Finding work of adiabatic compressor using ideal gas

AI Thread Summary
The discussion focuses on calculating the work done on an adiabatic compressor compressing methane from pressure P1 to P2 and determining the discharge temperature T2 using the ideal gas model. Participants clarify that the process is assumed to be reversible, which is essential for applying certain thermodynamic equations. The work done is expressed as W = ∫(P1 to P2) VdP, and there is a debate about whether dW can be represented as -PdV instead. It is emphasized that the calculation should focus on shaft work rather than total work, and an alternative method involving the energy balance of a closed system is suggested to find T2. The conversation concludes with a method to derive T2 using the relationship PV^γ = constant.
worryingchem
Messages
41
Reaction score
1

Homework Statement


Methane at ## P_1 ## and ## T_1 ## is compressed to a pressure of ## P_2 ## adiabatically at steady-state. Calculate the work done on the compressor and the temperature ## T_2 ## of the discharge gas. Use ideal gas model.
Given:
## T_1, P_1, P_2, C_p, \gamma = 1.4 ##

## PV^\gamma = constant ##

## W = \int_{P_1}^{P_2} VdP ##

## \Delta H = \int_{T_1}^{T_2} C_p dT = W ##

Homework Equations


## dH = TdS + VdP ##

## dQ = TdS ##

Steady-state energy balance: ## \Delta H = W_{shaft} ##

The Attempt at a Solution


Hello,
my first question is that is this process reversible?
Because if I started from ## dH = TdS + VdP ##, in order to apply ## 0 = dQ = TdS ## to get ## dH = VdP ##, won't the process have to be reversible?

Then, using the energy balance, I make ## dH = VdP = dW ## and got the given equation ## W = \int_{P_1}^{P_2} VdP ##. My second question here is that can we use ## dW = -PdV ## here instead of ## dW = VdP ##? And why can't or can we? I'm confused as to when we can use ## dW = -PdV ##.

Thank you for your help.
 
Physics news on Phys.org
worryingchem said:

Homework Statement


Methane at ## P_1 ## and ## T_1 ## is compressed to a pressure of ## P_2 ## adiabatically at steady-state. Calculate the work done on the compressor and the temperature ## T_2 ## of the discharge gas. Use ideal gas model.
Given:
## T_1, P_1, P_2, C_p, \gamma = 1.4 ##

## PV^\gamma = constant ##

## W = \int_{P_1}^{P_2} VdP ##

## \Delta H = \int_{T_1}^{T_2} C_p dT = W ##

Homework Equations


## dH = TdS + VdP ##

## dQ = TdS ##

Steady-state energy balance: ## \Delta H = W_{shaft} ##

The Attempt at a Solution


Hello,
my first question is that is this process reversible?
You're supposed to assume the process is reversible.
Because if I started from ## dH = TdS + VdP ##, in order to apply ## 0 = dQ = TdS ## to get ## dH = VdP ##, won't the process have to be reversible?
Yes.

Then, using the energy balance, I make ## dH = VdP = dW ## and got the given equation ## W = \int_{P_1}^{P_2} VdP ##. My second question here is that can we use ## dW = -PdV ## here instead of ## dW = VdP ##? And why can't or can we? I'm confused as to when we can use ## dW = -PdV ##.

Thank you for your help.
You are supposed to be calculating the shaft work, not the total work which, if you remember, also includes PV work pushing the gas into the compressor and it pushing gas ahead of it out of the compressor. For this situation, the equation gives you the shaft work.

But there is also another way to do this problem that gives the same answer for the shaft work. This is where you treat each parcel of gas passing through the compressor as a closed system experiencing an adiabatic reversible compression. On this basis, you can calculate the exit temperature from the compressor (as they ask you to do). Then you calculate ΔH from ##mC_p ΔT = W_s##. When you solve it this way, you should get the same answer as when you integrate VdP.
 
If I treat the parcel of gas as a closed system, will the energy balance be like the following?
## dU = -PdV ##
## C_v \Delta T = -\int_{P_1}^{P_2} PdV ##
From here, will I plug in ## PV^{\gamma} = constant ## to find ## T_2 ## ?
 
worryingchem said:
If I treat the parcel of gas as a closed system, will the energy balance be like the following?
## dU = -PdV ##
## C_v \Delta T = -\int_{P_1}^{P_2} PdV ##
From here, will I plug in ## PV^{\gamma} = constant ## to find ## T_2 ## ?
You write ## PV^{\gamma} = constant ## and you substitute ##V=RT/P##
 
So I get ## P_2(RT_2/P_2)^\gamma = constant ## and solve for ## T_2 ##.
Thank you for clearing this up for me.
 
  • Like
Likes Chestermiller
Back
Top