Solar Heat Engines

Low-Cost Tachometer

I bought the cheapest bicycle speedometer I could find at Kmart. I see Amazon has the same model. Amazon also has a Schwinn model for $10 that should work–but I haven’t tested it.

Tachometer Part 2 –I’ve tested the Schwinn Speedometer and some updates on the Bell Speedometer.

Copyright 2008 Doug Conner

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Engine model 3d is up and running!

Engine model 3d has actually been running for over a week now, but I finally am taking time to start getting this website up-to-date. You can see more photos in the photo pages.

I’m in the process of performance testing the engine to see how closely its power output agree with my simulation. There are many sinks for losing power including: bearing friction on the rotating shaft, sliding friction on the piston and displacer shaft, compression leakage around the piston and displacer shaft seal, air friction on the air moving around the displacer, and air friction on the spinning flywheel.

My basic and inexpensive instrumentation for this operation includes a stopwatch and a bicycle speedometer used as a tachometer. Using engineering computations and these two instruments I can derive the friction based on how long it takes the unpowered engine to spin down. Here’s how it works.

A flywheel stores energy proportional to the square of its speed (RPM). You can compute the actual energy stored in the flywheel at any RPM. Selecting two RPMs, a high and a low value, you compute the energy difference between the two RPMs. You spin the engine up unpowered to just above the high RPM and start your stopwatch when the RPM drops to the high RPM. The unpowered engine spins down and you stop the time when it reaches the low RPM. You already know the flywheel work difference between the two RPMs (ft-lb, joules, or whatever energy units you want). Dividing by the time you measured on the stop watch gives you the power (ft-lbs/sec, joules/sec, etc.) or rate at which flywheel work is being used up in friction.

A massive flywheel helps you make the measurements accurately and more easily because the flywheel will slow down more gradually. That’s the reason you see the club with massive bolts swinging on the flywheel above. This setup is for under 100 RPM use (don’t use it for high speeds or it could disintegrate with potential for bodily injury). You probably want the spin-down time to be at least 5 seconds and preferably 10 or more to minimize the timing measurement error. You’ll want to make enough spin down measurements so that you can see what the variation is due to timing and RPM measurement accuracy. You can also average the results to get a more accurate results.

The flywheel spin down test can be used with engine components starting with the flywheel and the main bearings it rides on. You add components one at a time, repeating tests to see the power loss due to each component. Using this method you can measure say the piston sliding friction instead of just knowing the friction of the entire engine.

I’ll be adding material for those without the technical background to make the flywheel computations. I’ll also post the method for using a bicycle speedometer as a tachometer. It’s inexpensive and really simple. I use it to measure the RPM on my variable-speed lathe too.

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Solar Heat

Solar Power—Radiation from the sun received on Earth

Here is some basic information for those interested in converting solar radiation into power. At the average earth-sun distance of 92.9 million miles the solar radiation intensity is:

435 BTU/(ft²-hr) or 1370 watts/meter².  

After going through the atmosphere and reaching the ground the values drop to:

340 BTU/(ft²-hr) or 1070 W/m². 

The above value is what you have to work with at noon on a clear day at the equator. As you depart from those conditions the values decrease. A good ballpark value to use for 40 degrees North latitude appears to be around:

300 BTU/(ft²-hr) or 944 W/m². 

Just so you know there is more to learn about solar radiation, there is the total radiation which is the sum of direct normal radiation (sunlight), diffuse or sky radiation (when you stand in the shade it isn’t dark and your solar calculator still works), and reflected solar radiation (you probably noticed it’s warmer in the sun by the south side of a building).

The above numbers will get you started if you know what to do with them. I’ll talk about heating things up with solar power on a future post so you can figure how much heat you need to change the temperature of things like air, water, or a piece of metal.  

By the way, typical solar cells have about 10% conversion efficiency so you get roughly 94 W/m² out of them. State-of-the-art solar cells are approaching 15-20% efficiency.

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How much heat to raise the temperature?

Everyone knows it takes heat to raise the temperature of something, whether it’s the air in your house or a kettle on the stove. If you want to design heat engines or use solar power to elevate the temperature of something, it helps if you can determine how much heat you need.

To begin with I’ll skip phase changes and just look at heating things that stay in the same phase—solid, liquid, or gas.

The basic equation is:

Q=mc(deltaT)

 where:

Q = heat (units: cal, kcal, Joules, ft-lb, Btu, or kWhr)

m= mass (units: kg or lb)

c= specific heat for the particular material and temperature (units:kJoules/(kg-degC) or Btu/(lb-degF)

deltaT = change in temperature (units: degC, or degF)

To use the above equation correctly you need to make sure you use consistent units.

Below is a table that contains specific heats for various materials at 20 deg C. If you need other materials you can probably find them on the web or in handbooks for engineering, physics, or chemistry. Note that the values change with temperature and may change greatly when the substance goes through a phase change. These values are also for constant pressure. If you heat or cool at constant volume there are other numbers to use and I’ll discuss that at a later date.

I’ll give a few examples:

You want to raise the temperature of 1.5 pounds of aluminum by 50 degF then you’ll need:

Q=1.5lb x 0.22 Btu/(lb-degF) x 50 degF = 16.5 Btu of heat

You want to raise the temperature of 0.1kg of air by 100 degC

Q=0.1kg x 1.02kJ/(kg-degC) x 100degC = 10.2kJ

A few useful conversions:

1kw-hr = 3412 Btu

1 kw-hr = 3600 kJ,

1Btu=1.055kJ

Sources:

Marks’ Handbook for Mechanical Engineers, tenth edition

General Physics, 1984, Giancoli, Douglas

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