Science, asked by anshucool193, 1 month ago

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

Answered by Rameshjangid
0

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,

qm \:  = qp\times ( \frac{nm }{np} ) \times ( \frac{dm}{dp})^{3}

Substitute the known values.

qm = (1) \times ( \frac{1450}{600} ) \times ( \frac{1}{4} )^{3}

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,

hm \:  = (hp) \times ( \frac{nm}{np} )^{2}  \times ( \frac{dm}{dp})^{2}

Substitute the known values.

hm \:  = (30) \times ( \frac{1450}{600} )^{2}  \times ( \frac{1}{4} ) ^{2}

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,

pp \:  =  \frac{γ \times qp \times hp}{ηo}

Substitute the known values.

pp \:  =  \frac{9790 \times 1 \times 30}{0.80}

pp = 367125 W

The expression for the power consumed by the model pump is,

pm \:  = pp \times ( \frac{nm}{np} )^{3}  \times ( \frac{dm}{dp})^{5}

Substitute the known values.

pp \:  = 367125  \times (\frac{1450}{600} )^{3}  \times ( \frac{1}{4} )^{5}

pp = 5060.16 W

Thus, the power consumed by the model pump is

5060.16 W.

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