# Homework Help: Polyprotic Expansion HELP

1. Nov 4, 2011

### ttb90

Polyprotic Expansion HELP!!!

1. The problem statement, all variables and given/known data

Nitrogen at 100°C and specific volume of 0.1846 m^3/kg undergoes a polyprotic expansion process such that pv^(1.2)=const and the final pressure is 100 kPa. A constant pressure compression is then performed to bring the volume back to 0.1846 m^3/kg. Finally a constant volume pressurisation occurs to bring the Nitrogen back to its initial pressure. Nitrogen is a perfect gas with R= 0.297 kJ/kg K and Cv=0.745 kJ/kg K.

i) Determine the initial pressure

ii) Determine T and v at the end of the polyprotic expansion

iii) Sketch 3 processes on a P-V diagram

iv) Calculate the net work done at the conclusion of all three processes

v) Calculate the net heat transfer at the conclusion of all three processes.

2. Relevant equations

p1v1^λ=p2v2^γ
w= Rx(T1-T2)/(n-1)

3. The attempt at a solution

So for the first section I am struggling to keep up with all the processes that are going on. My logic was since the first process is expanding as the volume increases the pressure must decrease therefore our expected initial pressure must be higher than 100 kPa. However I am not entirely sure on how to calculate it. I cannot use the ideal gas law as its not ideal to begin with and the pv^(1.2)=const has limited use as I need the final v for the expansion to be able to use it..

Can someone help me through this question.. I am really struggling with it. Thanks in advance.

2. Nov 4, 2011

### Simon Bridge

Re: Polyprotic Expansion HELP!!!

What is a polyprotic expansion? What does it mean? (Sounds like what I'd call an adiabatic expansion.)

You are missing a couple of relations:
P(V/m)=RT for each corner on the graph.
Work = area under the PV diagram.
... W = (V2-V1)P ... (isobaric)
... W = 0 ... (isochoric)
... W = $K\frac{V_2^{-0.2}-V_1^{-0.2}}{-0.2}$ ... (adiabatic)
... ... K=PV1.2

state1:
V1/m=0.1846 m^3/kg
P1=?
T1=373K

... get the idea?

Last edited: Nov 4, 2011
3. Nov 4, 2011

### ttb90

Re: Polyprotic Expansion HELP!!!

Sorry you are right. The question states the term polytropic expansion which would mean adiabatic in this case. I think polytropic is a general case where n=γ is adiabatic , n=1 is isothermal and n=0 is isobaric.

When you say P(V/m)=RT for each corner on the graph, do you mean that because they are just points on the corners. So we can use that relationship and assume it is ideal. I initially used another form of the same relationship but with specific volume and did the following:

T=100+273.15=373.15 K
v(specific volume) = 0.1846, R (nitrogen)= 0.297

Pv=RT
p(initial) =RT/v = (0.297 x 10^3 x 373.15)/0.1846 = 600 kPa

When it asks for the T at the end of the expansion, I used this relationship below

(P1/P2)^((γ-1)/γ)=T2/T1

(600/100)^((1.2-1)/1.2)= T2/373.15

T2= 503.04 K

Would this be correct ?

Not sure how i would continue with the v calculation but this is what i have done so far.

4. Nov 4, 2011

### Simon Bridge

Re: Polyprotic Expansion HELP!!!

Well V/m is the specific volume - that's why I parenthesized it off - Nitrogen is not an ideal gas so it is a good idea to adjust R for this, correct. Point is you can get an equation of state. What level is this aimed at?
looks like what I'd do in your place all right. Only I would have recorded the units for R - just for the record. (J/kg.K = kPa.m^2/kg.K)

If you looked up that relation - try deriving it.
The way to check processes and results is to see if they make sense in terms of the problem. Is the result consistent with what you know about adiabatic processes?
(Sooner or later you will have to deal with situation nobody knows the answer to - nobody to turn to to see if you are right - so you have to learn how to tell if you've done it right all by yourself while the problems are still known.)

Your graph usually tells you what to do next.
Usual process, list what you know for each corner.
Compare with the state equation.

(hint: you are given P, you have just calculated T, and you know R for Nitrogen.)

Last edited: Nov 4, 2011