Thermodynamics Compressor Work (1st law question)

[brk51]

Homework Statement

Compare the compressor work input required to compress water isentropically from 100kpa to 1MPa, assuming that the water exists as (a) saturated liquid and (b) saturated vapor at the inlet state

Homework Equations

Win = v(P2-P1) Win = (h2- h1)

The Attempt at a Solution

Thought I got this question completely right in the moment.. got most of it wrong actually. I'm confused on why i got it wrong. Here was my thought process:
(1) compressor has no heat input therefore, W(in) = change in internal energy = (h2-h1)

 

[rude man]

Win = v(P2-P1) Win = (h2- h1)

?? Maybe confusing p dV with V dp?

 

[brk51]

?? Maybe confusing p dV with V dp?

I'm considering the compressor to be a control volume so i use vdP

 

[rude man]

I'm considering the compressor to be a control volume so i use vdP

How do you propose to increase pressure at constant volume isentropically? Anyway, your reply doesn't make sense to me.

 

[brk51]

How do you propose to increase pressure at constant volume isentropically? Anyway, your reply doesn't make sense to me.

I'm looking at my book right now.

This is from the rankine cycle description.. : "1-2 isentropic compression in a pump. W_pump,in = h2-h1 or v(p2-p1)"

 

[brk51]

I'm looking at my book right now.

This is from the rankine cycle description.. : "1-2 isentropic compression in a pump. W_pump,in = h2-h1 or v(p2-p1)"

I think my mistake is treating a compressor and pump as basically the same thing only that they differ in that pumps deal with liquids and compressors deal with vapors. (My professor told us that)

 

[rude man]

I'm looking at my book right now.

This is from the rankine cycle description.. : "1-2 isentropic compression in a pump. W_pump,in = h2-h1 or v(p2-p1)"

I have to admit I don't understand this statement.
ΔH = V Δp looks right since V is approximately constant and we're compressing a liquid adiabatically.
But I can't get from that to ΔW = V Δp = ΔH. Hopefully someone else can. I get ΔW = p ΔV as usual. ΔV is indeed small in this phase of the cycle but can't be zero.

 

[brk51]

I have to admit I don't understand this statement.
ΔH = V Δp looks right since V is approximately constant and we're compressing a liquid adiabatically.
But I can't get from that to ΔW = V Δp = ΔH. Hopefully someone else can. I get ΔW = p ΔV as usual. ΔV is indeed small in this phase of the cycle but can't be zero.

If I was to use the first law on the compressor
Qin + mdot(h1) = Wout + mdot(h2)
Control Volume so mass flow rate into compressor equals mass flow rate out>>> Cancels the mdots. Then you are left with Qin + h1 = Wout + h2. Since there is heat exchange in the compressor you are left with Wout= h1-h2 ... which then you switch around to get Win.

That's what I have been taught atleast!

EDIT: Its really (h1 + KE + PE) but since the compressor is not moving physically or at a specificed elevation they also both cancel

 

[Chestermiller]

I see what you are trying to do in this problem, and it seems you have the right idea (even though, you called enthalpy internal energy in your original post). In your calculations, you seem to be using the steam tables. This is definitely the way to go. Do you want me to check your calculations and see what I get? Did you solve for the final state using the condition that the entropy is constant? What was your determination of the final state of the water (compressed liquid, saturated liquid, saturated vapor, superheated vapor) for the two cases?

 

[brk51]

I see what you are trying to do in this problem, and it seems you have the right idea (even though, you called enthalpy internal energy in your original post). In your calculations, you seem to be using the steam tables. This is definitely the way to go. Do you want me to check your calculations and see what I get? Did you solve for the final state using the condition that the entropy is constant? What was your determination of the final state of the water (compressed liquid, saturated liquid, saturated vapor, superheated vapor) for the two cases?

Yes that would be fantastic..

1) I don't believe I assumed entropy was constant when looking up values, that is probably why I lost most points. I think I got the correct answers now, let me know what you get.

