# Homework Help: Calculate Pressure in piston cylinder after heating

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1. May 5, 2016

### Nemo's

1. The problem statement, all variables and given/known data
Water is contained in a cylinder fitted with a frictionless piston (figure shows that atmospheric pressure acts on piston from above). The mass of the water is 0.45 kg, and the piston area is 0.186 m2. Initially, the water conditions are 110oC and 90% dryness fraction, and the spring just touches the piston but exerts no forces on it. Now, heat is added to the water and the piston begins to rise. During this process, the spring resisting force is proportional to the distance moved with a coefficient of 88 N/cm. Calculate the pressure in the cylinder when the temperature reaches 160oC.

2. Relevant equations
1)Fs = kΔx
Fs: spring force, k = coefficient, x = distance moved by piston

2)PA = PοA + Fs + mg
P = Water pressure

Water is not and ideal gas but just in case :
3)P1V1/T1 = P2V2/T2
3') PV=mRT

4) Q + W = m (cvΔT+gΔx)
W is the boundary work

3. The attempt at a solution
I can get P1 by letting x=0 in eq2 --> P1=18855.5 Pa
Now I'm stuck and I hope someone can help.

2. May 5, 2016

### Staff: Mentor

What is the saturation vapor pressure of water at 110 C? What is the weight of the piston? For a 90 % dryness fraction, what is the average specific volume of the fluid at the initial condition?

3. May 5, 2016

### Nemo's

saturated vapour pressure P= 143.24 k
Weight of the piston: at Δx=0 mg=143.24A-PοA so now I can get mg.
I actually understand what 90% dryness fraction means or how can I use it.

4. May 5, 2016

### Staff: Mentor

So, what is mg?
Are you saying you do understand what 90% dryness fraction means or you don't? If you know what it means, then you can calculate the initial average specific volume of the water. This, together with the initial mass of water, will give you the initial volume.

Let x = displacement of piston. In terms of x and A, what is the change in volume? In terms of x and A, what is the final volume? In terms of x and A, what is the final specific volume? In terms of x and A, what is the final pressure?

Last edited: May 5, 2016
5. May 5, 2016

6. May 5, 2016

### Staff: Mentor

90 % dryness fraction means that there is 90% water vapor and 10 % liquid water.

7. May 6, 2016

### Nemo's

So now that I know that there's 90% water vapour and 10% water How do I calculate the average specific volume ? Can I just use PV=mRT to get the volume?
Here's what I tried I used the saturated vapour pressure I'm not sure I can use it here:
143.24*V=0.45(90%)(461.5)(383) so V =0.5 m^3

8. May 6, 2016

### Staff: Mentor

No. You use your steam tables. You look up the specific volume of the saturated liquid and the saturated vapor at the temperature of the system, and take a weighted average of the two.
This would be what you would do if you wanted to approximate the answer using the ideal gas law. Of course, you would also have to include the volume of the liquid water as well. Why don't you see how there results compare with what you calculate using the steam tables?

9. May 6, 2016

### Nemo's

specific volume of saturated liquid = 0.0010516 m^3/kg
vapour =1.2093 m^3/kg
weighted average volume = (0.0010516*0.9*0.45+1.2093*0.1*0.45)/2 = 0.027422 m^3

change in volume in terms of A and x = AΔx
Final volume = AΔx+0.027422
Final specific volume = Final volume/0.45

So now I know T2 = 160. I can't just read the corresponding pressure from the steam table and solve the problem right? What should I do ?

10. May 6, 2016

### Staff: Mentor

This is not quite correct.

weighted average specific volume = (0.1)(0.0010516)+(0.9)(1.2093)=1.088 m^3/kg
mass of water = 0.45 kg
Initial volume of water in cylinder = 0.490 m^3

Another way of getting this same result is:
Initial mass of water vapor = (0.45)(0.9)=0.405 kg
Initial mass of liquid water = (0.45)(0.1)=0.045 kg
Initial volume of water vapor = (0.405)(1.2093) = 0.4897 m^3
Initial volume of liquid water = (0.045)(0.0010516) = 0.00005 m^3
Initial volume of water in cylinder = 0.4897 + 0.00005 = 0.490 m^3
Correcting this result: Final volume = 0.490 + AΔx
Final specific volume = (0.490 + AΔx)/(0.45)= 1.088 + 2.222 A Δx

You can't read the corresponding pressure from the steam table because the system may not saturated in the final state. It may be a superheated vapor.

11. May 6, 2016

### Nemo's

Thanks for correcting me I switched the values for 0.9 and 0.1.
Yes, I see. So in the final state can I use PV = mRT assuming all the liquid is turned into vapour? Or maybe the superheated vapour isn't an ideal gas so I can't do so. I still need to find the final pressure.

12. May 6, 2016

### Staff: Mentor

Your assignment expects you to use the steam tables. However, that should not prevent you from also using the ideal gas law to see how the results compare.

You still haven't answered my question from post #4: "In terms of Δx and A, what is the final pressure?" Even with the ideal gas law, you are going to need to have this relationship for the pressure. (You already have the equation for the volume in terms of A and Δx).

