Efficiency Calculation for a Stirling Engine Cycle

[CivilSigma]

Homework Statement

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:

Homework Equations

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

The Attempt at a Solution

I am pretty confident in my solution to part a ) of the problem which is this:

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

Thank you.

 

[Suraj M]

Whenever you're free try typing these equations use this

 

[CivilSigma]

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)
$$ efficiency = \frac{W}{Q_H} = \frac{2nRT_iLn(2)}{2nRT_iLn(2)} = 1 $$

 

[CivilSigma]

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?