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Testing Models

Testing of model steam engines and turbines.

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Turbine Guy23/01/2019 15:14:20
75 forum posts
32 photos

I estimated the following mechanical losses due to friction using a friction coefficient of 0.30. The crosshead power loss is estimated to be 0.51 watts. The crankshaft bearings power loss is estimated to be 1.31 watts. The propeller had a thrust force of 0.39 lbs at a speed of 700 rpm resulting in a estimated 0.23 watts of power loss. To estimate the piston rings power loss, I assumed the mean effective pressure pushed the rings out against the cylinder and used a coefficient of friction of 0.13. Since the Chiltern engine has two piston rings, I assumed half the pressure drop occurred across each piston ring. With these assumptions, I estimated the piston rings power loss to be 0.54 watts. The total of all the estimated mechanical losses is 2.59 watts. The indicated power with a clearance area of 0.028 in^3, a cutoff of 94%, an inlet pressure drop of 0.60 psi, a exhaust pressure drop of 0.68 psi, and a compression ratio of 1.36 is approximately 7.9 watts at 700 rpm. With an estimated indicated power of 7.9 watts and an estimated total mechanical loss of 2.59 watts, the estimated power is 5.31 watts. The actual power was 4.71 watts. This leaves 0.6 watts unaccounted for. Hopefully, this analysis gives an indication of the approximate magnitude of the losses and where they occur for the Chiltern engine tested. As previously stated, some of the losses are due to tradeoffs in performance versus ease of operation and setup, not poor design.

Tim Taylor 228/01/2019 17:01:32
66 forum posts
8 photos

TG,

Don't know if you've seen these, but links to a couple references I thought you might find of interest...

https://www.forgottenbooks.com/en/download/The_Steam_Turbine_1000777956.pdf

https://www.osti.gov/servlets/purl/1376860

Tim

Turbine Guy28/01/2019 22:03:30
75 forum posts
32 photos

Tim,

There is a lot of information in your first link, essentially a complete book on turbine design. It will take me a while to go through the PDF file and see what it adds to the several books on turbines I have. Quick scanning through the PDF showed it gets pretty deep in the technical aspects. Thanks for the link.

The second link was information on two phase nozzle design. I currently have no interest in that. When I made turbines with large enthalpy changes that got into the supersonic velocities, I was careful to use highly super heated steam to avoid moisture at the end of the expansion. It would be an interesting subject if change of phase could not be avoided. Thanks for showing the link.

Turbine Guy30/01/2019 21:00:33
75 forum posts
32 photos

After estimating the magnitude of the losses and what caused them for the Chiltern engine, I looked at ways to recover the losses. I started by seeing if the cutoff could be changed and what would be required to make the changes. I started with the best case. I estimated the indicated power without any pressure drop through the ports, no clearance volume, full expansion, and no compression of the air on the exhaust cycle. This is a perfect condition where the power produced equals the energy available. I used a 14mm bore, 18mm stroke, air with an inlet pressure of 24 psig, an exhaust pressure of 0.0 psig, and a mass flow of 1.74 lbm/hr. The cutoff for the ideal case is 48%. With these conditions, the power is approximately 17.5 watts at a speed of 1,620 rpm. I assumed a slightly higher inlet pressure than I was able to attain in the previous testing since I can get a propeller that keeps the maximum speed low enough that my airbrush compressor can maintain the higher pressure. I found by changing the piston valve and changing the ports from holes to slots I could get a cutoff of approximately 50% and still keep the pressure drop through the ports very low. I calculated the indicated power and mechanical (friction) losses by the methods I previously described plus added the 0.6 watts loss I was unable to account for. I estimated I could get a maximum power of 5.9 watts at a speed of 1,040 rpm. The test of the unmodified Chiltern engine resulted in a maximum power of 4.71 watts at 700 rpm. I’ll explain why this major change only has a potential gain of approximately 1.2 watts in my next post.

Turbine Guy01/02/2019 18:29:57
75 forum posts
32 photos

The following drawing shows the valve I designed to try to get approximately 50% cutoff with the existing eccentric and valve linkage. The existing valve has a 0.4mm inlet lap and a 0.6mm exhaust lap. My revised valve has an inlet lap of 2.75mm and no exhaust lap. It wasn’t until I made a series of drawings to check the valve operation that I learned the importance of exhaust lap. Valve Mid Travel Position

Turbine Guy01/02/2019 18:33:29
75 forum posts
32 photos

The following drawing shows the distance from the cylinder cover to the piston when the inlet valve closed. The distance traveled minus the clearance distance divided by the stroke is the cutoff. The actual clearance is 2.25mm, so the cutoff is approximately 58%. Because of the time it takes to close the valve, the effective cutoff is probably close to the 50% I was trying to get.

