Question

2. A 1.00 m3 rigid vessel is filled with steam with a quality of 98.0% at 180◦C. Energy is added to the vessel until the pressure reaches 3,000 kPa. Determine the following:

a) The initial pressure in the vessel

b) The mass of water in the vessel

c) The final temperature in the vessel

d) The change in enthalpy

Answer 1

Volume = 1 m³

Now here we will make use of steam table to simplify our problem

We have steam table of quality 98 % at 180 ^oC

From steam table

Hl = 763.1879 KJ/Kg

Hv = 2777.219 KJ/Kg

Vl = 0.001127 m³/Kg

Vv = 0.19387 m³/Kg

Pressure of saturated steam = 10.026 bar

Answer to part (A): pressure of the steam is 10.026 bar

Now specific volume of steam (V) = x*Vv + (1-x)*Vl

where x is the quality of steam, which is already provided to us as 0.98

H =x*Hv +(1-x)*Hl = 0.98* 2777.219 + (1-0.98)*763.1879 =2736.9388 Kj/Kg

V = 0.98*0.19387 + (1-0.98)*0.001127 =0.19 m³/Kg

The total volume of the rigid vessel is already given as 1 m³

mass of the steam = 1/0.19 =5.263 Kg

answer to part (B): The mass of water /steam = 5.263 Kg

Now the final pressure is 3000 KPa

Now the mass of water and the volume of vessel will remain constant

Hence specific volume = 1/5.263 =0.19 won't change, so we will find the state point corresponding to this specific volume at P = 3000 KPa

T = 950 ^oC

H at this temperature = 5409 KJ/Kg

answer to part (C). The final temperature of the system is 950 ^oC

Change in enthalpy = Enthalpy at 30000 KPa - Enthalpy of saturated steam at 180 ^oC = 5409 - 2736.9388 = 2672.06 KJ/Kg

Answer to part (D): The change in enthalpy is 2672.06 KJ/Kg