91Ó°ÊÓ

The first law of thermodynamics is a fundamental concept in thermodynamics. It states that energy cannot be created or destroyed, only transferred or converted from one form to another. This principle is crucial for understanding how energy flows in any thermodynamic process. In mathematical terms, this law can be expressed as: \[ \text{ΔU} = \text{Q} - \text{W} \] where:

• \( \text{ΔU} \) is the change in internal energy of the system
• \( \text{Q} \) is the heat added to the system
• \( \text{W} \) is the work done by the system

In the context of our problem, we used this law to calculate the heat transfer in process 2-3. Since the change in internal energy was known from other processes, we could find the heat transfer by rearranging the formula: \[ \text{Q}_{2–3} = \text{ΔU} + \text{W}_{2–3} \] This allowed us to determine that the heat added to the system during process 2-3 was 1120 kJ.

A constant pressure process, also known as isobaric process, is a thermodynamic process in which the pressure remains constant. This type of process frequently occurs in practical applications, such as heating or cooling a gas in a cylinder at atmospheric pressure. For the problem at hand, process 2-3 is a constant pressure process:

• The initial volume \( V_2 \) = 0.2 m³,
• The final volume \( V_3 \) = 1.6 m³,
• The initial and final pressures are both 8 bar, since the pressure remains constant.

To calculate the work done during a constant pressure process, you use the formula: \[ \text{W}_{2–3} = p \times ( V_3 - V_2 ) \] By substituting the known values: \[ \text{W}_{2–3} = 800\text{kPa} \times 1.4\text{m³} = 1120\text{kJ} \] This tells us that the work done by the gas as it expands from 0.2 m³ to 1.6 m³ under constant pressure is 1120 kJ.

Understanding how to calculate work and heat transfer is essential for analyzing any thermodynamic cycle. In our specific problem, we encountered multiple work and heat transfer scenarios across different processes:

• Process 1-2: Despite the compression, internal energy change \( ΔU = 0 \) meant no net work and heat transfer.
• Process 2-3: Involved calculating work done under constant pressure which we found to be 1120 kJ (as shown using the isobaric work formula).
• Process 3-1: From the internal energy relationship \( U_1 - U_3 = -3549\text{kJ} \), reflected heat loss and no work due to constant volume.

By combining these calculations with the first law of thermodynamics, you get a complete picture of the energy transferred and work done in the cycle. This cycle, as computed, confirms it's a power cycle, as it results in a net positive work done by the system.