Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 14, Problem 35EP

A water pump is used to pump water from one large reservoir to another large reservoir that is at a higher elevation. The free surfaces of both reservoirs are exposed to atmospheric pressure, as sketched in Fig. P14-35E. The dimensions and minor loss coefficients are provided in the figure. The pump's performance is approximated by the expression H a v a i l a b l e = H 0 a V ˙ 2 where the shutoff head H 0 = 125 f t of water column, coefficient a = 2.50 f t ( g p m ) 2 , available pump head H a v a i l a b l e is in units of feet of water column, and capacity is in units of gallons per minute (gpm). Estimate the capacity delivered by the pump.

   z 2 z 1 = 22.0 f t (elevation difference)

   D = 1.20 i n (Pipe diameter)

   K L , e n t r a n c e = 0.50 (Pipe entrance)

   K L , v a l u e 1 = 2.0 (value 1)

   K L , v a l u e 2 = 6.8 (value 2)

   K L , e l b o w = 0.34 (each elbow-the re are 3)

   K L , e x i t = 1.05 (pipe exit)

   L = 124 f t (total pipe length)

   ε = 0.0011 i n (pipe roughless)

Chapter 14, Problem 35EP, A water pump is used to pump water from one large reservoir to another large reservoir that is at a

Expert Solution & Answer
Check Mark
To determine

The capacity delivered by the pump.

Answer to Problem 35EP

The capacity delivered by the pump is 6.33gpm.

Explanation of Solution

Given information:

The shutoff head is 125ft, the coefficient ais 2.50ft/(gpm)2, the elevation difference is 22ft, the diameter of the pipe is 1.20in, the loss coefficient at pipe entrance is 0.50, the minor loss coefficient at valve 1 is 2.0, the minor loss coefficient at valve 2 is 6.8, the minor loss coefficient at each elbow is 0.34, the minor loss coefficient at pipe exit is 1.05, the total pipe length is 124ft, the pipe roughness is 0.0011in, the initial velocity of the water is zero, the final velocity of the water is zero and the pressure at exit is 1.01×105Paand inlet is pressure is 1.01×105Pa.

Write the expression for the required head using the energy balance equation.

  Hrequired=P2P1ρg+V22V122g+(z2z1)+hturbine+hLfriction (I)

Here, the initial pressure is P1, the final pressure is P2, the initial velocity is V1, the final velocity is V2, the density of the water is ρ, the acceleration due to gravity isg, the potential head of water at section 1 is z1, the potential head of the water at section 2 is z2, the turbine head is hturbine, the frictional loss is hLfriction.

Write the expression for the roughness factor.

  R=εD..... (II)

Here, the diameter of the pipe is D

Write the expression for the minor losses.

  FL=KL,entrance+KL,valve1+KL,valve2+KL,elbow+KL,exit..... (III)

Here, the minor loss coefficient at pipe entrance is KL,entrance, the minor loss coefficient at valve 1 is KL,valve1, the minor loss coefficient at valve 2 is KL,valve2, the minor loss coefficient at each elbow is KL,elbow, the minor loss coefficient at pipe exit is KL,exit.

Write the expression for the frictional loss head.

  hLfriction=(fLD+KL)V22g (IV)

Here, the friction factor is f, the length of the pipe is L.

Write the expression for the available head.

  Havailabe=HoaV˙2 (V)

Here, the shutoff head is Ho, the coefficient is athe capacity is V˙and the velocity of the water is V.

Write the expression for the capacity.

  V˙=πD24V (VI)

Substitute πD24Vfor V˙in Equation (V).

  Havailabe=Hoa( πD2 4V)2Havailabe=Hoa( π 2 D 416V2) (VII)

Write the expression for the Reynolds number.

  Re=ρVDμ (VIII)

Here, the density of the water is ρand the dynamic viscosity is μ.

Write the expression for the friction factor using the Colebrook equation.

  1f=2.0log(R3.7+2.51Ref) (IX)

Calculation:

Refer to table A-3E Properties of saturated water to obtain the density of water as 1.940slug/ft3and viscosity of water as 1.002×105lbs/ft2at room temperature.

Substitute 0.0011infor εand 1.20infor Din Equation (II).

  R=0.0011in1.20in=9.16×104

Substitute 0.5for KL,entrance, 2.0for KL,valve1, 6.8for KL,valve2, (0.34×3)for KL,elbow, 1.05for KL,exitin Equation (III).

