FLUID MECHANICS-PHYSICAL ACCESS CODE
FLUID MECHANICS-PHYSICAL ACCESS CODE
8th Edition
ISBN: 9781264005086
Author: White
Publisher: MCG
Question
Chapter 9, Problem 9.1P
To determine

(a)

The velocity V2 and change in entropy (s2− s1) at the exit of the duct for air.

Expert Solution
Check Mark

Answer to Problem 9.1P

At the exit of the duct for air.

The velocity V2=334.835m/s and change in entropy is s2s1=336.848J/kgK

Explanation of Solution

Given information:

At section 1,

Pressure p1 = 140 kPa

Temperature T1 = 260° C

Velocity V1 = 75 m/s

Downstream, at section 2, the values are

Pressure p2 = 30 kPa

Temperature T2 = 207° C

γair=1.4

Gas constant for air R = 287 J/kg-K

The specific heat at constant pressure cp = 1005 J/kg-K

Calculation:

For calculating the downstream velocity, the adiabatic steady-flow energy equation can be used

cpT1+12V12=cpT2+12V22

1005×260+12(75)2=1005×207+12V22

V2=334.835m/s

Now, for calculating the entropy change

s2s1=cpln(T2T1)Rln(p2p1)

s2s1=1005ln(207+273260+273)287ln(30140)

s2s1=336.848J/kgK

Conclusion:

Thus, at the exit of the duct for air, the velocity V2=334.835m/s and change in entropy is s2s1=336.848J/kgK.

To determine

(b)

The velocity V2 and change in entropy (s2− s1) at the exit of the duct for argon.

Expert Solution
Check Mark

Answer to Problem 9.1P

At the exit of the duct for argon, the velocity V2 = 246.034 m/s and change in entropy is s2s1=266.1596J/kgK

Explanation of Solution

Given information:

At section 1,

Pressure p1 = 140 kPa

Temperature T1 = 260° C

Velocity V1 = 75 m/s

Downstream, at section 2, the values are

Pressure p2 = 30 kPa

Temperature T2 = 207° C

γair=1.67

Gas constant for air R = 208 J/kg-K

The specific heat at constant pressure cp = 518 J/kg-K

Calculation:

For calculating the downstream velocity, the adiabatic steady-flow energy equation can be used

For Argon,

let us take

k = 1.67

The gas constant for Argon R = 208 J/kg.K

Specific heat at constant pressure cp = 518 J/kg.K

cpT1+12V12=cpT2+12V22

518×260+12(75)2=518×207+12V22

V2=246.034m/s

Now, for calculating the entropy change:

s2s1=cpln(T2T1)Rln(p2p1)

s2s1=518ln(207+273260+273)208ln(30140)

s2s1=266.1596J/kgK

Conclusion:

Thus, At the exit of the duct for argon, the velocity V2 = 246.034 m/s and change in entropy is s2s1=266.1596J/kgK..

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Chapter 9 Solutions

FLUID MECHANICS-PHYSICAL ACCESS CODE

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