Steam at 100 kPa and 8 percent quality is contained in a spr

Steam at 100 kPa and 8 percent quality is contained in a spring-loaded piston-cylinder device, with an initial volume of 2.5 m^3. Steam is now heated until its volume is 4.46 m^3 and its pressure is 300 kPa. Determine the heat transferred to and the work produced by the steam during the process.

Solution

Ans2

Given initial pressure P1= 100 KPa , initial volume V1=2.5m3 --(1)

final pressureP2= 300 KPa & final volume V2=4.46 m3 ---(2)

The change in pressure and volume happens due to addition of heat.

The total heat transferrred Q= h2-h1 ---(3)

where h2 = enthalpy of state 2 and h1= enthalpy of state 1

Now in state 1 steam is at 8% quality, i.e water is (100-8)% =92)%.

Using standard table for steam, h1 can be calculated: hf =417.51 KJ/kgand hg=2675.0KJ/Kg

h1=0.92 h_f + 0.08 h_g =0.92 X 417.51 +0.08 X2675.0=598.11 KJ/Kg ---(4)

For state 2, steam is saturated, hence h2=h_g

Substituting these values heat transferred can be calculated. From table: h2=2724.9KJ/Kg --(5)

Substituing (4) and (5) in (3)

Heat transferred Q=2125.89 KJ/Kg --(6)

(b) Internal energy

From giventable , at 100 Kpa internal energy of fluid and gas :uf=417.50Kj/ kg & ug=2505Kj/kg --(7)

Hence total internal energy of state u1= 0.92 uf + 0.08 ug =384.1+200.4=584.5---(8)

Similarly, the internal energy of state 2: u2=ug =2543.2 KJ/Kg--(9)

Hence , from (8) &(9) change internal energy=u2-u1=1958.7 KJ/Kg --(10)

from 1st law of thermodynamics, heat supplied= change in internal energy + work done.

Hence, from (6) & (10): work done= 2125.89-1958.7=167.19 KJ/Kg