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Testing Models

Testing of model steam engines and turbines.

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Turbine Guy02/01/2019 15:12:24
52 forum posts
23 photos

The preceding posts bring us up to date on all the testing I have completed so far. The descriptions of these tests were taken from Emails to Thor updating my progress. Thor helped me in my testing using his experience with the many steam engines he has made and run. These test results only show the results we were able to obtain and not some of the problems we had to overcome. You can see from these tests, it is not always easy to find what the real problems are. I was concentrating on the internal leakage of my ST steam engines and it took a long time to find that the biggest problem was the titling of the cylinder due to the loose fit of the cylinder support pin. I have machined a larger cylinder pivot pin and will drill and ream the standard of one of my ST steam engines to reduce the clearance. When I get the standard finished I will start the testing again.

Tim Taylor 202/01/2019 16:22:16
66 forum posts
8 photos
Posted by Turbine Guy on 02/01/2019 12:52:10:

You will need the maximum diameter of the nozzle for scaling the blades. You're right about using multiple nozzles and being able to isolate them individually. The supersonic nozzles are very sensitive to pressure drop and can become unstable with pressure ratios higher or lower than their design. For optimum performance the angle of the blades needs to take into account the speed difference between the stationary and moving parts. If you plan on only running at low speeds this won't be as important.

TG,

The way full size single stage Curtis turbines are made, the blade size is related to the wheel diameter and pretty much fixed. Instead of changing the blade size, they adjust the number and throat diameter of the nozzles to fit the application. As I'm sure you are aware, all full size turbines are sized for at least 10% reserve above rated capacity based on worst case (minimum inlet/maximum back pressure) specified steam conditions, and if dealing with a spec like API, you get an additional 10% on top of that. This is why the hand valves - isolating single or groups of nozzles allows you to adjust for optimum efficiency in actual operation - not really that relevant for a model turbine though......

If I have enough room, I'll probably use a nozzle block instead of machining the nozzles directly in the steam chest. What I am working on right now is how small I can make the buckets while maintaining a reasonably accurate profile. I plan on using a built up wheel - the buckets will be individually made and silver-soldered into slots in the wheel, then a shroud silver-soldered around the outside. Then the wheel will be chucked in the lathe and faced and turned to uniform OD.....at least that's my preliminary plan....

Now that the holidays are over, maybe I'll get some more time to work on it......then again, winter Steelhead season is just getting into swing here, so I'll likely be spending a day or two here and there on one of the local rivers with my fly rod.....

Tim

Turbine Guy06/01/2019 16:08:30
52 forum posts
23 photos

Tim,

It looks like you picked a very good turbine to base your design on. I searched on the internet for information on this turbine and found the following for a rebuilt turbine. They didn't give the temperature or the mass flow of the steam for this power. It surprised me the turbine would operate at such a low speed.

Specifications

Coppus Steam Turbine TF-9
RPM: 1250
HP: 0.75
Inlet PSIG: 75

Turbine Guy06/01/2019 18:52:33
52 forum posts
23 photos

I ran my Stuart ST with the tight packing and the increased diameter cylinder pivot pin. The tilting of the cylinder face against the standard face was almost completely eliminated with the tighter clearance. With 20 psig steam pressure, the leakage was about the same with the cylinder rotated in any position and the pressure didn’t drop when the engine was at its maximum speed. I stopped, oiled, and started the engine for about a half hour. The maximum speeds for 5, 10, 15, and 20 psig were 450, 655, 765, and 851 rpm respectively. The speeds kept increasing with each run and the leakage decreased slightly after all the running. My airbrush compressor ran the full time on the first run and was turning on an off on the last run. The engine ran much smoother and quieter without the cylinder tilting back and forth on the standard. I expect the engine will reach a higher speed after it is fully bedded in since it had a maximum speed of 1,160 rpm before these changes.

Tim Taylor 206/01/2019 20:11:25
66 forum posts
8 photos
Posted by Turbine Guy on 06/01/2019 16:08:30:

Tim,

It looks like you picked a very good turbine to base your design on. I searched on the internet for information on this turbine and found the following for a rebuilt turbine. They didn't give the temperature or the mass flow of the steam for this power. It surprised me the turbine would operate at such a low speed.

Specifications

Coppus Steam Turbine TF-9
RPM: 1250
HP: 0.75
Inlet PSIG: 75

TG,

The Coppus TF design dates back to the 1920's. 15 psig is on the low side of inlet pressure - depending on materials of construction they could operate up to 650psig/750F steam with back pressures up to 150psig, and could produce up to 1000bhp. They also had a vertical version of the design...

