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Modern Engineering Thermodynamics

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740 CHAPTER 18: Introduction to Statistical <strong>Thermodynamics</strong><br />

Table 18.3 Measured Values of the Specific Heats of Various Gases at 20.0ºC<br />

Gas c v /R c p /R k = c p /c v<br />

Monatomic<br />

He 1.50 2.50 1.67<br />

Ne 1.50 2.50 1.67<br />

Ar 1.51 2.52 1.67<br />

Kr 1.00 1.68 1.68<br />

Xe 1.52 2.51 1.65<br />

Diatomic<br />

CO 2.51 3.51 1.40<br />

NO 2.51 3.51 1.40<br />

H 2 2.44 3.42 1.40<br />

O 2 2.53 3.53 1.40<br />

N 2 2.50 3.50 1.40<br />

Triatomic<br />

CO 2 3.48 4.48 1.29<br />

SO 2 3.97 4.97 1.25<br />

H 2 O 3.05 4.05 1.33<br />

EXAMPLE 18.4<br />

Estimate the heat transfer rate required to heat low-pressure gaseous carbon tetrachloride (CCl 4 ) from 500. to 1200. K in a<br />

steady state, steady flow, single-inlet, single-outlet, aergonic (i.e., zero work) process at a flow rate of 1.00 kg/min.<br />

Solution<br />

The system here is just the gas in the heating zone. Neglecting the flow stream kinetic and potential energies, the energy rate<br />

balance for this system reduces to<br />

so that<br />

Q : + m : ðh in − h out Þ = 0<br />

Q :<br />

= m : ðh out − h in Þ = m : c p ðT out − T in Þ<br />

For CC1 4 , b = 5; consequently, F = 3b = 15. Then, Eq. (18.28) gives<br />

Now, the molecular mass of carbon tetrachloride is<br />

and its gas constant is<br />

so<br />

Therefore,<br />

R = R M<br />

c p = ð1 + 15=2ÞR = 8:5 × R<br />

M = 12:0 + 4ð35:5Þ = 154 kg=kgmole<br />

=<br />

8:3143 kJ/ðkgmole ⋅ KÞ<br />

154 kg/kgmole<br />

= 0:0540 kJ/ðkg ⋅ KÞ<br />

c p = 8:5½0:0540 kJ/ðkg⋅KÞŠ = 0:459 kJ/ðkg.KÞ<br />

_Q = ð1:00 kg/minÞ½0:459 kJ/ðkg⋅KÞŠð1200: − 500:KÞ = 321 kJ/min<br />

Exercises<br />

10. If the mass flow rate of gaseous carbon tetrachloride in Example 18.4 is suddenly increased from 1.00 to 3.50 kg/min,<br />

determine the new heat transfer rate, assuming all the other variables remain unchanged. Answer: _Q = 1120 kJ/min:<br />

11. The exit temperature of the gaseous carbon tetrachloride in Example 18.4 is increased from 1200. K to 2000. K.<br />

Determine the heat transfer rate, assuming all the other variables remain unchanged. Answer: _Q = 688 kJ/min:<br />

12. The gas in Example 18.4 is changed from carbon tetrachloride to gaseous dichlorodifluoromethane (CCl 2 F 2 ). Estimate the<br />

heat transfer rate for this gas, assuming all the other variables in Example 18.4 remain unchanged. Answer: _Q = 409 kJ/min:

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