A 01m3 rigid tank contains steam initially at 500 kPa and 20

A 0.1-m^3 rigid tank contains steam initially at 500 kPa and 200 degree Celsius The steam is now allowed to cool until its temperature drops to 50 degree Celsius. Determine the amount of heat transfer during this process and the final pressure in the tank.

Solution

Rigid tank: Constant vol\'m = 0.1 m^3

P1 = 500 kPa, T1 = 200+273 = 473K

P2 =?, T2 =50+273 = 323 K

From steam table @P = 500 kPa and T = 473K (200°C),

v1 = 0.4249 m^3/Kg, u1= 2642.9 KJ/Kg

At state 2, v2=v1 = 0.4249 m^3/Kg &

P2 = P_sat@50°C = 12.35 kPa & T2 = 50°C

From steam table,

v_2f= 0.001 m^3/Kg and V_2g = 12.032 m^3/Kg

u_2f=209.32 KJ/Kg & u_2g= 2443.5 KJ/Kg

From v= vf + x (vg-vf)

x= (0.4249-0.001)/ (12.032 -0.001) = 0.0352

So, u2= 209.32 + 0.0352 (2443.5-209.32)

u2= 287.96 KJ/Kg

Q_out = - m (u2-u1) = V/v (u1-u2)

= (0.1/0.4249) * (2642.9 - 287.96) = 554.233 KJ....Ans

= 323 * (500/473)

= 341.43 kPa