- Fri Aug 09, 2019 5:36 pm
#26002
I found a new thermodynamic process in which the energy you get out of the process can be more than the energy you put into the process. The process consist of the following four steps:
Step 1: Deflate a big spherical balloon, and squeeze the deflated balloon to a very small volume.
Step 2: Submerge the deflated balloon in a liquid and take it down to a depth where the pressure is at least 34atm. Because the volume of the deflated balloon is very small, the buoyant force that you have to push down against is negligible.
Step 3: When the balloon gets to the intended depth, hold the balloon stationary while you inflate it with air. Use an air compressor located outside the liquid to inflate the balloon. Use an air hose to connect the air compressor to the balloon.
Step 4: After the balloon gets inflated, release the balloon and let it ascend upward slowly. As the balloon ascends upward, it pulls upward on a rope that is attached to it. As the balloon pulls upward on the rope, the rope unwinds and spins a turbine that creates electricity.
After the balloon ascends up to the surface of the liquid, to start another cycle, all you have to do is deflate the balloon and perform the four steps again.
The energy output is maximum if the following three requirements are met:
1. The temperature of the hot air that came out of the air compressor and went into the balloon must remain constant. The temperature of the liquid must be equal to the temperature of the hot air that came out of the air compressor.
2. The balloon must ascend upward slowly enough so that the hydrodynamic drag force on the balloon is insignificant. Another reason why the balloon must ascend upward slowly is so that the balloon expands isothermally.
3. The balloon and the liquid must be in an insulated house so that heat energy is not wasted.
If those three requirements are met, the total energy output is approximately equal to the work done by the buoyant force as the balloon ascends to the surface: (P1V1)ln(P1÷P0). The total energy input is approximately equal to the energy needed to inflate the balloon adiabatically: 3.5P1V1 – 2.5P0V0. We can see that if ln(P1÷P0) is greater than or equal to 3.5, total energy output is greater than total energy input; and free energy is generated. P0 represents the atmospheric pressure where the air compressor is located; P1 represents the liquid pressure at the initial depth of the balloon; V0 represents the total volume of air that went into the air compressor; V1 represents the volume of the air after it has been adiabatically compressed from P0 to P1. If P0 = 1atm, for ln(P1÷P0) to be greater than or equal to 3.5, P1 must be greater than or equal to 34atm. In the next paragraphs, I will do the detailed calculations that led to my conclusions. Please read on.
Total Energy Input:
I will try to calculate the amount of energy you used to carry out each of the four steps (if the energy is negligible, I am going to ignore it). In step 1, you deflated the balloon and squeezed it to a very small volume. The energy you used to deflate and squeeze the balloon is negligible when compared to the energy output of the turbine. In step 2, you spent energy to submerge the deflated balloon in the liquid and push it down to the initial depth of the balloon. But because the volume of the deflated balloon was relatively very small, the buoyant force that you had to push down against was relatively very small. So the energy you spent in step 2 is negligible when compared to the energy output of the turbine. In step 3, the air compressor spent energy to inflate the balloon. This is a significant amount of energy. In step 4, you let the balloon ascend upward and spin the turbine. You did not spend any energy in carrying out step 4. So the only energy input that I will use to estimate the total energy input is the energy the air compressor used to inflate the balloon.
Total Energy Input ≈ Energy used to adiabatically compress the air from P0 to P1 + Energy used for the isobaric expansion of the balloon from a very small volume (approximately zero volume) to V1.
Total Energy Input ≈ [1/(γ−1)][P1V1 – P0V0] + P1[V1 – 0]
Total Energy Input ≈ [γ/(γ−1)] P1V1 – [1/(γ−1)] P0V0
For air, γ = 1.4:
Total Energy Input ≈ 3.5P1V1 – 2.5P0V0
Total Energy Output:
In step 4, you released the balloon and let it ascend upward. The upward force on the balloon as the balloon ascends upward is equal to the buoyant force minus the weight of the balloon minus the hydrodynamic drag force. For my calculations, I would estimate that the balloon is weightless (because the weight of the material from which the balloon is made plus the weight of the air inside the balloon is relatively very small). Also, I would estimate that the hydrodynamic drag force on the balloon is zero because I know that the balloon will ascend upward slowly. So the only force that I am going to use for my calculation of total energy output is the buoyant force.
