A room at temperature T2 loses heat to the outside at temperature T1 at a rate A(T2-T1). It is w

A room at temperature T2 loses heat to the outside at...

Key Concepts

Energy Balance in Thermal Systems

Energy balance involves applying the principles of the first law of thermodynamics to equate the energy entering a system with the energy leaving it, accounting for work done and heat transfer. In scenarios like a heated room interacting with its external environment, energy balance equations are used to derive relationships between the work input, heat transfer rates, and resulting equilibrium temperatures. This concept is central to solving problems involving steady-state and dynamic thermal behaviors.

Thermal Equilibrium

Thermal equilibrium refers to the state in which a system's temperature remains constant over time because the energy entering the system is balanced by the energy leaving it. In the context of heating a room, achieving thermal equilibrium means that the heat supplied by the heat pump equals the heat lost to the colder exterior. Understanding thermal equilibrium is key to analyzing steady-state conditions in thermal systems.

Coefficient of Performance (COP)

The coefficient of performance of a heat pump is a measure of its efficiency, defined as the ratio of useful heat delivered to the work input required. For a Carnot cycle, the COP is determined solely by the temperature reservoirs, providing the maximum possible gain in transferring heat. This concept is crucial in the design and evaluation of heating systems, as it quantifies the effectiveness of a heat pump working under ideal conditions.

Heat Transfer Rate

Heat transfer rate refers to the amount of heat energy that is exchanged per unit time between a system and its surroundings. In thermal systems, especially those involving heating or cooling, it is vital to quantify both the loss of heat and the supply of heat to maintain desired temperature conditions. This concept is fundamental in calculating equilibrium states where the heat input matches the heat loss.

Carnot Cycle

The Carnot cycle is a theoretical thermodynamic cycle that operates between two temperatures and is reversible. It provides the maximum possible efficiency for a heat engine or the maximum coefficient of performance for a heat pump, establishing an ideal benchmark for energy conversion processes. Understanding the Carnot cycle is essential for analyzing the theoretical limits of heat transfer and work conversion in thermodynamic systems.

Transcript

00:01 So here in this problem we are given that the carnot heat engine it received heat from a reservoir that is at 900 degrees celsius and the heat received the rate of heat received we are given as 800 kilojoule per minute and ambient condition we are given as the the lower temperature that is 27 degrees celsius if you convert this thing into kelvin we'll get 300 kelvin and we are given that the entire work output from the heat engine it is used by the refrigerator and the refrigerator space we have to maintain it at minus 5 degrees celsius and we have to calculate in the a part of the problem we have to calculate the maximum rate of heat removal from the refrigerated space that means we have to calculate this ql dot for refrigeration this thing if we see in the figure this thing we have to calculate so first of all to calculate this will be calculating the efficiency of the heat engine.

01:16 And we know that the sense the heat engine is a carnot cycle is based on the carnot cycle.

01:23 So we can say that the efficiency will be equals to 1 minus t lower divided by t higher.

01:30 And simply we'll plug the values 1 minus here t lower is 27 degrees celsius.

01:36 If you convert this thing into kelvin, we'll get 300 kelvin and here we'll get 100.

01:42 1 ,173 kelvin so 1 minus steel over is 300 kelvin and divide by 1 ,173 kelvin so we're going to solve this and we'll get the efficiency of the heat engine as 0 .744 now the maximum work output of this heat engine is determined from the definition of the thermal efficiency so we know that the work output the rate of work output so this is output is not input this is output and the same the work output is used by the refrigerator so it is equals to efficiency the thermal efficiency of the heat engine times qh we have all the values we can plug it here so this is 0 .744 times quh dot we have 800 kilojoule per minute so it comes out to be as nearly 595 .2 kilojoule per minute so this is the work output the rate of work output and the same work output is used by the refrigerator so we know that first of all it will be calculating the cop of the refrigerator so it is equals to 1 divided by th divided by tl minus 1 so you can plug all the values here 1 divided by th in this case is this 300 kelvin divide by minus 5 degree celsius that means it will be 268 kelvin minus 1 and it is equals to want to solve this and we get the cop as 8 .37 now we can calculate this ql and so again cop of refrigerator can also be written in this way also ql dot of this time is for a refrigerator divide by the rate of work input so this is the rate of work input and this thing we have to calculate cop we have already calculated it is 8 .37 is equals to ql dot for refrigerator we have to calculate and work input rate of work input is 595 .2 kilojoule per minute so from here the work this ql dot comes out to be as let me write here it comes out to be nearly 4 ,982 kilojoules j.

04:40 So this was the solution for the a part.

04:45 Next we'll be solving the b part of the problem and in the b part we have to calculate the total heat of the heat and the total rate of heat rejection to the ambient.

04:53 That means we have to calculate this heat...

Please give Ace some feedback