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Cylinder Heated by Induction

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A Cylinder Heated by Induction 

? 2009 Waterloo Maple Inc 



In this worksheet we consider a long metal cylinder heated by induction, as shown below. This has a magnetic field applied parallel to the axis of the cylinder and a constraint on the current density at a particular depth. We demonstrate how the temperature depends on time and depth from the surface in a very long cylinder. This is calculated using the current density, the power density and the partial differential equation of heat conduction. 




In induction, the heating is caused by eddy currents, which themselves give rise to alternating magnetic fields. Because of the skin effect and depending on the frequency of the magnetic field, the highest current density exists directly under the surface of the heated work piece. It decreases rapidly with increasing depth. We calculate the effective magnetic field intensity on the surface required to reach a given temperature in a given time. 



Problem Definition 

This example considers an aluminium cylinder of 6cm diameter, heated by induction. The problem we address is: 


What must the value of the effective magnetic field intensity on the surface (HS, Unit(`/`(`*`('A'), `*`('m')))) be to reach a temperature on the surface of 800Unit('K') in 60 seconds? Additionally, the current density (J, Unit(`/`(`*`('A'), `*`(`^`('m', 2))))) within `/`(1, 2) Unit('cm') distance of the surface must have a value of at least half its value on the surface 


The Material Properties of Aluminium 

Electric conductivity (sigma) 

`+`(`*`(3.82, `*`(`^`(10, 7), `*`(Unit(`/`(1, `*`('Omega', `*`('m'))))))))
`+`(`*`(3.82, `*`(`^`(10, 7), `*`(Unit(`/`(1, `*`('Omega', `*`('m'))))))))
`+`(`*`(3.82, `*`(`^`(10, 7), `*`(Unit(`/`(1, `*`('Omega', `*`('m'))))))))

Relative permeability (kappa) 


Thermal conductivity (lambda) 

`+`(`*`(237, `*`(Unit(`/`(`*`('W'), `*`('K', `*`('m'))))))) 

Heat capacity (c) 

`+`(`*`(888, `*`(Unit(`/`(`*`('J'), `*`('kg', `*`('K')))))))
`+`(`*`(888, `*`(Unit(`/`(`*`('J'), `*`('kg', `*`('K')))))))

Density (rho) 

`+`(`*`(2700, `*`(Unit(`/`(`*`('kg'), `*`(`^`('m', 3))))))) 

Also the radius of our cylinder (R) 

`+`(`*`(0.3e-1, `*`(Unit('m')))) 


In this worksheet we use the character 'A' at the end of names, to show when we are using the numerical values, rather than the symbolic parameter. 


Let's define the values in Maple: 





As the cylinder is long, we use the approximation of no end effects, ie equivalent to a cylinder with infinite length. Also, the magnetic field produced by a surrounding coil is assumed to be homogeneous along the entire length of the cylinder. In this case, the distribution of the effective current density (J) in the cylinder is given by the formula: 


`:=`(J, proc (r, delta) options operator, arrow; `/`(`*`(HS, `*`(`+`(`-`(KelvinBer(1, `/`(`*`(sqrt(2), `*`(r)), `*`(delta)))), `-`(KelvinBei(1, `/`(`*`(sqrt(2), `*`(r)), `*`(delta)))), `-`(`*`(`+`(I),...
`:=`(J, proc (r, delta) options operator, arrow; `/`(`*`(HS, `*`(`+`(`-`(KelvinBer(1, `/`(`*`(sqrt(2), `*`(r)), `*`(delta)))), `-`(KelvinBei(1, `/`(`*`(sqrt(2), `*`(r)), `*`(delta)))), `-`(`*`(`+`(I),...


The constant delta is usually known as "depth of penetration" and is defined as: 




where omega Unit(`/`(1, `*`('sec'))) is the circular frequency of the magnetic field. 

