CHEG232 Fluid Mechanics: Water Flow Rate Calculation in Pipe System

This assignment solution calculates the volume flow rate of water in a pipe system using Bernoulli's equation and considers factors like pipe diameter, length, roughness, and elevation differences. The solution addresses three scenarios: determining the initial flow rate, calculating the flow rate with a 20m elevation difference, and assessing the flow rate if the pipe is made of cast iron with the same elevation difference. Calculations include major and minor losses, friction factors derived from the Moody chart, and the application of the continuity equation. The final results provide the volume flow rate in cubic meters per second for each scenario, demonstrating the impact of elevation and pipe material on the flow characteristics within the system. This detailed solution, ideal for students studying fluid mechanics, is available on Desklib, which also offers a range of solved assignments and study resources.

Determine the volume flow rate of water (T = 40oC) flowing in the system below in (m3/min).
The information provided is as follows
Temperature = 400 C
Diameter of the smaller pipe = 4 in
Diameter of the larger pipe = 16 in
The length of the smaller pipe = 43 m
The length of the larger pipe = 25 m
Bernoulli’s equation states that
P 1
eg + V 12
2 g +Z 1 = constant
P 1
eg + V 12
2 g + Z1 = P 2
eg +V 22
2 g +Z2
0 + 0 + 15 = 0 + 0 +0
∑ hL = 15 m
For ductile pipe
Pipe roughness size = 0.15 mm
From equation of continuity
A1V1 = A2V2
V2 = A1/A2*V1
V2= 0.444V1
Major loss
Pipe 1

f 1 ( L1
D 1 ) ( V 2 1
2 g )
f*
43
0.1016 ∗v2
2 g
= 21.5713fV2
Pipe 2
f 2 ( L2
D 2 ) ( V 2 2
2 g )
f*
25
0.1016 ∗(0.444 V 1)2
2 g
= 2.4724fV12
Total major loss = 21.5713fV2 + 2.4724fV12 = 24.0437fV21
Minor loss
1. At the Entry
hentry = Kentry* v2 1
2 g
pipe 1
at sharp entry
Kentry = 0.5
hentry = 0.5* v2
2 g = 0.0255V21
Pipe 2

hentry = 0.5* (0.444 v 1)2
2 g = 0.005024V21
2. At exit
Pipe 1
Kexit = 1
hexit = 1.0* v 12
2 g = 0.051V21
Pipe 2
1.0*(0.444 v 1)2
2 g = 0.01005V21
3. Sudden expansion
hes = V 1−v 2
2 g
= v 1−0.444 v 1
2 g
= 0.02834V1
4. At valve
hvalve = 10∗V 2
29 = 0.51V21
hm for the two pipes
hm = (0.0255 + 0.005024 +0.051 + 0.01005+ 0.51 + 0.02834) = 0.629914V21
∑ hL = hf + 0.629914V21 = 15 m

e
D = 0.15
101.6 = 0.0015
From moody chart
f = 0.02075
Substituting to 24.0437*0.02075*V21= 0.49891V21
V21 = 15
0.49891+ 0.629914 = 13.288
V = 3.65 m/s
Flow rate = cross-section area of the pipe * velocity
Cross section area for 4 in diameter pipe
A = π D2
4 = π∗0.10162
4 = 0.0081073 m2
Q = 0.0081073 * 3.65 = 0.0296 m3/s
What would the flow if the elevation difference between A and B is 20 m?
Using the excel
Q = 0.034125429 m3/s
What would be the flow if the pipe is made of cast iron (for the 20m elevation)?
For cast iron e = 0.26 mm
e
D = 0.26
101.6 = 0.0026
From moody chart
f = 0.05075
Substituting to 24.0437*0.05075*V21= 1.22022V2

V2 = 20
1.22022+0.629914 = 10.81
V = 3.29 m/s
Flow rate = cross-section area of the pipe * velocity
Cross section area 6 in diameter pipe
A = π D2
4 = π∗0.10162
4 = 0.0081073 m2
Q = 0.0081073 * 3.29 = 0.02667 m3/s