Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
Question
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Chapter 6, Problem 6.92P
To determine

(a)

The air flow rate in m3h.

Expert Solution
Check Mark

Answer to Problem 6.92P

Q=1544.4m3/h

Explanation of Solution

Given information:

Fluid Mechanics, Chapter 6, Problem 6.92P , additional homework tip  1

The room pressure is 10Pa

The pressure drop is defined as,

ΔP=ρghf=ρgfLD V 2 2g

In above equation,

ρ - Density of the fluid

hf - Head loss

L - Length of pipe

D - Hydraulic diameter of the pipe

The volumetric flow rate Q can be defined as below,

Q=VA

In above equation,

V - Velocity of the flow

A -Area

The hydraulic diameter is defined as,

Dh=4AΡ

Dh - Hydraulic diameter of pipe

A - Area of the pipe

Ρ - Perimeter

Calculation:

First of all, to find the hydraulic diameter,

Dh=4AΡ=4900cm120cm=30cm=0.3m

To find volumetric flow rate,

Assume,

f=0.02

ΔP=ρfLD V 2 2 10N/m2=1.2kg/m3 0.02 12m 0.3m V 2 2 V=4.564m/s

To find the exact value of the velocity, further iterations needs to be done. For that we need to find the friction factor,

From the definitions, b/a=1.0,

Therefore the effective diameter will be equal to,

Deff=6456.91Dh=6456.910.3m=0.337m

To find the Reynolds’s number,

Re=VDeffν=4.564m/s0.337m1.5111×105m2s=101784.66

To find the roughness ratio,

εDeff=0.046×103m0.337m=1.22×104

To find the friction factor,

1f 1/21.8log 6.9 R e + /d 3.7 1.111.8log 6.9 101784.66+ 1.22× 10 4 3.7 1.110.0183

Therefore, to find the exact value of the velocity,

ΔP=ρfLD V 2 2 10N/m2=1.2kg/m3 0.0183 12m 0.3m V 2 2 V=4.772m/s

To find the flow rate,

Q=VA=4.772m/s0.09m2=0.429m3/s=1544.4m3/h

Conclusion:

The flow rate is equal to 1544.4m3/h

(b)

The room pressure if the flow rate is 1200m3/h.

ΔP=6.24N/m2

Given information:

Fluid Mechanics, Chapter 6, Problem 6.92P , additional homework tip  2

Assume the flow rate as 1200m3/h

The pressure drop is defined as,

ΔP=ρghf=ρgfLD V 2 2g

In above equation,

ρ - Density of the fluid

hf - Head loss

L - Length of pipe

D - Hydraulic diameter of the pipe

The volumetric flow rate Q can be defined as below,

Q=VA

In above equation,

V - Velocity of the flow

A -Area

The hydraulic diameter is defined as,

Dh=4AΡ

Dh - Hydraulic diameter of pipe

A - Area of the pipe

Ρ - Perimeter

Calculation:

Compute the mean velocity,

V=QA=0.333m3/s0.09m2=3.7m/s

To find the Reynolds’s number,

Re=VDeffν=3.7m/s0.337m1.5111×105m2s=82516.05

To find the friction factor,

1f 1/21.8log 6.9 R e + /d 3.7 1.111.8log 6.9 82516.05+ 1.22× 10 4 3.7 1.110.019

Therefore, according to the obtained results,

To find the room pressure,

ΔP=ρfLD V 2 2 =1.2kg/m3 0.019 12m 0.3m 3.7m/s 2 2 =6.24N/m2

Conclusion:

The room pressure is equal to 6.24N/m2.

To determine

(b)

The room pressure if the flow rate is 1200m3/h.

Expert Solution
Check Mark

Answer to Problem 6.92P

ΔP=6.24N/m2

Explanation of Solution

Given information:

Fluid Mechanics, Chapter 6, Problem 6.92P , additional homework tip  3

Assume the flow rate as 1200m3/h

The pressure drop is defined as,

ΔP=ρghf=ρgfLD V 2 2g

In above equation,

ρ - Density of the fluid

hf - Head loss

L - Length of pipe

D - Hydraulic diameter of the pipe

The volumetric flow rate Q can be defined as below,

Q=VA

In above equation,

V - Velocity of the flow

A -Area

The hydraulic diameter is defined as,

Dh=4AΡ

Dh - Hydraulic diameter of pipe

A - Area of the pipe

Ρ - Perimeter

Calculation:

Compute the mean velocity,

V=QA=0.333m3/s0.09m2=3.7m/s

To find the Reynolds’s number,

Re=VDeffν=3.7m/s0.337m1.5111×105m2s=82516.05

To find the friction factor,

1f 1/21.8log 6.9 R e + /d 3.7 1.111.8log 6.9 82516.05+ 1.22× 10 4 3.7 1.110.019

Therefore, according to the obtained results,

To find the room pressure,

ΔP=ρfLD V 2 2 =1.2kg/m3 0.019 12m 0.3m 3.7m/s 2 2 =6.24N/m2

Conclusion:

The room pressure is equal to 6.24N/m2.

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