TITLE.PM5

134 ENGINEERING THERMODYNAMICS

dharm

M-therm/th4-2.pm5

The change in internal energy of the fluid during the process

U 2 – U 1 = (42 + 3.6p 2 V 2 ) – (42 + 3.6p 1 V 1 )

= 3.6(p 2 V 2 – p 1 V 1 )

= 3.6(4.2 × 10^5 × 0.07 – 1.9 × 10^5 × 0.035) J

= 360(4.2 × 0.07 – 1.9 × 0.035) kJ

= 81.9 kJ

Now, p = a + bV

190 = a + b × 0.035 ...(i)

420 = a + b × 0.07 ...(ii)

Subtracting (i) from (ii), we get

230 = 0.035 b or b = 0 035.^230 = 6571 kN/m^5

and a = – 40 kN/m^2

Work transfer involved during the process

WpdVabVdVaVVbVV

V

V

V

V

12 2 1 2

2

1

2

1

2

1

2

− ==+=−+ 2

F −

HG

I

zz KJ

() ()

=−L + +

NM

O

QP

()()VVa 21 b 2 VV1 2

= (0.07 – 0.035) −+ +

L

NM

O

QP

40 kN / m^2565712 kN / m (.0 035 007. ) = 10.67 kJ
∴ Work done by the system = 10.67 kJ. (Ans.)
Heat transfer involved,
Q1–2 = (U 2 – U 1 ) + W1–2 = 81.9 + 10.67 = 92.57 kJ.
92.57 kJ of heat flow into the system during the process. (Ans.)
Example 4.21. 90 kJ of heat are supplied to a system at a constant volume. The system
rejects 95 kJ of heat at constant pressure and 18 kJ of work is done on it. The system is brought
to original state by adiabatic process. Determine :
(i)The adiabatic work ;
(ii)The values of internal energy at all end states if initial value is 105 kJ.
Solution. Refer Fig. 4.20.
Heat supplied at constant volume = 90 kJ
Heat rejected at constant pressure = – 95 kJ
Work done on the system = – 18 kJ
Initial value of internal energy, Ul = 105 kJ
Process l–m (constant volume) :
Wl–m = 0
Ql–m = 90 = Um – Ul
∴ Um = Ul + 90 = 105 + 90
= 195 kJ
Process m–n (constant pressure) :
Qm–n = (Un – Um) + Wm–n

• 95 = (Un – Um) – 18
  ∴ Un – Um = – 77 kJ