Answer:
The entropy change of the steam is 2.673 kJ/K
Explanation:
Mass of liquid-vapor mixture = 1.5 kg
Mass in liquid phase = 3/4 × 1.5 kg = 1.125 kg
Mass in vapor phase = 1.5 - 1.125 = 0.375 kg
From steam table
At 200 kPa (200/100 = 2 bar), specific entropy of steam = 7.127 kJ/kgK
Entropy of steam = specific entropy × mass = 7.127 × 0.375 = 2.673 kJ/K
Answer:
The change in entropy of steam during the process is found to be 5.55615 KJ/kg
Explanation:
First of all, we use steam tables to find the properties at both states:
At State 1:
Pressure = P1 = 200 KPa
Quality = X1 = 1 - 3/4 = 1/4 = 0.25
From the steam table at 200 KPa:
Liquid specific volume = Vf = 0.001061 m³/kg
Gas specific volume = Vg = 0.88578 m³/kg
Vfg = Vg - Vf = 0.88578 - 0.001061 = 0.884719 m³/kg
Sf = 1.5302 Kj/kg.k
Sfg = 5.5968 KJ/kg.k
Hence, the specific volume of water at state 1 will be:
V1 = Vf + X1 Vfg = 0.001061 m³/kg + (0.25)(0.884719 m³/kg)
V1 = 0.22224075 m³/kg
Similarly the entropy will be given as:
S1 = Sf + X1 Sfg = 1.5302 KJ/kg.k + (0.25)( 5.5968 KJ/kg.k)
S1 = 2.9294 KJ/kg.k
At State 2:
V2 = V1 = 0.22224075 m³/kg
X2 = 1 (Saturated Vapor)
So, from the steam table at these conditions, we find by interpolation:
S2 = 6.6335 KJ/kg.k
Now, the entropy change of steam is given as:
ΔS = m(S2 - S1)
ΔS = (1.5 kg)(6.6335 - 2.9294) KJ/kg.k
ΔS = 5.55615 KJ/K
