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//Example18.6//
a=8.9*10^4;//(amperes/m)(webers/m^2) // Area
mprintf("a= %e (amperes.webers)/m^3",a)
//one ampere weber is equal to 1joule. The area is then a volume density of energy ,or
e=8.9*10^4//J/m^3 //energy loss
b= 10^-3; //As 1Kilogram = 10^3 gram
e1=e*b
mprintf("\ne1 = %i kJ/m^3 (per cycle) (As 1Kilogram = 10^3 garm)",e1)
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