Page 114 - DJJ20063- Thermodynamics 1

DJJ20063- Thermodynamics 1

Example 3.4

The specific internal energy of a fluid is increased from 120 kJ/kg to 180 kJ/kg during

a constant volume process. Determine the amount of heat energy required to bring

about this increase for 2 kg of fluid.

Solution to Example 3.4

The non flow energy equation is

Q – W = U2 – U1

For a constant volume process
W = 0

and the equation becomes

Q = U2 – U1
 Q = 180 – 120

= 60 kJ/kg

Therefore for 2 kg of fluid

Q = 60 x 2 = 120 kJ

i.e. 120 kJ of heat energy would be required.

Example 3.5

3
2.25 kg of fluid having a volume of 0.1 m is in a cylinder at a constant pressure of 7

3
bar. Heat energy is supplied to the fluid until the volume becomes 0.2 m . If the initial

and final specific enthalpies of the fluid are 210 kJ/kg and 280 kJ/kg respectively,

determine

a) the quantity of heat energy supplied to the fluid

b) the change in internal energy of the fluid

Solution to Example 3.5

3
3
Data: p = 7.0 bar; V1 = 0.1 m ; V2 = 0.2 m

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