Concept of a new thermopiple array above 10% efficiency?

[manchild02]

Is is possible to design a more efficient thermopipe using a laminer flow heat source?

A. Test piece incorporates a length of 1.5 x 24 " black iron pipe to produce laminer flow.
B. A variable fan is used to input the air into one end of the pipe.
C. Fuel source is a 20 lb. propane bottle with regulator.
Propane feed consists of a 1/4 " clear vynal hose fitted with a 6 " length of 1/8th steel tubing.
D. Fan air is feed through a steel 1.5 x 1.5 T. The 6 " x 1/8th steel tube is embedded into a 1 x 1.5 steel nipple with refractory cement and threaded into a 1 x 1.5 steel coupling which is threaded into the 1.5 steel T. A gang valve is used to adjust the volume of propane into the steel pipe. This makes up the primary heat source.
E. The thermocouplers consist of 12-1/16th x 8 solid copper wire leads drilled into the steel pipe horizontally and spaced .75 " apart to form a thermopipe. 1/8th x 8 " threaded rod is used to make up the iron thermopile set at a 60 degree angle to the bank of 12 copper leads. Both sides of the multi-thermocoupler array are joined in series.
F. Two more copper-iron thermopile banks are spaced around the 1.5 " pipe at 60 degrees each to form three complete iron-copper thermopipe banks. All together you have 3 rows of copper thermocouplers and three rows of iron thermocouple leads, each joined in series. All six copper-iron banks make up the cold junction while the extended copper-irons leads fitted into the steel pipe make up the hot junction.
G. Heat is generated inside of the 1.5 x 24 " steel pipe by turning air on low and propane at 50 %. The mix is ignited at the end of the pipe and air-fuel pressure turned down to pull flame inside of the pipe. At this point both the fan-air and propane can be turned up to develope a high laminer flow producing core temperatures of 2, 300 F and sidewall temperature of 2, 000 F and exhaust heat at 2, 000 F. This should cause both the copper and iron junctions to heat up to above 1, 800 F.

Questions:

1. What will be the potential temp. difference of each copper and iron thermocoupler lead minus an ambient temp. of 72 degrees F?
2. How much dc voltage will be produced?
3. Can the potential temperature be increased by extending the indivisual length of each copper-iron thermocoupler?
4. What would be the optimun length of the copper and iron leads?
5. How much voltage will 1 thermopile bank produce?
6. What is the amperage output on 1 thermopiple bank?
7. What is the voltage and amperage of all 3 thermopile banks.
8. Using the cold side of the flame pipe, can this be used as the cold point for the copper side of the individual copper leads?
9. If so, how do the voltages and amperage compare using a iron cold point as a common junction?
10. What would be the combined dc voltage and amperage output if all three thermopile banks are joined in //?
11. Heating all 72 copper-iron juctions in union with ref. points at near ambient temperature, will this increase thermocoupler efficiencies above present 5 % conventional values?
12. If so, can improvements be made to increase efficiencies above 30 % Nasa standards?
13. Will the voltage and amperage produced have temperature effect of the flame?
14. Will the laminer flow of the flame be affected or increased thrust?
15. Will the thermocouler array induce a magnetic effect on the flame?

For further information of flame pipe set-up, please refer to u tube under 'joepipe.'

Using laminer flow as a primary energy souce for heating a thermocoupler array allows for near 100 % of the heat to be utilized, as oposed to using a external flame where much of the heat is absorded by the air.

Let see what we can do to combine our expertize to come up in theory to establish the said outputs prior to setting up for real time tests? As far as I know, no one to date is using this method to build a thermopile in this fashion.

 

[jbsteele]

1. To calculate the potential temperature difference between each copper and iron thermocoupler lead, we can use the standard formula for calculating Joule's Law: V = I x R, where V is voltage, I is current, and R is resistance. The resistance of the copper-iron leads can be determined from their length and diameter.2. We can also use the formula P = IV, where P is power, to determine the amount of DC voltage produced by each thermopile bank. The power will depend on the amount of current flowing through each bank.3. We can extend the individual length of each copper-iron thermocoupler to increase the potential temperature difference by increasing the resistance of the leads. This will require more voltage to be applied in order to create the same amount of current. 4. The optimum length of the copper and iron leads can be determined by determining the resistance of each thermocoupler based on its length and diameter. The greater the resistance, the lower the current.5. We can use the formula V=IR to calculate the voltage produced by each thermopile bank. The voltage will depend on the amount of current flowing through each bank.6. We can use the formula I=V/R to calculate the amperage output on each thermopile bank. The amperage will depend on the amount of resistance in the bank.7. We can use the formula P=IV to calculate the combined dc voltage and amperage output of all three thermopile banks when connected in series.8. We can use the cold side of the flame pipe as the cold point for the copper side of the individual copper leads by measuring the temperature difference between the hot and cold points.9. When using a iron cold point as a common junction, we can use the formula V=IR to calculate the voltage produced by each thermopile bank. The voltage will depend on the amount of current flowing through each bank.10. We can use the formula P=IV to calculate the combined dc voltage and amperage output if all three thermopile banks are joined in parallel.11. Heating all 72 copper-iron junctions in union with reference points at near ambient temperature may increase thermocoupler efficiencies above present 5% values, but this depends on how the thermopiles are arranged