`:=`(PD, proc (r, delta) options operator, arrow; `/`(`*`(evalc(`*`(`^`(abs(J(r, delta, R)), 2)))), `*`(sigma)) end proc); -1 

simplify(PD(r, delta)) 

`+`(`/`(`*`(2, `*`(`+`(`*`(`^`(KelvinBei(1, `/`(`*`(`^`(2, `/`(1, 2)), `*`(r)), `*`(delta))), 2)), `*`(`^`(KelvinBer(1, `/`(`*`(`^`(2, `/`(1, 2)), `*`(r)), `*`(delta))), 2))), `*`(`^`(HS, 2)))), `*`(s... (3.1)


For the transient temperature distribution, a solution of the partial differential equation for heat conduction in polar coordinates is needed.  


diff(T(r, t), t) = `/`(`*`(lambda, `*`(`+`(diff(T(r, t), r, r), `/`(`*`(diff(T(r, t), r)), `*`(r))))), `*`(rho, `*`(c))); -1 

T(r, t): temperature Unit('K') 


If the boundary conditions are assumed to approximate to adiabatic conditions (that is, no heat is dissipating from the surface of the cylinder to the environment) and if all the heat is entering via the surface of the cylinder, then an analytic solution can be found: 


`:=`(T, proc (t, r) options operator, arrow; `+`(`#msub(mi(
`:=`(T, proc (t, r) options operator, arrow; `+`(`#msub(mi(


`#msub(mi(: initial temperature at time = 0 seconds 

PA: heat entering on the surface Unit(`/`(`*`('W'), `*`(`^`('m', 2)))) 

R: radius of cylinder  


The sum should to be taken from i = 1..infinity, but as the exponential function becomes very small with increasing time, it is sufficient in most cases to restrict the sum to the first 50 or so elements (n = 50). The assumption of a surface heat source (PA) is reasonable, because the skin effect causes the heat to be produced in a shallow surface layer. To find PA, the power density must be integrated on the interval r = 0..R. 



First, the suitable frequency (`#msub(mi() has to be found, where the current density in 0.5cm depth is about half its value on the surface. 



1004.131532 (4.1)

Next, the surface power loss (PAA) is calculated: 



int(`+`(`*`(0.2290944224e-5, `*`(abs(`*`(`^`(HS, 2), `*`(`^`(`+`(`-`(`*`(1., `*`(KelvinBer(1., `+`(`*`(219.5492289, `*`(r))))))), `-`(`*`(1., `*`(KelvinBei(1., `+`(`*`(219.5492289, `*`(r))))))), `-`(`...
int(`+`(`*`(0.2290944224e-5, `*`(abs(`*`(`^`(HS, 2), `*`(`^`(`+`(`-`(`*`(1., `*`(KelvinBer(1., `+`(`*`(219.5492289, `*`(r))))))), `-`(`*`(1., `*`(KelvinBei(1., `+`(`*`(219.5492289, `*`(r))))))), `-`(`...


How big must the value of the applied magnetic field (HSA) be to heat the cylinder to a temperature of 800K in 60sec? This is calculated with 'fsolve': 


`:=`(HSA, fsolve(subs({PA = `#msub(mi(
`:=`(HSA, fsolve(subs({PA = `#msub(mi(

266406.5148 (4.3)

 The magnetic field has a value of `+`(`*`(2.67, `*`(`^`(10, 5)))) Unit(`/`(`*`('A'), `*`('m'))). This is equal to the amplitude of the magnetic induction B of about 0.48T 


The surface power loss evaluates to: 


`:=`(PAA, evalf(int(simplify(subs({HS = HSA, R = `#msub(mi( 

292701.7487 (4.4)



Graphical Display of the Solution 

Now the temperature increase is plotted as a function of time: 









The calculation of the temperature rise due to induction effects is important in many applications in industry. In this example, a short introduction is given on how such computations can be done with Maple. It is demonstrated how the temperature with respect to time and location in a very long cylinder is calculated using the current density, the power density and the partial differential equation of heat conduction. 



Carslaw, H.S.; Jaeger, J. C.: Conduction of heat in solids; 2nd. ed., Oxford University PressJoos, Georg: Lehrbuch der theoretischen Physik; 8. Auflage, Leipzig: Akademische Verlagsgesellschaft Geest & Portig KG 

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