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A mass of 5 kg of saturated liquid-vapor mixture of water is contained in a piston-cylinder device at 125 kPa. Initially, 2 kg of the water is in the liquid phase and the rest is in the vapor phase. Heat is now transferred to the water, and the piston, which is resting on a set of stops, starts moving when the pressure inside reaches 300 kPa. Heat transfer continues until the total volume increases by 20 percent.


Determine:

(a) the initial and final temperatures,

(b) the mass of liquid water when the piston first start moving

Respuesta :

a)  The initial and final temprature are 99.61°C and T3 = 259.07°C

b)   The mass of liquid water when the piston first start moving is 0g

Using the data provided:

Initial Pressure = P1 = 125kPa

Mass of Saturated Vapor Liquid = m = 5kg

Initial Mass of Liquid water = mf = 2kg

Initial Mass of Vapor = mg = 3kg

The temperature for a mixture of liquid steam saturated with water according to the thermodynamic table is:

T₁ = 99.61°C

The specific volume is:

vf = 0.001043m³/kg

And the saturated volume is:

vg = 1.6941m³/kg

Therefore, the volume in state 1 will be equal to:

V1 = mfvf + mgvg

V1 = (2kg)(0.001043m³/kg) + (3kg)(1.6941m³/kg)

V1 = 5.084m³

In state 3 then the volume will be equal to:

V3 = 1.2V1

V3 = 1.2(5.084m³)

V3 = 6.1008m³

Then the specific volume will be equal to:

v3 = V₃/m

v3 = 6.1008m³/5kg

v3 = 1.2202m³/kg

For a pressure of 300 kPa and the previous specific volume found through the thermodynamic table, the value of the temperature is:

T3 = 259.07°C

Question b) When the piston first stats moving at P₂ = 300kPa and V₂ = V₁ = 5.084m³

v₂ =V₂/m

v₂ = 5.084m³/5kg

v2 = 1.0168m³/kg

While at 300 kPa the value of the vapor saturated vg = 0.88578m³/kg This implies that the condition is met:

v2 > vg

Therefore, no liquid is there in-cylinder when piston stats moving, mf = 0kg

Learn more on solving problems on heat transfer from:

https://brainly.com/question/15289827?referrer=searchResults

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