Question

To defog therear window of an automobile, a very thin
transparent heating element is attached to the inner surface of the window. A Uniform
heat flux of 1300 W/m2 is provided to the heating element for defogging a rear window
with thickness of 5 mm. The interior temperature of the automobile is 22°C and the
convection heat transfer coefficient is 15 W/m2·K. The outside ambient temperature is
−5°C and the convection heat transfer coefficient is 100 W/m2·K. If the thermal
conductivity of the window is 1.2 W/m·K, determine the inner surface temperature of the
window. [Ans.: 14.9 0C ]

Answer 1

General guidance

Concepts and reason

The concepts required to solve the given problem are the heat transfer, conduction, convection, and thermal resistance.

Heat transfer: It is defined as the transfer of heat between two bodies by virtue of the temperature gradient in-between the two bodies. The heat transfer is always from higher temperature body to the lower temperature body.

Conduction:

It is defined as the transfer of heat from higher temperature body to lower temperature body due to direct contact in-between both the bodies.

Convection:

It is defined as the transfer of heat energy from the higher temperature body to lower temperature body due to the bulk fluid movement. The fluid medium is used here to transfer heat energy.

Thermal resistance: It is the resistance offered to the flow of heat by the medium through which the heat transfer takes place.

Consider the given steady state heat transfer process and draw its thermal circuit. Next, determine the thermal resistance for the associated convective and conductive heat transfer. Finally, use the concept of heat transfer through a thermal circuit and formulate a heat balance equation for the given steady state heat transfer process to compute ${T_1}$.

Fundamentals

For a $1$-D steady state heat transfer, write the expression to calculate the heat flux across the surface of wall through conduction.

${\\dot q_{cond}} = - k\\frac{{dT}}{{dx}}$

Here, ${\\dot q_{cond}}$ is heat flux across the wall through conduction, $k$ is thermal conductivity of wall, $dT$ is the temperature difference across the wall and $dx$ is the thickness of the wall.

For a given thickness of the slab, the above equation is rewritten as,

${\\dot q_{cond}} = k\\frac{{\\Delta T}}{L}$

Here, $\\Delta T$ is the temperature difference and $L$ is the thickness of the slab.

Write the expression to calculate the heat flux through convection.

${\\dot q_{conv}} = h\\Delta T$

Here, ${\\dot q_{conv}}$ is the heat flux from the surface through convection, $h$is the convective transfer coefficient and $\\Delta T$ is the temperature difference between the fluid and the surface of the wall.

Write the expression of thermal resistance for conduction.

${R_{cond}} = \\frac{L}{{KA}}$

Here, $A$ is the cross-sectional area normal to the direction of flow of heat energy and $K$ is the thermal conductivity of the material.

For a unit surface area the equation is written as,

${R_{cond}} = \\frac{L}{K}$

Write the expression of thermal resistance for convection.

${R_{conv}} = \\frac{1}{{hA}}$

For a unit surface area the above equation is written as,

${R_{conv}} = \\frac{1}{h}$

Write the expression of rate of heat transfer through a combined convection and conduction.

$\\dot Q = \\frac{{\\Delta T}}{{{R_{eq}}}}$

Here, ${R_{eq}}$ is the equivalent thermal resistance due to combined convection and conduction.

Step-by-step

Step 1 of 2

Consider the steady state one dimensional heat flow with negligible radiation effects and draw the thermal circuit for the given heat transfer process.

L=5mm Glass Outside air Inside air T 22\u00b0C window 2,0 Heater T. ond,1-1 tors, 2-0 \u0424.

In the above diagram, ${\\dot q_h}$ is the uniform heat flux to the heating element, ${T_{\\infty ,i}}$ is the air temperature inside the automobile, ${T_{\\infty ,o}}$ is the ambient temperature outside the automobile, ${T_2}$ is the outer surface temperature of the glass, ${R_{conv,i{\\rm{ - 1}}}}$ is the convective thermal resistance for the heat transfer in-between the air inside and the heating element.

Also, ${R_{conv,2{\\rm{ - }}o}}$ is the convective thermal resistance for the heat transfer in-between the air outside and the glass, and ${R_{cond,1{\\rm{ - 2}}}}$ is the conductive thermal resistance for the heat transfer through the glass of thickness $L$.

Calculate the thermal resistance for convective heat transfer in-between the air inside and the heating element for a unit surface area.

