Question

An air heater may be fabricated by coiling Nichrome wire and passing air in cross flow over the wire. Consider a heater fabricated from wire of diameter D=1 mm, electrical resistivity rhoe=10−6Ω⋅m, thermal conductivity k=25W/m⋅K, and emissivity ε=0.20. The heater is designed to deliver air at a temperature of T[infinity]=50∘C under flow conditions that provide a convection coefficient of h=250W/m2⋅K for the wire. The temperature of the housing that encloses the wire and through which the air flows is Tsur=50∘C. If the maximum allowable temperature of the wire is Tmax=1200∘C, what is the maximum allowable electric current I? If the maximum available voltage is ΔE=110V, what is the corresponding length L of wire that may be used in the heater and the power rating of the heater? Hint: In your solution, assume negligible temperature variations within the wire, but after obtaining the desired results, assess the validity of this assumption.

Answer 1

Assuming that the wire has an uniform temperature, the equivalent convective heat transfer coefficient is given as :

`$h_T= \epsilon \sigma (T_s+T_(surr))(T_s^2 +T^2_(surr))$`

`$h_T= 0.20 * 5.67 * 10^(-8) (1473+323)(1473^2 +323^2)$`

`$h_T=46.3 \ W/m^2 .K$`

The total heat transfer coefficient will be :

`$h_T=(250+46.3) \ W/m^2 .K$`

`$=296.3 \ W/m^2 .K$`

Now calculating the maximum volumetric heat generation :

`$\dot {q}_(max)=(2h_t)/(r_0)(T_s-T_(\infty))$`

`$\dot {q}_(max)=(2* 296.3)/(0.0005)(1200-50)$`

`$= 1.362 * 10^(9) \ W/m^3$`

The heat generation inside the wire is given as :

`$\dot{q} = (I^2R)/(V)$`

Here, R is the resistance of the wire

V is the volume of the wire

∴
`$\dot{q} = (I^2\left( \rho * (L)/(A) \right))/(A * L)$`

`$=(I^2 \rho)/(\left((\pi)/(4)D^2 \right))$`

where, ρ is the resistivity.

`$I_(max)= \left(\frac{\dot{q}_(max)}{\rho} \right)^(1/2) * (\pi)/(4)D^2$`

`$I_(max)= \left((1.36 * 10^9)/(10^(-6)) \right)^(1/2) * (3.14)/(4)(1 * 10^(-3))^2$`

= 28.96 A

Now considering the relation for the current flow through the finite potential difference.

`$E=I_(max) * R$`

`$E=I_(max) * \rho * (L)/(A)$`

`$L=(AE)/(I_(max) \ \rho)$`

`$L=((\pi)/(4) * (1 * 10^(-3))^2 * 110)/(28.96 * 10^(-6))$`

= 2.983 m

Now calculating the power rating of the heater:

`$P= E * I_(max)$`

`$P= 110 * 28.96}$`

= 3185.6 W

= 3.1856 kW