  FL=0.5+2.0+6.8+(0.34×3)+1.05=0.5+2.0+6.8+(1.02)+1.05=11.37

Substitute 1.20infor D, 2.50ft/(gpm)2for aand 125ftfor Hoin Equation (VII).

  Havailabe=125ft2.50ft/( gpm)2( π 2 ( 1.20in) 416V2)=125ft2.50ft/( gpm)2( π 2 ( 1.20in( 1ft12in )) 416V2)=125ft(1.54×104 ft5/ (gpm )2)V2

Substitute 0for hturbine

  (fLD+KL)V22gfor hL,friction, 0for V1, 0for V2, 1.01×105Pafor P1and 1.01×105Pafor P2in Equation (I).

  Hrequired=1.01×105Pa1.01×105Paρg+0+(z2z1)+0+((f L D+ KL) V 22g)=0+(z2z1)+((f L D+ KL) V 22g)=(z2z1)+((f L D+ KL) V 22g)(X)

Since, the required head is equal to the available head that is Havailable=Hrequired.

Substitute Havailablefor Hrequiredin Equation (X).

  Havailable=(z2z1)+((fLD+ K L)V22g) (XI)

Substitute 125ft(1.54×104ft5/( gpm)2)V2for Havailable, 22.0ftfor (z2z1), 124ftfor L, 1.20infor D, 32.2ft/s2for gand 11.37for KLin Equation (XI).

   [ 125ft ( 1.54× 10 4 ft 5 / ( gpm ) 2 ) V 2 ]=[ 22.0ft+ ( ( f 124ft 1.20in +( 11.37 ) ) V 2 2( 32.2 ft/ s 2 ) ) ]

   [ 125ft ( 1.54× 10 4 ft 5 / ( gpm ) 2 ) V 2 ]=[ 22.0ft+ ( ( f 124ft 1.20in( 1ft 12in ) +( 11.37 ) ) V 2 ( 64.4 ft/ s 2 ) ) ]

   V 2 [ ( 1.54× 10 4 ft 5 / ( gpm ) 2 )( ( f19.251/ ft/ s 2 +( 17.651/ ft/ s 2 ) ) ) ]=103ft

   V 2 = 103ft [ ( 1.54× 10 4 ft 5 / ( gpm ) 2 )( ( f19.251/ ft/ s 2 +( 17.651/ ft/ s 2 ) ) ) ]

  V2=103ft[1.611× 10 5 ft 4/ ( gpm) 2f19.251/ ft/ s2 ]f=5.53ftV20.0836×105ft4/( gpm)2

Substitute 1.940slug/ft3for ρ, 1.002×105lbs/ft2for μand 1.20infor Din Equation (VII).

  Re=(1.940 slug/ ft3 )(1.20in)V(1.002× 10 5 lbs/ ft2 )=(1.940 slug/ ft3 )(1.20in( 1ft 12in))V(1.002× 10 5 lbs/ ft2 )(1lbs1slug)=0.1941.002×105V=19361.27V

Substitute 19361.27Vfor Re, 9.16×104for R, 5.53ftV20.0836×105ft4/(gpm)2for fin Equation (IX).

  1( 5.53ft V2 0.0836×10 5 ft 4/ (gpm) 2 )=2.0log( ( 9.16× 10 4 ) 3.7+ 2.51 ( 19361.27V ) ( 5.53ft V 2 0.0836×105 ft4/ ( gpm )2 ) )12.35V9.143×104 ft2/( gpm)=2.0log(2.47× 10 4+ 2.51 45498.9817.70 ft 2/( gpm)V)V2.359.143×104 ft2/( gpm)V=2.0log(4.37× 10 3 ft2 / (gpm )V+13.7445498.9817.70 ft2 / (gpm )V)V=0.1798ft/s

Substitute 0.1798ft/sfor Vand 1.20infor Din Equation (VI).

  V˙=π( 1.20in)24(0.1798ft/s)=π( 1.20in( 1ft 12in ))24(0.1798ft/s)=0.056484ft3/s(448.83gpm ft3 /s)=6.33gpm

Conclusion:

The capacity delivered by the pump is 6.33gpm.

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Chapter 14 Solutions

Fluid Mechanics: Fundamentals and Applications

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