# Thermodynamics: Steam Powered Carnot Engine Question

• spinverted
In summary: Under 144 psi steam has a density of 3.11 ft^3/lb which is 285 mol/m^3. Using this information and the values for ɣ and P from the table, I was able to solve for V. My final answer is that the engine would require 0.9 m^3 of steam at 448K to reach its maximum efficiency.
spinverted

## Homework Statement

A steam engine has a cylinder 80cm in diameter with a piston that has a total travel length of 1.8m. Assume that there is no "dead space" between the cylinder head and the piston when the piston is farthest in the cylinder. Steam is piped into the cylinder head from an external boiler at 130psi above atmospheric pressure. The water in the boiler is boiling at the 130psi above atmospheric pressure. There is a "quick release" valve system that can shut off the flow of high pressure steam (at 130psi above atmosphere) at any point in the piston's 1.8m displacement down the cylinder. The displacement until the valve cuts off the flow of steam from the boiler is done at 130psi above atmosphere but after the cut off the steam already in the cylinder expands until the total 1.8m displacement is finished and another valve opens to allow the steam to be exhausted into the atmosphere. Assume the engine is running at 0.60 piston cycles per second.

What should be the piston position at steam cutoff (measured from the cylinder head) for the engine to operate at maximum efficiency?

NOTE: if you cool the steam such that it condenses into liquid water the pressure will decrease drastically, possibly below atmospheric!

## Homework Equations

1 (dP / dT) = L / (T * dV)
2 ln(P2/P1)= -(ΔH/R) * ((1/T2) - (1/T1))
3 PV=nRT
4 V=pi*r^2h
5 PV^ɣ=PV^ɣ
6 P^(ɣ-1) * T^(-ɣ) = P^(ɣ-1) * T^(-ɣ)
7 VT^α=VT^α

## The Attempt at a Solution

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My understanding of this problem is that I must find the volume of initial steam to add so that the adiabat is as large as possible without condensing the steam. Point B to Point C on the following image is that adiabatic transition that I am talking about.

http://img694.imageshack.us/img694/2987/carnotcycle.png
Using a steam table from http://www.energysolutionscenter.org/BoilerBurner/Eff_Improve/Primer/SteamTables.pdf" I got the following values. The ɣ is from the adiabatic index on Wikipedia.

Vgas=3.5*10^-3 m^3/mol
Vliq=2.0*10^-5 m^3/mol
Volume of full cylinder = 0.90735 m^3
P = 992736Pa
ɣ=1.13
T=453K
P = 992736Pa
ΔHvap=36260J/mol

The following is my attempt to solve for the initial amount of steam added:
http://img190.imageshack.us/img190/9198/mapleg.png
Equation 4 is the pressure and temperature of steam at which a phase change to liquid water will begin. Equation 5 is a rearranged form of equation 3 giving the final pressure after expansion. Equation 6 Is Equation 2 with pressure from equation 5 and the number of moles based on the amount of volume added. Equation 7 is equation 4 with equation 5 and equation 6 plugged into it. Volume is the only variable and I have attempted to solve for it, but Maple 11 will not solve for me. Here is a picture of what happens when I attempt to solve in Maple:
http://img19.imageshack.us/img19/3853/losti.png

Anyone know how to solve for V or whether or not I'm doing this right?

Last edited by a moderator:
I noticed I clearly made a mistake on my last post with my construction of equation 5. The denominator should bethe total volume of the cylinder fully expanded (0.90477m^3). The corrected formula for pressure is now found in equation 1 of this post. Using equation 2 I have the highest ratio of dP/dT possible without a phase change 23001.19256 Pa/K. The value I got from Maple when solving for the initial volume of steam added was 0.9073500003 m^3.
http://img44.imageshack.us/img44/5019/thermosolved.png
That's bigger than the volume of the cylinder! It seemed crazy so I plotted dP/dT for a volume change from 0 to 0.9 m^3 and sure enough it seems that it is impossible to have a conversion in this cylinder. This seems to mean that I would get the highest efficiency from the smallest addition of steam possible. I thought I might have screwed up the Clapeyron equation so I plotted T vs V and P vs V for initial volumes of steam ranging from 0 to 0.9 m^3.

From the looks of the PV plot an initial volume of steam of 0.1 m^3 would result in a final pressure of 100,000 Pa (roughly atmospheric). Looking at the temperature for this point I find around 50 K. Steam has definitely condensed by this point.

http://img521.imageshack.us/img521/9647/pvtv.png
What am I doing wrong?

Last edited by a moderator:

## 1. How does a steam powered Carnot engine work?

A steam powered Carnot engine works by using the expansion and compression of steam to convert thermal energy into mechanical energy. The engine consists of two chambers, one hot and one cold, and a piston that moves between the two chambers. The steam is heated in the hot chamber, causing it to expand and push the piston to the cold chamber. As the steam cools and condenses, the piston moves back to the hot chamber, completing the cycle.

## 2. What is the Carnot cycle?

The Carnot cycle is a theoretical thermodynamic cycle that describes the most efficient way to convert heat into work. It consists of four processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. The Carnot cycle is used as a standard for comparing the efficiency of real-world engines.

## 3. What is the maximum efficiency of a steam powered Carnot engine?

The maximum efficiency of a steam powered Carnot engine is determined by the temperature difference between the hot and cold chambers. It is given by the equation efficiency = (T1 - T2)/T1, where T1 is the temperature of the hot chamber and T2 is the temperature of the cold chamber. The Carnot cycle is the only cycle that can achieve this maximum efficiency.

## 4. How does a steam powered Carnot engine differ from other types of engines?

A steam powered Carnot engine differs from other types of engines in that it operates on a theoretical cycle and has the highest possible efficiency. Other types of engines, such as internal combustion engines, operate on real-world cycles and have lower efficiencies. Additionally, steam powered Carnot engines use steam as the working fluid, while other engines may use gases, liquids, or even solids.

## 5. What are the applications of a steam powered Carnot engine?

Although the Carnot cycle is used as a theoretical standard, steam powered Carnot engines are not commonly used in practical applications. However, the principles of the Carnot cycle are used in the design of more efficient real-world engines, such as steam turbines and gas turbines. The Carnot cycle is also used in the analysis of thermodynamic systems and processes.

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