Question

Q1- A pipe with D1= 100 mm carries water with a flow of 1415 Litre/min. With a sudden enlargement, it is connected to another pipe with D2 = 260 mm. a) Calculate the head loss in this connection. Express your answer in [m]. b) If a pump with efficiency of 75% is used to compensate this energy loss only, how much mechanical power is needed to run the pump. Express the answer in [kW].

Answer 1

Given

d1 = 100 mm =0.1m

d2 = 260 mm = 0.26 m

Flow = 1415 litre/min =1415* 0.001 m3 /min = 0.02358333333 m3/sec

A1 V1 =  $\pi /4$ d1 2 V 1 = 0.0078539816 m 2 V 1 = 0.02358333333 m3/sec

V1 = 3.0235 m/sec

  

From Continuity Equation , we have

A1V1=A2V2

$\pi /4$  d1 2 V 1 2 =  $\pi /4$ d2 2 V 22

1002 V 1  2 = 2602  V 22

( V1/V2 )2 = ( 260/100)2

( V1/V2 ) = 13/5

V 1 = 13/5 V 2

V 2 = 5/13 V1

head loss due to sudden enlargement is given by following formula written below

head loss = (V1- V2)2/2g

= (13/5 V 2 - V 2)2 /2g   

= (V1 - 5/13 V1)2 /2 g

= (8/13 V 1)2 /2g

= (8/13 * 3.0235 ) 2/2g

=   0.176m

efficiency = power output/power input

75/100 = power output /power input

power input = power output *100/75 =100/75* q ρ g h = 100/75 * 0.0235833 * 1000*9.81* 0.176m

= 52.947 KW

where,

q = flow = 0.0235833 m3/sec

ρ = density of fluid =1000kg/m3

g = acceleration of gravity = 9.8 m/ s 2

h = head = 0.176 m