Question

Water is circulated from a large swimming pool, through a filter, and back to the pool. The power added to the water by the pump is 200 lb-ft/s. The flowrate (in ft/min) through the filter is most likely: Req. pump power = m-gehpumppug.Qhpump = y.Qehpump For obtaining the Darcy friction factor (1), use the following Colebrook equation. 1 eD -2.0log 2.51 3.7 * Revf + K2 exit = 1.0 KZ elbow = 1.5 KĽ valve = 6.0 Kz filter = 12.0 Ky ent 0.8 200 ft. of 0.1-ft-diameter pipe with e/D = 0.01 Filter Pump

Answer 1

NOTE:

We must use Engineering Equations Solver (EES) software for this sum.

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In this sum, we are not given the height of pool, the extraction height and other elevation heads. So, we will assume that the extraction of water and the release of water is occuring at nearly the same height. So, the head added by the pump can be assumed to not overcome any elevation head but solely the major head and minor head losses.

H(pump) = H(major) + H(minor)................(1)

Major head loss =

L V2 X H major = fx D 29

where

• f = friction factor = ?
• L = length of piping system = 200 ft
• D = Diameter of pipe = 0.1 ft
• V = Velocity of flow =
  Q Area 127.324Q m/s 0.007854
  
  • Q = Volumetric flow rate = ? (our answer)
  • Area =
    ΠD? Π0.12 4 = 0.007854 ft-
    
    • D = Diameter of pipe = 0.1 ft
• g = 32.185 ft/s2

Friction factor = f = ?

1 1 e/D -210910 + 2.51 Rex vf 3.7

where

• Re = Reynold's Number =
  VD Re 127.324 Q x 0.1 1.1337e – 5 1123083.708 Q
  ​​​​​​​
  • V = Velocity of flow = 127.324 Q m/s
  • D = 0.1 ft
  • $1590620841381_blob.png$ = Kinematic viscosity of water = 1.1337e-5 ft2/s (from the internet)
• e/D = 0.01 (given)

1 1 e/D -210910 + 2.51 Rex vf 3.7

........................(2)

1 1 -2log10 0.01 3.7 + 2.51 1123083.708 Q x V)

So,

L V2 H major = fx х D 29

Hmajor = fx 200 0.1 (127.324 Q4 2 x 32.185 503694.2978 X Q? x fft

Minor head loss =

v2 Hminor = kminor X 29

where

• k(minor) = Total minor loss coefficients = ?
• V = 127.324 Q m/s

Total minor loss coefficients:

• K(L-ent) = 0.8
• K(L-filter) = 12
• K(L-elbow) (x2) = 1.5 (x2)*
• K(L-valve) = 6
• K(L-exit) = 1
• K(L-90(deg) flanged bends) (x3) = 0.3 (x3)**

*In the picture only one elbow bend is given, although, there is a second bend in the question as shown below (point A)

**In the picture shown, no 90(deg) bends are mentioned, however, in the diagram, there are 3 such bends as shown below. (points X, Y, Z)

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K2 exit = 1.0 Kz elbow = 1.5 Flanged Kz valve = 6.0 Kz filter = 12.0 Y. Lent = 0.8 ΑΙ Filter Z Pump X 200 ft. of 0.1-ft-diameter pipe with e/D = 0.01

Total minor loss coeff = 0.8 + 12 + ( 1.5 x 2 ) + 6 + 1 + (0.3 x 3 ) = 23.7

v2 Hminor = kminor X 29

H minor 23.7 x (127.324 Q4 2 x 32.185 5968.777 x Q2 ft

Now, using eqn 1, we get,

H(pump) = H(major) + H(minor)

.................................(3)

Hритр 503694.2978 x Q? xf+ 5968.777 x Q-

Also, we know,

Power wXQ XH pump

where

• Power = 200 lb-ft/s
• $1590620844235_blob.png$ = Specific weight of water = 62.4 lb(f)/ft3
• Q = ?
• 
  Hритр 503694.2978 x Q? xf+ 5968.777 x Q-
  

Power wXQ XH pump

.......................(4)

200 = 62.4 x Q x 503694.2978 x 22 x f +5968.777 X Q2

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Solving these 4 equations simultaneously in EES software, we get,

Flow rate =

Q = 0.05003 ft/s = 0.05003 cfs = 3.0018ft/min

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