TITLE.PM5

322 ENGINEERING THERMODYNAMICS

DHARM
M-therm\Th6-2.PM5

= 85.4 – 288 × 0.061 = 67.8 kJ/kg

∴ Minimum work required

25 678

60

×. = 28.25 kJ/s = 28.25 kW. (Ans.)

Example 6.10. 1 kg of oxygen at 1 bar and 450 K is mixed with 1 kg of hydrogen at the

same temperature and pressure by removing the diaphragm as shown in Fig. 6.11. Determine the

loss in availability if the surrounding temperature is 290 K.

Assume that the system is fully isolated.

Fig. 6.11

Solution. Mass of oxygen, mO 2 = 1 kg

Mass of hydrogen, mH 2 = 1 kg

Pressure, p = 1 bar = 1 × 10^5 N/m^2

Temperature, TTOH 22 = = 450 K

Surrounding temperature = 290 K

Characteristic gas constant of O 2 ,

RO 2 = R

M

0

O 2

=^831432 = 259.6 J/kg K

Now to find volume of O 2 using the relation,

pv = mRT

v = mRTp = 1 259 6 450

1105

××

×

. = 1.168 m 3

i.e., vO 2 = 2.336 m^3

Characteristic gas constant of H 2 ,

RH 2 = R

MH

0

2

=^83142 = 4157 J/kg K

Volume of H 2 , vH 2 = 1 4157 450

1105

××
×
= 18.706 m^3
Total volume after mixing
= vvOH 22 + = 1.168 + 18.706
= 19.874 m^3
The partial pressure of each gas changes after the mixing even though the temperature is
the same due to increase in volume.