Coventry University: Thermodynamics and Fluid Mechanics Report

This assignment solution focuses on the analysis of fully developed turbulent pipe flow, addressing the objective of investigating and comparing the accuracy of various mathematical approaches in determining mass flow. It involves examining relations for the pipe friction factor and calibrating an orifice plate. Key calculations include shear force, frictional velocity, average velocity, friction factor, Reynolds number, and mass flow rate. The solution also covers the relationship between flow rate and Reynolds number, determination of ideal pipe size, and calculation of major and minor head losses. A schematic diagram of an additional pipe is provided, and the report includes references to relevant fluid mechanics literature. Desklib is a platform where students can find similar assignments and study resources.

Running Head
1 | P a g e
Thermodynamics and fluid mechanics

Running Head
Contents
Solution 1.........................................................................................................................................5
Solution 2.........................................................................................................................................5
Solution 3.........................................................................................................................................6
Solution 4 (Students ID is 4416).....................................................................................................6
References........................................................................................................................................8
2 | P a g e

Running Head
As given in question,
Dia. 95.6
L 10.4
Orifice 95.6 47.8
S no Inverter Freq(Hz) 8 10 24 32
Velocity 2.67 3.33 8.00 10.67
1 Orifice ΔP(Pa) 52 230 535 975
2 ΔP1(Pa) 4.5 15.5 32.5 53.5
3 ΔP2(Pa) 5.21 16.2 33 54
4 ΔP3(Pa) 4.1 12 24 37.5
S no
R(mm)From 0 to
46 Pitot Velocity (m/s)
AVERAG
E
1 20 -5.03 -4.03 -3.89 -4.38 -4.3325
2 15 -5.12 -4.14 -3.63 -4.03 -4.23
3 10 -4.83 -4.05 -3.47 -3.89 -4.06
4 5 -4.92 -4.14 -3.47 -3.63 -4.04
5 0 -4.83 -4.3 -3.47 -3.63 -4.0575
6 5 -5.03 -4.03 -3.47 -4 -4.1325
7 10 -4.6 -4.14 3.47 -4.14 -2.3525
8 15 -4.5 -4.2 3.63 -3.47 -2.135
9 20 -4.27 -4.14 -3.89 -4.03 -4.0825
Figure 1-The pipe network
3 | P a g e

Running Head
From the above equation,
We must consider, maximum frequency which is given as 32, and its sum Δp for different
component = 1120, the distance covered by fluid is 10.4 m.
In this condition,
dp
dz = 1120
10.4 =107.69
We can calculate shear force with the help of equation
τ o=−Rh
2
dP
dz Putting the value from the given problem
τ o=−107.69
2 x 47.8=−¿2573.85/1000 = -2.573 N/m2
The calculated shear force for the given pipe is = 2.573 N/m2,
The frictional velocity can be calculated as
Ur = √ τ
ρ = √ 2.573
1.225 =1.449 m/sec
The velocity profile obtained from graph
4 | P a g e

Running Head
0 1 2 3 4 5 6
0
0.01
0.02
0.03
0.04
0.05
0.06
Velocity Profile
Axial Velocity (m/s)
Radial Position (m)
Figure 2-Velocity profile of given pipe system
Solution 1
The velocity at centre line = 15.22 m/sec
The average velocity can be calculated by
V =U 1−4.07 U r=15.22−4.07 x 1.449=9.12 m/sec
Solution 2
The frictional factor can be calculated by formula
f = 8 to
ρV 2 = 8∗2.573
1.225∗9.122 =0.025253061
The practical value of frictional factor = 0.02525,
As per moody chart the roughness for Aluminium pipe = 0.0015 mm, Then its relative roughness
= 0.0015/47.8 = 0.000538 = 5.38x10-4
Now the Reynolds number can be calculated as follows
5 | P a g e

Running Head
= Re = ρVd
μ = 1.225∗9.12∗0.0478
0.00001534 =348123.6
The from the roughness and Reynolds number as per Moody’s chart, the friction is around 0.017
Ans
Solution 3
The mass flow rate is given by
m=ρπ R2 V Putting the value from above
M = 1.225 x 3.14x (0.0478)2* 9.12 = 0.08015 kg/sec Ans
Solution 4 (Students ID is 4416)
The Reynolds number = 4416,
Horizontal distance = 16 m
Vertical distance = 8 m
a) The relation of flow rate and Reynolds number = ℜ= m∗d
A∗μ
Or m= ℜ∗A∗μ
d Putting the value from above
M = 4416*0.0071744*0.00001534/0.0956 = 0.005084 kg/sec
b) As per the flow rate calculated above, the ideal pipe size will be 7.8 mm outer dia. and 5.4 mm
inner diameter is suitable for above flow rate with tolerance +- 0.005 mm.
c) The major head loss for pipe
6 | P a g e

Running Head
h=−fl V 2
2 gD = 0.017∗24∗9.122
2∗9.81∗0054 =0.3203± 0.05 m
The K value of bend and valve = 0.3
Minor losses in pipes = K∗V 2
2 g = 0.3∗9.122
2∗9.81 =1.27 ± 0.05 m
Total losses = 0.3203 + 1.27 = 1.59 ± 0.05 m Ans
d) The diagram is as given below
7 | P a g e
Figure 3-Schmematic diagram of additional pipe
16 m
08 m

Running Head
References
Cengel, Y, 2017, Fluid Mechanics, 3rd ed, New York: Mc Graw Hill.
Guha, H, a, 2009, The design of a test rig and study of the performance and efficiency of a Tesla
disc turbine, Journal of Power and Energy, 1(1), pp, 451-455.
Philip Pritchard, J, L, 2011, Introduction to fluid Mechanics, 8th ed, New York: John Wiley &
Sons.
White, F, 2013, Fluid Mechanics, 7th ed, New York: Mc Graw Hill.
8 | P a g e