Question

an engine working on carnot cycle absorves heat from 3 blocks at 1000 kelvin ,800 kelvin and 600 kelvin.The engine developes 600 kj per min of work and reject 400 kj per min of heat to the block at 300 kelvin.if the heat supplied to the block at 1000 kelvin is 60% of the heat supplied by the block at 600 kelvin.find the quantity of heat absorves by 3 blocks.

Answer 1

an engine working on Carnot cycle absorves heat from 3 blocks at 1000 kelvin, 800 kelvin and 600 kelvin. The engine develops $600\mathrm{kJ}$ per min of work and reject $400\mathrm{kJ}$ per min of heat to the block at 300 kelvin. If the heat supplied to the block at 1000 kelvin is $60\%$ of the heat supplied by the block at 600 kelvin. Find the quantity of heat absorves by 3 blocks.

Solution

We define several quantities for a cycle:

$Q_{A1}$ is the heat absorbed by block at $T_{1} = 1000$ kelvin,

$Q_{A2}$ is the heat absorbed by block at $T_{2} = 800$ kelvin,

$Q_{A3}$ is the heat absorbed by block at $T_{3} = 600$ kelvin,

$Q_{R}$ is the heat rejected by the system,

$W$ is the net work done by the system,

$T_{4} = 300$ kelvin.

The thermal efficiencies of the cycles:

$$\eta_ {1} = \frac {W _ {1}}{Q _ {A 1}}, \eta_ {2} = \frac {W _ {2}}{Q _ {A 2}}, \eta_ {3} = \frac {W _ {3}}{Q _ {A 3}},$$

where $W_{1}, W_{2}, W_{3}$ - work done by 3 blocks ($W_{1} + W_{2} + W_{3} = W$).

Then

$$W _ {1} = \eta_ {1} Q _ {A 1}, W _ {2} = \eta_ {2} Q _ {A 2}, W _ {3} = \eta_ {3} Q _ {A 3} \text{ and } (\eta_ {1} Q _ {A 1} + \eta_ {2} Q _ {A 2} + \eta_ {3} Q _ {A 3}) = W.$$

For Carnot cycle thermal efficiency is done by formula:

$$\eta = 1 - \frac {T _ {c}}{T _ {h}},$$

where $T_{c}$ - is the temperature of the cold reservoir and $T_{h}$ - is the absolute temperature of the hot reservoir.

The thermal efficiencies of the cycles:

$$\eta_ {1} = 1 - \frac {T _ {4}}{T _ {1}}, \eta_ {2} = 1 - \frac {T _ {4}}{T _ {2}}, \eta_ {3} = 1 - \frac {T _ {4}}{T _ {3}}.$$

And now

$$\left(\left(1 - \frac {T _ {4}}{T _ {1}}\right) Q _ {A 1} + \left(1 - \frac {T _ {4}}{T _ {2}}\right) Q _ {A 2} + \left(1 - \frac {T _ {4}}{T _ {3}}\right) Q _ {A 3}\right) = W.$$

But we know that

$$W = Q _ {A} - Q _ {R} = Q _ {A 1} + Q _ {A 2} + Q _ {A 3} - Q _ {R},$$

and $Q_{A1} = 0.6 * Q_{A3}$.

So $W = 0.6 * Q_{A3} + Q_{A2} + Q_{A3} - Q_R = Q_{A2} + 1.6 * Q_{A3} - Q_R$

or $Q_{A2} = W + Q_R - 1.6 * Q_{A3}$.

Substituting $Q_{A2}$ and $Q_{A1}$ in equation for work we get

$$\left(\left(1 - \frac{T_4}{T_1}\right) * 0.6 * Q_{A3} + \left(1 - \frac{T_4}{T_2}\right) (W + Q_R - 1.6 * Q_{A3}) + \left(1 - \frac{T_4}{T_3}\right) Q_{A3}\right) = W.$$ $$Q_{A3} = \frac{W - \left(1 - \frac{T_4}{T_2}\right) (W + Q_R)}{\left(1 - \frac{T_4}{T_1}\right) * 0.6 + \left(1 - \frac{T_4}{T_3}\right) - 1.6 * \left(1 - \frac{T_4}{T_2}\right)}.$$ $$Q_{A3} = \frac{600 - \left(1 - \frac{300}{800}\right) * (600 + 400)}{\left(1 - \frac{300}{1000}\right) * 0.6 + \left(1 - \frac{300}{600}\right) - 1.6 * \left(1 - \frac{300}{800}\right)} = 312.5 \frac{\text{kg}}{\text{min}}.$$ $$Q_{A1} = 0.6 * 312.5 = 187.5 \frac{\text{kg}}{\text{min}}.$$ $$Q_{A2} = 600 + 400 - 1.6 * 312.5 = 500 \frac{\text{kg}}{\text{min}}.$$

Answer: 187.5 $\frac{\text{kg}}{\text{min}}$, 500 $\frac{\text{kg}}{\text{min}}$ and 312.5 $\frac{\text{kg}}{\text{min}}$ respectively.