arnot’s engine with air as working substance is initially at 3270C and pressure of 12 atmospheres. The volume is one litre to start with. The expansion or compression ratio is 1:6. Find the lowest temperature and efficiency. (γ= 1.4)
To find the lowest temperature and efficiency of Arnot’s engine, we can make use of the First Law of Thermodynamics and the equations for adiabatic processes. Here's how you can proceed:
Step 1: Determine the final volume of the working substance:
The compression ratio of 1:6 means that the final volume will be 1/6th of the initial volume. Therefore, the final volume (V2) can be calculated as:
V2 = (1/6) * 1 L = 1/6 L
Step 2: Calculate the final temperature (T2) using the adiabatic process equation:
For an adiabatic process, the relation between temperature, pressure, and volume is given by:
T1 * (V1/V2)^(γ-1) = T2
Given:
Initial temperature (T1) = 3270 °C = 3270 + 273 = 3543 Kelvin
γ = 1.4 (for air)
Substituting the given values in the equation, we can solve for T2:
3543 * (1/6)^(1.4-1) = T2
Step 3: Calculate the lowest temperature (T3) using the adiabatic process equation:
For an isothermal expansion or compression, the temperature remains constant. Therefore, the lowest temperature (T3) will be equal to T2.
Step 4: Calculate the efficiency (η) of the engine:
The efficiency of the engine is given by the equation:
η = 1 - (T3 / T1)
Substitute the values of T1 and T3 to find the efficiency.
Now, let's calculate the values:
Step 2:
T2 = 3543 * (1/6)^(1.4-1)
Step 3:
T3 = T2
Step 4:
η = 1 - (T3 / T1)
Solve the above equations to find the values of T2, T3, and η.