Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 3, Problem 3.177P
To determine

(a)

The formula for the viscosity of the liquid.

Expert Solution
Check Mark

Answer to Problem 3.177P

The formula for the viscosity of the liquid is (H+L)(πρgd4128LQ)(ρQα216πL).

Explanation of Solution

Given information:

The flow rate is Q . The flow is laminar. The head loss in the pipe is hf=32μLVρgd2 . The density is ρ.

The Figure-(1) shows the section (1) and the section (2) for the given system.

Fluid Mechanics, Chapter 3, Problem 3.177P

Figure-(1)

Write the expression for the Bernoulli’s equation between the section (1) and section (2) of the given system.

p1ρg+α1V122g+z1=p2ρg+α2V222g+z2+hf( p 1 p 2ρg)+(α1 V 1 22gα2 V 2 22g)+(z1z2)=hf …… (I)

Here, the pressure at inlet is p1, the velocity at inlet is V1, the height of the inlet from the datum is z1, the pressure at outlet is p2, the velocity at outlet is V2, the height of the outlet from the datum is z2, the density of water is ρ, the head loss due to friction is hf, the kinetic energy correction factor at section (1) is α1, the kinetic energy correction factor at section (2) is α2 and the acceleration due to gravity is g.

Write the expression for the head loss due to friction in the pipe region.

hf=32μLV2ρgd2

Here, the lead loss due to friction is hf, dynamic viscosity is μ, the length of the pipe is L, the flow velocity is V2, density is ρ, the gravitational acceleration is g, the diameter of the pipe is d.

Since the pressure at the inlet and outlet are same so,

p1=p2

Since the fluid is stationary at section (1), so the velocity of fluid at section (1) is zero.

V1=0

Write the expression for the height of section (2) from datum.

z1=H+L

Substitute p2 for p1, 0 for V1, 0 for z2, 32μLV2ρgd2 for hf and H+L for z1 in Equation (I).

( p 2 p 2ρg)+(α102gα2 V 2 22g)+{(H+L)0}=32μLV2ρgd20+0+(H+L)α2V222g=32μLV2ρgd2α2V222g+32μLV2ρgd2(H+L)=0 …… (II)

Write the expression for the flow rate at section (2).

Q=A2V2 …… (III)

Here, the flow rate is Q and the area is A2.

Write the expression for the area at section (2).

A2=π4d2

Substitute π4d2 for A2 in Equation (III).

Q=π4d2V2V2=Qπ4d2V2=4Qπd2

Substitute 4Qπd2 for V2 in Equation (II).

α2( 4Qπd2 )22g+32μL( 4Q πd2 )ρgd2(H+L)=08Q2α2π2d4g+128μLQπρgd4(H+L)=0128μLQπρgd4=[(H+L)8Q2α2π2d4g]μ=πρgd4128LQ[(H+L)8Q2α2π2d4g]

Further solve the above expression.

μ=[( πρg d 4 128LQ)(H+L)( πρg d 4 128LQ)( 8 Q 2 α 2 π 2 d 4 g)]=[(H+L)( πρg d 4 128LQ)(ρ 16L)( Q α 2 π)]=(H+L)(πρg d 4128LQ)(ρQ α 216πL)

Conclusion:

The formula for the viscosity of the liquid is (H+L)(πρgd4128LQ)(ρQα216πL).

To determine

(b)

The viscosity of the fluid.

Expert Solution
Check Mark

Answer to Problem 3.177P

The viscosity of the fluid is 0.001925kg/ms.

Explanation of Solution

Given information:

The diameter of the pipe is 2mm . The density is 800kg/m3 . The length of the pipe is 95cm . The height of the water level is 30cm . The flow rate is 760cm3/h.

Write the expression for the viscosity.

μ=(H+L)(πρgd4128LQ)(ρQα216πL) …… (IV)

Calculation:

Substitute 2mm for d, 800kg/m3 for ρ, 95cm for L, 760cm3/h for Q, 9.81m/s2 for g and 30cm for H in Equation (IV).

Further solve the above expression.

μ=1.925395×103kg/ms=0.001925395kg/ms0.001925kg/ms

Conclusion:

The viscosity of the fluid is 0.001925kg/ms.

To determine

(c)

The name of the fluid having viscosity 0.001925kg/ms.

Expert Solution
Check Mark

Answer to Problem 3.177P

The name of the fluid having viscosity 0.001925kg/ms is

Explanation of Solution

Given information:

The viscosity of the fluid is 0.001925kg/ms.

Refer to the viscosity table for liquids to obtain the fluid having viscosity 0.001925kg/ms as propyl alcohol.

μp=0.00192kg/ms

Here, the dynamic viscosity of propyl alcohol is μp.

μpμ

Thus, the fluid could be propyl alcohol.

Conclusion:

The name of the fluid having viscosity 0.001925kg/ms is propyl alcohol.

To determine

(d)

Whether the Reynolds number is less than 2000 or not.

Expert Solution
Check Mark

Answer to Problem 3.177P

Yes, the Reynolds number is less than 2000.

Explanation of Solution

Given information:

The diameter of the pipe is 2mm . The density is 800kg/m3 . The length of the pipe is 95cm . The height of the water level is 30cm . The flow rate is 760cm3/h.

Write the expression for the Reynolds number.

Re=ρV2dμ …… (V)

Here, the Reynolds number is Re.

Write the expression for the velocity at section (2).

V2=4Qπd2

Substitute 4Qπd2 for V2 in Equation (V).

Re=ρ( 4Q πd2 )dμ=4ρQπμd …… (VI)

Calculation:

Substitute 2mm for d, 800kg/m3 for ρ, 760cm3/h for Q, 9.81m/s2 for g and 0.001925kg/ms for μ Equation (VI).

Re=4(800 kg/ m 3 )(760 cm 3 /h)π(0.001925 kg/ ms)(2mm)=4(800 kg/ m 3 )(760 cm 3 /h)( 1m3 1×106 cm3 )( 1h 3600s)π(0.001925 kg/ ms)(2mm)( 1m 1000mm)=55.8556

Thus, the Reynolds number is 56 which is less than 2000.

Re<2000

Conclusion:

Yes, the Reynolds number is less than 2000.

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Chapter 3 Solutions

Fluid Mechanics

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