an engine operates on the carnot cycle between the temperature limits of 250c and 50c. the pressure and the volume at the start of the isothermal expansion are 6 bar and 0.1 m^3 respectively. pressure after isothermal expansion is 3 bar. assume gamma=1.4,R=0.287kj/kg.k

calculate the efficiency of cycle
,output work and change in entropy of each process?

To calculate the efficiency of the Carnot cycle, we need to find the temperatures of the hot and cold reservoirs. We are given the temperature limits of the cycle, which are 250°C and 50°C.

Step 1: Convert the temperatures from Celsius to Kelvin by adding 273.15 to each value.
Hot reservoir temperature (Th) = 250 + 273.15 = 523.15 K
Cold reservoir temperature (Tc) = 50 + 273.15 = 323.15 K

Step 2: We can use the formula for Carnot efficiency to calculate the efficiency (η):
η = 1 - (Tc/Th)

Plug in the values:
η = 1 - (323.15/523.15)
η ≈ 0.383 or 38.3%

So, the efficiency of the Carnot cycle is approximately 38.3%.

Now let's calculate the output work and change in entropy for each process in the Carnot cycle.

Process 1: Isothermal Expansion
We are given the initial pressure (P1) as 6 bar and initial volume (V1) as 0.1 m^3. The pressure after expansion (P2) is 3 bar.

To calculate the output work (W), we can use the formula:
W = P1 * V1 * ln(P2/P1)

Plug in the values:
W = 6 bar * 0.1 m^3 * ln(3/6)
W ≈ -0.115 kJ (negative because work is done on the system)

To calculate the change in entropy (ΔS), we can use the formula:
ΔS = C * ln(P2/P1)

Where C is the heat capacity at constant pressure, which is given as R/gamma.

Plug in the values:
ΔS = (0.287 kJ/kg.K / 1.4) * ln(3/6)
ΔS ≈ -0.035 kJ/K

Process 2: Adiabatic Compression
Since the process is adiabatic, there is no heat transfer (Q = 0). Therefore, the change in entropy (ΔS) is zero.

Process 3: Isothermal Compression
We know the initial pressure (P3) is 3 bar and the final pressure (P4) is 6 bar.

To calculate the output work (W), we can use the same formula as before since the process is reversible:
W = P3 * V2 * ln(P4/P3)

But we need to find the final volume (V2). This can be calculated using the ideal gas law:
P3 * V2 = P4 * V3 (V3 is given as V1)

Solving for V2:
V2 = (P4 * V3) / P3

Plug in the values:
V2 = (6 bar * 0.1 m^3) / 3 bar
V2 ≈ 0.2 m^3

Now let's calculate the work and change in entropy:
W = 3 bar * 0.2 m^3 * ln(6/3)
W ≈ 0.058 kJ

ΔS = (0.287 kJ/kg.K / 1.4) * ln(6/3)
ΔS ≈ 0.035 kJ/K

Process 4: Adiabatic Expansion
Similar to Process 2, this is an adiabatic process with no heat transfer (Q = 0), so the change in entropy (ΔS) is zero.

So, the output work for process 1 is approximately -0.115 kJ (negative value indicates work done on the system), and the change in entropy is approximately -0.035 kJ/K.
The output work for process 3 is approximately 0.058 kJ, and the change in entropy is approximately 0.035 kJ/K.

Note: The change in entropy for the adiabatic processes (2 and 4) is zero because there is no heat transfer and thus no change in entropy.