Thermodynamics friction less piston problem

[annnoyyying]

Homework Statement

Consider an ideal friction-less cylinder-piston arrangement, in which the piston is free to move between two sets of stops. When the piston rests on the lower stops, the volume available for the contained fluid is 0.5 m3; when the piston reaches the upper stops, the volume is 0.7 m3. The piston has a mass such that an internal pressure of 3.5 bar is required to lift it.

Initially (State 1), the system contains a liquid-vapour mixture of water at 1 bar and 14% quality. The system is heated until the water is fully evaporated (i.e., to saturated vapour; State 2).

a) Determine the initial temperature in the cylinder.

b) Determine the final pressure and temperature in the cylinder.

c) Calculate the work done by the fluid.

d) What is the heat transferred during the process?

If the system in State 1 was cooled, instead of heated, to ambient temperature (25ºC; State 3) what would be the final pressure, and what would be the quality?

Homework Equations

for closed system, du = q+w
w = -PdV

The Attempt at a Solution

The initial temperature is 99.6 degrees c according to the steam tables. (because the temperature has to be the temperature of saturation at 1 bar or else there would be no liquid-vapour coexistence?)
at quality = 0.14
specific volume of saturated liquid = 0.001043 m3/kg
specific volume of saturated vapour = 1.6939 m3/kg
using x=(v-vl)/(vg-vl)
(v being specific volume of vapour liquid, vl being specific volume of saturated liquid, vg being specific volume of saturated vapour)
v = 0.2380 m3/kg
taking 1kg water basis,
V = 0.2380 m3
but initial volume available in the cylinder for the water is 0.5m3
so actual mass of water present to begin with is 0.5/0.2380 = 2.1kg

I'm thinking that at state 2 the specific volume is going to be 0.3333m3/kg as the volume will now be 0.7m3 instead of 0.5m3, so 0.7/2.1 = 0.3333 The pressure is 5.67 bar with a temperature of 156.6 degrees c?

If the above is correct, then comes the question of how the process is going to be like (on a pv diagram)...
Please shed some light on this.

 

[Chestermiller]

Your results are correct so far. Here's a hint: You now have enough information without doing any more calculations to precisely establish the entire P vs V variation from State 1 to State 2.

Chet

 

[annnoyyying]

How should the full process look like though? Say if it is isochoric from state 1 to 8.1 bar with heating, does it turn into saturated vapour? How does it proceed to state 2? Does the saturated vapour at 8.1 bar get isobarically heated to superheated steam to the specific volume in state 2 then back down to saturated vapour at the reduced pressure at state 2? So basically up to intersect saturated vapour, across to superheated region, then down again to saturated vapour?

 

[Chestermiller]

How should the full process look like though? Say if it is isochoric from state 1 to 8.1 bar with heating, does it turn into saturated vapour? How does it proceed to state 2? Does the saturated vapour at 8.1 bar get isobarically heated to superheated steam to the specific volume in state 2 then back down to saturated vapour at the reduced pressure at state 2? So basically up to intersect saturated vapour, across to superheated region, then down again to saturated vapour?

No. None of this. You start out at 1 bar and 0.5 m³. The piston starts to rise when you get to 3.5 bar and 0.5 m³. During the time that the volume is between 0.5 m³ and 0.7 m³, the piston holds the pressure in the cylinder constant at 3.5 bar. After the piston hits the stop, the volume stops increasing, and the pressure begins rising again, reaching 5.67 Bars at State 2. So, how much work is done?

Chet

 

[annnoyyying]

So on the pv diagram there should be a straight line going up from state one to 3.5 bar, then a horizontal line at 3.5 bar from 0.2380 to 0.3333, then another straight line up to 5.67 bar to intersect with the saturated steam line?
Work can only be done at the horizontal stage?

 

[Chestermiller]

So on the pv diagram there should be a straight line going up from state one to 3.5 bar, then a horizontal line at 3.5 bar from 0.2380 to 0.3333, then another straight line up to 5.67 bar to intersect with the saturated steam line?
Work can only be done at the horizontal stage?

Yes and yes. Don't forget to multiply by the mass, or use the actual volumes.

Chet

 

[annnoyyying]

So i calculated the work done to be -70 kj and the total heat transferred to be 3969 kj
What happens to the water if it was cooled to 25 degrees c from state 1? The volume stays at0

 

[annnoyyying]

What happens to the water if it was cooled to 25 degrees from the initial state. Does the specific volume remain the same? If the specific volume stays the same then pressure decreases and quality increases?

 

[Chestermiller]

So i calculated the work done to be -70 kj and the total heat transferred to be 3969 kj
What happens to the water if it was cooled to 25 degrees c from state 1? The volume stays at0

I get +70 kJ for the work done by the gas. Please show the details of how you got the heat.

 

[Chestermiller]

What happens to the water if it was cooled to 25 degrees from the initial state. Does the specific volume remain the same? If the specific volume stays the same then pressure decreases and quality increases?

The combined weighted average specific volume stays the same. The quality decreases.

Chet

 

[annnoyyying]

Oh yes the work is +70kj by the gas being -70kj done on the fluid.
Working to find heat:
Closed system, du=q+w
No work done at vertical stages 1 bar to 3.5bar and 3.5bar to 5.67 bar so du=q.
For isobaric evapouration stage, du=q+w, W is -70kj as calculated before, do some algebra and cancel out, only need u values for initial and final states to find q as u is a state variable. So Q = U final - U initial - W

 

[Chestermiller]

Oh yes the work is +70kj by the gas being -70kj done on the fluid.
Working to find heat:
Closed system, du=q+w
No work done at vertical stages 1 bar to 3.5bar and 3.5bar to 5.67 bar so du=q.
For isobaric evapouration stage, du=q+w, W is -70kj as calculated before, do some algebra and cancel out, only need u values for initial and final states to find q as u is a state variable. So Q = U final - U initial - W

You're confident that you determined Uinitial correctly, yes?

Chet

 

[annnoyyying]

according to my steam tables (rogers and mayhew)
at state 1
u liquid = 417 kJ/kg
u vapour = 2506 kJ/kg
quality = 0.14
u initial = 417 + 0.14(2506-417) = 709.46 kJ/kg

at state 2
u vapour = u final = 2566 kJ/kg

du = 1856.54 kJ/kg
2.1 kg water present = dU = 3898.7 kJ
Q = dU - W = 3898.7- (-70) = 3968.7 kJ

 

[Chestermiller]

according to my steam tables (rogers and mayhew)
at state 1
u liquid = 417 kJ/kg
u vapour = 2506 kJ/kg
quality = 0.14
u initial = 417 + 0.14(2506-417) = 709.46 kJ/kg

at state 2
u vapour = u final = 2566 kJ/kg

du = 1856.54 kJ/kg
2.1 kg water present = dU = 3898.7 kJ
Q = dU - W = 3898.7- (-70) = 3968.7 kJ

Looks good. Very nice job.

Chet

 

[annnoyyying]

Thank you very much for your help. I would not have been able to do this without you.

 

[Chestermiller]

Thank you very much for your help. I would not have been able to do this without you.

Don't be so hard on yourself. All you needed was a few hints. My impression was that you had a feel for this, but needed some experience.

Chet

 

[annnoyyying]

Yes I was struggling to start the problem. But I am still extremely grateful for your help.