${R_{conv,i{\\rm{ - 1}}}} = \\frac{1}{{{h_i}}}$

Here, ${h_i}$ is the heat transfer coefficient for the air inside the automobile.

Calculate the thermal resistance for convective heat transfer in-between the air outside and the glass for a unit surface area.

${R_{conv,2{\\rm{ - }}o}} = \\frac{1}{{{h_o}}}$

Here, ${h_o}$ is the heat transfer coefficient for the air outside the automobile.

Calculate the thermal resistance for conductive heat transfer in-between the air outside and the glass for a unit surface area.

${R_{cond,1{\\rm{ - 2}}}} = \\frac{L}{K}$

Here, $K$ is the thermal conductivity of the glass.

Explanation | Common mistakes | Hint for next step

The heat transfer takes place in-between the air inside the automobile and the heating element through convection; also heat transfer takes places through the glass due to conduction, while another heat transfer process is carried out in-between the ambient air and the glass through convection.

Consider the resistance of the thin heating element is negligible, such that the inner surface temperature of the glass is ${T_1}$. Thus, draw the thermal circuit for the flow of heat energy from inside air to the outside air and compute the thermal resistance for the above heat transfer processes.

Step 2 of 2

Write the expression for heat flux from the inside air to the heating element.

${\\dot q_{i{\\rm{ - 1}}}} = \\frac{{{T_{\\infty ,i}} - {T_1}}}{{{R_{conv,i{\\rm{ - 1}}}}}}$

Write the heat flux from the heating element to the outside air.

${\\dot q_{{\\rm{1 - }}o}} = \\frac{{{T_1} - {T_{\\infty ,o}}}}{{{R_{conv,2{\\rm{ - }}o}} + {R_{cond,1{\\rm{ - 2}}}}}}$

Write the heat balance equation for the given heat transfer process.

${\\dot q_{i{\\rm{ - 1}}}} + {\\dot q_h} = {\\dot q_{{\\rm{1 - }}o}}$

Here, ${\\dot q_{i{\\rm{ - 1}}}}$ is the heat flux from the inside air to the heating element and ${\\dot q_{{\\rm{1 - }}o}}$ is the heat flux from the heating element to the outside air.

Substitute $\\left( {{T_{\\infty ,i}} - {T_1}} \\right)/\\left( {{R_{conv,i{\\rm{ - 1}}}}} \\right)$ for ${\\dot q_{i{\\rm{ - 1}}}}$ and $\\left( {{T_1} - {T_{\\infty ,o}}} \\right)/\\left( {{R_{conv,2{\\rm{ - }}o}} + {R_{cond,1{\\rm{ - 2}}}}} \\right)$ for ${\\dot q_{{\\rm{1 - }}o}}$.

$\\frac{{{T_{\\infty ,i}} - {T_1}}}{{{R_{conv,i{\\rm{ - 1}}}}}} + {\\dot q_h} = \\frac{{{T_1} - {T_{\\infty ,o}}}}{{{R_{conv,2{\\rm{ - }}o}} + {R_{cond,1{\\rm{ - 2}}}}}}$

Substitute $\\left( {1/{h_i}} \\right)$ for ${R_{conv,i{\\rm{ - 1}}}}$, $\\left( {L/K} \\right)$ for ${R_{cond,1{\\rm{ - 2}}}}$, and $\\left( {1/{h_o}} \\right)$ for ${R_{conv,2{\\rm{ - }}o}}$.

$\\begin{array}{l}\\\\\\frac{{{T_{\\infty ,i}} - {T_1}}}{{\\left( {1/{h_i}} \\right)}} + {{\\dot q}_h} = \\frac{{{T_1} - {T_{\\infty ,o}}}}{{\\left( {1/{h_o}} \\right) + \\left( {L/K} \\right)}}\\\\\\\\{h_i}{T_{\\infty ,i}} - {h_i}{T_1} + {{\\dot q}_h} = \\frac{{{T_1}}}{{\\left( {1/{h_o}} \\right) + \\left( {L/K} \\right)}} - \\frac{{{T_{\\infty ,o}}}}{{\\left( {1/{h_o}} \\right) + \\left( {L/K} \\right)}}\\\\\\\\{h_i}{T_{\\infty ,i}} + {{\\dot q}_h} + \\frac{{{T_{\\infty ,o}}}}{{\\left( {1/{h_o}} \\right) + \\left( {L/K} \\right)}} = \\frac{{{T_1}}}{{\\left( {1/{h_o}} \\right) + \\left( {L/K} \\right)}} + {h_i}{T_1}\\\\\\\\{h_i}{T_{\\infty ,i}} + {{\\dot q}_h} + \\frac{{{T_{\\infty ,o}}}}{{\\left( {1/{h_o}} \\right) + \\left( {L/K} \\right)}} = {T_1}\\left( {\\frac{1}{{\\left( {1/{h_o}} \\right) + \\left( {L/K} \\right)}} + {h_i}} \\right)\\\\\\end{array}$

