It is proposed to design a homologous model for a centrifugal pump. The prototype pump is to run at 600 r.p.m and develop 30 m head, the flow rate being 1 m /s. The model of scale is to run at 1450 r.p.m. Determine the head developed discharge and power required for the model. Overall efficiency = 80%.
Answers
Answer:
The head developed of the model pump is 10.95m.
The power consumed by the model pump is
5060.16 W.
Explanation:
Given:
- The discharge of the prototype pump is (qp) = 1 metre cube per second
- The head developed of the prototype pump is (hp) = 30m
- The overall efficiency of the pump is (ηo) = 0.80
- The speed of the prototype is (np) = 600rpm
- The speed of the model is (nm) = 1450rpm
- The scale factor is (dm/dp) = 1/4
To find:
- The head developed discharge
- Power required for the model
Solution:
Assume, the working fluid is water and its specific weight is (γ) = 9790 N/m^3.
The expression for the discharge of the model is,
Substitute the known values.
qm = 0.377 meter cube per second
Thus, the discharge of the model pump is
0.377 meter cube per second.
The expression for the discharge of the model is,
Substitute the known values.
hm = 10.95 m
Thus, the head developed of the model pump is
10.95m.
The expression for the power consumed by the prototype pump is,
Substitute the known values.
pp = 367125 W
The expression for the power consumed by the model pump is,
Substitute the known values.
pp = 5060.16 W
Thus, the power consumed by the model pump is
5060.16 W.
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