Problem #3:
N
e
t momen
tum rate transferred across yplane is zero
For xmomentum rate transferred across xplane at x = x (influxes) and at x+
∆
x (outfluxes)
()
(
)
( ) ( )
;
x
xx
x
x
x
x
x
WL
p
u u
WL p
WL p
τρ
+Δ
++
=
xmomentum balance
(
)
( )( ) ( )
sin
0
x
g
x
x
WL
p
p
F
WL
p
p
WL x
g
ρβ
−+
=
Δ
=
This equation just shows that pressure is a unique function of x which is not needed for
determining velocity profile. Therefore, based on the assumptions discussed in the lecture, z
momentum balance is sufficient to determine velocity profile.
Problem #4:
Problem 2B.3 at the end of chapter 2 of BSL, parts (a), (b) and (c) only.
Follow the terminology
in the diagram in Figure 2B.3 on p. 63.
Note that x = 0 in the center of the slit in this figure.
Note also that the answers are given; what is desired is the derivation.
For convenience in
grading, perform the derivation in the following steps:
a)
Define a control volume
b)
State assumptions
c)
Perform a momentum balance on zmomentum, and explain why there is no need to
perform respective balances on x and y momentum.
d)
Obtain a differential equation for
τ
xz.
e)
Solve this differential equation but do not attempt to determine the constant of
integration yet.
f)
Insert Newton's law to obtain a differential equation for uz.
g)
Solve this equation without evaluating the constants of integration yet.
h)
State necessary boundary conditions that apply to this problem.
i)
Evaluate the two constants of integration and obtain the equation for u
z
.
j)
Using the constants of integration, evaluate
τ
xz
as a function of x.
k)
Integrate u
z
across the area of the slit to obtain Q.
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 Spring '08
 dicarlo
 Fluid Dynamics, τxz, uz

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