TITLE.PM5

(Ann) #1
FIRST LAW OF THERMODYNAMICS 143

dharm
/M-therm/Th4-3.pm5

or Taking log on both sides, we get
loge (0.6357) = (γ – 1) loge (0.333)


  • 0.453 = (γ – 1) × (– 1.0996)
    ∴γ =
    0
    0996


.453
1.
+ 1 = 1.41

Also,

c
c

p
v = γ = 1.41
Work done during adiabatic process,

W

mR T T
12 − =^121



()
γ

∴ 52.5

0 453 288
41 1
=


.44
1.

R()
()

∴ R= −

52.5 41 1
44 453 288

()
()

1.
0.
= 0.296

∴ cp – cv = 0.296 [Q R = cp – cv]

Also

c
c

p
v
= 1.41 or cp = 1.41 cv

∴ 1.41 cv – cp = 0.296
or cv = 0.722 kJ/kg K. (Ans.)
and cp = 1.018 kJ/kg K. (Ans.)



  • Example 4.30. 1 kg of ethane (perfect) gas is compressed from 1.1 bar, 27°C according
    to a law pV1.3 = constant, until the pressure is 6.6 bar. Calculate the heat flow to or from the
    cylinder walls.
    Given : Molecular weight of ethane = 30, cp = 1.75 kJ/kg K.
    Solution. Mass of ethane gas, m = 1 kg
    Initial pressure, p 1 = 1.1 bar
    Initial temperature, T 1 = 27 + 273 = 300 K
    Final pressure, p 2 = 6.6 bar
    Law of compression, pV1.3 = C
    Quantity of heat transferred, Q :
    Now, characteristic gas constant,


R R
M

=Universal gas constart ( )
Molecular weight ( )

0

=^831430 = 277.13 N-m/kg K = 277.31 J/kg K
= 0.277 kJ/kg K
Also cp – cv = R
∴ cv = cp– R = 1.75 – 0.277 = 1.473 kJ/kg K

γ =
c
c

p
v

=
1.
1.

75
473
= 1.188
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