# 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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gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

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

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.

bennyaus | Student, Graduate | eNotes Newbie

Posted on

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