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Stirling Engine Efficiency

  1. Feb 21, 2015 #1
    1. The problem statement, all variables and given/known data

    Figure 2 represents a model for the thermodynamic
    cycle of the Stirling engine, patented by Scottish clergyman Robert Stirling in
    1816. The engine operates by burning fuel externally to warm one of its two
    cylinders. A xed quantity of inert gas moves cyclically between the cylinders,
    expanding in the hot one and contracting in the cold one.

    (a) Consider n mol of an ideal monatomic gas being taken
    once through the cycle in Fig. 2, consisting of two isothermal processes
    at temperatures 3Ti and Ti and two isochoric processes. In terms of n,
    R, and Ti, determine Q for the complete cycle.

    (b) What is the eciency of the engine? (Hint: The heat QH
    transferred into the system happens during steps 1 and 4).

    Figure 2:
    f5PXbGE.png

    2. Relevant equations

    Work = nRT ln(v2/v1)
    U = n Cv T
    efficiency = W/Q

    3. The attempt at a solution
    I am pretty confident in my solution to part a ) of the problem which is this:
    z2N7EPE.jpg
    OP9XCo7.jpg


    However when it comes to calculating efficiency, I am getting an answer of 1 . Is this possible?

    b53uY45.jpg

    Thank you.
     
  2. jcsd
  3. Feb 22, 2015 #2

    Suraj M

    User Avatar
    Gold Member

    Whenever you're free try typing these equations use this
     
  4. Feb 22, 2015 #3
    Okay,
    So for a)

    $$ Step \, 1-2\,$$
    $$ Isothermal \, \therefore \delta U=0 \, , W=Q$$
    $$W=nR3T_iLn(2V/V) \implies Q= nR\cdot3T_iLn(2)$$

    $$Step \, 2-3 $$
    $$Isochoric, \, W=0 $$
    $$\delta U=n \frac{3}{2}R (T_i-3T_i) \implies Q=-2T_in\frac{3}{2}R $$

    $$Step \, 3-4 $$
    $$ Isothermal \, \therefore \delta U=0 \, , W=Q$$
    $$W=nR3T_iLn(1V/2V) \implies Q= nR\cdot3T_iLn(1/2) = -nR\cdot3T_iln(2)$$

    $$Step \,4-1$$
    $$Isochoric, \, W=0 $$
    $$\delta U=n \frac{3}{2}R (3T_i- T_i) \implies Q= 2T_in\frac{3}{2}R $$

    $$Q_{total} = 2T_in\frac{3}{2}R + -nR\cdot3T_iln(2) + -2T_in\frac{3}{2}R + nR\cdot3T_iLn(2) = 2nRT_iLn(2) $$

    $$W_{total} = nR3T_iLn(2) + 0 + nR3T_iLn(1/2) + 0 = 2nRT_iLn(2)$$

    Part b)
    $$ efficiency = \frac{W}{Q_H} = \frac{2nRT_iLn(2)}{2nRT_iLn(2)} = 1 $$
     
  5. Feb 22, 2015 #4
    I would also like to add that this does make sense since for a cyclic process, $$\delta U = 0 = Q-W $$ and since Q=w, this holds. So,the question now is did I use the correct efficiency formula?
     
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