EBK FLUID MECHANICS
EBK FLUID MECHANICS
2nd Edition
ISBN: 9780134626055
Author: HIBBELER
Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)
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Chapter 13, Problem 1P
To determine

The changes in pressure and entropy.

Expert Solution & Answer
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Explanation of Solution

Given:

Initial  pressure (p1) is 120psi.

Initial temperature (T1) is 40°F.

Final temperature (T2) is 120°F.

Calculation:

Consider oxygen as compressible.

Write an equation for pressure ratio (p1p2) and calculate final pressure (p2).

 p1p2=ρRT1ρRT2p1p2=T1T2

120psip2=460+40°F460+120°Fp2=139.2psi

Calculate the change in pressure (Δp).

Δp=p2p1

Δp=(139.2120)psi=19.2psi

Thus, the change is pressure is 19.2psi_.

From Appendix A table of “Physical properties of Gases”, for Oxygen, obtain the values for the gas constant (R) as 1,554ftlb/slug°R and specific heat ratio (k) as 1.4.

Write an equation for change in enthalpy (Δh) and calculate specific heat at constant pressure (cp).

Δh=kRk1(T2T1)cpΔT=kRk1ΔTcp=kRk1

cp=1,554ftlb/slug°R ×1.41.41=5439 ftlb/slug°R

Calculate the change in entropy (Δs).

(Δs)=cplnT2T1Rlnp2p1={(5439ftlb/slug°R)[ln(580°R500°R)](1554ftlb/slug°R)[ln(139.2psi120psi)]}=576.61ftlb/slug°R

Thus, entropy change (Δs) is 576.61ftlb/slug°R_.

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Chapter 13 Solutions

EBK FLUID MECHANICS

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