Let the initial depth of the balloon be D meters deep. Let the volume of the balloon, at depth
(D – x) meters, be Vx. Let the pressure of the liquid at depth (D – x) meters be Px. Let the Buoyant force on the balloon, at depth (D – x) meters, be Bx. Let ρ represent the density of the liquid.
D
Total Energy Output ≈ ∫ Bx dx
0
Bx = Vx × ρ × g (g is equal to 9.8m/s^2)
Because the temperature of the hot air inside the balloon remains constant,
(P1)(V1) = (Px) (Vx). So therefore, Vx = P1×V1 ÷ Px
Bx = (P1×V1 ÷ Px)× ρ × g
Px = P0 + ρ(g)(D – x)
Bx = (P1×V1 ÷ [P0 + ρ(g)(D – x)]) × ρ × g
Bx = P1×V1× ρ × g ÷ [P0 + ρ(g)(D – x)]
D
Total Energy Output ≈ ∫ P1×V1× ρ × g ÷ [P0 + ρ(g)(D – x)]dx
0
D
Total Energy Output ≈ P1×V1× ∫ ρ(g) ÷ [P0 + ρ(g)(D) – ρ(g)(x)]dx
0
D
Total Energy Output ≈ P1×V1× – ln[P0 + ρ(g)(D) – ρ(g)(x)]|
0
Total Energy Output ≈ P1×V1×– [ln(P0) – ln(P0 + ρ(g)D)]
Total Energy Output ≈ P1×V1×– [ln(P0) – ln(P1)]
Total Energy Output ≈ P1×V1×ln(P1÷P0)
Now that we have the Total Energy Output and we also have the Total Energy Input, I can calculate the free energy generated:
Free Energy ≈ Total Energy Output – Total Energy input
Free Energy ≈ P1V1 ×ln(P1÷P0) – (3.5P1V1 – 2.5P0V0)
Free Energy ≈ P1V1 ×[ln(P1÷P0) – 3.5] + 2.5P0V0
If ln(P1÷P0) is greater than or equal to 3.5, Free Energy is positive, and free energy is actually generated.
Example:
We carried out the four steps. We submerged a spherical balloon in liquid aluminum and took it down to a depth of 486.21m. After the balloon got to depth 486.21m, we held the balloon stationary while the air compressor inflated it with (4/3)π(100)^3 m^3 of air. After the balloon got inflated, we released it and let it ascend upward with a constant speed of 0.25m/s. The atmospheric pressure where the air compressor is located is 1atm; the temperature where the air compressor is located is 300K; the density of liquid aluminum is 2311kg/m^3. The spherical balloon is made out of copper material. Calculate the free energy generated.
Let T0 be the temperature of the air before it goes into the compressor, and let T1 be the temperature of the air after it comes out of the compressor.
So therefore we have:
P0 =1atm =101300pa
P1 = P0 + ρ(g)D = 101300pa + (2311kg/m^3)(9.8m/s^2)(486.21m) = 11112886.84pa = 109.7atm
V0 = (4/3)π(100)^3 m^3 = 4188790.205m^3
V1 = V0×[(P0/ P1)^(1/γ)] (for air, γ = 1.4).
V1 =(4188790.205m^3)×(101300pa/11112886.84pa)^(1/1.4)
= (4188790.205m^3)(1/109.7)^(1/1.4) =146149.7784m^3
T0 = 300K
T1 = T0× (V0/V1)^(γ-1) = 300K × (4188790.205m^3/ 146149.7784m^3)^(0.4) = 1148.253K.
So the temperature of the hot air that came out of the air compressor is 1148.253K. The temperature of the liquid aluminum is also 1148.253K. Aluminum is actually liquid at 1148.253K. Copper is solid at 1148.253K. So the copper balloon will be solid.
Free Energy ≈ P1V1 ×[ln(P1÷P0) – 3.5] + 2.5P0V0
Free Energy ≈ (11112886.84pa)(146149.7784m^3) ×[ln(11112886.84pa ÷ 101300pa)–3.5] + 2.5(101300pa)(4188790.205m^3)
Free Energy ≈ (3.005×10^12)J
So free energy generated is approximately (3.005×10^12)J.
