A heat engine operates between two thermal reservoirs at 800 and 20"C. One-half of the work output of the heat engine is used to drive a Camot heat pump that removes heat from the cold surroundings at 2°C and transfers it to a house maintained at 22 C. If the house
is losing heat at a rate of 95,000 kJ/h.
a) Define the following
i. Heat
Thermal reservoir
[2]
Heat engine iv. Heat pump
(2)
b) Determine
12)
[2]
i. The COP for the Carnot heat pump ii. The rate of work output for the heat engine.
[3]
The minimum rate at which heat must required to keep the house at 22°C.
[51 be
a) Definitions:
i. Heat: Heat is a form of energy that is transferred between two substances or systems as a result of a temperature difference.
Thermal reservoir: A thermal reservoir is a large body or system that can provide an infinite amount of heat or absorb an infinite amount of heat without changing its own temperature significantly.
Heat engine: A heat engine is a device that transforms heat energy into mechanical work in a cyclic process.
Heat pump: A heat pump is a device that uses mechanical work to transfer heat from a low-temperature reservoir to a high-temperature reservoir.
b) Calculations:
i. The coefficient of performance (COP) for a Carnot heat pump can be calculated using the formula COP = Th/(Th - Tc), where Th is the temperature of the hot reservoir and Tc is the temperature of the cold reservoir. In this case, Th = 22°C and Tc = 2°C. Plugging in the values into the formula, we get COP = 22/(22 - 2) = 22/20 = 1.1.
ii. The rate of work output for the heat engine can be calculated using the formula W = Qh - Qc, where Qh is the heat absorbed from the hot reservoir and Qc is the heat rejected to the cold reservoir. Since we know that one-half of the work output is used to drive the heat pump, the work output of the heat engine is 2 times the work input of the heat pump. Therefore, W = 2 * (Qh - Qc).
iii. The minimum rate of heat required to keep the house at 22°C can be calculated using the formula Qh = W + Qc, where Qh is the heat absorbed from the hot reservoir, W is the work input to the heat pump, and Qc is the heat rejected to the cold reservoir. In this case, Qh = 2 * Qc.
Please provide the inputs for Qh, Qc, or any other relevant information to further calculate the answers.
a) Definitions:
i. Heat: Heat is a form of energy that is transferred between two objects or systems due to a temperature difference. It flows from a higher temperature to a lower temperature until thermal equilibrium is reached.
Thermal reservoir: A thermal reservoir is an idealized system that can supply or absorb an unlimited amount of heat without undergoing any change in temperature. It is assumed to have a constant temperature, allowing it to serve as a source or sink of heat in heat engines and heat pumps.
Heat engine: A heat engine is a device that converts heat energy into mechanical work. It operates by taking heat from a high-temperature reservoir, converting part of it into work, and rejecting the remaining heat to a low-temperature reservoir.
Heat pump: A heat pump is a device that uses mechanical work to transfer heat from a low-temperature reservoir to a high-temperature reservoir. It operates in the reverse mode of a heat engine. Unlike a heat engine, which aims to convert heat into work, a heat pump aims to transfer heat against the natural direction of heat flow.
b) Calculations:
i. The coefficient of performance (COP) for a Carnot heat pump is defined as the ratio of the heat transfer to the high-temperature reservoir to the work input. The COP is given by:
COP = (Th - Tc) / Th,
where Th is the temperature of the high-temperature reservoir and Tc is the temperature of the low-temperature reservoir.
In this case, Th = 22°C + 273 = 295K and Tc = 2°C + 273 = 275K.
COP = (295 - 275) / 295 = 0.068
ii. The rate of work output for the heat engine can be determined using the first law of thermodynamics, which states that the net work output is equal to the difference between the heat absorbed and the heat rejected.
The net work output is given by:
W_net = Qh - Qc,
where Qh is the heat absorbed from the high-temperature reservoir and Qc is the heat rejected to the low-temperature reservoir.
Given that half of the work output of the heat engine is used to drive the Carnot heat pump, the net work output is equal to half of the total work output.
W_net = (1/2) * W_total
Also, from the definition of the Carnot heat engine:
Qh = Th * (1 - Tc/Th) * W_total,
Qc = Tc/Th * Qh.
Substituting the values:
Qh = 295 * (1 - 275/295) * W_total = 20 * W_total,
Qc = (275/295) * 20 * W_total = (22/295) * W_total.
So, the net work output is:
W_net = 20 * W_total - (22/295) * W_total = (591 - 22) / 295 * W_total = 569/295 * W_total.
iii. The minimum rate at which heat must be supplied to keep the house at 22°C can be determined using the equation:
Qh = Qc + Q_house,
where Qh is the heat absorbed from the high-temperature reservoir, Qc is the heat rejected to the low-temperature reservoir, and Q_house is the heat lost by the house.
Given that Q_house = 95,000 kJ/h and Th = 295K, we can substitute the known values:
Qh = (295/275) * Qc + Q_house.
Substituting the value of Qc from part ii:
Qh = (295/275) * (22/295) * W_total + 95,000 kJ/h.
Now we can solve for the minimum rate at which heat must be supplied (Qh) to keep the house at 22°C.