Question

1. Water at 100°F is pumped from a sea-level reservoir to a tank open to the atmosphere through a new 14-in ID galvanized iron pipe as shown below. The minor loss coefficients in the suction and discharge piping are EK = 1.5 and Eduk = 7.2. Perform the following: a. derive the equation for the system curve as hy = a + b where he and a have units of feet and b has units of ft/gpm); b. plot the system curve derived in Part (a) on the attached pump curve; c. if the desired flow rate is 22,000 gpm, identify i. the required impeller diameter (in.) assuming that the impeller can be trimmedto the nearest "/-inch; ii. the pump efficiency %); and iii. the required NPSH (ft). 95 100 Pune S' 156 n = 1170 r/min 800 50 40 NPSH 700 — 36-in dia. 65% 729 30 NPSH, 78% 82% 20 185% = 600 87% 88% Total head, ft 32-in dia. 87% 500 400 28-in dia. 2500 bhp 3000 bhp 3500 bhp 300 1500 bhp 2000 bhp 200 0 . 8 12 16 20 24 28 U.S. gallons per minute x 1000

Answer 1

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Ans) Consider point 1 at water surface at tank and point 2 at water surface sea level reservoir and apply Bernoulli equation between these two points :

  P1/$\dpi{100} \rho$g + V1²/2g + Z1 + H_p = P2/$\dpi{100} \rho$g + V2²/2g + Z2 + H_f + H_m

Since, both reservoir are open to atmosphere, the pressure is only atmospheric pressure so gauge pressure is zero. Also, velocity at surface 1 and 2 is also zero ,therefore, V1 = V2 = 0

  Let point 2 be datum, then, Z1 = 100 ft and Z2 =0

100 + H_p= H_f + H_{m ................(1)}

H_f = f L V² / (2gD)

We have to determine Darcy friction factor 'f' . To determine 'f' , lets use Colebrook equation,

1 / $\dpi{80} \sqrt{f}$ = -2.0 Log (e/3.7D + 2.51/R$\dpi{80} \sqrt{f}$)

where, e = Roughness (0.006 in for GI pipe)

D = Diameter of pipe

R = Reynold number = $\dpi{100} \rho$VD/$\dpi{100} \mu$

R = 62.4 V (14/12) / 1.42 x 10^-5

R = 51.26 x 10⁵ V

Lets assume V = 10ft/s , then f = 0.016

From equation 1,

100 + Hp = f L V² / (2gD) + (K1 +K2)V² /2g

100 + Hp= 0.016 x 100 V² / (2 x 32.2 x1.167) + 8.7 V² / (2 x 32.2)

100 + Hp = 0.02 V² + 0.135 V²

=> Hp = 0.155V² - 100 .............(2)

Also, Q = A V

=> V = Q/A

V = Q / (0.785 x 1.167²)

V² = 0.875 Q²

Putting value in equation 2,

=> Hp = 0.135 Q² - 100

Ans b) From part a,

Hp = 0.135 Q² - 100

where, discharge is in ft³/s and total head is in ft. Therefore, required curve is as follows :

^ 303 Pony Polo 6oo 400 200 24 3o 36 42 US gpm x 1000

Ans c) Desired flow rate = 22000 gpm or 49 ft³/s

Lets assume diamter of impeller = 'd' with 3600 rpm

Impeller circumference (C)= 3.14d

Velocity as it exit vanes, V = C x rpm

V = 3.14 x 3600 d

V = 11309.73 d ft/min or 188.5 d ft/s

Head = V²/2g = 551.74d²

Pump head at Q = 49ft³/s ,

Hp = 0.135(49)² - 100

= 224.135 ft

Hence, 551.74 d² = 224.135

d = 0.637 ft or 7.56 or 8 inches

b) According to given curve for Q= 22000 gpm and head = 225 ft, Power (P) consumed =  3000 bpm or 2237 kW

Theoratical Power consumed = $\dpi{100} \rho$ g QH

= 1000 x 9.81 x 1.38 x 68.58

= 928.42 kW

Efficiency = 928.42 / 2237

= 41.50 %

c) NPSH = H_abs + H_s + h_s - H_vap

= 12.8 + 20 + 0 - 5.60

= 27.20 m or 81.60 ft