However, on the exam I went to the Saturated Water - Pressure table and got values for hf at both 1000 KPa and 100KPa, (762.51 kJ/kg and 417.51 kJ/kg). But looking at it now, I should have used vdP since I knew I could define state 1>>> with v1 = .0001043 and the change in pressure 900 kpa. Win should be .9387 kJ

2) For the saturated vapor at the inlet state: At 100 KPa, s_g = 7.3589 kJ/kg*K..which equals the entropy at 1MPa...Looking in the superheated water table I found that at .1MPa, h = (approximately) 2675 kJ/kg and at 1MPa h =(approximately) 3165 kJ/kg*K .. I didn't interpolate just for the sake of time. Thus Win = h2-h1 = 3165-2675 = 525 kJ.

 

[Chestermiller]

Yes that would be fantastic..

1) I don't believe I assumed entropy was constant when looking up values, that is probably why I lost most points. I think I got the correct answers now, let me know what you get.

However, on the exam I went to the Saturated Water - Pressure table and got values for hf at both 1000 KPa and 100KPa, (762.51 kJ/kg and 417.51 kJ/kg). But looking at it now, I should have used vdP since I knew I could define state 1>>> with v1 = .0001043 and the change in pressure 900 kpa. Win should be .9387 kJ

2) For the saturated vapor at the inlet state: At 100 KPa, s_g = 7.3589 kJ/kg*K..which equals the entropy at 1MPa...Looking in the superheated water table I found that at .1MPa, h = (approximately) 2675 kJ/kg and at 1MPa h =(approximately) 3165 kJ/kg*K .. I didn't interpolate just for the sake of time. Thus Win = h2-h1 = 3165-2675 = 525 kJ.

I agree with your answer to item (2).

In item 1, there is a big difference. Did you check the compressed water tables, rather than using the saturated liquid water data (which is obviously the wrong approach because the temperatures are so different, so that the entropies are not constant)? The answer from the tables should be very close to the vdP result.

 

[brk51]

I agree with your answer to item (2).

In item 1, there is a big difference. Did you check the compressed water tables, rather than using the saturated liquid water data (which is obviously the wrong approach because the temperatures are so different, so that the entropies are not constant)? The answer from the tables should be very close to the vdP result.

That's where I'm confused... are you saying there is another approach other than vdP to get the correct answer?

The compressed water tables start out at a pressure of 5MPa

 

[Chestermiller]

That's where I'm confused... are you saying there is another approach other than vdP to get the correct answer?

The compressed water tables start out at a pressure of 5MPa

Oh. I don't have the compressed water tables at my fingertips right now. If 5 MPa is the lowest that your table goes, then I guess doing vdP is the only option. There must be tables that go down to lower pressures. Anyway, you can at least use the tables to go from 0.1 MPa to 5 MPa, and scale it to going from 0.1 MPa to 1 MPa to see how the answer compares to the vdP answer.

For the case of the vapor, you might also try to compare what you get by integrating vdP, using an approximate value of $\gamma$ for water vapor over the range of interest (assuming ideal gas). A pressure of 10 bars is not too far out of the range of ideal gas behavior for water vapor.

 

[brk51]

Oh. I don't have the compressed water tables at my fingertips right now. If 5 MPa is the lowest that your table goes, then I guess doing vdP is the only option. There must be tables that go down to lower pressures. Anyway, you can at least use the tables to go from 0.1 MPa to 5 MPa, and scale it to going from 0.1 MPa to 1 MPa to see how the answer compares to the vdP answer.

For the case of the vapor, you might also try to compare what you get by integrating vdP, using an approximate value of $\gamma$ for water vapor over the range of interest (assuming ideal gas). A pressure of 10 bars is not too far out of the range of ideal gas behavior for water vapor.

So, I'm trying to bring it all together. I use enthalpies for saturated vapor because I can use the constant entropy value to find corresponding enthalpy values in the superheated water table, thus giving work input. My initial thought is that I can use vdP the same way I use it for item 1 except that v is now it's the specific volume of vapor at 100 kPa instead of liquid.

 

[Chestermiller]

So, I'm trying to bring it all together. I use enthalpies for saturated vapor because I can use the constant entropy value to find corresponding enthalpy values in the superheated water table, thus giving work input.

Yes, for the vapor.

My initial thought is that I can use vdP the same way I use it for item 1 except that v is now it's the specific volume of vapor at 100 kPa instead of liquid.

No. To approximate it this way, you would have to integrate vdP, subject to $Pv^{\gamma}=const$. You should get pretty close to the same result as from the steam tables. Try using your steam tables solution to estimate an approximate value of $\gamma$ to use.