Once you have the equation for the final pressure in terms of A and Δx, you can combine it with your equation for the specific volume in terms of A and Δx to express the final specific volume as a function of the final pressure. This will enable you to find the final state in the steam tables.

Chet

13. May 6, 2016

### Nemo's

P=Pο+mg/A+kΔx/A
=143240+8800Δx/A
Now I have the pressure in terms of A and Δx.
Vf = 1.008+2.222AΔx

P=Pο+mg/A+km(Vf-Vi)/(A^2)
=143240+8800AΔx
How can I get the final state from steam tables when Δx is still unknown? Am I doing something wrong?

14. May 6, 2016

### Staff: Mentor

This equation should read $$v_f=1.088+2.222A\Delta x=1.088+0.41329\Delta x$$
This equation should read $P_f=143240+8800\Delta x /A=143240+47312\Delta x$

So, combining these two equations gives:
$$\Delta x=2.4196v_f-2.6325=2.11364 \times 10^{-5}p_f-3.0275$$
or$$p_f=114472v_f+18691$$
You need to find the combination of pf and vf in the steam tables that satisfy this equation at 160 C.

Last edited: May 7, 2016
15. May 7, 2016

### Nemo's

I'm still confused about one thing though, the saturated steam table gives specific values for the pressure and specific volume at a certain temperature. e.g at 160οC pressure is 6.1823 bar and specific volume is 0.30678 for saturated vapour. So what other combinations could there be?

16. May 7, 2016

### Staff: Mentor

It's not saturated in the final state. There is no liquid water remaining. You need to use the part of the steam tables for superheated steam. At 160 C and 1.5 bars, my superheated steam tables gives a value of 1.317 m^3 for the specific volume, and at 3 bars, it gives 0.6506 m^3. So at 1.5 bars, Pv= 197550 J/kg, and at 3 bars, Pv = 195180 J/kg. From the ideal gas equation, what do you get?

17. May 7, 2016

### Nemo's

I'm afraid I don't know the value for gas constant R in case of the superheated steam. Can I even use the ideal gas equation in this case or is it just an approximation like you said earlier?
So now I'm supposed to select a value for vf right? e.g: Pf=114472vf+18691 so if vf=1.317 Pf=169450 Is that correct?

18. May 7, 2016

### Staff: Mentor

$$PV=nRT=\frac{m}{M}RT$$where m is the mass of vapor and M is the molecular mass (M=18). So$$Pv=\frac{RT}{M}$$where v is the specific volume (V/m). Thus, if we treat water vapor as an ideal gas, at T=160 C, we have:$$Pv=\frac{8314(273.2+160)}{18}=200090\tag{ J/kg}$$This is pretty close to the more accurate values for water vapor we obtained from the steam tables.
This predicts that the final pressure is 1.695 Bars even though the steam tables indicate that the specific volume value of 1.317 corresponds to a pressure of 1.5 Bars. So that is not quite the solution.

If we take the equation $p_f=114472v_f+18691$ and multiply by $v_f$, we obtain:
$$p_fv_f=114472v_f^2+18691v_f$$
For the case of an ideal gas, we know that the left hand side is 200090. Thus,
$$114472v_f^2+18691v_f-200090=0$$
For an ideal gas, all you need to do now is solve this quadratic equation for $v_f$. Once you know $v_f$, you can then determine the final pressure.

Now it's your turn. Do the same type of procedure to get the final values using the data from the steam tables.

19. May 7, 2016

### Nemo's

1)114472vf^2+18691vf-197550=0 so vf=1.2346 and pf=160063
2)114472vf^2+18691vf-195180=0 so vf=1.243 and pf=159113.8
and for the ideal gas pf=160979

These values for pf all seem pretty close to me the pressure lies around 160kPa so why did we calculate twice, once at 1.5 bars and the other at 3 bars ?

20. May 7, 2016

### Staff: Mentor

Because there is really only one combination of P and v that satisfies both our derived equation and the values in the steam tables. If we linearly interpolate the values of P and v in the steam tables between 1.5 and 3.0 Bars, capitalizing on the fact that we expect P to vary nearly linearly with 1/v, we obtain:
$$\frac{(p-1.5)}{(3.0-1.5)}=\frac{((1/v)-(1/1.317))}{((1/0.6506)-(1/1.317))}$$
This reduces to $$pv=0.03557v+1.92866$$
or, in terms of pressures in Pascals,
$$pv=192866+3557v$$
So this relationship is the best we can do by interpolating linearly in the steam tables. See what you get for p and v when you combine this with the equation $$pv=114472v^2+18691v$$ (by eliminating pv).

21. May 8, 2016

### Staff: Mentor

What we are saying here is that the behavior of the vapor is very close to that of an ideal gas (constant pv), but not quite. We didn't know this in advance, so we had to use the steam tables to investigate it. If the behavior were exactly that of an ideal gas, then it wouldn't matter if we used the values in the steam tables at 1.5 Bars or at 3.0 Bars. However, there is a slight non-ideal gas effect that we are taking into account. If the pressures were somewhat higher, say 10 Bars, there would be more variation of pv with pressure (or specific volume), and the effect would have been more significant.