Question

A well-insulated rigid tank contains 1.5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters of the mass is in the liquid phase. An electric resistance heater placed in the tank is now turned on and kept on until all the liquid in the tank is vaporized. Determine the entropy change of the steam during this process. Use steam tables.

Answer 1

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 2

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