Further solve as,

${T_1} = \\frac{{{h_i}{T_{\\infty ,i}} + {{\\dot q}_h} + \\frac{{{T_{\\infty ,o}}}}{{\\left( {1/{h_o}} \\right) + \\left( {L/K} \\right)}}}}{{\\left( {\\frac{1}{{\\left( {1/{h_o}} \\right) + \\left( {L/K} \\right)}} + {h_i}} \\right)}}$

Substitute $22^\\circ {\\rm{C}}$ for ${T_{\\infty ,i}}$, $- 5^\\circ {\\rm{C}}$ for ${T_{\\infty ,o}}$, $5{\\rm{ mm}}$ for $L$, $1.2{\\rm{ W/m}} \\cdot {\\rm{K}}$ for $K$, $15{\\rm{ W/}}{{\\rm{m}}^2} \\cdot {\\rm{K}}$ for ${h_i}$, $100{\\rm{ W/}}{{\\rm{m}}^2} \\cdot {\\rm{K}}$ for ${h_o}$, and $1300{\\rm{ W/}}{{\\rm{m}}^2}$ for ${\\dot q_h}$.

$\\begin{array}{c}\\\\{T_1} = \\frac{{\\left( {15{\\rm{ W/}}{{\\rm{m}}^2} \\cdot {\\rm{K}}} \\right)\\left( {22 + 273} \\right){\\rm{K}} + 1300{\\rm{ W/}}{{\\rm{m}}^2} + \\left( {\\frac{{\\left( { - 5 + 273} \\right){\\rm{K}}}}{{\\frac{1}{{100{\\rm{ W/}}{{\\rm{m}}^2} \\cdot {\\rm{K}}}} + \\frac{{5{\\rm{ mm}}\\left( {1{\\rm{ m}}/{{10}^3}{\\rm{ mm}}} \\right)}}{{1.2{\\rm{ W/m}} \\cdot {\\rm{K}}}}}}} \\right)}}{{\\left( {\\frac{1}{{\\frac{1}{{100{\\rm{ W/}}{{\\rm{m}}^2} \\cdot {\\rm{K}}}} + \\frac{{5{\\rm{ mm}}\\left( {1{\\rm{ m}}/{{10}^3}{\\rm{ mm}}} \\right)}}{{1.2{\\rm{ W/m}} \\cdot {\\rm{K}}}}}} + 15{\\rm{ W/}}{{\\rm{m}}^2} \\cdot {\\rm{K}}} \\right)}}\\\\\\\\ = {\\rm{287}}{\\rm{.92 K}}\\\\\\\\ = \\left( {{\\rm{287}}{\\rm{.92}} - 273} \\right)^\\circ {\\rm{C}}\\\\\\\\ = 14.92^\\circ {\\rm{C}}\\\\\\end{array}$

The temperature on the inner surface of the glass is $14.92^\\circ {\\rm{C}}$.

The heat energy flows from the inside air at $22^\\circ {\\rm{C}}$ to the inner surface temperature ${T_1}$ of the glass. Further, through the heat flux provided to the heating element, heat is conducted into the glass and then the heat is further dissipated into the outside atmosphere.

The heat transfer from inner surface temperature to the outside air is due to combined conduction through the glass and the convection heat transfer in-between the outer surface of the glass and the outside air. Thus, formulate the heat transfer equation for the above thermal circuit for the given steady state heat transfer and solve for ${T_1}$.

Answer

The temperature on the inner surface of the glass is $14.92^\\circ {\\rm{C}}$.

Answer only

The temperature on the inner surface of the glass is $14.92^\\circ {\\rm{C}}$.