Thermodynamics: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684
Thermodynamics: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684
9th Edition
ISBN: 9781260048667
Author: Yunus A. Cengel Dr.; Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 12.6, Problem 77P
To determine

The velocity of the oxygen at nozzle exit by treating the oxygen as an ideal gas and using enthalpy departure charts.

Expert Solution & Answer
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Answer to Problem 77P

The exit velocity of the nozzle by treating the oxygen as an ideal gas is 1738ft/s.

The exit velocity of the nozzle by using enthalpy departure charts is 1740ft/s.

Explanation of Solution

Write the general formula energy balance equation for closed system.

E˙inE˙out=ΔE˙system                                                                                           (I)

Here, rate of energy transfer into the system is E˙in, rate of energy transfer from the system is E˙out, and rate of change in net energy of system is ΔE˙system.

At the steady state, the rate of change of net energy of the system is zero.

ΔE˙system=0

Since the inlet velocity is negligible and the Equation (I) is rewritten as follows for the nozzle.

[h1+(V122)][h2+(V222)]=0h1+(V122)h2(V222)V2=2(h1h2)                                                                            (II)

Here, inlet velocity is V1, exit velocity is V2, enthalpy at initial state is h1 and enthalpy at final state is h2.

Write the formula for change in entropy equation (s2°s1°).

s2°s1°=Ruln(P2P1) (III)

Here, universal gas constant is Ru, final pressure is P2, and initial pressure is P1.

Write the change in enthalpy equation per mole basis.

(h¯2h¯1)Ideal=h¯2,Idealh¯1,Ideal (IV)

Here, Ideal enthalpy at final state is h¯2,Ideal and Ideal enthalpy at initial state is h¯1,Ideal.

Write the formula for change in enthalpy equation in mass basis.

(h¯2h¯1)Ideal=(h¯2h¯1)IdealM (V)

Here, molar mass is M.

Calculate the reduced temperature (TR1) at initial state.

TR1=T1Tcr (VI)

Here, critical temperature is Tcr and initial temperature is T1.

Calculate the reduced pressure (PR1) at initial state.

PR1=P1Pcr (VII)

Here, critical pressure is Pcr and initial pressure is P1.

Calculate the reduced temperature (TR2) at final state.

TR2=T2Tcr (VIII)

Here, critical temperature is Tcr and final temperature is T2.

Calculate the reduced pressure (PR2) at final state.

PR2=P2Pcr (IX)

Here, critical pressure is Pcr and final pressure is P1.

Write the formula for change in enthalpy (h2h1) using generalized enthalpy departure chart relation.

h2h1=RTcr(Zh1Zh2)+(h2h1)ideal (X)

Here, change in enthalpy of ideal gas is (h2h1)ideal and gas constant is R.

Refer Table A-19E, “Ideal properties of oxygen”.

The inlet enthalpy (h1) and entropy (s1) corresponding to the temperature of 1060R is 7543.1Btu/lbmol and 53.921Btu/lbmolR respectively.

h¯1,Ideal=7543.6Btu/lbmols1°=53.921Btu/lbmolR

Refer table A-1E, “Molar mass, gas constant and critical properties table”.

The molar mass of oxygen is 31.999 lbm/lbmol.

The critical temperature of oxygen is 278.6 R.

The critical pressure of oxygen is 736psia.

The gas constant of oxygen is 0.06206 Btu/lbmolR.

Conclusion:

Substitute 53.921Btu/lbmolR for s1°, 1.9858 Btu/lbmolR for Ru, 70psia for P2, and 200psia for P1 in Equation (III).

s2°53.921Btu/lbmolR=1.9858 Btu/lbmolR×ln(70psia200psia)s2°=53.921Btu/lbmolR2.085 Btu/lbmolRs2°=51.836 Btu/lbmolR

Refer Table A-19E, “Ideal properties of oxygen”.

The exit enthalpy (h2) and exit temperature (T2) at final entropy of 51.836 Btu/lbmolR is 5614.1Btu/lbmol and 802R respectively.

h¯2,Ideal=5614.1Btu/lbmolT2=802R

Substitute 5614.1Btu/lbmol  h¯2,Ideal and 7543.6Btu/lbmol for h¯1,Ideal in Equation (IV).

(h¯2h¯1)Ideal=5614.1Btu/lbmol7543.6Btu/lbmol=1929.5Btu/lbmol

Substitute 31.999lbm/lbmol for M and 1929.5Btu/lbmol for (h¯2h¯1)Ideal in

Equation (V).

(h¯2h¯1)Ideal=1929.5Btu/lbmol31.999lbm/lbmol=60.30Btu/lbm

Substitute 60.30Btu/lbm for (h2h1) in Equation (II).

V2=2×60.30Btu/lbm=2×60.30Btu/lbm×(25037ft2/s21Btu/lbm)=3019462.2ft2/s2=1737.6899ft/s

1738ft/s

Thus, exit velocity of the nozzle by treating the oxygen as an ideal gas is 1738ft/s.

Substitute 1060 R for T1 and 278.6 R for Tcr in Equation (VI).

TR1=1060R278.6R=3.805

Substitute 200psia for P1 and 736psia for Pcr in Equation (VII).

PR1=200psia736psia=0.272

Refer the table A-15E, “Nelson-Obert generalized compressibility chart”.

Obtain the enthalpy departure factor (Zh1) at initial state corresponding to the reduced pressure (PR1) and temperature (TR1) is 0.000759.

Zh1=0.000759

Substitute 802 R for T2 and 278.6 R for Tcr in Equation (VIII).

TR2=802 R278.6 R=2.879

Substitute 70psia for P2 and 736psia for Pcr in Equation (IX).

PR2=70psia736psia=0.0951

Refer the table A-15E, “Nelson-Obert generalized compressibility chart”.

Obtain the enthalpy departure factor (Zh2) at initial state corresponding to the reduced pressure (PR2) and temperature (TR2) is 0.00894.

Zh2=0.00894

Substitute 0.00894 for Zh2, 0.000759 for Zh1, 0.06206 Btu/lbmolR for R, 60.30Btu/lbm (h2h1)Ideal, and 278.6 R for Tcr in Equation (X).

h2h1=RTcr(Zh1Zh2)+(h2h1)ideal

(h2h1)={60.30Btu/lbm+[(0.06206 Btu/lbmolR)(278.6R)(0.008940.000759)]}=60.30Btu/lbm+0.1414Btu/lbm=60.44Btu/lbm

Substitute 60.44Btu/lbm for (h2h1) in Equation (II).

V2=2×60.44Btu/lbm=2×60.44Btu/lbm×(25037ft2/s21Btu/lbm)=3026472.56ft2/s2=1739.6759ft/s

1740ft/s

Thus, the exit velocity of the nozzle by using enthalpy departure charts is 1740ft/s.

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Chapter 12 Solutions

Thermodynamics: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684

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