Sadi Carnot and the Power of Fire, the 1824 Edition Part 2

DOFPro Team

Introduction

These two videos present two examples of how you might actually implement a Carnot cycle in the physical world.

• Part 1 presents the air-standard Carnot cycle.
• Part 2 presents the steam Carnot cycle.

We will use these two examples as the basis for all of the power and refrigeration cycles we will study.

Carnot Steam Cycle

\(a\)

\(b\)

\(c\)

\(d\)

\(100\ ^\circ\mathrm{C}\)

\(200\ ^\circ\mathrm{C}\)

\(a\)

\(b\)

\(c\)

\(d\)

\(100\ ^\circ\mathrm{C}\)

\(200\ ^\circ\mathrm{C}\)

Boiler

1^st Law

\(\Delta \dot{H} = \dot{Q} = \dot{m} \Delta \hat{H} = \dot{m}(2792.1 - 852.33) = 1939.77 \dot{m}\)

Turbine

1^st Law

\(\Delta \dot{H} = \dot{W}_s = \dot{m} \Delta \hat{H} =\ ?\)

2^nd Law

\(\Delta \dot{S} = 0 = \dot{m} \Delta \hat{S}\)

\(a\)

\(b\)

\(c\)

\(d\)

\(100\ ^\circ\mathrm{C}\)

\(200\ ^\circ\mathrm{C}\)

so

\(\hat{S}_d = \hat{S}_c\)

\(\hat{S}_c = \hat{S}(\mathrm{vapor,\ 200\ ^\circ C}) = 6.4303\)

\(\hat{S}(\mathrm{liquid,\ 100\ ^\circ C}) = 1.3070\)

\(\hat{S}(\mathrm{vapor,\ 100\ ^\circ C}) = 7.3541\)

What is \(\hat{S}\) of a mixture of steam and water?

\(a\)

\(b\)

\(c\)

\(d\)

\(100\ ^\circ\mathrm{C}\)

\(200\ ^\circ\mathrm{C}\)

Reminder: The mole- or mass-fraction of steam is the quality.

\(\mathrm{quality} = x = \dfrac{m_\mathrm{vapor}}{m_\mathrm{total}}\)

Then the specific entropy of a mixture is

\(\hat{S}_\mathrm{mixture} = x \hat{S}_v + (1-x) \hat{S}_l\)

In our case

\(6.4303 = x (7.3541) + (1-x) (1.3070)\)

\(x = \dfrac{6.4303 - 1.3070}{7.3541 - 1.3070} = 0.8472\)

\(a\)

\(b\)

\(c\)

\(d\)

\(100\ ^\circ\mathrm{C}\)

\(200\ ^\circ\mathrm{C}\)

To calculate the enthalpy at point \(d\)

\(\hat{H} = x \hat{H}_v + (1-x) \hat{H}_l\)

\(\hat{H}_d = 2330.9\)

By a similar process

\(x(\mathrm{at\ point\ }a) = 0.1693\)

\(\hat{H}_a = 801.0\)

\(\eta = \dfrac{|Q_\mathrm{hot}| - |Q_\mathrm{cold}|}{|Q_\mathrm{hot}|}\)

\(= \dfrac{(2792.1 - 852.33) - (2330.9-801.0)}{(2792.1 - 852.33)} = 0.2113\)

\(a\)

\(b\)

\(c\)

\(d\)

\(100\ ^\circ\mathrm{C}\)

\(200\ ^\circ\mathrm{C}\)

From the formula for Carnot efficiency

\(\eta = 1 - \dfrac{T_\mathrm{cold}}{T_\mathrm{hot}} = 1 - \dfrac{373.15}{473.15} = 0.2113\)

\(-W_\mathrm{net} = 410.0\ \mathrm{kJ/kg}\)

\(Q_\mathrm{hot} = 1939.8\ \mathrm{kJ/kg}\)

\(Q_\mathrm{cold} = -1529.8\ \mathrm{kJ/kg}\)

\(T_\mathrm{hot}\)

\(T_\mathrm{cold}\)

\(|Q_\mathrm{hot}|\)

\(|Q_\mathrm{cold}|\)

\(|W|\)

What happens if we run a Carnot engine backwards?

\(\dfrac{|W|}{|Q_\mathrm{hot}|} = 1 - \dfrac{T_\mathrm{cold}}{T_\mathrm{hot}}\)

\(\dfrac{|W|}{|W| + |Q_\mathrm{cold}|} = \dfrac{T_\mathrm{hot} - T_\mathrm{cold}}{T_\mathrm{hot}}\)

\(\dfrac{|Q_\mathrm{cold}|}{|W|} = \dfrac{T_\mathrm{cold}}{T_\mathrm{hot} - T_\mathrm{cold}}\)

In general

\(\dfrac{|Q_\mathrm{cold}|}{|W|} = \omega = \text{C.O.P.}\)

What do we call a device that removes heat at a low temperature and rejects it at a high temperature?

The videos, Boil, Expand, Condense, Repeat: The Rankine Cycle in Action Part 1 and Part 2 explains the Rankine cycle, which is based on the steam Carnot cycle.

The video, Revenge of the Fridge: Vapor Compression Strikes Back examines refrigeration cycles.

The Takeaways

1. The Carnot cycle consists of an isentropic temperature rise, an isothermal heat addition, an isothermal temperature drop, and an isothermal heat removal.
2. The steam Carnot cycle implements the four steps with an isentropic compressor, an isothermal boiler, an isentropic turbine, and an isothermal condenser.
3. A Carnot engine running in reverse is a refrigerator.

Thanks for watching!
The previous video in the series is in the link in the upper left. The next video in the series is in the upper right. To learn more about Chemical and Thermal Processes, visit the website linked in the description.

The DOFPro Team