A piston cylinder device initially contains 0.6 kg of steam with a volume of 0.1 m3. The mass of the piston is such that it maintains a constant pressure of 800 kPa. The cylinder is connected through a valve to a supply line that carries steam at 5 MPa and 5000C. Now the valve is opened and steam is allowed to flow into the cylinder until the volume of the cylinder doubles and the temperature in the cylinder reaches 2500C, at which point the valve is closed. Determine (a) the mass of steam that has entered and (b) the amount of heat transfer.

Qin = 448.23 kJ

Explanation:

given data

mass m1 = 0.6 kg

volume v1 = 0.1 m³

pressure P1 = P2 = 800 kPa

pressure Pi = 5 MPa

temperature Ti = 500°C

temperature T2 = 250°C

volume V2 = 2V1 = 2 (0.1) = 0.2 m³

solution

we get here some value from steam table

as we get first for initial state at pressure 800 kPA

v1 =    ...........1

v1 =  

v1 = 0.1667 m³/kg

and u1 = 2004.4 kJ/kg

and

for final state at pressure 800 kPa and temperature 250

u2 = 2715.9 kJ/kg

and

v2 = 0.29321 m³/kg

now we get for stream in supply line that is

hi = 3434.7 kJ/kg

for the preesure 5 MPa and temperature 500

and

now we get mass m2 =      ............2

m2 =    

m2 = 0.6821 kg

so

now we consider here tank as control system

so that steam cross control surface is

as mass balance

mi = m2-m1

mi = 0.6821 - 0.6

mi = 0.0821 kg

and

now we neglect microscopic for energies

so energy balance is

Qin - W (b) + mihi = m2u2 - m1u1  ............3

Qin - P(V2-V1) + mihi = m2u2 - m1u1

Qin - 800 (0.2-0.1) + 0.082 (3434.7) = 0.6821 × 2715.9 - 0.6 × 2004.4

Qin = 448.23 kJ

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation: