# Efficiency of heating up aluminum vs steel by induction

Karl Karlsson
Summary: In a heating by induction experiment performed, the idea was to determine the efficiency of heating up a small steel cylinder, an aluminum cylinder and then compare the two efficiencies. The cylinder was surrounded by a metall coil that alternating current was going through, alternating with the resonance frequency. The surprising thing about the result of the lab was that according to our measurements, the efficiency of aluminum was more than twice as high as the efficiency of aluminum.

In a heating by induction experiment performed, the idea was to determine the efficiency of heating up a small steel cylinder, an aluminum cylinder and then compare the two efficiencies. The cylinder was surrounded by a metall coil that was connected to a power supply which generated an alternating current, alternating with the resonance frequency for the materials. The surprising thing about the result of the lab was that according to our measurements, the heating efficiency for aluminum (about 20%) was more than twice as high as the heating efficiency for steel (about 10%). Due to the steel cylinder used being a ferromagnetic (higher permeability?) material, should not the heating efficiency for steel be higher than that of aluminum?

Mentor
The surprising thing about the result of the lab was that according to our measurements, the efficiency of aluminum was more than twice as high as the efficiency of aluminum.
Wow, that really is surprising! Homework Helper
Gold Member
2022 Award
according to our measurements
Please explain what measurements were taken and how you calculated the efficiencies.

Karl Karlsson
Wow, that really is surprising! Good that I am not the only one who thinks that! Do you know why? Of course there are sources of error, but that big of a difference in efficiency makes me doubt that it is the reason

Last edited:
Karl Karlsson
Please explain what measurements were taken and how you calculated the efficiencies.

The specific heat capacities for the volumes of the cylinders of the different materials were given and the power was set to be about 20W which i got by adjusting the voltage and current from the power supply. The efficiency is equal to: μ = c*V*ΔT/(Δt*I*U), the volume was exactly the same for all the cylinders, U*I is the avarage effect from the power supply, c is the specific heat capacities for the different materials (with the unit J/(K*m^3)). From steel I got the efficiency to about c*V*ΔT/(Δt*I*U)=0.1, using a power of almost U*I=20W, whereas for aluminum i got the efficiency to about 0.2 also using a power of almost 20W. What could be the reason fot this?

Last edited:
Homework Helper
Gold Member
2022 Award
Good that I am not the only one who thinks that! Do you know why? Of course there are sources of error, but that big of a difference in efficiency makes me doubt that it is the reason
You missed the sarcasm. Read carefully your text as quoted in post #2.

Homework Helper
Gold Member
2022 Award
The specific heat capacities for the volumes of the cylinders of the different materials were given and the power was set to be about 20W which i got by adjusting the voltage and current from the power supply. The efficiency is equal to: μ = c*V*ΔT/(Δt*I*U), the volume was exactly the same for all the cylinders, U*I is the avarage effect from the power supply, c is the specific heat capacities for the different materials (with the unit J/(K*m^3)). From steel I got the efficiency to about c*V*ΔT/(Δt*I*U)=0.1, using a power of almost U*I=20W, whereas for aluminum i got the efficiency to about 0.2 also using a power of almost 20W. What could be the reason fot this?
Nothing jumps out, but take a look at https://en.m.wikipedia.org/wiki/Induction_heating#Details. Maybe the frequency was not high enough for the a/d ratio.

Karl Karlsson
Nothing jumps out, but take a look at https://en.m.wikipedia.org/wiki/Induction_heating#Details. Maybe the frequency was not high enough for the a/d ratio.
They are calculating reference depth in centimetres in the linked wikipedia page so i will do that as well.I have already checked that but using that formula gives considering:

Relative depth d = 5000*(ρ/(μ*f))^(1/2)
diameter, a=10^(-2)m

Steel:
Resistivity: 6.9*10^(-9)Ω*cm
Relative permeability: 100
Frequency: 118000Hz
a/d=8271

Aluminum:
Resistivity: 2.65*10^(-10)Ω*cm
Relative permeability: 5000
Frequency: 116000Hz
a/d=4184

The ratio relative depth divide by diameter is higher for steel and should therefore have higher efficiency, right?

Last edited:
Mentor
Good that I am not the only one who thinks that!
Did you read the text of yours that I quoted...? Karl Karlsson
Did you read the text of yours that I quoted...? Haha, what I meant to write was: "The surprising thing about the result of the lab was that according to our measurements, the efficiency of aluminum was more than twice as high as the efficiency of steel."

• berkeman