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A pump having an efficiency of 90%, lifts water to a heightof 155m at the rate of 7.5 m^3/s the friction head loss in the pipe is 13m the required pump power in kw will be?
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Problem Statement:

A pump having an efficiency of 90%, lifts water to a height of 155m at the rate of 7.5 m^3/s the friction head loss in the pipe is 13m the required pump power in kW will be?


Solution:

Given Parameters:


  • Efficiency of Pump (η) = 90%

  • Height of Water Lift (H) = 155m

  • Rate of Water Flow (Q) = 7.5 m^3/s

  • Friction Head Loss (hf) = 13m



Step 1: Calculate Total Head (Ht)

Total Head (Ht) is the sum of height lift and head loss due to friction in the pipe.

Ht = H + hf = 155m + 13m = 168m


Step 2: Calculate the Pump Power (P)

The formula for pump power is:

P = (Q x Ht x ρ) / η


  • Q = Rate of Water Flow = 7.5 m^3/s

  • Ht = Total Head = 168m

  • ρ = Density of Water = 1000 kg/m^3

  • η = Efficiency of Pump = 0.9


Substituting the given values, we get:

P = (7.5 x 168 x 1000) / 0.9 = 1400000 W = 1400 kW


Step 3: Final Answer

Therefore, the required pump power in kW is 1400 kW.


Explanation:

The given problem required us to calculate the required pump power in kW for a pump which lifts water to a height of 155m at the rate of 7.5 m^3/s. The friction head loss in the pipe is also given as 13m. We first calculated the total head which is the sum of height lift and head loss due to friction in the pipe. Then we used the formula for pump power which is the product of rate of water flow, total head, density of water and reciprocal of pump efficiency. Finally, we substituted the given values in the formula and arrived at the answer.
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