V) Solar Augmented Heat-of-Compression Fluid Power Systems
Tesla's Magnificent Mechanical Oscillator Fluid Heat Engine
.... and some things to do with it
V) Solar Augmented Heat-of-Compression Fluid Power Systems
Background: This design approach was developed over the past year or so due to an interest in providing a practical alternative to the traditional high temperature solar heat engine designs with their dependence on reflective surfaces, tracking mechanisms, special thermal transport fluids, and other temperamental fallible costly high maintenance schemes and their generally abysmal overall thermal efficiencies which are not well suited for small scale non-commercial power applications.
The primary philosophical principle here is that our atmosphere provides the most practical dependable solar heat reservoir available to earth bound inhabitants, and since for all practical purposes air is still free and can't be permanently broken, frozen, or evaporated it might be a sensible idea to make use of what Mother Nature in her bounty has so generously given us.
Variations Three on a Theme: With that in mind what follows is pretty much just variations on the concepts discussed in the previous chapter, with the addition of passive solar heating units, and the expansion from the small mobile utility device design of the Mark-10 to that of stationary thermal solar power installations serving farms or small communities.
Example 1: Simple Air Compressor using air as the motive and suction fluids - or - How things get difficult real fast. This example is mainly to clarify why hydraulic motive injectors are better suited for this type of application than pneumatic. Compressed air serves the external load.
A: Pre-pressurize by compressing 2.4kg of air (@ 14.7psia @ 288K @ density = 1.2kg/cubic meter), into a one (1) cubic meter volume which is considered constant. Then exhaust one (1) cubic meter of pressurized air through the Mechanical Oscillator. Assume _no_ heat losses. Solve for Pf, Tf, and Ef.
B: To begin the pressure tank is at the above ambient conditions. The intake air during the Oscillator's operation will be heated to 338 degrees K by the solar panels.
C: The Mechanical Oscillator and Pump are affixed to a common shaft so the stroke of each is the same. The working surface area of the Oscillator's power piston is double that of the Pump's piston, so the Pump's output is always double the tank pressure driving the Oscillator - normal losses for such systems are ignored here.
D: The Injector's ratio of motive fluid to entrained suction intake fluid = 1:5.
From the above conditions it can seen that for every cubic meter of compressed tank air that powers the Oscillator and exhausts to the atmosphere, half a cubic meter of compressed tank air will be drawn into the compressor and delivered to the Injector as a quarter cubic meter of air at twice the pressure. At the ratio of 1:5 the Injector will then draw in (.25 * 5) 1.25 cubic meters of ambient air.
While from a volumetric view point things don't look all that bad, from a mass perspective this is a disaster, for the cubic meter of pressurized air exhausted driving the Oscillator had a mass of 2.4 kg, while the 1.25 cubic meters of intake ambient air only had a mass of 1.5 kg. In this scenario the pressure will fall from 38.79 psia to 18.97psia.
Where the sign 'y' = gamma = cp/cv = 1.4; gas constant 'R' = cp - cv = 0.287; rho = density. Since the original 2.4kg air mass was entirely replaced by the 1.5kg intake mass we can solve this just using the intake values. Solve for Pf, Tf, and Ef:
Pf = P1(V1/V2)^y = 14.7psia(1.25/1)^1.4 = 14.7(1.291) = 18.97psia
Tf = T1(P2/P1)^(y-1)/y = 288K(18.97/14.7)^0.286 = 288K(1.29)^0.286 = 309.77K
Ef = cv*m*Tf = 0.718*1.5*309.77K = 333.62kJ
*(I got a Tf of 314.87K when using the formula Tf = T1(rho2/rho1)^y-1; don't know why but it certainly indicates I did something wrong...)
This design fails because of the variable relationship between volume and mass the compressibility of air engenders. As the following examples demonstrate this problem is eliminated with the substitution of basically non-compressible water as the motive fluid.
Example 2: Air Compressor using water as the motive fluid, with air as the primary suction fluid & water as the secondary, or, the Mark-10's big brother. Compressed air serves the external load.
A: Pre-pressurize by compressing 2.4kg of air (@ 14.7psia @ 288K @ density = 1.2kg/cubic meter), into a one (1) cubic meter volume which is considered constant. The water volume is also considered constant as it is replaced nearly as fast as it is used. Then exhaust one (1) cubic meter of pressurized water through the Mechanical Oscillator. Assume _no_ heat losses.
B: To begin the pressure tank is at the above ambient conditions. The intake air during the Oscillator's operation will be heated to 338 degrees K by the solar panels.
C: The Mechanical Oscillator and Pump are affixed to a common shaft so the stroke of each is the same. The working surface area of the Oscillator's power piston is double that of the Pump's piston, so the Pump's output is always double the tank pressure driving the Oscillator - normal losses for such systems are ignored here.
D: The Injector's ratio of motive fluid to entrained suction intake fluid = 1:5.
In this example the former compressor is now a pump delivering 0.5 cubic meters of pressurized tank water to the Injectors instead of the 0.25 cubic meters of air of Example 1. The Injector's performance ratio stays at 1:5, so for every cubic meter exhausted by the Oscillator the Injector will draw in (0.5 * 5) 2.5 cubic meters of intake fluid. With water replaced at a 1:1 volumetric ratio, 1 cubic meter of the intake is dedicated to make-up water leaving 1.5 cubic meters for fresh intake air, or an additional 1.8 kg of air added to the 2.4 kg already present for a total air mass of 4.2 kg in the cubic meter volume.
Including the pre-pressurization stage, this results in an 'Ideal' pressure increase to 89.32 psia, and a temperature increase to 511 K (238 C | 460.4 F).
Where the sign 'y' = gamma = cp/cv = 1.4; gas constant 'R' = cp - cv = 0.287; rho = density. Solve for Pf and Tf.
Step One: Compress 2.4kg of air @ 14.7psia @ 288K @ density = 1.2kg/cubic meter into 0.5714285 cubic meter. Assume _no_ heat losses. Solve for P2, T2, and E1:
P2=P1(V1/V2)^y = 14.7(2m3/0.57m3)^1.4 = 14.7(5.78) = 84.92psia
T2=T1(P2/P1)^(y-1)/y = 288K(84.92/14.7)^0.286 = 288K(1.651) = 475.59K
E1=cv*m*T2 = 0.718*2.4*475.59K = 819.54 kJ
Step Two: Compress 1.8kg of air @338K into a 0.4285714 cubic meter volume. Solve for P2, T2, and E2:
P2=P1(V1/V2)^y = 14.7(1.5/0.4285714)^1.4 = 14.7(5.78) = 84.92psia
T2=T1(P2/P1)^y-1)/y = 338K(84.92/14.7)^0.286 = 338K(1.651) = 558K
E2=cv*m*T2 = 0.718*1.8kg*558K = 721.16kJ
Step Three: Combine the above results in a 1 cubic meter volume with rho3 = 4.2kg. Solve for Pf & Tf:
Tf=(E1+E2)/(cv*rho3)=(819.54kJ+721.16kJ)/0.718*4.2=(1,0540.70)/3.0156 = 511K
Pf=(rho3*R*Tf)= 4.2*287*511 = 615,852.33Pa / 6895Pa/psi = 89.32psia
(If all air is solar boosted final temp = 558K @ 97.55 psi)
(continued)