Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
9th Edition
ISBN: 9781259822674
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 17.7, Problem 108P

Air is heated as it flows through a 6 in × 6 in square duct with negligible friction. At the inlet, air is at T1 = 700 R, P1 = 80 psia, and V1 = 260 ft/s. Determine the rate at which heat must be transferred to the air to choke the flow at the duct exit, and the entropy change of air during this process.

Expert Solution & Answer
Check Mark
To determine

The rate of heat transfer in the duct.

The entropy change in the duct.

Answer to Problem 108P

The rate of heat transfer in the duct is 16,090Btu/s_.

The entropy change in the duct is 0.434Btu/lbmR_.

Explanation of Solution

Determine the inlet density of air.

ρ1=P1RT1 (I)

Here, the inlet pressure of air is P1, the universal gas constant of air is R, and the inlet temperature of air is T1.

Determine the mass flow rate of the duct.

m˙air=ρ1Ac1V (II)

Here, the inlet velocity of air is V1 and the density of the air is ρ1.

Determine the inlet stagnation temperature of air.

T01=T1+V122cp (III)

Here, the inlet static temperature of ideal gas is T3, the specific heat of pressure for ideal gas is cp, and the inlet velocity of the ideal gas flow is V1.

Determine the relation of ideal gas speed of sound at the inlet.

c1=kRT1 (IV)

Here, the specific heat ratio of air is k, the gas constant of the air is R, and the inlet temperature of the air is T1.

Determine the speed of sound at the inlet.

Ma1=V1c1 (V)

The inlet velocity of the air flow in the device is V1 and the inlet Mach number is Ma1.

Determine the static temperature in the duct.

T2T1=T2/TT1/T (VI)

Here, the ratio of Rayleigh flow for inlet temperature is T1/T and the ratio of Rayleigh flow for outlet temperature is T2/T.

Determine the static pressure in the duct.

P2P1=P2/PP1/P (VII)

Here, the ratio of Rayleigh flow for inlet pressure is P1/P and the ratio of Rayleigh flow for outlet pressure is P2/P.

Determine the stagnation temperature in the duct.

T02T01=T02/TT01/T (VIII)

Here, the ratio of Rayleigh flow for exit stagnation temperature is T01/T and the ratio of Rayleigh flow for outlet stagnation temperature is T02/T.

Determine the rate of heat transfer of the duct.

Q˙=m˙aircp(T02T01) (IX)

Determine the entropy change of the duct.

Δs=cplnT2T1RlnP2P1 (X)

Conclusion:

From the Table A-2E, “Ideal-gas specific heats of various common gases” to obtain value of universal gas constant, specific heat of pressure, and the specific heat ratio of air at 80°F temperature as 0.06855Btu/lbmR, 0.2400Btu/lbmR and 1.4.

Substitute 80psia for P1, 0.06855Btu/lbmR for R, and 700R for T1 in Equation (I).

ρ1=80psia(0.06855Btu/lbmR)(700R)=80psia(0.06855Btu/lbmR×(5.40395psiaft31Btu))(700R)=80psia(0.3704psiaft3/lbmR)(700R)=0.3085lbm/ft3

Substitute 260ft/s for V1, 0.3085lbm/ft3 for ρ1, and 6in×6in for Ac1 in Equation (II).

m˙air=(0.3085lbm/ft3)(6in×6in)(260ft/s)=(0.3085lbm/ft3)(6in×6in)×(1ft2144in)(260ft/s)=20.052lbm/s

Substitute 700R for T1, 260ft/s for V1, and 0.2400Btu/lbmR for cp in Equation (III).

T01=(700R)+(260ft/s)22×(0.2400Btu/lbmR)=(700R)+(67600ft2/s2)(0.48Btu/lbmR)=(700R)+(67600ft2/s2)×(1Btu/lbm25,037ft2/s2)(0.48Btu/lbmR)=(700R)+(5.625R)

=705.6R

Substitute 1.4 for k, 0.06855Btu/lbmR for R, and 700R for T1 in Equation (IV).

c1=(1.4)(0.06855Btu/lbmR)×(700R)=(1.4)(0.06855Btu/lbmR)×(25,037ft2/s21Btu/lbm)×(700R)=1681960.6ft2/s2=1296.9ft/s

   =1297ft/s

Substitute 260ft/s for c1 and 1297ft/s for V1 in Equation (V).

Ma1=(260ft/s)(1297ft/s)=0.200460.2005

Refer to Table A-34, “Rayleigh flow function for an ideal gas with k=1.4”, to obtain the value ratio of static temperature, pressure, and stagnation temperature at 0.2005 inlet Mach number using interpolation method of two variables.

Write the formula of interpolation method of two variables.

y2=(x2x1)(y3y1)(x3x1)+y1 (XI)

Here, the variables denote by x and y is ratio of stagnation temperature and Mach number.

Show the Mach number at 0.2 and 0.3 as in Table (1).

S. No

Mach number

(x)

ratio of stagnation temperature

(y)

10.20.1736
20.2005y2=?
30.30.3469

Calculate ratio of static temperature, pressure, and stagnation temperature at 0.2005 inlet Mach number using interpolation method.

Substitute 0.2 for x1, 0.2005 for x2, 0.3 for x3, 0.1736 for y1, and 0.3469 for y3 in Equation (XI).

y2=(0.20050.2)(0.34690.1736)(0.30.2)+0.1736=0.1744

From above calculation the ratio of stagnation temperature at 0.2005 is inlet Mach number 0.1744.

Repeat the Equation (XII), to obtain the value of inlet ratio of static temperature and pressure at 0.2005 inlet Mach number as:

T1/T=0.2075P1/P=2.272

From the Table A-34, “Rayleigh flow function for an ideal gas with k=1.4”, to obtain the value of the outlet ratio of temperature, pressure, and velocity at 1 outlet Mach number as:

T2/T=1P2/P=1V2/V=1

Substitute 700R for T1, 1 for T2/T, and 0.2075 for T1/T in Equation (VI).

T2(700R)=10.2075T2=(700R)0.2075T2=3373.49R

Substitute 80 psia for P1, 1 for P2/P, and 2.272 for P1/P in Equation (VII).

P2(80psia)=12.272P2=(80psia)2.272P2=35.21psia

Substitute 705.6R for T01, 1 for T02/T, and 0.1743 for T01/T in Equation (VIII).

T02(705.6R)=10.1743T02=(705.6R)0.1743T02=4048R

Substitute 20.052lbm/s for m˙air, 0.2400Btu/lbmR for cp, 4048R for T02 and 705.6R for T01 in Equation (IX).

Q˙=(20.052lbm/s)(0.2400Btu/lbmR)(4048705.6)R=(20.052lbm/s)(0.2400Btu/lbmR)(3342.4R)=16085Btu/s16090Btu/s

Thus, the rate of heat transfer in the duct is 16,090Btu/s_.

Substitute 0.2400Btu/lbmR for cp, 3373.49R for T2, 700R for T1, 0.06855Btu/lbmR for R, 35.21psia for P2,and 80psia for P1 in Equation (X).

Δs=[(0.2400Btu/lbmR)(ln(3373.49R)(700R))(0.06855Btu/lbmR)(ln(35.21psia)(80psia))]=[(0.2400Btu/lbmR)(1.572)(0.06855Btu/lbmR)(0.8207)]=0.4336Btu/lbmR0.434Btu/lbmR

Thus, the entropy change in the duct is 0.434Btu/lbmR_.

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Chapter 17 Solutions

Thermodynamics: An Engineering Approach

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