
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.
Want to see more full solutions like this?
Chapter 4 Solutions
Fluid Mechanics
- Solve, use engineering economic tablesarrow_forwardSolve, use engineering economic tablesarrow_forwardA pinion has a pressure angle of 20 degrees a module of 3mm and 20 teeth. It is meshed with a gear having 32 teeth. The center distance between the shafts is 81mm. Determine the gear ratio and diametral pitch .arrow_forward
- USE MATHLAB WITH CODES Estimate the damping ratio, stiffness, natural frequency, and mass of the SDOF system. Please use a MATHLAB with CODES and no negative damping ratio. Data Set 1:Time(s) Data Set 1:top1(g) Data Set 1:bottom(g)0 0.002593181 0.007262860.01 0.011367107528507709 -0.0015110660.02 0.007467585 -0.0058980290.029999999999999999 0.004542943 0.0028758970.040000000000000001 0.018678712689042091 -0.0019985060.050000000000000003 0.004542943 0.0009261360.059999999999999998 0.014779189431130886 -0.0068729090.070000000000000007 0.004055502 -0.0088226710.080000000000000002 0.008442465 -0.0015110660.089999999999999997 0.011854547366917134 -0.0039482670.10000000000000001 0.007467585 0.0058005390.11 0.004055502 0.0043382180.12 0.010392226334810257 0.0019010160.13 0.010392226334810257 -0.001998506% 0.14000000000000001 0.016728950301647186 0.0048256580.14999999999999999 0.007955025…arrow_forwardProvide an example of at least five features produced by a certain machining process (for example, a keyway to accommodate a key iarrow_forwardHow to draw a gam from the data of the subject's readings three times and difficulties in drawing a gam Material Name: Machinery Theory I'm a vehicle engineering student. Please describe details about gam in addition the law gam: 1-tangent cam with reciprocating roller follower. 2-circular arc cam with flat-faced follower.arrow_forward
- Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University PressMechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSONThermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
- Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEYMechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage LearningEngineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY





