FLUID MECHANICS-PHYSICAL ACCESS CODE
FLUID MECHANICS-PHYSICAL ACCESS CODE
8th Edition
ISBN: 9781264005086
Author: White
Publisher: MCG
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Question
Chapter 1, Problem 1.31P
To determine

i.

The new density and temperature.

Expert Solution
Check Mark

Answer to Problem 1.31P

The new density is 4.99 kg/m3

The new temperature is 578 K

Explanation of Solution

Given:

Initial temperature = 10℃

Initial Pressure = 1atm ⇒ 101.350 kPa

New Pressure = 600kPa

Concept Used:

P=ρRT

ρ=PRT

For isentropic process:

ρ1ρ2=( P 1 P 2 )1/kwherekistheconstant( P 2 P 1 )=( T 2 T 1 )k/(k1)T2=( P 2 P 1 )(k1)/k*T1

Calculation:

From Table A.4, for argon gas R=208m2/s2Kandk=1.67

ρ1=P1RT1

ρ1=101350N/m2(208m2/s2K)*283Kρ1=1.72kg/m3

ρ1ρ2=( P 1 P 2 )1/kρ2=( P 2 P 1 )1/k*ρ1ρ2=( 600kPa 101.350kPa)1/1.67*1.72ρ2=4.99kg/m3

For isentropic process:

ρ1ρ2=( P 1 P 2 )1/kwherekistheconstant( P 2 P 1 )=( T 2 T 1 )k/(k1)T2=( P 2 P 1 )(k1)/k*T1T2=( 600kPa 101.350kPa)(1.671)/1.67*283T2=578K

Conclusion:

The new density is 4.99 kg/m3

The new temperature is 578 K.

To determine

ii.

The final pressure if it is allowed to cool back to 10℃ at the new volume.

Expert Solution
Check Mark

Answer to Problem 1.31P

The final pressure is 294kPa

Explanation of Solution

Given:

Final temperature = 10℃

Initial Pressure = 1atm ⇒ 101.350 kPa

New Pressure = 600kPa

Concept Used:

P=ρRT

Calculation:

Since the cooling process takes place at constant volume, the density will remain constant i.e. ρ=4.99 kg/ m3.

Therefore

Pfinal=ρRTPfinal=4.99*208*283Pfinal=294kPa

Conclusion:

The final pressure is 294kPa.

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Chapter 1 Solutions

FLUID MECHANICS-PHYSICAL ACCESS CODE

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