How much work?
Using relative measurements makes understanding and computing results much simpler. Pick a volume, density, heat and temperature. Any at all. Those are all 1. Volume is 1, density is 1, heat is 1, temperature is 1. For a vapor (air, steam, any gas), heat is proportional to temperature. So in this model, that constant is also 1, so we can treat temperature and heat interchangeable.
It is important to realize, this is only true because of the ideal gas law (gas as in vapor). One cannot continue to use these formulas through state changes. The heat energy change from water to steam is quite large, temperature change quite small (under 1 degree), compared to raising steam 1 Degree. Degree on which scale? All of them.
Also because of the ideal gas law, pressure = density * heat, we can substitute temperature, so pressure = density times temperature. Which scale? The scale where 100% of the starting heat is 1, and 100% of starting temperature is also 1.
So Work(density curve) or W(dc) = Volume change times Pressure.
For the Density curve, Pressure = Density = Volume -1, when temperature = 1. To calculate the area of Pressure times Volume change, it is Integral (V=A,V=B) Volume-1 or 1/Volume. Get out the old definite integral table, that's Ln(B) - Ln(A).
V = .5 to 1,W= Ln(1) - Ln(.5) = 0 - (-.69) = .69
V = 1 to 2,W= Ln(2) - Ln(1) = .69 - 0 = .69
V = 2 to 4,W= Ln(4) - Ln(2) = (.69+.69) - .69 = .69
If we went the opposite direction, density curve work for a compression factor of 2 is -.69. The work to change the density from A to K*A is always the same. Double the volume always yields the same Work due to density expansion. Half the Volume always takes the same work.
Note that for infinite expansion, there is infinite work.
V= 1 to infinity, W = Ln(infinity)-Ln(1) = infinity
(Requires available heat energy is also infinite). This would mean constantly adding heat to the air to maintain a fixed temperature
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Can one compare Work expressed as area (Volume*Pressure) directly to a quantity of Heat or Temperature?
Measuring Work area under the Work Curve.
Work(work curve) or W(wc)= Volume change times area under Pressure curve.
Pressure = Density * Temperature
Pressure = Volume-Gamma
and
Density = Volume-1
Temperature = Volume-Beta, where Beta = Gamma - 1.
For good old air, gamma = 1.4 which means Work is area under (Pressure = Density * Temperature = Volume-1.4) curve. The amount of work changes based on the amount of work accumulated on compression, or exported on expansion. But start from any place, the relative changes are still the same. Check the definite integral table and find, W(wc) = ∫(V=a,V=b) (Volume-1.4)
or B1-1.4/(1-1.4) - A1-1.4/(1-1.4)
or B-.4/(-.4) - A-.4/(-.4)
Simplifying to remove negative denominators...
W(wc) = A-.4/(.4) - B-.4/(.4)
W(wc) = (A-.4 - B-.4)/0.4
V= 1 to 2, W=(1-.4 - 2-.4)/0.4 = (1-.77)/0.4=(0.23)/0.4=0.61 Volume-Pressure units
To calculate relative work from other points, we multiply to normalize the beginning heat to 1. Since below points are picked spans at half and double, we multiply or divide by the Heat Ratio of 2, Hr(2), or 0.77.
V= 2 to 4, W=(2-.4 - 4-.4)/0.4/Hr(2)=((0.77-.4 - (0.77*0.77)-.4)/0.4)/0.77= .61 units
V= .5 to 1, W=((0.5-.4 - 1-.4)/0.4)*Hr(2) =((0.5-.4 - 1)/0.4)*.77=.61 units
Infinite expansion, is now a fixed amount of work, as no heat is being added. How much work? The Energy we gave the system to begin with, 1 normalized heat unit. The area is
W= (1-.4 - infinity-.4)/0.4 =(1-0)/.4 = 2.5 volume-pressure units.
Since The heat at infinity is 0, and the heat at 1 is 1, we can also assign the value 1/2.63 volume-pressure area units to 1 unit of heat on our relative vertical scale. So 1 unit of area = .4 units of relative temperature or heat, or 38% of the heat.
So in a complex graph, we can take the calculated volume-pressure area times 0.4 to get an equivalence with temperature (Which is an equivalence for heat energy also, because the fluid is in gaseous state).
Work(density curve) from 1 to 2 = .69 VP work units = .4*.69 = .33 T temperature units.
Work(work transfer curve) from 1 to 2
= .61 VP work units = .61*.4 = .23 T Temperature units
On the relative scale, we equated both heat and temperature to 1, so its easy to confuse the two. The amount of heat is also proportional to mass, or volume x density. And specific heat, or the amount of heat it takes to raise the temperature 1 degree, which varies by material. In the gas phase, specific heat is a constant, so the
total heat = density x volume x specific heat x temperature
In heat engines, when you want to compare the change in heat, and mass and specific heat are constant, and heat is not being added or subtracted, then you can compare temperatures.
However, you cannot compare temperatures. of different materials, like air and steam, to compare their heat energy, or even relative heat energy. So using the maximum temperature of steam and the ambient temperature of air in the "Work ratio" or "Carnot Ratio" is doubly silly.
Note for a gas with exponent 1.7, the total possible work, from 1 to infinity is still finite, but a smaller total area than gas with exponent 1.4.
W= (1-.7 - infinity-.7)/0.7 =(1-0)/.7 = 1.67 volume-pressure units. When comparing gasses the comparable units are volume - pressure units. If the exponent gets higher, the total work relative to the unit temperature scale is less. This is really less energy for a given temperature, which is to be expected, because the specific heat is less, a smaller amount of heat raises the exponent 1.7 gas to the same maximum temperature
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