Problem 3.105
Extend Prob. 3.104 to the case where the rocket has a linear air drag force F = cV, where c is a
constant. Assuming no burnout, solve for
(t) and find the terminal angular velocity, — that is,
the final motion when the angular acceleration is zero. Apply to the case Mo = 6 kg, R = 3 m,
m = 0.05 kg/s, Ve = 1100 m/s, and c = 0.075 N·s/m to find the angular velocity after 12 s of
burning.
Solution 3.105
If linear resistive drag is added to Prob. 3.104, the equation of motion becomes
Problem 3.106
Actual air flow past a parachute creates a variable distribution of velocities and directions. Let us
model this as a circular air jet, of diameter half the parachute diameter, which is turned
completely around by the parachute, as in Fig. P3.106. (a) Find the force F required to support
the chute. (b) Express this force as a dimensionless drag coefficient, CD = F/[(1/2)
V2(4)D2]
and compare with Table 7.3.
Solution 3.106
This model is crude, compared to velocity-field theory, but gives the right order of magnitude.
(a) Let the control volume surround the parachute and cut through the oncoming and leaving air
streams:
Problem 3.107
The cart in Fig. P3.107 moves at constant velocity Vo = 12 m/s and takes on water with a scoop
80 cm wide which dips h = 2.5 cm into a pond. Neglect air drag and wheel friction. Estimate the
force required to keep the cart moving.
Solution 3.107
The CV surrounds the cart and scoop and moves to the left at cart speed Vo. Momentum within
Problem 3.108*
A rocket sled of mass M is to be decelerated by a scoop, as in Fig. P3.108, which has width b
into the paper and dips into the water a depth h, creating an upward jet at 60°. The rocket thrust
is T to the left. Let the initial velocity be Vo, and neglect air drag and wheel friction. Find an
expression for V(t) of the sled for (a) T = 0 and (b) finite T
Solution 3.108
The CV surrounds the sled and scoop and moves to the left at sled speed V(t).
Problem 3.109
For the boundary layer flow in Fig. 3.10, let the exit velocity profile, at x = L , simulate turbulent
flow, u Uo(y/
)1/7. (a) Find a relation between h and
. (b) Find an expression for the drag
force F on the plate between 0 and L.
Solution 3.109
(a) Since the upper and lower boundaries of the control volume are streamlines, the mass flow
in, at x=0, must equal the mass flow out, at x=L.
Problem 3.110
Repeat Prob. P3.49 by assuming that p1 is unknown and using Bernoulli’s equation with no
losses. Compute the new bolt force for this assumption. What is the head loss between 1 and 2
for the data of Prob. P3.49?
Solution 3.110
Use one-dimensional, incompressible continuity to find V1:
22
1 1 1 2 2 1
( / 4)(1 ) (56 / )(0.5 ) , or 14 /V A V ft V A ft s ft V ft s
= = = =
Problem 3.111
As a simpler approach to Prob. P3.96, apply the unsteady Bernoulli equation between 1 and 2 to
derive a differential equation for the motion z(t). Neglect friction and compressibility.
Solution 3.111
The flow from 1 to 2 is along a “streamline”, so we can use Eq. (3.54):
222
21
2 1 2 1
1
1( ) ( ) 0
2
pp
Vds V V g z z
t
−
+ + − + − =
Problem 3.112
A jet of alcohol strikes the vertical plate in Fig. P3.112. A force F 425 N is required to hold the
plate stationary. Assuming there are no losses in the nozzle, estimate (a) the mass flow rate of
alcohol and (b) the absolute pressure at section 1.
Solution 3.112
A momentum analysis of the plate (e.g. Prob. 3.40) will give