Question about a thermal cycle

In summary, Leg 4 and Leg 1 can be expressed in terms of change in pressure and volume, and the ratio of work to heat can be simplified to 1/(5/2 + 5/2 p-1/ delta p + 3/2 * V_0/ delta V). The maximum value of this ratio is 2/5 when both p_1 and V_0 approach 0.
  • #1
Clara Chung
304
14
Homework Statement
Attached below
Relevant Equations
Attached below
1554051481000.png

Attempt:
a) Leg 4 and Leg 1 W=P_0(V_1-V_0)+P_1(V_0-V_1)
b) Leg 4: Q= C_v*ΔT=C_v*ΔP*V_0/nR = 3/2*(V_0*P_0-V_0*P_1)
Leg 1: Q=C_p*ΔT=C_p*P_0*ΔV/nR = 5/2*(P_0*V_1-P_0*V_0)
Q=3/2*(V_0*P_0-V_0*P_1)+ 5/2*(P_0*V_1-P_0*V_0)
c) W/Q = [P_0(V_1-V_0)+P_1(V_0-V_1)] / [3/2*(V_0*P_0-V_0*P_1)+ 5/2*(P_0*V_1-P_0*V_0)]
No idea how to get 2/5
 

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  • #2
Please try to factor your expressions and also to express as much as possible in terms of ##\Delta p## and ##\Delta V##.
 
  • #3
Chestermiller said:
Please try to factor your expressions and also to express as much as possible in terms of ##\Delta p## and ##\Delta V##.
Ok..
I got W/Q = ΔpΔV / ((3/2)*Δp*V0 + (5/2)*ΔV*P0). I don't know how to continue because I can't see any constraint on the variables...
 
  • #4
Clara Chung said:
Ok..
I got W/Q = ΔpΔV / ((3/2)*Δp*V0 + (5/2)*ΔV*P0). I don't know how to continue because I can't see any constraint on the variables...
Much better. Now, if I write ##p_0=p_1+\Delta p##, I obtain:
$$\frac{W}{Q}=\frac{\Delta p\Delta V}{[\frac{5}{2}\Delta p\Delta V+\frac{5}{2}p_1\Delta V+\frac{3}{2}V_0\Delta p]}$$
Does this give you any ideas?
 
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  • #5
Chestermiller said:
Much better. Now, if I write ##p_0=p_1+\Delta p##, I obtain:
$$\frac{W}{Q}=\frac{\Delta p\Delta V}{[\frac{5}{2}\Delta p\Delta V+\frac{5}{2}p_1\Delta V+\frac{3}{2}V_0\Delta p]}$$
Does this give you any ideas?
I try to divide the numerator and denominator by \Delta p\Delta V, so 1/ (5/2 + 5/2 p-1/ \Delta p + 3/2 * V_0/\Delta V) , because \Delta p and \Delta V are positive, the maximum value 2/5 occurs when p_1 and V_0 = 0?
 
  • #6
Clara Chung said:
I try to divide the numerator and denominator by \Delta p\Delta V, so 1/ (5/2 + 5/2 p-1/ \Delta p + 3/2 * V_0/\Delta V) , because \Delta p and \Delta V are positive, the maximum value 2/5 occurs when p_1 and V_0 = 0?
Yes. They can both approach zero. See your diagram.
 
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Related to Question about a thermal cycle

1. What is a thermal cycle?

A thermal cycle is a process in which a substance or system undergoes a series of changes in temperature, often resulting in expansion and contraction.

2. What is the purpose of a thermal cycle in scientific research?

Thermal cycles are often used in scientific research to study the effects of temperature on materials or to simulate real-world conditions, such as weather patterns or industrial processes.

3. What is an example of a thermal cycle?

An example of a thermal cycle is the process of heating and cooling a metal rod to study its expansion and contraction properties.

4. How does a thermal cycle affect the properties of a material?

A thermal cycle can cause changes in the physical and chemical properties of a material, such as its strength, conductivity, and structure.

5. Are there any potential dangers associated with thermal cycles?

Yes, thermal cycles can cause stress and potential damage to materials, especially if the temperature changes are extreme or rapid. It is important to carefully design and monitor thermal cycles in scientific research to ensure safety and accurate results.

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