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An idealized centrifugal pump (blade outer radius of 50 mm) consumes 2 kW power while running at 3000 rpm. The entry of the liquid into the pump is axial and exit from the pump is radial with respect to impeller. If the losses are neglected, then the mass flow rate of the liquid through the pump is ________ kg/s (round off to two decimal places).
Correct answer is '8.106'. Can you explain this answer?
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Verified Answer
An idealized centrifugal pump (blade outer radius of 50 mm) consumes 2...
Diameter = 2×radius of blade=2×50=100mm=0.1m
Now, power = mass flow rate × u2
⇒2000=m×(15.71)2
Or, m=8.106
Now, power = mass flow rate × u2
⇒2000=m×(15.71)2
Or, m=8.106
Most Upvoted Answer
An idealized centrifugal pump (blade outer radius of 50 mm) consumes 2...
Given:
- Power consumed by the pump: 2 kW
- Pump speed: 3000 rpm
- Blade outer radius: 50 mm
To find:
The mass flow rate of the liquid through the pump.
Assumptions:
- Negligible losses in the pump
- Steady-state operation
- Incompressible flow
Approach:
The mass flow rate through the pump can be determined using the power consumption and pump speed. The power consumed by the pump is equal to the product of the mass flow rate, head, and acceleration due to gravity. Since the losses are neglected, the head can be assumed to be constant.
Calculation:
Given that the power consumed by the pump is 2 kW, we can convert it to watts:
Power = 2 kW = 2000 W
The pump speed is given as 3000 rpm. To convert it to radians per second, we can use the following formula:
Angular speed (ω) = 2πN/60, where N is the speed in rpm
ω = (2π * 3000) / 60 = 100π rad/s
The head (H) can be assumed to be constant since losses are neglected.
The power consumed by the pump is given as the product of mass flow rate (ṁ), head (H), and acceleration due to gravity (g):
Power = ṁ * H * g
Rearranging the equation, we can solve for the mass flow rate:
ṁ = Power / (H * g)
Since the head is constant and the acceleration due to gravity is known, we can substitute the given values:
ṁ = 2000 / (H * 9.81)
To find the head (H), we can use the following equation for a centrifugal pump:
H = (Vout^2 - Vin^2) / (2g)
Given that the entry of the liquid into the pump is axial and the exit is radial, we can assume that the velocity at the entry (Vin) is zero. Hence, the head simplifies to:
H = Vout^2 / (2g)
The velocity at the exit (Vout) can be calculated using the equation for the tangential velocity of the impeller:
Vout = ω * R, where R is the blade outer radius
Substituting the known values, we get:
Vout = 100π * 0.05 = 5π m/s
Finally, substituting the values of H and g into the equation for mass flow rate, we get:
ṁ = 2000 / ((5π)^2 / (2 * 9.81)) = 8.106 kg/s (rounded to two decimal places)
Hence, the mass flow rate of the liquid through the idealized centrifugal pump is 8.106 kg/s.
- Power consumed by the pump: 2 kW
- Pump speed: 3000 rpm
- Blade outer radius: 50 mm
To find:
The mass flow rate of the liquid through the pump.
Assumptions:
- Negligible losses in the pump
- Steady-state operation
- Incompressible flow
Approach:
The mass flow rate through the pump can be determined using the power consumption and pump speed. The power consumed by the pump is equal to the product of the mass flow rate, head, and acceleration due to gravity. Since the losses are neglected, the head can be assumed to be constant.
Calculation:
Given that the power consumed by the pump is 2 kW, we can convert it to watts:
Power = 2 kW = 2000 W
The pump speed is given as 3000 rpm. To convert it to radians per second, we can use the following formula:
Angular speed (ω) = 2πN/60, where N is the speed in rpm
ω = (2π * 3000) / 60 = 100π rad/s
The head (H) can be assumed to be constant since losses are neglected.
The power consumed by the pump is given as the product of mass flow rate (ṁ), head (H), and acceleration due to gravity (g):
Power = ṁ * H * g
Rearranging the equation, we can solve for the mass flow rate:
ṁ = Power / (H * g)
Since the head is constant and the acceleration due to gravity is known, we can substitute the given values:
ṁ = 2000 / (H * 9.81)
To find the head (H), we can use the following equation for a centrifugal pump:
H = (Vout^2 - Vin^2) / (2g)
Given that the entry of the liquid into the pump is axial and the exit is radial, we can assume that the velocity at the entry (Vin) is zero. Hence, the head simplifies to:
H = Vout^2 / (2g)
The velocity at the exit (Vout) can be calculated using the equation for the tangential velocity of the impeller:
Vout = ω * R, where R is the blade outer radius
Substituting the known values, we get:
Vout = 100π * 0.05 = 5π m/s
Finally, substituting the values of H and g into the equation for mass flow rate, we get:
ṁ = 2000 / ((5π)^2 / (2 * 9.81)) = 8.106 kg/s (rounded to two decimal places)
Hence, the mass flow rate of the liquid through the idealized centrifugal pump is 8.106 kg/s.
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An idealized centrifugal pump (blade outer radius of 50 mm) consumes 2 kW power while running at 3000 rpm. The entry of the liquid into the pump is axial and exit from the pump is radial with respect to impeller. If the losses are neglected, then the mass flow rate of the liquid through the pump is ________ kg/s (round off to two decimal places).Correct answer is '8.106'. Can you explain this answer?
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An idealized centrifugal pump (blade outer radius of 50 mm) consumes 2 kW power while running at 3000 rpm. The entry of the liquid into the pump is axial and exit from the pump is radial with respect to impeller. If the losses are neglected, then the mass flow rate of the liquid through the pump is ________ kg/s (round off to two decimal places).Correct answer is '8.106'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about An idealized centrifugal pump (blade outer radius of 50 mm) consumes 2 kW power while running at 3000 rpm. The entry of the liquid into the pump is axial and exit from the pump is radial with respect to impeller. If the losses are neglected, then the mass flow rate of the liquid through the pump is ________ kg/s (round off to two decimal places).Correct answer is '8.106'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An idealized centrifugal pump (blade outer radius of 50 mm) consumes 2 kW power while running at 3000 rpm. The entry of the liquid into the pump is axial and exit from the pump is radial with respect to impeller. If the losses are neglected, then the mass flow rate of the liquid through the pump is ________ kg/s (round off to two decimal places).Correct answer is '8.106'. Can you explain this answer?.
An idealized centrifugal pump (blade outer radius of 50 mm) consumes 2 kW power while running at 3000 rpm. The entry of the liquid into the pump is axial and exit from the pump is radial with respect to impeller. If the losses are neglected, then the mass flow rate of the liquid through the pump is ________ kg/s (round off to two decimal places).Correct answer is '8.106'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about An idealized centrifugal pump (blade outer radius of 50 mm) consumes 2 kW power while running at 3000 rpm. The entry of the liquid into the pump is axial and exit from the pump is radial with respect to impeller. If the losses are neglected, then the mass flow rate of the liquid through the pump is ________ kg/s (round off to two decimal places).Correct answer is '8.106'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An idealized centrifugal pump (blade outer radius of 50 mm) consumes 2 kW power while running at 3000 rpm. The entry of the liquid into the pump is axial and exit from the pump is radial with respect to impeller. If the losses are neglected, then the mass flow rate of the liquid through the pump is ________ kg/s (round off to two decimal places).Correct answer is '8.106'. Can you explain this answer?.
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