Question

The following heat engines produce power of 95,000 kW. Determine in each case the rates at which heat is absorbed from the hot reservoir and discarded to the cold reservoir. a. A carnot engine operates between heat reservoir at 750K and 300K. b. A practical engine operates between the same heat reservoirs but with a thermal efficiency n = 0.35

Answer 1

a. For the Carnot engine operating between heat reservoirs at 750K and 300K, the rate at which heat is absorbed from the hot reservoir (Qh) is approximately 142,857 kW, and the rate at which heat is discarded to the cold reservoir (Qc) is approximately 47,857 kW.

b. For the practical engine with a thermal efficiency (η) of 0.35 operating between the same heat reservoirs, the rate at which heat is absorbed from the hot reservoir is approximately 271,429 kW, and the rate at which heat is discarded to the cold reservoir is approximately 176,429 kW.

Step-by-step explanation:

a. The Carnot engine's efficiency is given by the Carnot efficiency formula: η_c = 1 - (Tc/Th), where Th and Tc are the temperatures of the hot and cold reservoirs, respectively. Using the given temperatures, we can calculate η_c, and then use it to find Qh and Qc.

b. For the practical engine, the thermal efficiency (η) is given by the equation η = 1 - (Qc/Qh). Using the provided efficiency, we can find Qh and then determine Qc. Note that in a practical engine, the efficiency is less than the Carnot efficiency due to irreversibilities and losses.

Answer 2

For a Carnot engine operating between heat reservoirs at 300 K and 750 K, the rates at which heat is absorbed and discarded are 95,000,000 J/s and 0 J/s respectively. For a practical engine with an efficiency of 0.35 operating between the same heat reservoirs, the rates at which heat is absorbed and discarded are 95,000,000 J/s and 61,750,000 J/s respectively.

Step-by-step explanation:

In each case, you can use the formula for the efficiency of a heat engine:

Efficiency (η) = 1 - (Tc/Th)

Where:

η is the efficiency, Tc is the temperature of the cold reservoir, and Th is the temperature of the hot reservoir.

a. For the Carnot engine, the efficiency is given as 0.60. Solving for Th:

0.60 = 1 - (300/Th)

Th = 750 K

Since the power produced is 95,000 kW, the rate at which heat is absorbed and discarded can be determined using the formula:

Power = (Qh - Qc)/t

Where:

Power is the rate at which work is done in watts, Qh is the rate at which heat is absorbed from the hot reservoir, Qc is the rate at which heat is discarded to the cold reservoir, and t is the time in seconds.

Solving for Qh:

95,000,000 = (Qh - 0)/1

Qh = 95,000,000 J/s

Solving for Qc:

95,000,000 = (95,000,000 - Qc)/1

Qc = 0 J/s

b. For the practical engine with an efficiency of 0.35:

0.35 = 1 - (300/Th)

Th = 462.86 K

Using the same formula, we can determine the rates at which heat is absorbed and discarded:

Solving for Qh:

95,000,000 = (Qh - 0)/1

Qh = 95,000,000 J/s

Solving for Qc:

95,000,000 = (95,000,000 - Qc)/1

Qc = 61,750,000 J/s