TITLE.PM5

FIRST LAW OF THERMODYNAMICS 189

dharm

/M-therm/Th4-5.pm5

2 × 10^512

p (N/m )^2

v = 0.885 1 v = 1.316 2

v (m /kg)^3

Fig. 4.60
(i)Heat supplied :
Heat supplied, Q = H 2 – H 1 = m(h 2 – h 1 )
= 0.08(3071.8 – 2706.3)
= 29.24 kJ. (Ans.)
(ii)Work done :
The process is shown on a p-v diagram in Fig. 4.60. The work done is given by the shaded
area
i.e., W = p(v 2 – v 1 ) Nm/kg
Here v 1 = vg at 2 bar = 0.885 m^3 /kg
and v 2 = 1.316 m^3 /kg
∴ W = 2 × 10^5 (1.316 – 0.885) = 2 × 10^5 × 0.431 Nm/kg
Now work done by the total mass of steam (0.08 kg) present
= 0.08 × 2 × 10^5 × 0.431 × 10–3 kJ
= 6.896 kJ. (Ans.)
Example 4.53. 1 kg of steam at 8 bar, entropy 6.55 kJ/kg K, is heated reversibly at con-
stant pressure until the temperature is 200°C. Calculate the heat supplied, and show on a T-s
diagram the area which represents the heat flow.

Solution. Mass of steam, m = 1 kg

Pressure of steam, p = 8 bar

Entropy of steam (at 8 bar), s = 6.55 kJ/kg K

Temperature after heating = 200°C

At 8 bar, sg = 6.66 kJ/kg K, hence steam is wet, since the actual entropy, s, is less than sg.

To find the dryness fraction x 1 , using the relation,

ss xs 11 =+f 1 fg 1