Sharing information about model turbines
|Martin Johnson 1||13/03/2019 16:53:46|
|106 forum posts|
Thanks for making me think, chaps.
You are right for constant momentum, but greater mass flow, the TORQUE would remain constant for a given design of blades, but at the same time the optimum running speed would fall in the ratio of the mass flow increase (same as velocity decrease). Hence power falls as the ratio of mass flow increase or velocity decrease. All assuming motive and entrained fluids of identical density.
BUT earlier in this thread TG is reporting a blade speed to fluid speed ratio of 0.022 - way down on where it should be for a two velocity staged machine. AND even if he can get the running speed up to where it should be, how long will the bearings stand up to it? Or, indeed, how long will the blades stay attached to the hub? AND what losses would there be in a wee gearbox to reduce the speed to something manageable?
So, the question remains for tiny turbines, would an ejector lose more than it gains?
For all that, I am in awe of the tiny rotor and blade cutting that TG is undertaking. Also, the methodical way it is being thought through. Keep the reports on progress coming, please.
|Turbine Guy||13/03/2019 18:54:24|
|68 forum posts|
The following is a similar analysis as in the 13/03/2019 posts for the GH ½ B model jet pump
Air inlet pressure of 0 psig
Air inlet temperature of 70 F
Air inlet enthalpy of 126.7 btu/lb
Steam pressure of 120 psig,
Steam temperature is 351 F
Steam enthalpy is 1192 btu/lb.
Suction capacity is approximately 54 SCFM.
½ B Capacity factor is 0.030.
1½ model steam consumption is 390 lb/hr
½ B model steam consumption is 11.7 lb/hr
½ B air pumped = 1.6 SCFM
½ B mass flow of air is 7.2 lb/hr
Discharge pressure is 12 psig,
Discharge temperature is 300 F
Discharge enthalpy is 1192 btu/lb
Discharge mass flow is 18.9 lb/hr.
The mixture isentropic enthalpy drop to a pressure of 0 psig is 1192 – 1144 = 48 btu/lb. The exit velocity from a nozzle with this enthalpy drop is 1550 ft/sec. The energy of the mixture is 48 btu/lb x 18.9 lb/hr = 907 btu/hr
The isentropic enthalpy drop from the 120 psig and 351 F to a pressure of 0 psig is 1192 – 1045 = 147 btu/lbm. The exit velocity from a nozzle with this enthalpy drop is 2,713 ft/sec. The energy of the steam is 147 btu/lb x 11.7 lb/hr = 1720 btu/hr.
This model jet pump was designed to move higher flows with less steam and works better. The extra mass flow and lower velocity increases the nozzle size or quantity which can make a substantial difference on small turbines.
I'll read the discussions on momentum exchange and respond.
Edited By Turbine Guy on 13/03/2019 19:00:55
|Turbine Guy||13/03/2019 23:13:27|
|68 forum posts|
The equation I use for the power of an ideal impulse turbine might help show the effect of changes.
Increasing the mass flow or spouting velocity without any other changes will always increase the power. Increasing the rotor speed up to a maximum of ½ the spouting velocity will also increase power in the ideal case but rotational losses have to be dealt with in real turbines.
What I found in the last post is starting to convince me using an ejector might actually help. I'm going to look at this a little more.
I appreciate you comments on my methodical approach and machining the tiny blades. I enjoy the engineering and design as much as making the models. Because of the cost of the tools required and my limited machining skills, It will be a while before I try making the model Terry turbine. I have an improved version of the open pocket rotor I plan to make next.
|Turbine Guy||14/03/2019 18:14:21|
|68 forum posts|
I finally have enough information to give an indication of how effective using an ejector could be for small turbines. A Penberthy GL ½ A ejector has the following performance.
With this information I can use the following chart to estimate the performance with and without the ejector. This chart estimates the maximum performance for the given specific speed Ns and specific diameter Ds. It is intended to be used by first deciding the speed you would like to run and then calculating the specific speed Ns. You then use Ns to find the Ds that gives the best efficiency. You can then find the optimum rotor diameter from the Ds. I am using a turbine speed of 10,000 rpm for this comparison.
For the ejector, Q3 = 0.095 ft^3/sec, Had = 24,896 ft-lb/lb, and Ns = 1.56
Without the ejector, Q3 = 0.055 ft^3/sec, Had = 108,142 ft-lb/lb, and Ns = 0.39
Using the ejector, the efficiency of the turbine is double that of the turbine without the ejector. The ejector definitely improves the transfer of energy in the turbine. However, the discharge mixture from the ejector has so much less energy that the power ends up almost the same.
|Turbine Guy||15/03/2019 12:05:30|
|68 forum posts|
The heavy solid lines in the chart of the preceding post are for axial turbines. The dashed lines are for Terry turbines. The thin solid lines are for Drag turbines. The drag turbines are like turbine pumps. The blades circulate the flow in a way that increases the drag force on the rotor. The units for the volume flow are ft*3/sec and for the gas density lb/ft^3.
Edited By Turbine Guy on 15/03/2019 12:08:55
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