Question

A piston-cylinder device initially contains 0.35-kg of steam at 3.5 MPa, superheated by 7.4°C. Now the steam loses heat to the surroundings and the piston moves down, hitting a set of stops at which point the cylinder contains saturated liquid water. The cooling continues until the cylinder contains water at 200°C.

Determine (a) the final pressure and the quality (if mixture), (b) the boundary work, (c) the amount of heat transfer when the piston first hits the stops, (d) and the total heat transfer.

Answer 1

Answer:

a) The final pressure is 1.5538 kPa, the quality is 0.062%

b) The boundary work is W=--707.42kJ

c) The amount of heat transfer when the piston first hits the stops is: $Q_{1-2}=-622.8kJ$

d) The total heat transfer is: Q=-689.9kJ

Explanation:

Consider that the piston initially moves down until saturated liquid is obtained at 3.5 MPa (The pressure is constant until this point because the piston is the only and constant weight that is under the system), so let's name three states:

State 1: Initial state

State 2: State where saturated liquid is obtained and the piston hit the stops.

State 3: The final state.

The volume changes but the pressure is constant in the 1-2 process, but the  opposite for the 2-3 process, where the volume remains constant and the pressure changes because the system is not under the piston weight (the piston weight is supported by the stops, not by the fluid).

So, considering that the pressure is constant in 1-2, it is possible to calculate the volume at state 2. The state 2 has the properties:(P=3.5MPa, and v=vf(3.5MPa) i.e. the volume of saturated liquid at this pressure. According to the tables I attach, that volume is 0.001235 $\frac{m^{3} }{kg}$.

So, the volume at the final state, which is constant between 2-3 is that one. Looking for the third state (T=200ºC, v=0.001235 $\frac{m^{3} }{kg}$), we realize that it is a vapor-liquid mixture, because the specific volume is between the saturated vapor and liquid volume. So, the pressure has to be the saturation pressure at 200ºC, which is 1.5538 kPa, according to the tables I attach.

The quality can be calculated according to:

$v=v_{f}+x v_{fg}\\x=\frac{v-v_{f}}{v_{fg}}=\frac{0.001235-0.001157}{0.12736-0.001157}=0.000618=0.0618$ %

In order to answer the other required data, it is just necessary to make the energy balance and read the enthalpies:

b) The boundary work

The work is done just by the piston displacement, so in the 2-3 process there is no work done because the piston does not move. The work in the process 1-2 is a work done under constant pressure, so it is given by:

$W=m*P(v_{2}-v_{1})\\0.35*3500kPa*(0.001235-0.05872)=-70.42kJ$

This work is negative because it is being done over the fluid, and it is the surroundings which make the work.

c) This is the heat transfer in the 1-2 process; the energy balance for this is:

$Q-W=m(u_{2}-u_{1})$

And applying the work equation above and the enthalpy definition:

$Q=m*(u_{2}-u_{1})+m*P(v_{2}-v_{1})\\Q=m*(u_{2}+P*v_{2}-(u_{1}+P*v_{1}))=m*(h_{2}-h_{1})$

The enthalpy of initial state (3.5MPa, 250ºC) is 2829.2 kJ/kg. (The temperature can be calculated looking for the saturation temperature at 3.5 MPa: 242.6 and adding 7.4ºC). The enthalpy of 2 state is the enthalpy of saturated liquid at 3.5MPa (1049.75kJ/kg)

$Q=0.35*(1049.75-2829.2)=-622.8kJ$

This quantity is also negative because the system is losing energy.

d) To the previous amount of heat sum the heat transfer in the 2-3 process, where the energy balance is:

$Q=m(u_{3}-u_{2})$

The internal energy of 3 is calculated with the quality: 853.65 kJ/kg. The state 2 internal energy is: 1045.43kJ/kg; so:

$Q=0.35*(853.65-1045.43)=-67.123kJ$.

So, the total heat transfer is:

$Q=-67.123-622.8=-689.9kJ$

I will be happy to solve any doubt you have.

Answer 2

1) The final pressure of steam will be 1.5538 kPa, the quality is 0.062%

2) The boundary work is W=--707.42kJ

3) The amount of heat transfer when the piston first hits the stops $Q_{1-2} =-622.8KJ$

4) The total heat transfer will be  $Q=-689.9kJ$

What will be the heat transfer and pressure of the steam?

There are three stages at the initial stage the condition of steam is mass is 0.35kg pressure 3.5Mpa and steam is superheated to 7.4 C

At the initial stage, the volume changes suppose in process 1-2 and in process 2-3 the volume remains constant, and the pressure of steam changes. because the system is not under the piston weight (the piston weight is supported by the stops, not by the steam).

For processes, 1-2 given conditions are  (P=3.5MPa, and v=vf(3.5MPa) i.e. the volume of saturated liquid at this pressure. According to the tables I attach, that volume is 0.001235.

So, the volume at the final state, which is constant between 2-3 is that one. Looking for the third state (T=200ºC, v=0.001235 ), we realize that it is a vapor-liquid mixture because the specific volume is between the saturated vapor and liquid volume. So, the pressure has to be the saturation pressure at 200ºC, which is 1.5538 kPa, according to the tables I attach.

1) The quality of steam will be calculated according to:

$v=v_{f} + xv_{fg}$

$x=\dfrac{v-v_{f} }{v_{f} } =\dfrac{0.001235-0.001137}{0.12736-0.00157}$

$x=0.000618=0.0618$

2) The boundary work

In process 2-3 it is a constant volume process so there is not any change in the volume the work done will be zero but in process 2-3 it is a constant pressure process so work done will be calculated as

The work is done just by the piston displacement, so in the 2-3 process there is no work done because the piston does not move. The work in process 1-2 is work done under constant pressure, so it is given by:

$W=m\timesp\times(v_{2}-v_{1} )$

$W=0.35\times3500\times(0.001235-0.05872 )=-70.42KJ$

This work is negative because work is done on the steam the surroundings that make the work.

3) This is the heat transfer in the 1-2 process; the energy balance for this is:

$Q-w=m\times(u_{2} -u_{1} )$

And applying the work equation above and the enthalpy definition:

$Q=m\times(u_{2} -u_{1} )+m\times p(v_{2} -v_{1} )$

$Q=m\times(u_{2} +p\timesv_{2} -(u_{1} +p\times v_{1} )=m\times(h_{1} -h_{2} )$

The enthalpy of the initial state (3.5MPa, 250ºC) is 2829.2 kJ/kg.

(The temperature can be calculated by looking for the saturation temperature at 3.5 MPa: 242.6 and adding 7.4ºC).

The enthalpy of 2 state is the enthalpy of saturated liquid at 3.5MPa (1049.75kJ/kg)

$Q=0.35\times (1049.75-2829.2)=-622.8KJ$

This quantity is also negative because heat is flowing out of the system

4) To the previous amount of heat sum the heat transfer in the 2-3 process, where the energy balance is

$Q=m\times(u_{3} -u_{2})$

The internal energy of 3 is calculated with 853.65 kJ/kg. The state 2 internal energy is 1045.43kJ/kg; so:

$Q=0.385\times(853.65-1045.43)=-67.123KJ.$

So, the total heat transfer is:

$Q=-67.123-622.8=-689.9KJ$

Thus,

1) The final pressure of steam will be 1.5538 kPa, the quality is 0.062%

2) The boundary work is W=--707.42kJ

3) The amount of heat transfer when the piston first hits the stops $Q_{1-2} =-622.8KJ$

4) The total heat transfer will be  $Q=-689.9kJ$

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