Question

6. Dr. Li boils water using a kettle with a 1.5 kW Nichrome (80% Ni and 20% Cr) heating element (resister heater). The diameter and length of this heating element is 8 mm and 20 cm, respectively. When exposed to liquid water, the convection heat transfer between water and heating element is 800 W m2K . Please find: (25 Points) (1) The surface temperature and maximum temperature of the heating element when water is boiling in the kettle in Dr. Li’s office, ESB 751. (2) The surface temperature and maximum temperature when all liquid water has been evaporated and the heating element is exposed completely to the superheated vapor. You can assume the convection heat transfer between superheated water vapor and heating element is 24 W m2K . You can assume the temperature of the superheated water vapor is 100 °C. (3) Please justify if the heating element can survive in case (2) without melting.

Answer 1

Assuming that:

The heat generation is uniform throughout the heating element, Nichrome wire.

The cross-sections are insulated and heat transfer is taking place only in the radial direction.

All the heat generated is conducted, there is energy storage.

Now, from the properties of Nichrome:

The melting temperature of Nichrome,
`T_m=1400 ^(\circ)C`

Thermal conductivity,
`K=11.3 W/m^(\circ)C`

Given that:

The power generated by the heating element,
`Q_g=1.5kW=1500W`

Diameter,
`D=8mm=8*10^(-3)m`

So, radius,
`R=4*10^(-3)m`

Length,
`L=20cm=0.2m`

The volume of the Nichrome wire,

`V=\pi R^2L=\pi* (4*10^(-3))^2*0.2=1.005*10^(-5)m^3`

Heat generation rate per unit volume,

`q_g=(Q_g)/(V)=(1500)/(1.005*10^(-5))=149.2*10^(6)W/m^3`

With the assumptions made above, this is the case of heat transfer in one direction.

Let
`T` be the temperature at the radius
`r`.

Now, the heat generated within the cylinder of radius
`r` is conducted in a radially outward direction. i.e

`q_g(\pi r^2)L=-K(2\pi r L)(dT)/(dr)` [where K is the thermal conductivity oof Nichrome]

`\Rightarrow -(q_g)/(2K)rdr=dT`

`\Rightarrow T=-(q_g)/(4K)r^2+ C_0` , where
`C_0` is constant.

Let the surface temperature is
`T_s`, i.e at
`r=R, T=T_s`.

Putting this boundary condition to get
`C_0`, we have

`T=T_s+(q_g)/(4K)(R^2-r^2)`

This is the temperature profile within the Nichrome wire, which is maximum at
`r=0`.

So, the maximum temperature,

`T_(max)=T_s+(q_g)/(4K)R^2\;\cdots(i)`

(1) The water is boiling, to the temperature of the water is,
`T_b=100^(\circ)C`.

The total heat generated within the heating element is convected to the water from the surface. i.e

`Q_g=h_w(2\pi RL)(T_s-T_b)`

where,
`h_w=800W/m^2K=800W/m^2^(\circ)C` is the convective heat transfer constant (Given).

`\Rightarrow 1500=800* (2\pi*4*10^(-3)* 0.2(T_s-100)`

`\Rightarrow T_s=100+373=473^(\circ)C`

So, the surface temperature is
`473^(\circ)C`.

From equation (i), the maximum temperature is at the center of the wire which is

`T_(max)=473+(149.2*10^(6))/(4*11.3)(4*10^(-3))^2`

`\Rightarrow T_(max)=473+53=526^(\circ)C`

(2) In this case, the temperature of the superheated water vapor,
`T_b = 100^(\circ)C` (Given)

The heat transfer coefficient between the superheated water vapor and the heating surface is,
`h_v=24W/mK=24W/m^(\circ)C`.

Similarly,
`Q_g=h_v(2\pi RL)(T_s-T_b)`

`\Rightarrow 1500=24* (2\pi*4*10^(-3)* 0.2(T_s-100)`

`\Rightarrow T_s=100+12434=12534^(\circ)C`

So, the surface temperature is
`12534^(\circ)C`.

From equation (i), the maximum temperature is at the center of the wire which is

`T_(max)=12534+(149.2*10^(6))/(4*11.3)(4*10^(-3))^2`

`\Rightarrow T_(max)=12534+53=12587^(\circ)C`

(3) The melting temperature of Chromium is
`1400 ^(\circ)C`.

So, the 1st case when the heating element is surrounded by water is safe as the maximum temperature within the element is below the melting temperature.

But, it the 2nd case, the heating element will melt out as the maximum temperature is much higher than the melting temperature of the element.