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A pump is running at a speed of 4800 rpm and delivers 2.5 m3/s of water under a head of 20 m. The power input to a pump (in kW) at a shaft speed of 1600 rpm is (assume, pump efficiency = 85 %)
Correct answer is between '21.35,21.39'. Can you explain this answer?
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A pump is running at a speed of 4800 rpm and delivers 2.5 m3/s of wate...
Given, η0 = 0.85, Q = 2.5 m3/s, H = 20 m, N1 = 4800 rpm, N
Power at 4800 rpm,
2
= 1600 rpmPower at 4800 rpm,
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A pump is running at a speed of 4800 rpm and delivers 2.5 m3/s of wate...
Given data:
- Pump speed (N1) = 4800 rpm
- Water flow rate (Q1) = 2.5 m^3/s
- Head (H1) = 20 m
- Pump efficiency (η) = 85%
Conversion of pump speed:
To calculate the power input at a different speed, we need to convert the pump speed N1 to the desired speed N2. This can be done using the affinity laws for pumps:
N1/N2 = (Q1/Q2) * (H1/H2)^0.5
Given:
- N1 = 4800 rpm
- N2 = 1600 rpm
Rearranging the formula, we get:
N2 = N1 * (Q2/Q1) * (H2/H1)^0.5
Calculating the flow rate:
Using the affinity laws, we can calculate the flow rate at the desired speed:
Q2 = Q1 * (N2/N1) * (H2/H1)^0.5
Given:
- Q1 = 2.5 m^3/s
- N1 = 4800 rpm
- N2 = 1600 rpm
- H1 = 20 m
Substituting the values:
Q2 = 2.5 * (1600/4800) * (H2/20)^0.5
Calculating the head:
Given:
- H1 = 20 m
Using the affinity laws, we can calculate the head at the desired speed:
H2 = H1 * (N2/N1)^2
Substituting the values:
H2 = 20 * (1600/4800)^2
Calculating the power input:
The power input to the pump can be calculated using the following formula:
P = (ρ * g * Q * H) / η
Given:
- ρ = density of water (assumed to be 1000 kg/m^3)
- g = acceleration due to gravity (9.81 m/s^2)
- Q = flow rate (Q2)
- H = head (H2)
- η = pump efficiency (85%)
Substituting the values:
P = (1000 * 9.81 * Q2 * H2) / η
Final calculation:
Substituting the calculated values of Q2 and H2 into the power formula, we can find the power input:
P = (1000 * 9.81 * (2.5 * (1600/4800) * (H2/20)^0.5) * (20 * (1600/4800)^2)) / 0.85
Calculating the above expression will give the power input in watts. To convert it to kilowatts, divide the answer by 1000.
After performing the calculations, the result should be between 21.35 kW and 21.39 kW.
- Pump speed (N1) = 4800 rpm
- Water flow rate (Q1) = 2.5 m^3/s
- Head (H1) = 20 m
- Pump efficiency (η) = 85%
Conversion of pump speed:
To calculate the power input at a different speed, we need to convert the pump speed N1 to the desired speed N2. This can be done using the affinity laws for pumps:
N1/N2 = (Q1/Q2) * (H1/H2)^0.5
Given:
- N1 = 4800 rpm
- N2 = 1600 rpm
Rearranging the formula, we get:
N2 = N1 * (Q2/Q1) * (H2/H1)^0.5
Calculating the flow rate:
Using the affinity laws, we can calculate the flow rate at the desired speed:
Q2 = Q1 * (N2/N1) * (H2/H1)^0.5
Given:
- Q1 = 2.5 m^3/s
- N1 = 4800 rpm
- N2 = 1600 rpm
- H1 = 20 m
Substituting the values:
Q2 = 2.5 * (1600/4800) * (H2/20)^0.5
Calculating the head:
Given:
- H1 = 20 m
Using the affinity laws, we can calculate the head at the desired speed:
H2 = H1 * (N2/N1)^2
Substituting the values:
H2 = 20 * (1600/4800)^2
Calculating the power input:
The power input to the pump can be calculated using the following formula:
P = (ρ * g * Q * H) / η
Given:
- ρ = density of water (assumed to be 1000 kg/m^3)
- g = acceleration due to gravity (9.81 m/s^2)
- Q = flow rate (Q2)
- H = head (H2)
- η = pump efficiency (85%)
Substituting the values:
P = (1000 * 9.81 * Q2 * H2) / η
Final calculation:
Substituting the calculated values of Q2 and H2 into the power formula, we can find the power input:
P = (1000 * 9.81 * (2.5 * (1600/4800) * (H2/20)^0.5) * (20 * (1600/4800)^2)) / 0.85
Calculating the above expression will give the power input in watts. To convert it to kilowatts, divide the answer by 1000.
After performing the calculations, the result should be between 21.35 kW and 21.39 kW.
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A pump is running at a speed of 4800 rpm and delivers 2.5 m3/s of water under a head of 20 m. The power input to a pump (in kW) at a shaft speed of 1600 rpm is (assume, pump efficiency = 85 %)Correct answer is between '21.35,21.39'. Can you explain this answer?
Question Description
A pump is running at a speed of 4800 rpm and delivers 2.5 m3/s of water under a head of 20 m. The power input to a pump (in kW) at a shaft speed of 1600 rpm is (assume, pump efficiency = 85 %)Correct answer is between '21.35,21.39'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A pump is running at a speed of 4800 rpm and delivers 2.5 m3/s of water under a head of 20 m. The power input to a pump (in kW) at a shaft speed of 1600 rpm is (assume, pump efficiency = 85 %)Correct answer is between '21.35,21.39'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A pump is running at a speed of 4800 rpm and delivers 2.5 m3/s of water under a head of 20 m. The power input to a pump (in kW) at a shaft speed of 1600 rpm is (assume, pump efficiency = 85 %)Correct answer is between '21.35,21.39'. Can you explain this answer?.
A pump is running at a speed of 4800 rpm and delivers 2.5 m3/s of water under a head of 20 m. The power input to a pump (in kW) at a shaft speed of 1600 rpm is (assume, pump efficiency = 85 %)Correct answer is between '21.35,21.39'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A pump is running at a speed of 4800 rpm and delivers 2.5 m3/s of water under a head of 20 m. The power input to a pump (in kW) at a shaft speed of 1600 rpm is (assume, pump efficiency = 85 %)Correct answer is between '21.35,21.39'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A pump is running at a speed of 4800 rpm and delivers 2.5 m3/s of water under a head of 20 m. The power input to a pump (in kW) at a shaft speed of 1600 rpm is (assume, pump efficiency = 85 %)Correct answer is between '21.35,21.39'. Can you explain this answer?.
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A pump is running at a speed of 4800 rpm and delivers 2.5 m3/s of water under a head of 20 m. The power input to a pump (in kW) at a shaft speed of 1600 rpm is (assume, pump efficiency = 85 %)Correct answer is between '21.35,21.39'. Can you explain this answer?, a detailed solution for A pump is running at a speed of 4800 rpm and delivers 2.5 m3/s of water under a head of 20 m. The power input to a pump (in kW) at a shaft speed of 1600 rpm is (assume, pump efficiency = 85 %)Correct answer is between '21.35,21.39'. Can you explain this answer? has been provided alongside types of A pump is running at a speed of 4800 rpm and delivers 2.5 m3/s of water under a head of 20 m. The power input to a pump (in kW) at a shaft speed of 1600 rpm is (assume, pump efficiency = 85 %)Correct answer is between '21.35,21.39'. Can you explain this answer? theory, EduRev gives you an
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