EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220102809444
Author: CENGEL
Publisher: YUZU
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Chapter 12.6, Problem 75P

a)

To determine

The exit velocity of the nozzle.

a)

Expert Solution
Check Mark

Answer to Problem 75P

The exit velocity of the nozzle is 1738ft/s.

Explanation of Solution

Write the 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 change in internal energy of system is ΔE˙system.

Substitute h1+(V122) for Ein, h2+(V222) for Eout, 0 for V1 and 0 for ΔEsystem in Equation (VI).

h1+(V122)h2(V222)=0V2=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 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 change in enthalpy equation in mass basis.

(h¯2h¯1)Ideal=(h¯2h¯1)IdealM (V)

Here, molar mass is M.

Conclusion:

Refer table A-19E, “Ideal properties of oxygen”, obtain the enthalpy of inlet and entropy at initial temperature of 1060R as 7543.1Btu/lbmol and 53.921Btu/lbmolR.

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

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/lbmolR=51.836 Btu/lbmolR

Refer table A-19E, “Ideal properties of oxygen”, obtain the enthalpy of inlet and final temperature at final entropy of 51.836 Btu/lbmolR as 5614.1Btu/lbmol and 802R.

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)=1738ft/s

Thus, the exit velocity of the nozzle is 1738ft/s.

b)

To determine

The exit velocity of the nozzle.

b)

Expert Solution
Check Mark

Answer to Problem 75P

The exit velocity of the nozzle is 1740ft/s.

Explanation of Solution

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.

Calculate the change in enthalpy (h2h1) using generalized chart relation.

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

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

Conclusion:

Refer table A-1E, “Molar mass, gas constant and critical properties table”, obtain the molar mass, critical temperature, critical pressure, and gas constant of oxygen as 31.999 lbm/lbmol, 278.6 R, 736psia, and 0.06206 Btu/lbmolR.

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”, select the compressibility factor Zh1 at initial state of reduced pressure and temperature of 3.805 and 0.272 as 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”, select the compressibility factor Zh2 at final state of reduced pressure and temperature of 2.879 and 0.0951 as 0 and 0.00894.

Zh2=0.00894

Substitute cp(T2T1) for (h2h1)ideal.

h2h1=RTcr(Zh1Zh2)(h2h1)Ideal

Here, specific heat at constant pressure is cp, final temperature is T2, and initial temperature is T1.

Substitute 0.00894 for Zh2, 0.000759 for Zh1, 0.06206 Btu/lbmolR for R, 60Btu/lbm (h2h1)Ideal, and 278.6 R for Tcr in Equation (X).

(h2h1)={0.06206 Btu/lbmolR(278.6R)(0.0007590.00894)60Btu/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)=1740ft/s

Thus, the exit velocity of the nozzle is 1740ft/s.

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Chapter 12 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

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