
Concept explainers
(a)
Value of nonzero velocity,

Answer to Problem 4.2CP
The value of nonzero velocity is,
Explanation of Solution
Assume the flow to be parallel,
Due to this, the momentum equation along x and z directions can be neglected.
Consider the y momentum equation,
Substitute 0 for u, 0 for w, 0 for
Integrate the second-order linear differential equation with respect to y, it gives:
Δ
Integrate again gives:
Δ
By applying the 1st boundary condition,
Δ
By applying the2nd boundary conditions,
Δ
Substitute,
(b)
The average velocity

Answer to Problem 4.2CP
The average velocity is,
Explanation of Solution
Based on the knowledge of calculus, the average value on an interval [a,b] of an integrable function f is given by:
Likewise, the average velocity,
Therefore,
(c)
Velocity,

Answer to Problem 4.2CP
Velocity,
Explanation of Solution
With the relationship of the velocity distribution obtained in part (a), the equation of flow-rate per unit width, Q, can be formulated from relation,
Since, there is no net flow either up or down, therefore,
(d)
Sketch the

Explanation of Solution
The equation which describe the velocity distribution, u(y), which derived in part (a) can be expressed in dimensionless form by dividing both sides of the equation by the velocity, V.
Where,
The velocity distribution of the film at various y-location which has the domain range from
The results are tabulated below:
x/h | u/V | ||||
A=0 | A=0.5 | A=1 | A=1.5 | A=2 | |
0 | 1 | 1 | 1 | 1 | 1 |
0.05 | 1 | 0.95125 | 0.9025 | 0.85375 | 0.805 |
0.1 | 1 | 0.905 | 0.81 | 0.715 | 0.62 |
0.15 | 1 | 0.86125 | 0.7225 | 0.58375 | 0.445 |
0.2 | 1 | 0.82 | 0.64 | 0.46 | 0.28 |
0.25 | 1 | 0.78125 | 0.5625 | 0.34375 | 0.125 |
0.3 | 1 | 0.745 | 0.49 | 0.235 | -0.02 |
0.35 | 1 | 0.71125 | 0.4225 | 0.13375 | -0.155 |
0.4 | 1 | 0.68 | 0.36 | 0.04 | -0.28 |
0.45 | 1 | 0.65125 | 0.3025 | -0.04625 | -0.395 |
0.5 | 1 | 0.625 | 0.25 | -0.125 | -0.5 |
0.55 | 1 | 0.60125 | 0.2025 | -0.19625 | -0.595 |
0.6 | 1 | 0.58 | 0.16 | -0.26 | -0.68 |
0.65 | 1 | 0.56125 | 0.1225 | -0.31625 | -0.755 |
0.7 | 1 | 0.545 | 0.09 | -0.365 | -0.82 |
0.75 | 1 | 0.53125 | 0.0625 | -0.40625 | -0.875 |
0.8 | 1 | 0.52 | 0.04 | -0.44 | -0.92 |
0.85 | 1 | 0.51125 | 0.0225 | -0.46625 | -0.955 |
0.9 | 1 | 0.505 | 0.01 | -0.485 | -0.98 |
0.95 | 1 | 0.50125 | 0.0025 | -0.49625 | -0.995 |
1 | 1 | 0.5 | 0 | -0.5 | -1 |
Dimensionless Velocity Distribution (Velocity Profile):
As illustrated on the plot of velocity profile of fluid flow shown above, there are still some portions of the fluid flow downward (
Conclusion:
It can be concluded that in order to lift a fluid with small viscosity, a relatively large belt speed is required.
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Chapter 4 Solutions
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