# Water is pumped at a rate of 60L/s through the system. head loss in the suction line is 0.8m and in the delivery line is 4.5m. the pump has an efficiency of 72%. determine a) the pressure at inlet...

Water is pumped at a rate of 60L/s through the system. head loss in the suction line is 0.8m and in the delivery line is 4.5m. the pump has an efficiency of 72%.

determine a) the pressure at inlet p1

b) the pressure at outlet p2

c) power required to drive the pump.

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This numerical is based on the application of Bernoulli's Theorem.

For part 1)

Applying the Bernoulli's Theorem at left hand tank and the pump inlet, we get

`P_3/(rhog) + v_3^2/(2g) +z_3 = P_1/(rhog) + v_1^2/(2g) + z_1 + h_l`

Here both the sections are at the same datum and hence datum head will cancel out

v3 = 0 (no flow, stationary condition)

v1 = Q/A = `(60*10^-3 )/ (pi/4 (0.2)^2)= 1.9 m/s`

and head loss is given as 0.8 m.

Substituting all the values in the equation, we get

**P1 = 9.95 kPa**

Similarly applying the Bernoulli's Theorem on pump outlet and right side tank,

`P_2/(rhog) + v_2^2/(2g) + Z_2 = P_4/(rhog) + v_4^2/(2g) + Z_4 +h_l`

Here, Z2 = 3m, Z4 = 7 m, P4 = 35 kPa and hl = 4.5 m

v2 = Q/A = `(60*10^-3)/(pi/4 0.15^2) = 3.39 m/s`

Substituting all the values, we get **P2 = 142.05 kPa**

The pump capacity can be determined by the head that it has to provide. This head will include, suction head, delivery head and head loss.

Here, total head loss = 4.5 + 0.8 = 5.3 m.

delivery head = 35 kPa = 3.57 m

and the difference in datum = 10.5 m-5 m = 5.5 m

Pump capacity = `(rhoQh)/(eff) = (9.81*0.06*(5.5+5.3+3.57))/0.72 = 11.75 kW`

Hope this helps.

I am not getting different answer for p1 and p2 ,when solving it

so how you find P3, then only we can solve p1 in first equation