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 41P

Suppose that the free surface of the inlet reservoir in Prob. 13-39 is 3.0 m higher in elevation, such that Z2- z1= 4.85 m. All the constants and parameters are identical to those of Prob. 14-39 except for the elevation difference. Calculate the volume flow rate for this case and compare with the result of Prob. 14-39. Discuss.

Expert Solution & Answer
Check Mark
To determine

The volume flow rate by the pump.

Answer to Problem 41P

The volume flow rate by the pump is 14.58755Lpm.

Explanation of Solution

Given information:

The diameter of pipe is 2.03cm, elevation difference is 4.85m, total length of pipe is 176.5m, shutoff head is 24.4m, coefficient ais 0.0678m/Lpm2, pipe roughness is 0.25mm, pipe roughness is 0.25mm, coefficient of entrance loss is 0.5, coefficient of valve loss is 17.5, coefficient of elbow loss is 0.92and coefficient of exit loss is 1.05.

Write the expression for head required.

  Hrequired=P2P1ρg+V22V122g+(z2z1)+hL   ....(I)

Here, pressure at exit is P2, pressure at inlet is P1, velocity at exit is V2, velocity inlet is V1, elevation difference is z2z1, head loss is hL.

Write the expression for coefficient of total minor head loss.

  K=K1+K2+K3+K4  ....(II)

Here, coefficient of entrance loss is K1, coefficient ofvalve loss is K2, coefficient of elbow loss is K3, coefficient of exit loss is K4.

Write the expression for head loss.

  hL=(fLD+K)V22g   ....(III)

Here, length of pipe is L, diameter of pipe is D, coefficient of friction loss is f, velocity of flow is Vand acceleration due to gravity is g.

Write the expression for Havailable.

  Havailable=H0aV˙2  ....(IV)

Here, shutoff head is H0, coefficient is aand volume flow rate is V˙.

Write the expression for Reynolds number.

  Re=ρVDμ.  ....(V)

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

Write the expression for Colebrook equation.

  1f=2.0log(ε/D3.7+2.51Ref)  ....(VI)

Here, coefficient of friction is fand pipe roughness factor is ε.

Write the expression for volume flow rate.

  V˙=π4D2V   ....(VII)

Calculation:

Substitute 0.5for K1, 17.5for K2, 5×0.92for K3, 1.05for K4in Equation (II).

  K=0.5+17.5+(5×0.92)+1.05=0.5+17.5+4.6+1.05=23.65

Substitute 23.65for K, 176.5mfor L, 2.03cmfor Dand 9.81m/s2for gin Equation (III).

  hL=(f×176.5m2.03cm+23.65)V22×9.81m/s2=(f×176.5m2.03cm× 1m 100cm+23.65)V22×9.81m/s2=(f×176.5m0.0203m+23.65)V22×9.81m/s2=(8694.58f+23.65)V219.62m/s2

Substitute 0for P1, 0for P2, 0for V1, 0for V2, 9.81m/s2for g, 7.85mfor z2z1, (8694.58f+23.65)V219.62for hLin Equation (I).

  Hrequired=00ρg+002g+(4.85m)+(8694.58f+23.65)V219.62m/s2=(4.85m)+(8694.58f+23.65)V219.62m/s2

Substitute 2.03cmfor Din Equation (VII).

  V˙=π4(2.03cm)2V=π4(2.03cm× 1m 100cm)2V=π4(0.0203m)2V=3.2365×104m2×V

Refer Table-A-3 "Properties of saturated water" at 20°Cto obtain density of water as 998kg/m3and dynamic viscosity as 1.002×103kg/ms

Substitute 3.2365×104m2×Vfor V˙, 24.4mfor H0, 0.0678m/Lpm2for aand (4.85m)+(8694.58f+23.65)V219.62for Havailablein Equation (IV).

  [( 4.85m)+( 8694.58f+23.65) V2 19.62m/s2 ]=[24.4m0.0678m/ Lpm 2( 3.2365× 10 4m2×V)2][( 8694.58f +23.65)V219.62m/ s 2]=[19.55m( 0.0678m/ Lpm2 × 3.6× 10 8 s2 / m5 1m/ Lpm2 )( 3.2365× 10 4m2×V)2][(8694.58f+23.65)V219.62m/ s 2]=[19.55m( 244080000s2/m5 )( 3.2365× 10 4m2×V)2](443.148s2/mf+1.205s2/m)V2=19.55m25.56s2/m×V2

  f=0.044116m2/s2V20.06041

Substitute 998kg/m3for ρ, 2.03cmfor Dand 1.002×103kg/msfor μin Equation (V).

  Re=998kg/m3×2.03cm×V1.002×103kg/ms=998kg/m3×2.03cm×1m100cm×V1.002×103kg/ms=20218.96s/m×V

Substitute 0.25mmfor ε, 2.03cmfor D, 20218.96s/m×Vfor Reand 0.044116m2/s2V20.06041for fin Equation (VI).

  [1 0.044116 m2 / s2 V 20.06041 =2.0log( 0.25mm/ 2.03cm 3.7 + 2.51 20218.96s/m×V 0.044116 m2 / s2 V 20.06041 )][1 0.044116 m2 / s2 V 20.06041 =2.0log( 0.25mm/ 2.03cm× 10mm 1cm 3.7 + 2.51 20218.96s/m×V 0.044116 m2 / s2 V 20.06041 )][1 0.044116 m2 / s2 V 20.06041 =2.0log( 0.00332432 + 2.51 20218.96s/m×V 0.044116 m2 / s2 V 20.06041 )]  ....(VIII)

Here, apply hit and trial to obtain velocity of flow.

Trial-(1)

Substitute 1m/sfor Vin Equation (VIIII).

  [1 0.044116 m2 / s2 (1m/s) 20.06041 =2.0log( 0.00332432 + 2.51 20218.96s/m×1m/s 0.044116 m2 / s2 ( 1m/s ) 20.06041 )]10.016294=2.0log(0.00332432+ 2.51 20218.96 0.016294)

Since the values on both sides are not equal, so assumption V=1m/sis not correct.

Trial-(2)

Substitute 0.7512m/sfor Vin Equation (VIII).

  [1 0.03734 m2 / s2 (0.7512m/s) 20.06041 =2.0log( 0.00332432 + 2.51 20218.96s/m×0.7512m/s 0.03734 m2 / s2 0.7512m/s0.06041 )]10.0057074=2.0log(0.00332432+ 2.51 20218.96s/m×0.7512m/s 0.0057074)13.2366=13.2363

Here, the values on both sides are equal, so velocity of flow is 0.7512m/s

Substitute 0.7512m/sfor Vin Equation (VII).

  V˙=3.2365×104m2×0.7512m/s=2.43125×104m3/s×60000Lpm1m3/s=14.58755Lpm

Here, the volume flow rate for elevation difference of 3mis 14.58755Lpm.

Since the volume flow rate for the for the elevation difference of 7mis 11.576Lpm

Hence on decreasing elevation difference between the reservoir the volume flow rate increases.

Conclusion:

The volume capacity delivered by the pump is 14.58755Lpm.

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Chapter 14 Solutions

Fluid Mechanics: Fundamentals and Applications

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