Reward: No reward
Step 1: Deflate a big spherical balloon, and squeeze the deflated balloon to a very small volume.
Step 2: Submerge the deflated balloon in a liquid and take it down to a depth where the pressure is at least 34atm. Because the volume of the deflated balloon is very small, the buoyant force that you have to push down against is negligible.
Step 3: When the balloon gets to the intended depth, hold the balloon stationary while you inflate it with air. Use an air compressor located outside the liquid to inflate the balloon. Use an air hose to connect the air compressor to the balloon.
Step 4: After the balloon gets inflated, release the balloon and let it ascend upward slowly. As the balloon ascends upward, it pulls upward on a rope that is attached to it. As the balloon pulls upward on the rope, the rope unwinds and spins a turbine that creates electricity.
After the balloon ascends up to the surface of the liquid, to start another cycle, all you have to do is deflate the balloon and perform the four steps again.
The energy output is maximum if the following three requirements are met:
1. The temperature of the hot air that came out of the air compressor and went into the balloon must remain constant. The temperature of the liquid must be equal to the temperature of the hot air that came out of the air compressor.
2. The balloon must ascend upward slowly enough so that the hydrodynamic drag force on the balloon is insignificant. Another reason why the balloon must ascend upward slowly is so that the balloon expands isothermally.
3. The balloon and the liquid must be in an insulated house so that heat energy is not wasted.
If those three requirements are met, the total energy output is approximately equal to the work done by the buoyant force as the balloon ascends to the surface: (P1V1)ln(P1÷P0). The total energy input is approximately equal to the energy needed to inflate the balloon adiabatically: 3.5P1V1 – 2.5P0V0. We can see that if ln(P1÷P0) is greater than or equal to 3.5, total energy output is greater than total energy input; and free energy is generated. P0 represents the atmospheric pressure where the air compressor is located; P1 represents the liquid pressure at the initial depth of the balloon; V0 represents the total volume of air that went into the air compressor; V1 represents the volume of the air after it has been adiabatically compressed from P0 to P1. If P0 = 1atm, for ln(P1÷P0) to be greater than or equal to 3.5, P1 must be greater than or equal to 34atm. In the next paragraphs, I will do the detailed calculations that led to my conclusions. Please read on.
Total Energy Input:
I will try to calculate the amount of energy you used to carry out each of the four steps (if the energy is negligible, I am going to ignore it). In step 1, you deflated the balloon and squeezed it to a very small volume. The energy you used to deflate and squeeze the balloon is negligible when compared to the energy output of the turbine. In step 2, you spent energy to submerge the deflated balloon in the liquid and push it down to the initial depth of the balloon. But because the volume of the deflated balloon was relatively very small, the buoyant force that you had to push down against was relatively very small. So the energy you spent in step 2 is negligible when compared to the energy output of the turbine. In step 3, the air compressor spent energy to inflate the balloon. This is a significant amount of energy. In step 4, you let the balloon ascend upward and spin the turbine. You did not spend any energy in carrying out step 4. So the only energy input that I will use to estimate the total energy input is the energy the air compressor used to inflate the balloon.
Total Energy Input ≈ Energy used to adiabatically compress the air from P0 to P1 + Energy used for the isobaric expansion of the balloon from a very small volume (approximately zero volume) to V1.
Total Energy Input ≈ [1/(γ−1)][P1V1 – P0V0] + P1[V1 – 0]
Total Energy Input ≈ [γ/(γ−1)] P1V1 – [1/(γ−1)] P0V0
For air, γ = 1.4:
Total Energy Input ≈ 3.5P1V1 – 2.5P0V0
Total Energy Output:
In step 4, you released the balloon and let it ascend upward. The upward force on the balloon as the balloon ascends upward is equal to the buoyant force minus the weight of the balloon minus the hydrodynamic drag force. For my calculations, I would estimate that the balloon is weightless (because the weight of the material from which the balloon is made plus the weight of the air inside the balloon is relatively very small). Also, I would estimate that the hydrodynamic drag force on the balloon is zero because I know that the balloon will ascend upward slowly. So the only force that I am going to use for my calculation of total energy output is the buoyant force.
