Question

A well-insulated tank of 50 m³ volume initially contains 16,000 kg of water distributed between liquid and vapor phases at 25°C. Saturated steam at 1500 kPa is admitted to the tank until the pressure reaches 800 kPa. What mass of steam is added (kg)?

Answer 1

As the liquid and vapor are distributed at 25 C, implies that the initial state of water is saturated steam.

For saturated steam at 25 C, from the steam table

specific enthalpy of saturated steam Hv = 2546.54 kJ/kg

specifc enthalpy of saturated water Hl = 104.838 kJ/kg

specific volume of saturated steam Vv = 43.3414 m³/kg

specific volume of saturated water Vl = 0.00100301 m³/kg

Given: Volume = 50 m³

mass of water m1 = 16000 kg

specific volume V1 = volume / mass = 50 / 16000 = 0.003125 m³/kg

x = quality of steam (mass fraction of vapor in the mixture)

V1 = xVv + (1-x)Vl

.: 0.003125 = x(43.3414) + (1-x)*0.00100301

.: x = 0.000048961

Now,

specifc enthlapy of the initial mixture H1 = xHv + (1-x)Hl = 0.000048961*2546.54 + (1-0.000048961)* 104.838

.: H1 = 104.957 kJ/kg

Now, saturated steam at 1500 kPa is admitted to the tank.

For saturated steam at 1500 kPa,

Hv = 2791.01 kJ/kg

Hl = 844.717 kJ/kg

As the steam is saturated steam, specific enthalpy if the inlet stream H2 = Hv = 2791.01 kJ/kg

V2 = Vv = 0.131702 m³/kg

Final state will be a wet steam of certain quality x.

Therefore, from the saturated steam table at 800 kPa,

Hv = 2768.3 kJ/kg

Hl = 721.018 kJ/kg

Vv = 0.240328 m³/kg

Vl = 0.00111479 m³/kg

Let m2 be the mass of the saturated steam added.

final mass m3 = (m2 + 16000) kg

final specific volume Vf = volume of tank / final mass = [50 / (m2 + 16000) ] m³/kg

.: By energy balance,

m1H1 + m2H2 = m3H3

16000*104.957 + m2*2791.01 = (16000 + m2)*H3

But H3 = xHv + (1-x)Hl = x( 2768.3) + (1-x)*721.018

.: 16000*104.957 + m2*2791.01 = (16000 + m2) * [x( 2768.3) + (1-x)*721.018].................................(1)

V = xVv + (1-x)Vl

[50 / (16000 + m2) ] = x*0.240328 + ( 1- x)*0.00111479

...............................................(2)

substitute in (1) and solve for m2

we get,

m2 = 0.000040908835 kg.........................................Answer