Actual Efficiency

I pre-heated the furnace and crucible to 750 °C and weighed out 1.005 Kg of aluminum to charge into the hot crucible.  I placed the pieces into the crucible one at a time and then replaced the lid.

As the furnace heated back up, I noticed smoke coming from the lid.  I checked on it and there was a burned circle on top of the firebricks and burned stains on the fiberboard.  The gasket material is supposed to be good to 1800 °F continuously, and 2300 °F for brief periods.  When I set the lid down on bricks some liquid dripped from it.  I suspect the sodium silicate was not completely dried.

After about 15 minutes I began to lift the lid and peek in on the heating charge to determine when it had melted.  I found it melted at the 40 minute mark, but it may have melted as early as 35 minutes.

So, I plugged the mass and the time into my equations and found an efficiency between 12% and 14%.  So, we will just call it 13% for now.  The true efficiency is probably a little higher - my lifting the lid every 5 minutes did not help with efficiency, and the furnace does not use 3200 watts continuously, it modulates on and off as it approaches the setpoint temperature.

I really don't have any other furnaces to compare this one to, so the efficiency is acceptable to me.

Calculating Efficiency

To calculate how efficient the electric resistance foundry furnace is, we need to measure how much (electrical) energy is supplied to the furnace and compare it to how much energy (heat) is required to melt a known mass of aluminum.  The furnace will never be 100% efficient because part of the supplied energy goes into heating the furnace materials, and some of it will escape from the furnace.

The furnace consumes 3200 watts of electrical energy.  The electricity is converted to heat with 100% efficiency.  So the furnace generates 3200 watts of electrical energy, or 3200 Joules per second.

Before we can melt the aluminum, we will have to heat it from room temperature (T1) to the melting point (T2).  Assume that the room temperature is 20 °C.  Aluminum melts at 660 °C.   We will assume that we have 1 kilogram of aluminum (the mass, m) in the crucible.

The specific heat capacity of SOLID aluminum is 900 Joules/kilogram °C (c).  We will use the following equation to calculate the heat required to bring the 1 kilogram of aluminum to the melting point (Q1):  

Q1 = mc(T2 - T1)

Q1 = 1 kilogram (900 Joules/kilogram °C) (660 °C - 20 °C)

Q1 = 576,000 Joules or 576 kiloJoules

After the metal reaches the melting point, additional heat must be added to it make it change phase from solid to liquid.  This heat (Q2) is called the latent heat of fusion (Lf), and for aluminum is 390 kiloJoules per kilogram.

Q2 = mLf 

Q2 = 1 kilogram (390,000 Joules/kilogram)

Q2 = 390,000 Joules or 390 kiloJoules

The total amount of heat needed to melt 1 kilogram of aluminum is the heat needed to raise its temperature to the melting point (Q1), plus the latent heat of fusion (Q2) needed to make it melt and change phase to a liquid.

QT = Q1 + Q2

QT = 576 kiloJoules + 390 kiloJoules

QT = 966 kiloJoules

If additional heat is supplied to the aluminum to raise its temperature above the melting point for pouring (Q3), we can use the first equation again to find out how much heat is required.  The specific heat capacity of LIQUID aluminum is 1180 Joules/kilogram °C (c).  Assume we heated the aluminum to 700 °C.

Q3 = mc(T2 - T1)

Q3 = 1 kilogram (1180 Joules/kilogram °C) (700 °C - 660 °C)

Q3 = 47,200 Joules or 47.2 kiloJoules

So, to bring 1 kilogram of solid aluminum from room temperature to liquid at 700 °C, you add the heat required to raise the aluminum to the melting point (Q1), plus the latent heat required to cause the phase change (Q2), plus the heat added to the liquid to bring the temperature to 700 °C (Q3).

QT = Q1 + Q2 + Q3

QT = 576 kiloJoules + 390 kiloJoules + 47.2 kiloJoules

QT = 1,013.2 kiloJoules or 1.013 megaJoules

The first melt will require more energy than subsequent melts because the furnace materials and the crucible will have to be heated up for that first melt.  It will be more accurate to pre-heat the furnace and crucible, then add the mass of room temperature aluminum for the test.

If the furnace were 100% efficient, all of the energy supplied to it would go into melting the aluminum.  The furnace supplies 3200 Joules per second.  966,000 Joules are required to completely melt 1 kilogram of aluminum.  It would take 966,000 Joules divided by 3200 Joules per second, or 302 seconds, about 5 minutes, to melt 1 kilogram of aluminum.

I am sure that it will take much more time to melt the aluminum.  Assuming it takes 40 minutes to melt 1 kilogram of aluminum, we will supply 3200 Joules/second times 2,400 seconds, or 7,680,000 Joules (7,680 kiloJoules) of energy.

The efficiency of the furnace (Eff) is amount of energy required to melt the aluminum (Q1) divided by the amount of energy supplied to the furnace (Q2), times 100%.

Eff = (Q1/Q2)x100%

Eff =(966 kiloJoules / 7,680 kiloJoules) x 100%

Eff = 12.6% 

I intend to pre-heat the furnace and crucible, add 1 kilogram of room temperature aluminum to the crucible, and then time how long it takes for the charge to completely melt. There will some uncertainty in this measurement, but it will give me an idea of how of the supplied energy escapes or is wasted.

Early testing was encouraging - the steel shell remained cool enough to comfortably touch while a test charge was melted.  So hopefully this furnace will be more than 13% efficient.

Finished Lid

I completed the lid.  I placed a piece of 1/2-inch-thick fiberboard on top of the firebrick and ceramic wool insulation.  On top of that I layed the firebrick.  Then I placed a ring of fiberboard on top of the brick to cushion it and fastened steel clips to the steel drum via 5/16-inch rivnuts and bolts.

Next I painted sodium silicate on to the brick in a circle next to the fiberboard and I painted the end of the fiberboard to seal it.  This dried to a shiny, glossy, finish.  I can see why they call it "water glass".

After the sodium silicate was dry, I applied another layer of wet sodium silicate and then pressed my braided ceramic rope gasket into the circle formed by the fiberboard.  I went around the outside of the gasket applying more sodium silicate to bond the rope gasket to the fiberboard securely.

Now that it is dry it does not pull out.

To check the fit, I rubbed sidewalk chalk on top of the bricks that form the heating chamber, then I layed the lid on top and twisted it from side-to-side.  I lifted the lid off and confirmed that I had a continuous circle of colored chalk transferred to the gasket.  Yes, it will seal.

Now I am ready to begin testing for efficiency.

Change Of Plans

Originally I had planned to use a lid lifting mechanism to lift the lid and then pivot the lid off of the furnace body, but I was not able to make this work in practice because the lid sagged too much.

So, I have decided that the lid will be a lift-off lid for now.

 

I cut four pieces of fire brick so that the firebrick, plus a sheet of fiberboard, plus the firebrick top, plus another layer of fiberboard, will come up to the beginning of the rolling ring where the drum was cut.

 

Between the firebrick I have packed in ceramic wool insulation.

 

I marked the tops of the eight steel clips that hold the fiberboard and firebrick in the lid, then I pulled the fiberboard and brick back out so that I may mark and drill the holes in the shell.  5/16" x 3/4" long bolts will screw into the rivnuts expanded into the clips through holes drilled in the shell.

Once this is done, I will cover the jagged cut edge of the steel shell with vinyl edge trim.