Let the initial depth of the balloon be D meters deep. Let the volume of the balloon, at depth
(D – x) meters, be Vx. Let the pressure of the liquid at depth (D – x) meters be Px. Let the Buoyant force on the balloon, at depth (D – x) meters, be Bx. Let ρ represent the density of the liquid.
D
Total Energy Output ≈ ∫ Bx dx
0
Bx = Vx × ρ × g (g is equal to 9.8m/s^2)
Because the temperature of the hot air inside the balloon remains constant,
(P1)(V1) = (Px) (Vx). So therefore, Vx = P1×V1 ÷ Px
Bx = (P1×V1 ÷ Px)× ρ × g
Px = P0 + ρ(g)(D – x)
Bx = (P1×V1 ÷ [P0 + ρ(g)(D – x)]) × ρ × g
Bx = P1×V1× ρ × g ÷ [P0 + ρ(g)(D – x)]
D
Total Energy Output ≈ ∫ P1×V1× ρ × g ÷ [P0 + ρ(g)(D – x)]dx
0
D
Total Energy Output ≈ P1×V1× ∫ ρ(g) ÷ [P0 + ρ(g)(D) – ρ(g)(x)]dx
0
D
Total Energy Output ≈ P1×V1× – ln[P0 + ρ(g)(D) – ρ(g)(x)]|
0
Total Energy Output ≈ P1×V1×– [ln(P0) – ln(P0 + ρ(g)D)]
Total Energy Output ≈ P1×V1×– [ln(P0) – ln(P1)]
Total Energy Output ≈ P1×V1×ln(P1÷P0)
Now that we have the Total Energy Output and we also have the Total Energy Input, I can calculate the free energy generated:
Free Energy ≈ Total Energy Output – Total Energy input
Free Energy ≈ P1V1 ×ln(P1÷P0) – (3.5P1V1 – 2.5P0V0)
Free Energy ≈ P1V1 ×[ln(P1÷P0) – 3.5] + 2.5P0V0
If ln(P1÷P0) is greater than or equal to 3.5, Free Energy is positive, and free energy is actually generated.
Example:
We carried out the four steps. We submerged a spherical balloon in liquid aluminum and took it down to a depth of 486.21m. After the balloon got to depth 486.21m, we held the balloon stationary while the air compressor inflated it with (4/3)π(100)^3 m^3 of air. After the balloon got inflated, we released it and let it ascend upward with a constant speed of 0.25m/s. The atmospheric pressure where the air compressor is located is 1atm; the temperature where the air compressor is located is 300K; the density of liquid aluminum is 2311kg/m^3. The spherical balloon is made out of copper material. Calculate the free energy generated.
Let T0 be the temperature of the air before it goes into the compressor, and let T1 be the temperature of the air after it comes out of the compressor.
So therefore we have:
P0 =1atm =101300pa
P1 = P0 + ρ(g)D = 101300pa + (2311kg/m^3)(9.8m/s^2)(486.21m) = 11112886.84pa = 109.7atm
V0 = (4/3)π(100)^3 m^3 = 4188790.205m^3
V1 = V0×[(P0/ P1)^(1/γ)] (for air, γ = 1.4).
V1 =(4188790.205m^3)×(101300pa/11112886.84pa)^(1/1.4)
= (4188790.205m^3)(1/109.7)^(1/1.4) =146149.7784m^3
T0 = 300K
T1 = T0× (V0/V1)^(γ-1) = 300K × (4188790.205m^3/ 146149.7784m^3)^(0.4) = 1148.253K.
So the temperature of the hot air that came out of the air compressor is 1148.253K. The temperature of the liquid aluminum is also 1148.253K. Aluminum is actually liquid at 1148.253K. Copper is solid at 1148.253K. So the copper balloon will be solid.
Free Energy ≈ P1V1 ×[ln(P1÷P0) – 3.5] + 2.5P0V0
Free Energy ≈ (11112886.84pa)(146149.7784m^3) ×[ln(11112886.84pa ÷ 101300pa)–3.5] + 2.5(101300pa)(4188790.205m^3)
Free Energy ≈ (3.005×10^12)J
So free energy generated is approximately (3.005×10^12)J.
Reward: No reward
- By AaronBurns
- By Uolevi Kattun