Efficiency of a heat engine

  • #1
402
14

Homework Statement


I am having an issue trying to decipher this question, as I am not sure if it a lack of knowledge on my half or there is an assumption I have to make.

pic2018.png

Homework Equations


##\epsilon=\frac{W_{total}}{Q_{in}}##

The Attempt at a Solution


My issue is calculating the heat along the path ##Q_{A \rightarrow B}## my confusion stem from that second law there is no working being done so then heat is equal to ##Q_{A \rightarrow B}=C_v\Delta T##.

Now in the question I have no information on if this a monotonic gas or diatomic ideal gas, if I have to make the assumption it is fine I can calculate the efficiency just fine, but I was just wondering if someone could take a look as I may have overlooked something.
 

Attachments

Answers and Replies

  • #2
20,523
4,396
Just do it algebraically, and see what the efficiency looks like.

Also, does 5x higher mean 5x as high, or does it mean 6x as high?
 
Last edited:
  • #3
402
14
I have attempted to solve algebraically and seem to have gotten a result which, dose not simplify easily.

Here is my current working

Step 1 : What I know.

$$\Delta E_{int}=Q_in+Wd_{on} \ [1]$$

$$\Delta E_{int}=C_v n\Delta T \ [2]$$

$$\Delta Wd_{on}=-\Delta Wd_{by}=-P\Delta V \ [3]$$

In the PV diagram displayed I have three processes
  1. Isotherm path ##B \rightarrow C##
  2. Isobaric path ##C \rightarrow A##
  3. Isochoric paths ##A \rightarrow B##
Step 2 : Solving Isotherm path
$$\Delta E_{int}=0$$

$$0=Q_{in}+Wd_{on}=Q_{in}-Wd_{by}$$

$$Q_{in}=Wd_{by}=Pdv$$

$$P=\frac{nR5T_1}{V}$$

$$Qin=\int^{V_2}_{V_1} \frac{nR5T_1}{V} dv=nR5T_1 ln(\frac{V_2}{V_1})$$

finding ##\frac{V_2}{V_1}##

$$P_1V_1=nRT_1$$

$$PV_2=nR5T_1$$

$$\frac{V_2}{V_1}=5$$

therefore

$$Q_{in}=nR5ln(5)T_1=Q_H$$

Step 3 : Solving Isobaric path

$$\Delta E_{int}=Q_{in}-Wd_{by}$$

$$C_vn\Delta T=Q_{in}-P(\Delta T)$$

$$Q_{in}=C_vn\Delta T + nR\Delta T $$

$$(C_v+R)n\Delta T=Q_{in}$$

$$C_p n\Delta T = Q_{in}$$

$$\Delta T=-4T_1$$

$$Q_{in}=-4C_pnT_1$$

Step 4 : Solving Isobaric path

$$\Delta E_{int}=Q_{in}$$

$$C_v n \Delta T=Q_{in}$$

$$4C_v n T_1=Q_{in}$$

Step 5 : Solving for Efficiency

$$\epsilon = \frac{W}{Q_in}$$

As system is a complet cycle then ##\Delta E_{int}=0## therefore ##W_T=\sum Q_in##

$$\epsilon = \frac{(5ln5)R+4C_v-4C_p}{(5ln5)R+4C_v}$$

$$\frac{(5ln5)R-4R}{(5ln5)R+4C_v}=\frac{(5ln5-4)R}{(5ln5)R+4C_v}$$

And this is were I am stuck, I have I missed a step in my workings? Surley I should be able to simplyfy this in to an actual percentage
 
  • #4
20,523
4,396
You almost have it.

$$\frac{(5ln5)R-4R}{(5ln5)R+4C_v}=\frac{(\ln{5}-0.8)R}{(\ln{5}R+0.8C_v}=\frac{\left(1-\frac{0.8}{\ln{5}}\right)}{\left(1+\frac{0.8}{\ln{5}(\gamma-1)}\right)}$$
 
  • Like
Likes Taylor_1989

Related Threads on Efficiency of a heat engine

  • Last Post
2
Replies
37
Views
5K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
4
Views
231
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
523
  • Last Post
Replies
8
Views
10K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
9K
Top