In the 1970's the TF design was superseded by the RLA design, which changed from a flyball governor to a woodward oil relay governor and a new safety trip design. In the 1980's a horizontally split design called the RLH was introduced. Coppus is now part of Dresser. The following link is to a brochure describing the various designs - you might find it of interest.

http://www.mercado-ideal.com/catalogosd/DRESSER-RAND%20COPPUS%20STEAM%20TURBINES.pdf

The one I am going to model is actually similar to one of the horizontally split designs - both have similar wheels.

Tim

Turbine Guy07/01/2019 19:19:46
52 forum posts
23 photos

I’ve run the ST engine with the tight packing and the increased cylinder shaft size off and on for about a half an hour and the maximum speed is up to about 900 rpm now. Since my airbrush compressor is able to maintain a pressure of 20 psig at the maximum speed the engine is running and the power output is very low, I tried to estimate the power losses. I looked at the friction power loss at each of the motion points and estimated the total power loss due to friction is approximately 2.68 watts. I use a friction coefficient of 0.3 for the analysis. I estimated the pressure drop in the inlet port to be 1 psi. I estimated the pressure drop in the exhaust port to be 4 psi. With these pressure drops, the existing clearance volumes, and the cut off point of the ports I calculated the mean effective pressure. I estimated the mean effective pressure to be approximately 15 psi and the corresponding indicated power to be 3.25 watts. Subtracting the friction power loss from the indicated power gave a power output of 0.57 watts. The power required to turn the 8” propeller 900 rpm is 0.09 watts. My calculations don’t quite show where all the power was lost but confirm that the power loss due to friction and pressure drop through the ports is very high for the Stuart ST engine. The lower speed with the increased cylinder shaft size is partially explained by the engine not being fully broken in. In addition, the friction loss of this shaft goes up in direct proportion with the increase in shaft diameter. The low power output is finally starting to make sense.

Turbine Guy08/01/2019 22:46:12
52 forum posts
23 photos

I found an error in my calculations for the power output of my ST engine. The corrected total power loss due to friction is approximately 2.55 watts and the resulting calculated power output is 0.7 watts. This leaves a little more loss unaccounted for, but the result is about the same. In my calculations I used the friction given by Parker for a floating 0-ring even though I ran the last tests with the tight packing. I needed some method to calculate the friction when the seal gets pressed against the cylinder bore by pressure. Since I got almost identical performance with the 0-ring and the tight packing in earlier tests I thought this method was valid for either. The only unknown in the rest of the calculations is the friction coefficient. This varies with the materials used, the finish, and whether dry or lubricated. The value of 0.3 I used for the friction coefficient is what I have seen used before for estimating the friction. Because of these unknowns the calculations are only indicative of the approximate size of these losses.

Turbine Guy18/01/2019 21:34:12
52 forum posts
23 photos

I purchased a Chiltern Vertical Single Marine Engine. This engine has a 14mm bore, 18mm stroke, a piston valve, and Teflon piston rings. I ran my Chiltern engine for a short time and very quickly it was able to run on 5 psig (0.34 bar) air pressure. I checked the leakage holding the valve in various positions with maximum pressure. When the valve was in the position that both ports were closed there was very little leakage after the engine had been run for a few minutes and not reoiled. After I oiled the engine the leakage completely stopped in this position. This was with a pressure of 24 psig (1.63 bar), the maximum I plan to use. When I moved the valve to a position that the ports were open, the leakage started but did not seem excessive. The leakage was about the same in all positions of the valve accept at each end of its travel where it increased quite a bit. These positions are where the distance from the edge of the port to the edges of the inlet and outlet are shortest. Apparently, the valve clearance is tight enough that oil can seal it in most positions. Even after running a while the valve leakage only increased a little. The piston rings have some leakage but it doesn't appear to be excessive. The minimum pressure the engine would continue to run stayed at about 5 psig (0.34 bar). Each time I shut the compressor off when the engine was running, the speed increased before slowing down and stopping. I'm still trying to figure out what causes this. I'll add a photo of this engine in the next post.

Turbine Guy18/01/2019 22:10:30
52 forum posts
23 photos

I like using the propellers for a load since I can get the amount of power they require for a given speed. For a small air compressor and medium size engine, I have to use a very large propeller to keep the speed low. If I use a small propeller, my compressor can't keep up with the flow required and the pressure drops. With a 14mm bore and 18mm stroke, my Chiltern engine requires the maximum output of my airbrush compressor to maintain a pressure of 24 psig (1.63 bar) at a speed of 675 rpm with no leakage. With leakage, the speed will be even lower to maintain this pressure. I bought a 22" (560mm) diameter propeller for testing this engine. The propeller is a APC 22 x 12 WE It is 22" diameter propeller with a 12" pitch and requires about 14 watts to run at 1000 rpm. I show this propeller mounted to the Chiltern steam engine below. There is also a couple of pictures of the Chiltern steam engine without the propeller mounted in my album. I would appreciate it if someone could tell me how rotate the pictures in the album to the position they should be viewed.Chiltern with propeller

