Carnot Cycle Analysis: Thermal Efficiency, Heat Added, and Heat Rejected

What is the thermal efficiency, heat added, and heat rejected in a Carnot cycle operating between 1000°F and 60°F while developing 10 hp?

Determine the values based on the given information.

Thermal Efficiency, Heat Added, and Heat Rejected Calculation:

Given data: Carnot cycle develops 10 hp, operates between 1000°F and 60°F.

The Carnot cycle in this scenario has a thermal efficiency of 64.4% and a heat input and heat rejection of 660 Btu/min. The Carnot cycle is a theoretical thermodynamic cycle that represents the most efficient heat engine possible. The thermal efficiency of the Carnot cycle is calculated using the formula:

Efficiency = 1 - (T_c / T_h)

where T_c is the absolute temperature at the cold reservoir (60°F converted to absolute temperature) and T_h is the absolute temperature at the hot reservoir (1000°F converted to absolute temperature). By plugging in the values and calculating, the thermal efficiency is approximately 64.4%.

To determine the heat added and heat rejected, we first convert the power output of 10 hp to Btu per minute (Btu/min) using the conversion factor of 2544 Btu/hp. The power output is calculated as 25440 Btu/min.

Since the Carnot cycle is reversible, the heat added and heat rejected are equal in magnitude but opposite in sign. Therefore, both the heat added and heat rejected are half of the power output, which is equal to 12720 Btu/min each.

Therefore, the thermal efficiency of the Carnot cycle is approximately 64.4%, with both heat added and heat rejected equal to 660 Btu/min.