Finding area from work, pressure and volume

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SUMMARY

The discussion focuses on solving the equation for area (A) derived from the work-energy principle, specifically using the integral W=PVf-A(Vf^3)/3-PVi+A(Vi^3)/3. The user initially calculated A as 3.179, while the correct answer is 5.05. The confusion arises from the misinterpretation of pressure variables P0 and P1, which are crucial for setting up the equations correctly. Participants clarify that two equations involving P0, P1, P2, V1, V2, and A are necessary to solve for A accurately.

PREREQUISITES
  • Understanding of the work-energy principle in thermodynamics
  • Familiarity with integral calculus and algebraic manipulation
  • Knowledge of pressure-volume relationships in fluid mechanics
  • Ability to set up and solve systems of equations
NEXT STEPS
  • Review the derivation of the work-energy equation W=∫PdV
  • Study the relationship between pressure and volume in ideal gases
  • Learn how to solve systems of equations involving multiple variables
  • Practice solving problems involving integrals in thermodynamic contexts
USEFUL FOR

Students and professionals in engineering, particularly those specializing in thermodynamics and fluid mechanics, will benefit from this discussion as it addresses common pitfalls in solving pressure-volume work problems.

JoeyBob
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Homework Statement
see attached
Relevant Equations
W=integral(PdV)
So I basically took the integral and ended up with W=PVf-A(Vf^3)/3-PVi+A(Vf^3)/3

so 65.7=72*5.3-A(5.3)^3/3-72(2.4)+A(2.4)^3/3

But when I solve for A I get the wrong answer of 3.179 when the answer is suppose to be 5.05. I've checked my calculation with an algebra calculator too...
 

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JoeyBob said:
Homework Statement:: see attached
Relevant Equations:: W=integral(PdV)

So I basically took the integral and ended up with W=PVf-A(Vf^3)/3-PVi+A(Vf^3)/3

so 65.7=72*5.3-A(5.3)^3/3-72(2.4)+A(2.4)^3/3

But when I solve for A I get the wrong answer of 3.179 when the answer is suppose to be 5.05. I've checked my calculation with an algebra calculator too...
You seem to have mixed up P0 and P1.
 
haruspex said:
You seem to have mixed up P0 and P1.
Wait wouldn't tha mean there's two unknowns then? How would I find P0?
 
JoeyBob said:
Wait wouldn't tha mean there's two unknowns then? How would I find P0?
You have two equations relating P0, P1, P2, V1, V2 and A.
 
haruspex said:
You have two equations relating P0, P1, P2, V1, V2 and A.

Wait so is it P1=P0+AV and then the integral eqn?
 
JoeyBob said:
Wait so is it P1=P0+AV and then the integral eqn?
No.
You are given P=P0-AV2 as a general fact. In particular this will be true at the initial and final states. Plug in the values/variables for those states to get two equations.
 
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You have $$P_1=P_0-AV_1^2$$ so $$P_0=P_1+AV_1^2$$So you can eliminate ##P_0##
 
Chestermiller said:
Well the problem statement says: $$P=P_0-AV^2$$so at the initial condition, $$P_1=P_0-AV_1^2$$So my algebra tells me that $$P_0=P_1+AV_1^2$$
What am I missing?
Sorry, I read what I was expecting, not what you wrote.
I read your second line as $$P_2=P_0+AV_2^2$$, reinforced by your comment about eliminating P0. So what I should have written is $$P_2=P_0-AV_2^2$$.
 
haruspex said:
Sorry, I read what I was expecting, not what you wrote.
I read your second line as $$P_2=P_0+AV_2^2$$, reinforced by your comment about eliminating P0. So what I should have written is $$P_2=P_0-AV_2^2$$.
What do you need to know P2 for?
 
  • #10
Chestermiller said:
What do you need to know P2 for?
I'm not saying you do. I'm explaining what I thought you had written and what I thought you had intended.
 

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