A Carnot engine operating on air accepts 50 kJ / kg of heat and rejects 20 kJ / kg. Calculate

step 1 Hello and welcome to this video solution of numeric. Here is given that a Carnot engine operates

A Carnot engine operating on air accepts 50 kJ / kg of heat...

A Carnot engine operating on air accepts $50 \mathrm{~kJ} / \mathrm{kg}$ of heat and rejects $20 \mathrm{~kJ} / \mathrm{kg}$. Calculate the high and low reservoir temperatures if the maximum specific volume is $10 \mathrm{~m}^{3} / \mathrm{kg}$ and the pressure after the isothermal expansion is $200 \mathrm{kPa}$. The thermal efficiency is $$ \eta=1-\frac{q_{L}}{q_{H}}=1-\frac{20}{50}=0.6 $$ Hence, $T_{L} / T_{H}=0.4$. For the adiabatic processes we know that (see Fig. 5-8) $$ \frac{T_{L}}{T_{H}}=\left(\frac{v_{2}}{v_{3}}\right)^{k-1} \quad \therefore \frac{v_{2}}{v_{3}}=0.4^{2.5}=0.1012 $$ The maximum specific volume is $v_{3}$; thus, $v_{2}=0.1012 v_{3}=(0.1012)(10)=1.012 \mathrm{~m}^{3} / \mathrm{kg}$. Now, the high temperature is $$ T_{H}=\frac{P_{2} v_{2}}{R}=\frac{(200)(1.012)}{0.287}=705.2 \mathrm{~K} \quad \text { or } 432.2^{\circ} \mathrm{C} . $$ The low temperature is then $T_{L}=0.4 T_{H}=(0.4)(705.2)=282.1 \mathrm{~K}$ or $9.1^{\circ} \mathrm{C}$.

A Carnot engine operating on air accepts $50 \mathrm{~kJ} / \mathrm{kg}$ of heat and rejects $20 \mathrm{~kJ} / \mathrm{kg}$. Calculate the high and low reservoir temperatures if the maximum specific volume is $10 \mathrm{~m}^{3} / \mathrm{kg}$ and the pressure after the isothermal expansion is $200 \mathrm{kPa}$.
The thermal efficiency is
$$
\eta=1-\frac{q_{L}}{q_{H}}=1-\frac{20}{50}=0.6
$$
Hence, $T_{L} / T_{H}=0.4$. For the adiabatic processes we know that (see Fig. 5-8)
$$
\frac{T_{L}}{T_{H}}=\left(\frac{v_{2}}{v_{3}}\right)^{k-1} \quad \therefore \frac{v_{2}}{v_{3}}=0.4^{2.5}=0.1012
$$
The maximum specific volume is $v_{3}$; thus, $v_{2}=0.1012 v_{3}=(0.1012)(10)=1.012 \mathrm{~m}^{3} / \mathrm{kg}$. Now, the high temperature is
$$
T_{H}=\frac{P_{2} v_{2}}{R}=\frac{(200)(1.012)}{0.287}=705.2 \mathrm{~K} \quad \text { or } 432.2^{\circ} \mathrm{C} .
$$
The low temperature is then $T_{L}=0.4 T_{H}=(0.4)(705.2)=282.1 \mathrm{~K}$ or $9.1^{\circ} \mathrm{C}$.

Key Concepts

Ideal Gas Law

The ideal gas law, expressed as PV = RT (or in specific terms as P*v = R*T), relates pressure, specific volume, and temperature for an ideal gas. This foundational concept in thermodynamics is used to calculate state properties during various processes in the cycle, such as determining the high reservoir temperature after isothermal expansion. Understanding the ideal gas law is crucial for analyzing real and idealized cycles involving gaseous working fluids.

Adiabatic Process

An adiabatic process involves a change in the state of a system in which no heat is transferred to or from the system. For an ideal gas undergoing an adiabatic process, the relationship between temperature and volume is given by T*v^(?-1) = constant, where ? is the specific heat ratio. This concept is essential in understanding portions of the Carnot cycle, as it allows the determination of state variables during the compression and expansion phases without heat transfer.

Isothermal Process

An isothermal process is one in which the temperature remains constant during the change of state. For an ideal gas, this process implies that the product of pressure and specific volume remains constant (as per the ideal gas law). In the context of the Carnot cycle, the isothermal expansion and compression phases are critical for absorbing and rejecting heat at fixed temperatures, thereby determining the engine's overall efficiency.

Carnot Cycle

The Carnot cycle represents an idealized thermodynamic cycle that defines the maximum efficiency any heat engine can achieve when operating between two heat reservoirs. It consists of two isothermal processes (heat addition and rejection at constant temperature) and two adiabatic processes (where no heat transfer occurs). This theoretical model provides a benchmark for the performance of real engines, emphasizing the importance of operating reversibly to maximize efficiency.

Thermal Efficiency

Thermal efficiency in the Carnot engine is defined as the ratio of net work output to the heat input, and it is expressed by the formula ? = 1 - (T_L/T_H), where T_H and T_L are the absolute temperatures of the high- and low-temperature reservoirs, respectively. This concept illustrates how the efficiency of any heat engine is fundamentally limited by the temperature difference between its heat reservoirs, with greater differences yielding higher potential efficiencies.

Transcript

00:01 Hello and welcome to this video solution of numeric.

00:04 Here is given that a carnot engine operates on air accepts 50 kilojoule per kg of heat and rejects 20 kilojoule per kg.

00:15 Right.

00:16 So here we have got the heat that is rejected to the cold reservoir or the lower temperature is of a ql i'm taking to be 20 kilojoule per kg.

00:26 Right.

00:27 And the heat coming from the hot reservoir is 50 kilojoule per kg.

00:34 Right.

00:34 So these are the two parameters.

00:37 Now you have to calculate the high and low temperature of the temperatures if the maximum specific volume is 10 meter cube per kg.

00:47 Right.

00:48 So v dot is 10 meter cube per kg.

00:55 Right.

00:56 And the pressure after the isothermal expansion is 200 kilo pascal.

01:05 Now here what we have is the efficiency we can calculate to be at the carnot efficiency 1 minus the heat from the lower reservoir over the heat from the high reservoir.

01:17 Right.

01:18 So 1 minus 20 over 50.

01:21 This gives you 0 .6 rate is the efficiency.

01:24 Right.

01:26 Now as per this diagram we are given that this is tl over th that is 0 .4.

01:34 Right.

01:35 And based on adiabatic process tl over th that is equal to v2 over v3 to the power of gamma minus 1.

01:47 Right.

01:53 And from here you will be having v2 over v3 if you just plug in then this th over tl is 0 .4 and here we will be having gamma minus 1 as 2 .5.

02:04 This is equal to 0 .1012...