Question

The radiator of a steam system has a volume of 20 L and is filled with superheated water vapor at 200 kPa and 150 oC. At t = 0, both the inlet and outlet valves to the radiator are closed. After a while the temperature of the steam drops to 40 oC as a result of heat transfer to the room air. Determine the entropy change of the steam during this process.

Answer 1

Answer:

Δs = 0.133 KJ/.K

Explanation:

Given that

V= 20 L = 0.02 m³

At P₁= 200KPa and T₁=150 C :

v₁=0.93 m³/kg

s₁=7.2 KJ/kg.K

At  T₂=40 C :

v₁ = v₂

vf= 0.001 m³/kg

vg = 19.51 m³/kg

v₂ = vf + x(vg-vf)

0.93 = 0.001 + x(19.51 - 0.001)

x=0.047

s₂ = sf + x(sg-sf)

s₂ = 0.57 + 0.047(7.638) KJ/kg.K

s₂ = 0.93 KJ/kg.K

Δs= s₁ - s₂

Δs= 7.2 - 0.93 =6.2 KJ/kg.K

m=V/v₁

m=0.02/0.93 = 0.021 kg

Δs = 0.021 x 6.2 = 0.133 KJ/.K

Δs = 0.133 KJ/.K