Turbine Guy19/01/2019 18:22:12
52 forum posts
23 photos

I attached the propeller to the engine, mounted the engine on a plate, clamped the plate to my workbench, and turned on the airbrush compressor. The propeller spun at a maximum speed of 700 rpm with a pressure of 20 psig (1.36 bar). The power required by the propeller at this speed is approximately 4.71 watts. The amount of air required by the engine to fill the cylinder at this speed and pressure is approximately 1.52 lbm/hr. Since my airbrush compressor is only capable of about 1.74 lbm/hr, the total leakage was very small. I was very pleased with the Chiltern engine. It did better than I thought it would. The energy available from the airbrush compressor is approximately 16 watts, so I would like to determine where the losses come from. With steam engines, friction, pressure loss through the ports, and leakage can all be major parts of the loss and are difficult to estimate. In addition, using the full pressure for almost the entire stroke (almost no cutoff) wastes quite a bit of energy. Also, the Chiltern engine has relatively large clearance volumes that waste some energy. I’ll see if I can estimate where the major losses came from.

Turbine Guy19/01/2019 20:39:54
52 forum posts
23 photos

To estimate where the losses came from, 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 steam 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 20 psig, an exhaust pressure of 0.0 psig, and a mass flow of 1.74 lbm/hr. With these conditions, the power is approximately 16 watts at a speed of 1,660 rpm. I kept everything the same except adding the Chiltern’s actual clearance volume of 0.028 in^3. With only the clearance volume added, the power dropped to 11.7 watts at a speed of 1,250 rpm. The actual clearance volume causes a reduction in power of approximately 4.3 watts. The Chiltern’s clearance volume is so large primarily due to the clearance between the piston and cylinders at each end of travel being 2.25mm (0.089 in). I used the Onshape solid model shown below to find the clearances and other dimensions needed for analysis. The solid model has the parts mated with the same degrees of freedom as the actual engine. Turning the flywheel in the solid model moves the parts the same as they would in the Chiltern engine.Solid Model of Chiltern

Tim Taylor 220/01/2019 19:36:28
66 forum posts
8 photos

TG,

Might be a stupid question, but is your 20 watt estimate based on compressor input power or net output? Compressors are notoriously inefficient, typically 50% or less - a lot of the energy is lost as low grade heat from the heat of compression.

Tim

Tim Taylor 220/01/2019 20:10:44
66 forum posts
8 photos

sorry....16 watts.....

Turbine Guy20/01/2019 20:43:22
52 forum posts
23 photos

Tim,

When I decided to use the airbrush compressor for my testing, I connected the compressor to my turbine to find the maximum pressure I could run with all the air going through the turbine nozzle. I found the maximum pressure was 24 psig and the turbine temperature was approximately 70 F. For this pressure, temperature, and .028” nozzle throat diameter the theoretical mass flow is approximately 1.74 lbm/hr. For this temperature, pressure, and mass flow the energy available to the turbine is approximately 18 watts. This is my assumed maximum output for my airbrush compressor when running the turbine.

Turbine Guy20/01/2019 20:53:04
52 forum posts
23 photos

The 16 watts maximum for the engines was found the same way. 20 psig was the maximum pressure the airbrush compressor could maintain with the engines which lowered the available power to 16 watts.

Tim Taylor 220/01/2019 22:11:55
66 forum posts
8 photos

TG,

That answers my question.

Regarding efficiency, I think you have a couple things going on that may be throwing you a curve.

First, there is a big difference between D&S steam and compressed air - steam has roughly 8 times the enthalpy of compressed air at the same pressure & temperature. Steam is not an ideal gas, and steam engines are designed to take advantage of the isentropic enthalpy drop and the expansion characteristics of steam. The valve timing for steam, for example, is not what would be the most efficient for compressed air. A steam engine running on air will not be as efficient as if it were running on steam.

Second are the leakage paths you have already mentioned. Design clearances take into account thermal expansion at operating temperature, and what may be a significant leak at room temp, will not be at designed operating temperature.

Was your clearance volume in in^3 or cm^3? Based on your 14mm bore by 18mm stroke, I calculated the swept volume as 2.77 cm^3. The optimum clearance volume is typically around 10% of the total volume, so if your clearance volume is in cm^3, then it's in the ball park.

I find your progress quite interesting - please keep us posted.........

Tim

Edited By Tim Taylor 2 on 20/01/2019 22:13:48

Turbine Guy20/01/2019 23:32:28
52 forum posts
23 photos

Tim,

I agree with your comments about the difference in using steam and air and testing the engine with air when it was probably designed for steam. What I am trying to determine is where the losses come from for the tests I've made with air. The actual amount of these losses would be different for steam, but I chose air since it eliminates one of the major difficulties. With steam, condensing can occur if the steam does not have enough superheat or the engine hasn't been run very long. The amount of moisture in the steam is very difficult to determine and has a big effect on the performance. With air and a compressor with very low output, the change in temperature is very small and the process is basically isothermal rather that isentropic. I can't tell the difference in temperature between the outlet of my airbrush compressor and the outlet of my engine even after a long run. I use the formulas for isothermal expansion of air to determine the enthalpies. I agree my findings are only valid for running the Chiltern engine on air, but should give a reasonable indication of the magnitude of the losses.

