Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 9, Problem 9.42P
To determine

(a)

To calculate:

The Mach number, velocity and temperature at exit plane.

Expert Solution
Check Mark

Answer to Problem 9.42P

Mae=0.9

Te=260.75K

Ve=291.31m/s

Explanation of Solution

Given information:

At stagnation point,

p0=169.12kPaT0=30°C

The absolute pressure is equal to pa=100kPa and Ta=20°C

The valve exit diameter is equal to 2mm

The pressure ratio is defined as,

p0p=[1+12(k1)Ma2]k/k1

Assume, for air,

k=1.4R=287m2/s2.K

The temperature ratio is defined as,

T0T=[1+12(k1)Ma2]

Speed of sound is defined as,

a=kRT

Where,

R - Gas constant

k - Specific heat capacity

The Mach number is defined as,

Ma=Va

Where,

V - Air velocity

Calculation:

Calculate the Mach number,

p0pa=[1+12(k1)Mae2]k/k1

Substitute for known values,

169.12kPa100kPa=[1+12(1.41)Mae2]1.4/1.41

Solve to find Mach number,

Mae=0.9

Calculate the exit temperature,

T0Te=[1+12(k1)Mae2]

Convert,

T0=30°C=303K

Substitute for known values,

303KTe=[1+12(1.41)(0.9)2]

Therefore,

Te=260.75K

Calculate the exit velocity,

Ve=Mae(a)=MaekRTe

Substitute for known values,

Ve=(0.9)1.4(287m2/s2.K)(260.75K)

Therefore,

Ve=291.31m/s

Conclusion:

The exit Mach number is equal to Mae=0.9

Exit temperature is equal to Te=260.75K

Exit velocity is equal to Ve=291.31m/s.

To determine

(b)

To calculate:

Initial mass flow rate.

Expert Solution
Check Mark

Answer to Problem 9.42P

m=1.223×103kg/s

Explanation of Solution

Given information:

At stagnation point,

p0=169.12kPaT0=30°C

The absolute pressure is equal to pa=100kPa and Ta=20°C

The valve exit diameter is equal to 2mm

The density at section 1 is defined as,

ρ1=p1RT1

The mass flow is defined as,

m=ρ1A1V1

Where,

A1 - Area at section 1,

For ideal gas,

R=287m2/s2.Kk=1.4

Calculation:

Calculate the exit density,

ρe=peRTe=100000kPa(287m2/s2.K)(260.75K)=1.336kg/m3

Calculate the mass flow rate,

m=ρeAeVe=(1.336kg/m3)(π4 ( 0.002m )2)(291.31m/s)=1.223×103kg/s

Conclusion:

The mass flow is equal to m=1.223×103kg/s.

To determine

(c)

To calculate:

The exit velocity using incompressible Bernoulli’s equation.

Expert Solution
Check Mark

Answer to Problem 9.42P

Ve=266.59m/s

The above obtained value of 266.59m/s is almost 8% lower than the velocity found in sup-part a Ve=291.31m/s

Explanation of Solution

Given information:

At stagnation point,

p0=169.12kPaT0=30°C

The absolute pressure is equal to pa=100kPa and Ta=20°C

The valve exit diameter is equal to 2mm

The density at section 1 is defined as,

ρ1=p1RT1

According to incompressible Bernoulli’s equation,

The exit velocity is defined as,

Ve=2Δpρ0

Calculation:

Calculate the density,

ρ0=p0RT0

Assume,

p0=ptire=169.12kPa

Therefore,

ρ0=p0RT0=169120Pa(287m2/s2.K)(303K)=1.945kg/m3

Calculate the exit velocity

According to incompressible Bernoulli’s equation,

Ve=2Δpρ0

Substitute for known values,

Ve=2Δpρ0=2( 169120Pa100000Pa)1.945kg/m3=266.59m/s

The above obtained value of 266.59m/s is almost 8% lower than the velocity found in sup-part a Ve=291.31m/s

Conclusion:

The exit velocity is equal to 266.59m/s.

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