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Author Topic: Net potential energy gain - please point out mistake  (Read 3308 times)

Offline Positron360

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Net potential energy gain - please point out mistake
« on: September 04, 2010, 09:33:15 PM »
Consider two gas-filled pistons connected to each other by a tube, such that the pressure in one piston must always be equal to the pressure in the other.

One piston is cylindrical, while the other is a circular cone, such that its piston head area is a function of displacement.

Let the formula describing the shape of the conical piston be given by:

x^2 + y^2 = cz^2, where c is a scalar (1)

At any specific height above the XY-plane, the piston head area will be given by

A = pi*h^2*c (2)

This can be derived easily by noticing that the radius is equal to (z^2*c)^(1/2) from (1).

The volume of the cone at a specific vertical displacement h is given by:

V = (pi/3)*h^3*c (3)

This was determined by the use of a triple integral. I can give the details if required, but I will leave it for now.

For simplicity, I will consider the case where c = 1.

I will refer to the conical piston with subscript 1, and to the cylindrical piston with subscript 2.

Consider the following process to occur adiabatically: (Isothermal analysis also work, give slightly smaller result, but adiabatic is more realistic since the process is considered to happen instantaneously.)

(Subscript i refers to an initial value (before the masses where added) and subscript f refers to a final value.)

Define the following constants for use:

g: 9.81 [N / kg]
R: 8.314 [J / (mol*K)]
cp: (7/2)R [J / (mol*K)]
cv: (5/2)R [J / (mol*K)]

n: 100 [mol]
Ti: 298 [K]
Patm: 1*10^5 [Pa]
A2: 0.1 [m^2]

m1: 1000 [kg]
m2: 1000 [kg]

Assume ideal gas behavior:

PiVi = n*R*Ti

Therefore Vi = (n*R*Ti)/Pi [m3] (4)

But also, Vi = pi*h1i^2 + A2*h2i (5)

Pi = Patm

Values for h1i and h2i can be solved by equating (4) and (5). There are an infinite number of solutions. Ideally, the starting position will be chosen such that the net potential energy gain is the greatest. However, as an aid in demonstrating the scenario of gain in potential energy more clearly, I will solve such that h1i = h2i.

Using the previously defined data and constraint, solving for h1i = h2i gives

h1i = h2i = 1.3086... [m]

In the next step, masses m1 and m2 are placed on the piston heads. Again, the masses would normally be chosen such that they provide for the most efficient process cycle, but for this example, let's choose m1 = m2 = 1000 kg.

Since the pressure in the pistons must remain equal;

Pf = Patm + (m1*g)/(A1f) = Patm + (m2*g)/A2 (6)

must hold.

This simplifies to A1f = (A2*m1)/m2 (7)

Calculate Pf by using the second part of equation (6) gives:

Pf = 198100 [Pa]

Knowing this value allows for the calculation of the final temperature by the following equation: (Derivation can be provided if necessary)

Tf = (cp/R)^-1*Ti*((Pf/Pi) + (cv/R)) (8)


Tf = 381.5... [K]

If the process was considered to be isothermal, Tf  = Ti = 298 [K]. As stated earlier, adiabatic conditions are not required to achieve a net increase in potential energy, so you can just substitute Tf = 298 [K] if you prefer. It will still work for this specific example.

Again, by the ideal gas law:

Vf = (n*R*Tf)/Pf (9)

and also,

Vf = (pi/3)*h1f^3 + A2*h2f (10)

Solving (7), (9) and (10) simultaneously gives a unique answer. According to my calculations:

h1f = 0.1784... [m]

h2f = 15.953... [m]

Calculating the net change in potential energy:

delta Ep = m*g*delta h

For m1:

delta Ep1 = m1*g*(0.1784... - 1.3086...) = -11 087.2... [J]

For m2:

delta Ep2 = m2*g*(15.9526... - 1.3086...) = 143 657.9... [J]

Adding, delta Ep(net) = 132 570.7... [J]

This means that the mass placed on the circular conic piston has lost less Ep than the one placed on the cylindrical piston gained.

I could not get a positive answer when using two cylindrical pistons, or a liquid working fluid.

Therefore I think that the concept of having the piston head area as a function of the displacement of the piston is the reason for this effect.

It is important to note that for each scenario there is a specific critical area for the cylindrical piston head. Anything greater than this results in a negative result, and anything smaller results in a positive result.

If someone can point out the mistake in my calculations or assumptions, I will be very thankful.

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Offline Positron360

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Re: Net potential energy gain - please point out mistake
« Reply #1 on: September 05, 2010, 10:16:51 AM »
I have found the mistake in my reasoning. I assumed that the masses will move so as to balance the system, while in reality the system starts out of balance and will only become more out of balance. A simple entropy calculation would probably have helped me see this error.