The actual clearance volume of the Chiltern engine I gave was 0.028 cubic inches or 0.459 cubic centimeters. This is 16.5% of the swept volume.

Sorry I use different measuring systems. All my calculations use English units. The Chiltern engine is metric so I made my solid model metric and when I pull dimensions off the model they are in millimeters which I usually convert to inches

Thanks for your feedback..

Turbine Guy21/01/2019 14:17:33
52 forum posts
23 photos

To estimate the loss due to the long cutoff for the Chiltern engine, I started again with the ideal 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. I used a 14mm bore, 18mm stroke, air with an inlet pressure of 20 psig, an exhaust pressure of 0.0 psig, and a mass flow of 1.74 lbm/hr. With these conditions, the power is approximately 16 watts at a speed of 1,660 rpm. I kept everything the same except using the Chiltern’s actual cutoff of 94% of the stroke. The cutoff for full expansion is 52% for the ideal case. With only the cutoff changed, the power dropped to 11.3 watts at a speed of 910 rpm. The actual cutoff causes a reduction in power of approximately 4.7 watts. The previously found power loss due to the clearance volume of 0.028 in^3 was 4.3 watts when it was the only thing changed in the ideal indicated power. The total power loss for these found independently is 9 watts. Since the clearance volume losses are not independent from cutoff losses, I added both changes to the ideal case at the same time and the power dropped to 9.6 watts at a speed of 772 rpm. The power lost for both run in combination is 6.4 watts and should be used as a total amount to account for both these losses. It should be noted that the long cutoff and large clearance make the Chiltern engine very forgiving in assembly and operation. Since maximum power is not as important in the model engine, Manufacturers may choose to accept more losses to insure the model can be assembled easily and be less sensitive to changes in pressure.

Turbine Guy21/01/2019 21:59:30
52 forum posts
23 photos

The next lost I tried to estimate is the pressure drop through the ports. I used a 14mm bore, 18mm stroke, air with an inlet pressure of 20 psig, an exhaust pressure of 0.0 psig, a mass flow of 1.74 lbm/hr, and a speed of 772 rpm. With these conditions, I calculated a pressure drop through the inlet port of 0.60 psi and the exhaust port of 0.68 psi. I looked at the length of the airbrush compressor hose, all the lengths and direction changes of the ports, the viscosity of the air, abrupt changes in cross sectional area, and the friction of the air moving through the ports in my calculations. I was surprised by the low pressure drop. The reason appeared to be the low velocity (67 ft/sec maximum) in the ports as a result of the 3mm diameter port that is fully open through most of the intake and exhaust cycles. Since the previously found losses are not independent from the pressure drops, I included the actual clearance area of 0.028 in^3 and the actual cutoff of 94% with the inlet pressure drop of 0.60 psi and the exhaust pressure drop of 0.68 psi and made a new calculation of the indicated power. This resulted in a indicated power of 9.1 watts at 785 rpm. The estimated indicated power without any pressure drop through the ports, no clearance volume, full expansion, and no compression of the air on the exhaust cycle is 16 watts. The total of all the losses calculated up to now for the Chiltern engine is 6.9 watts.

Turbine Guy22/01/2019 16:06:21
52 forum posts
23 photos

The final loss for the indicated power is the compression of the air when the exhaust closes before the end of the stroke. For high speed engines the exhaust is closed early so the compression slows the piston before it changes direction. The Chiltern engine closes the exhaust at approximate .043 inches before the end of travel, so there is some compression. The total volume of air including the clearance volume is 0.0381 in^3. With a clearance volume of 0.028 in^3 there is a compression ratio of 1.36. The resulting pressure at the end of travel is 7.3 psig. The power lost due to this compression is approximately 0.3 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 8.8 watts at 785 rpm. The Chiltern engine obtained a power of 4.7 watts at 700 rpm. The indicated power at 700 rpm with the same clearance volume, cutoff, port pressure drops, and exhaust compression is 7.9 watts. The estimated indicated power without any pressure drop through the ports, no clearance volume, full expansion, and no compression of the air on the exhaust cycle is 16 watts. The total of all the indicated power losses for the Chiltern engine running at 700 rpm is 8.1 watts. The actual power is approximately 59% of the indicated power. K.N. Harris in his book Model Stationary and Marine Engines stated ‘the useful horse-power the engine can give out at its shaft is not likely to be more than two-thirds of the calculated indicated horse-power’.

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