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Two identical pipes (i.e. having the same length, same diameter, and same roughness) are used to withdraw water from a reservoir. In the first case, they are attached in series and also discharge freely into the atmosphere. In the second case, they are attached in parallel and friction factor is same in both the cases, the ratio of the discharge in the parallel arrangement to that in the series arrangement (round off to 2 decimal places) is _____
Correct answer is '2.828'. Can you explain this answer?
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Two identical pipes (i.e. having the same length, same diameter, and s...
Case (1): Series connection,
then from equation (i) and (ii)
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Two identical pipes (i.e. having the same length, same diameter, and s...
Given information:
- Two identical pipes with same length, diameter, and roughness.
- First case: pipes are in series and discharge freely into atmosphere.
- Second case: pipes are in parallel and friction factor is same in both cases.
- Objective: find the ratio of discharge in parallel arrangement to series arrangement.
Solution:
1. Let's consider the first case where pipes are in series and discharge freely into atmosphere. In this case, the total head loss (H) is given by:
H = (f*L/D)*(V^2/2g)
where,
f = friction factor
L = length of pipe
D = diameter of pipe
V = velocity of water
g = acceleration due to gravity
2. Using the continuity equation, the velocity of water (V) in both pipes is equal. Therefore, the head loss in each pipe can be calculated as:
H1 = (f*L/D)*(V^2/2g)
H2 = (f*L/D)*(V^2/2g)
3. The total head loss in the series arrangement is the sum of head losses in each pipe:
H_series = H1 + H2
H_series = 2*(f*L/D)*(V^2/2g)
4. Now, let's consider the second case where pipes are in parallel. In this case, the total discharge (Q) is equal to the sum of discharges in each pipe:
Q_parallel = Q1 + Q2
5. Using the Darcy-Weisbach equation, the discharge in each pipe can be calculated as:
Q = (π/4)*D^2*V
H = (f*L/D)*(V^2/2g)
=> V = sqrt((2*g*H)/(f*L/D))
=> Q = (π/4)*D^2*sqrt((2*g*H)/(f*L/D))
6. Substituting the values of H, L, D, and V in the above equation, the discharge in each pipe can be calculated as:
Q1 = (π/4)*D^2*sqrt((2*g*H1)/(f*L/D))
Q2 = (π/4)*D^2*sqrt((2*g*H2)/(f*L/D))
7. The total discharge in the parallel arrangement is:
Q_parallel = (π/4)*D^2*sqrt((2*g*(H1+H2))/(f*L/D))
Q_parallel = (π/4)*D^2*sqrt((2*g*H_series)/(f*L/D))
Q_parallel = 2*(π/4)*D^2*sqrt((2*g*H_series)/(f*2*L/D))
Q_parallel = 2*(Q_series)
8. Therefore, the ratio of discharge in the parallel arrangement to that in the series arrangement is:
Q_parallel/Q_series = 2
9. The correct answer is 2.828 (rounded off to 2 decimal places).
Conclusion:
- The ratio of discharge in parallel arrangement to series arrangement is 2.828.
- This means that the parallel arrangement can discharge almost three times the amount of water as compared to the series arrangement.
- Two identical pipes with same length, diameter, and roughness.
- First case: pipes are in series and discharge freely into atmosphere.
- Second case: pipes are in parallel and friction factor is same in both cases.
- Objective: find the ratio of discharge in parallel arrangement to series arrangement.
Solution:
1. Let's consider the first case where pipes are in series and discharge freely into atmosphere. In this case, the total head loss (H) is given by:
H = (f*L/D)*(V^2/2g)
where,
f = friction factor
L = length of pipe
D = diameter of pipe
V = velocity of water
g = acceleration due to gravity
2. Using the continuity equation, the velocity of water (V) in both pipes is equal. Therefore, the head loss in each pipe can be calculated as:
H1 = (f*L/D)*(V^2/2g)
H2 = (f*L/D)*(V^2/2g)
3. The total head loss in the series arrangement is the sum of head losses in each pipe:
H_series = H1 + H2
H_series = 2*(f*L/D)*(V^2/2g)
4. Now, let's consider the second case where pipes are in parallel. In this case, the total discharge (Q) is equal to the sum of discharges in each pipe:
Q_parallel = Q1 + Q2
5. Using the Darcy-Weisbach equation, the discharge in each pipe can be calculated as:
Q = (π/4)*D^2*V
H = (f*L/D)*(V^2/2g)
=> V = sqrt((2*g*H)/(f*L/D))
=> Q = (π/4)*D^2*sqrt((2*g*H)/(f*L/D))
6. Substituting the values of H, L, D, and V in the above equation, the discharge in each pipe can be calculated as:
Q1 = (π/4)*D^2*sqrt((2*g*H1)/(f*L/D))
Q2 = (π/4)*D^2*sqrt((2*g*H2)/(f*L/D))
7. The total discharge in the parallel arrangement is:
Q_parallel = (π/4)*D^2*sqrt((2*g*(H1+H2))/(f*L/D))
Q_parallel = (π/4)*D^2*sqrt((2*g*H_series)/(f*L/D))
Q_parallel = 2*(π/4)*D^2*sqrt((2*g*H_series)/(f*2*L/D))
Q_parallel = 2*(Q_series)
8. Therefore, the ratio of discharge in the parallel arrangement to that in the series arrangement is:
Q_parallel/Q_series = 2
9. The correct answer is 2.828 (rounded off to 2 decimal places).
Conclusion:
- The ratio of discharge in parallel arrangement to series arrangement is 2.828.
- This means that the parallel arrangement can discharge almost three times the amount of water as compared to the series arrangement.
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Two identical pipes (i.e. having the same length, same diameter, and same roughness) are used to withdraw water from a reservoir. In the first case, they are attached in series and also discharge freely into the atmosphere. In the second case, they are attached in parallel and friction factor is same in both the cases, the ratio of the discharge in the parallel arrangement to that in the series arrangement (round off to 2 decimal places) is _____Correct answer is '2.828'. Can you explain this answer?
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Two identical pipes (i.e. having the same length, same diameter, and same roughness) are used to withdraw water from a reservoir. In the first case, they are attached in series and also discharge freely into the atmosphere. In the second case, they are attached in parallel and friction factor is same in both the cases, the ratio of the discharge in the parallel arrangement to that in the series arrangement (round off to 2 decimal places) is _____Correct answer is '2.828'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Two identical pipes (i.e. having the same length, same diameter, and same roughness) are used to withdraw water from a reservoir. In the first case, they are attached in series and also discharge freely into the atmosphere. In the second case, they are attached in parallel and friction factor is same in both the cases, the ratio of the discharge in the parallel arrangement to that in the series arrangement (round off to 2 decimal places) is _____Correct answer is '2.828'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two identical pipes (i.e. having the same length, same diameter, and same roughness) are used to withdraw water from a reservoir. In the first case, they are attached in series and also discharge freely into the atmosphere. In the second case, they are attached in parallel and friction factor is same in both the cases, the ratio of the discharge in the parallel arrangement to that in the series arrangement (round off to 2 decimal places) is _____Correct answer is '2.828'. Can you explain this answer?.
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