Question

1. A fire engine develops a head of 35 m, i.e., it increases the energy per unit weight of the water
passing through it by 35 m. The pump draws water from a sump at A through a 125 mm diameter
pipe in which there is a loss of energy per unit weight due to friction h1 = 6V1
2
/2g varying with
the mean velocity V1 in the pipe, and discharges it through a 75 mm nozzle at C, 28 m above the
pump, at the end of a 100 mm diameter delivery pipe in which there is a loss of energy per unit
weight h2 = 8V2
2
/2g. Calculate,
a) The velocity of the jet issuing from the nozzle at C,
b) The pressure in the suction pipe at the inlet to the pump at B.

Answer 1

Ans a) Apply Bernoulli equation between point A and C respectively,

/ + ^2 / 2 g + + Hp = / + ^2 / 2 g + + Hf

Since, both point 1 and 2 are open to atmosphere, pressure is only atmospheric hence gauge pressure = = 0

Also, there is no velocity at surface so = 0

Elevation , = 0 m and = 28 + 2 = 30 m

Hf is Friction head loss

Hp is head added by pump = 35 m

Hence,above equation reduce to,

0 + 0 + 35 = ^2 / 2 g + Hf + 30

=>   / 2 g + Hf = 5 .....................................(1)

Also, Hf = 6 /2g + 8 / 2g

Now,according to contunity equation,

A1 V1 = A2 V2 = Ac Vc

=> (/4) (125)^2 x V1 = (/4) (100)^2 x V2 = (/4) (75)^2 x Vc

=> V1 = 0.64 V2 = 0.36 Vc

=> V1 = 0.36 Vc and V2 = 0.5625 Vc

Hence, Hf = 6 /2g + 8 / 2g

=> Hf = 0.1686

Putting value in equation 1,

=>   / 2 g + 0.1686 = 5

=> 0.1296 = 5

=> = 38.58

=> = 6.21 m/s

Also, V1 = 0.36(6.21) = 2.235 m/s

V2 = 0.5625(6.21) = 3.49 m/s

Ans b) To determine pressure in suction pipe B, apply Bernoulli equation between point A and B,

    / + ^2 / 2 g +   = / + ^2 / 2 g + + 6 V1^2 / 2 g

Velocity at point B = V1 = 2.235 m/s

=> 0 + 0 + 0 = / +  [/ 2 g] + 2 + [6 / 2 g]

=> / + / (2 x 9.81) + 2 + 6 / (2 x 9.81) = 0

   => / + 0.2545 + 2 + 1.5276 = 0

=>  / = - 3.7821

We know, unit weight of water, = 9.81 kN/

=> = 9810 x (-3.7821)

=> =  - 37102.4 Pa or -37.10 kPa