Valve Cutoff Position

Turbine Guy01/02/2019 18:36:32
75 forum posts
32 photos

The following drawing shows the distances from the end plates when the exhaust closes on each side of the piston. The distance below the piston is used to calculate the volume of trapped steam or air that is compressed. I estimated the amount of power to compress the trapped air to be approximately 0.67 watts. The distance above the piston is used to see the point in the stroke the exhaust is released. The distance moved by the piston 18.08mm minus the clearance distance of 2.25mm and is approximately 15.83 mm. Since the stroke is 18mm, the air is released 2.17mm before the piston has moved the full stroke. I estimated the power lost due to the early release to be approximately.0.85 watts. In my original estimate of the power with the revised valve, I missed the early release. With this loss, the gain in power would be very little. The cure to this problem would be to add exhaust lap. The exhaust lap would almost eliminate the early release but increase the compression. If I could add the required exhaust lap, I would probably get the performance I first estimated. There isn’t enough room to add the required exhaust lap so I will have to look for a better way to improve the performance.

Exhaust closed and release points

Edited By Turbine Guy on 01/02/2019 18:37:39

Turbine Guy12/02/2019 19:52:00
75 forum posts
32 photos

I decided I would have to make separate exhaust and inlet valves to substantially increase the efficiency. The link **LINK** confirmed the increase in efficiency with separate valves. This article also gave me enough information that I could check my formulas and methods of estimating performance with some actual test data. I used the 1.27 in. bore, 2.00 in. stroke, 15% clearance volume, 38% cutoff, early release distance of 0.30 in., exhaust closed distance of 0.44 in., inlet pressure of 38 psig, superheat of 81.2 C, and 502 rpm engine speed shown in fig. 6 for the conventional valve. With these inputs and using my equations I calculated an indicated power of 116 watts. The test result was 119 watts so my equations and methods appear to be valid. The following drawing shows a concept to add a second valve to my existing engine. Separate Valves TDCAdding the separate valves requires 1 each of the following new parts, a valve chest, inlet valve, exhaust valve, eccentric, and a valve connecting rod. To get maximum efficiency I would also add new top and bottom cylinder covers to the preceding new parts. These covers would extend further into the cylinder to reduce the piston clearance from 2.25mm to 1mm at each end of travel.

Turbine Guy13/02/2019 19:50:55
75 forum posts
32 photos

With the preceding new parts, 14mm bore, 18mm stroke, air with an inlet pressure of 24 psig, a exhaust pressure of 0.0 psig, and a mass flow of 1.74 lbm/hr. the input power is approximately 17.5 watts. The average cutoff for each side of the piston is approximately 52%. The estimated output power is 7.52 watts. The estimated hydraulic efficiency is 66.9% and the estimated overall efficiency is 41.9%. Adding separate valves and reducing the clearance volume increases the power by approximately 2.8 watts, a 60% improvement, but requires 7 new parts.

Turbine Guy13/02/2019 21:17:40
75 forum posts
32 photos

I tried various combinations of inlet and exhaust lap for a single valve and found that an inlet lap of 2mm and no exhaust lap that gave an average cutoff of approximately 75% was the best combination for air with an available energy 17.5 watts. The air has an inlet pressure of 24 psig, a exhaust pressure of 0.0 psig, and a mass flow of 1.74 lbm/hr. The estimated output power is 5.6 watts. The estimated hydraulic efficiency is 51.8% and the estimated overall efficiency is 32.3%. Making a new valve without any other changes could increase the power by approximately 0.9 watts, a 19% improvement, and only require 1 new part.

Valve 2 Mid Travel

Turbine Guy10/04/2019 22:09:06
75 forum posts
32 photos

In the post of 31/12/2018 I stated: ‘The velocity coefficient for the existing rotor geometry, nozzle position and throat diameter, a power of 1.92 watts, an air flow of 1.74 lbm/hr, a inlet pressure of 24 psig, a inlet temperature of 70 F, and a speed of 17,012 rpm is 0.34. The velocity coefficient for the optimum Terry turbine rotor for these same conditions is 0.53.’ I built a rotor with twice the number of pockets of my first turbine rotor. The pockets of the new rotor overlap eliminating most of the edge blockage of the first design. I thought this design could possibly approach the performance of the optimum Terry turbine mentioned above. The following photo shows the new rotor (aluminum) next to the original rotor (brass). I tested this rotor with the same air pressure and flow used in the test of my original rotor that had a maximum speed of 17,000 rpm. The maximum speed with the new rotor was 18,250 rpm. The required power of the EP2508 propeller used in these tests is approximately 1.9 watts at 17,000 rpm and 2.4 watts at 18,250 rpm. The average rotor velocity coefficients for these output powers are 0.34 for 1.9 watts and 0.53 for 2.4 watts. I got the maximum I thought was possible, so this is a very significant increase. The original rotor has 24 pockets and the new rotor has 48 pockets. In Dr. Balje's study of high energy level, low output turbines the highest average rotor velocity coefficient for a Terry turbine with a single nozzle, 45 blades, and an admission length to rotor pitch length ratio of 1.72 was 0.53. The open pockets appear to be as efficient as the Terry turbine blades for this very small pocket size.

Turbine Rotors

Turbine Guy10/04/2019 22:21:08
75 forum posts
32 photos

The following drawing shows the details of the Tangential turbine of the previous post with the brass rotor.Tangential Turbine BB

Turbine Guy10/04/2019 22:23:34
75 forum posts
32 photos

The following drawing shows the details of the latest Tangential turbine with the aluminum rotor.

Tangential Turbine 2

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