Question

Question a. Two reservoirs are connected together by a pipeline which is 200 mm in diameter and 2000 m long for which the friction factor f, is constant and equal to 0.0045. The head difference between the reservoirs is 25 m and remains constant. Under these conditions, determine the flow in the pipeline. Minor losses can be ignored, however the rejected kinetic energy should be taken into account. b. In this case, is the rejected kinetic energy worth taking into account? c. Halfway along the pipeline there is a small fire hydrant which requires a minimum flow of 7.5 ls and a minimum pressure of 10 kN/m2 above the atmospheric. Determine whether or not the pipeline is able to satisfy these conditions and if so what will be the flow downstream of the hydrant when it is in use. Assume that the level of the hydrant is 15 m below water level of the upstream reservoir. In this case all minor losses and rejected kinetic energy can be ignored. u2 Hint: Rejected kinetic energy ke 29 2942

Answer 1

Ans a)  Apply Bernoulli equation between point 1 and 2 located at water surface elevation of upper and lower reservoir respectively,

P1/$\gamma$ +
V1
/2g + Z1 = P2/$\gamma$ +
V22
/2g +Z2 + Hf

Since, both points are open to atmosphere, pressure is only atmospheric , hence gauge pressure P1 = P2 = 0

Velocity at surface is negligible so V1 = 0

Elevation difference (Z1- Z2) = 25 m

Hf is head loss due to friction

Putting values,

0 + 0 + 25 = 0 + $V^2$ / 2g + Hf

=> Also, Hf =  f L $V^2$ / 2 g D

where, f = friction factor = 0.0045

L = Pipe length = 2000 m

V = Flow velocity

D = Pipe diameter = 200 mm or 0.20 m

=> [0.0045 x 2000 x $V^2$ / (2 x 9.81 x 0.20)] + [$V^2$ / (2 x 9.81)] = 25

=> (9 $V^2$ / 0.20)  + $V^2$   = 25 x 2 x 9.81

=> 46 $V^2$ = 490.5

=> $V^2$ = 10.66

=> V = 3.26 m/s

We know, flow rate (Q) = area x velocity

=> Q = ($\pi$/4)$(0.2)^2$ x 3.26

=> Q = 0.103 $m^3$ /s

Hence, flow rate in pipeline is 0.102 $m^3$ /s

.

Ans b) Rejected kinetic energy = $V^2$ / 2 g

=> $V^2$ / 2 g = $3.3^2$ / (2 x 9.81)

=> $V^2$ /2g = 0.55 m

Since, kinetic energy head (0.55 m) is very less as compared to static elevation head (25 m), kinetic energy head not worth to taken into account

.

Ans c) Apply Bernoulli equation between point 1 and 2 located at water surface elevation of upper reservoir and just befor fire hydrant respectively,

P1/$\gamma$ +
V1
/2g + Z1 = P2/$\gamma$ +
V22
/2g +Z2 + Hf

Again, pressure at point 1, P1 = 0

According to question, neglect Velocity head

Elevation, Z1 = 15 m and Z2 = 0

Hf = (0.0045 x 1000 x $3.26^2$ / (2 x 9.81 x 0.20) = 12.18 m

Putting values,

0 + 0 + 15 = P2/$\gamma$ + 0 + 12.18

=> P2/$\gamma$ = 2.82 m

=> P2 = 9810 x 2.82 = 27664.2 Pa or 27.66 kPa > 10 kPa (OK)

Flow rate at hydrant = 0.102 m3 /s or 102 L/s > 7.5 L/s (OK)

Since, both flow rate and pressure at fire hydrant is more than minimum required values, pipeline is able to satisfy required conditions with downstream flowrate of 102 L/s