Question

The figure below shows a portion of a fire protection system in which a pump draws water from a reservoir and deliver it to a point Befowo 150 lei pe .) Calculate the required height of the water level in the turke in order to maintain - 38.5 kPa pressure at polit A035.5 kPa below mewhere) Prictions in b) Calculate the power delivered by the pop to the water in order to maintain a pressure increase of 551 ks between points And B. market 792.5 m long in schedule 40 steel pipe 762 13.7 m-long 10 in schedule 40 steel pipe

Answer 1

Ans a) Apply Bernoulli equation between point C located at water surface elevation of tank and at point A respectively,

$P_c$ / $\gamma$ + $V_c^2$ /2g + $Z_c$ = $P_A$ / $\gamma$ + $V_A^2$ /2g + $Z_A$ + Hf

Since, point C is open to atmosphere, pressure is only atmospheric hence, gauge pressure $P_c$ = 0

Velocity at surface is negligible so, Vc = 0

   Elevation from datum, $Z_c$ = h and $Z_A$ = 0

Velocity at point A = Flow rate / Area

   Flow rate = 1500 gal/min or 3.33 cfs

Area = ($\pi$/4)$D^2$

Diameter at point A = 10 in or 0.833 ft

=> Area = ($\pi$/4) = 0.5447 sq.ft

(0.833}=

=> Velocity at point A,$V_A$ = 3.33 / 0.5447 = 6.11 ft/s or 1.86 m/s

Frictional head loss, Hf = 6.94 m

Putting values in Bernoulli equation,

0 + 0 + h = (-35.5/9.81) + $(1.86)^2$ / (2 x 9.81) + 0 + 6.94

=> h = -3.62 + 0.176 + 6.94

=> h = 3.5 m  

Hence, required height 'h' = 3.5 m

Ans b) Apply Bernoulli equation between point A and B respectively,

$P_A$ / $\gamma$ + $V_A^2$ /2g + $Z_A$ + Hp = $P_c$ / $\gamma$ + $V_B^2$ /2g + $Z_B$ + $H_{f(A-B)}$

Given, pressure difference, $P_c$ - $P_A$ = 551 kPa

Velocity at point B = Flow rate / Area

Diameter at point B = 8 in or 0.75 ft

=> Area = ($\pi$/4)$(0.75)^2$ = 0.441 sq.ft

=> Velocity at point A,$V_A$ = 3.33 / 0.441 = 7.55 ft/s or 2.3 m/s

Hp is Head added by pump

Elevation , $Z_A$ = 0 and $Z_B$ = 7.62 m

  $H_{f(A-B)}$ is frictional head loss between A and B

Also, $H_{f(A-B)}$ = f L $V_B^2$ / 2 g D

To determine friction factor, calculate Reynold number (Re)

Re = V D / $\nu$

Pipe diameter = 8 in or 0.20 m

Kinematic viscosity of water = $10^{-6}$ m2/s

=> Re = 2.3 x 0.20 / $10^{-6}$

=> Re = 460000

Roughness of steel pipe = 0.045 mm

Relative roughness,e/D = 0.045 / 200 = 0.000225

According to Moody diagram for Re = 460000 and e/D = 0.0000225, friction factor (f) = 0.016

=> $H_{f(A-B)}$ = 0.016(792.5)$(2.3)^2$/(2 x 9.81 x 0.20)

=> $H_{f(A-B)}$ = 17.1 m

Putting values in Bernoulli equation,

=> $(1.86)^2$ / (2 x 9.81) + 0 + Hp = (551 / 9.81) +  $(2.3)^2$ / (2 x 9.81) + 7.62 + 17.1

=> 0.176 + Hp = 56.16 + 0.27 + 24.72

=> Hp = 81 m or 265.75 ft

We know, power (P) = $\gamma$ Q Hp

=> P = 62.4 x 3.33 x 265.75

=> P = 55221 lb ft/s or 100.4 hp

Hence, power delivered by pump to water is 100.4 hp