Question

1. A house has a composite wall of wood, fiberglass insulation, and plaster board, as indicated below. On a cold winter day the convection heat transfer coefficients are ho = 60 W/m2K and hi = 30 W/m2K. The total wall surface area is 350 m2.

1. (a) Determine the heat transfer rate through the wall in Watts.

2. (b) If the wind were blowing violently, raising ho to 300 W/m2K, determine the percentage

increase in the heat transfer rate.

(c) What is the controlling resistance that determines the heat transfer through the wall?

Glass fiber blanket (28 kg/m3), kp Plaster board, k Plaster board, k, Plywood siding, k, Inside Outside hi, T”, i = 20°C A,, T,,o =-15°C 10 mm ?+I - 100mm-?-20mm

Answer 1

Concepts and reason

The concepts used to solve this problem are heat transfer, equivalent thermal resistance and heat transfer.

Heat transfer, in general, is defined as the flow of thermal energy from one object to another having different temperatures. Basically, there are three modes of heat transfer.

1. Conduction: Heat transfer takes place due to a temperature difference by the mechanism of intermolecular interactions. It can take place without the presence of bulk matter.

2. Convection: Heat transfer takes place due to a temperature difference but requires combined mechanisms of intermolecular interactions and bulk matter transport.

3. Radiation: Heat transfer takes place due to electromagnetic radiation which occurs because of the temperature of the body. Radiation does not need medium of transport.

Thermal resistance: It is a property of a material by which the material is able to resist the heat flow.

Equivalent thermal resistance: It is the sum of all the thermal resistances. The expression for the net thermal resistance is different for series and parallel combination.

Initially, the value of total thermal resistance is calculated by using the concept of thermal resistance. Then, by using the concept of heat transfer, the value of heat transfer rate is calculated. Similarly, the value of heat transfer rate is calculated for the new value of thermal coefficient by using the concept of thermal resistance and the heat transfer rate.

Finally, by using the known value of thermal resistance, find the controlling resistance.

Fundamentals

Write the expression for resistance due to convection.

${R_{{\rm{convection}}}} = \frac{1}{{{h_i}A}}$

Here, ${h_i}$ is the convective heat transfer coefficient and A is the area of heat transfer.

Write the expression for resistance due to conduction through a slab.

${R_{{\rm{conduction}}}} = \frac{{{l_p}}}{{{k_p}A}}$

Here, ${l_p}$ is the length of the slab and ${k_p}$ is the conduction heat transfer coefficient.

Write the expression for heat transfer in the tube.

$q = \frac{{\Delta T}}{{{R_{th}}}}$

Here, $\Delta T$ is the change in temperature, and ${R_{th}}$ is the total thermal resistance.

Write the expression for the total resistance when the resistances are connected in series.

${R_{th}} = {R_1} + {R_2} + {R_3}$

Here, ${R_1}$ , ${R_2}$ and ${R_3}$ are the thermal resistances of different surfaces.

Write the expression for the percentage increase of heat transfer rate.

$\% {\rm{ increase}} = \frac{{{Q_n} - Q}}{Q} \times 100$

Here, ${Q_n}$ and $Q$ are the value of new and old heat transfer rate.

(A)

Consider the inner surface of the plasterboard as $A$ , the outer surface of the plasterboard as $B$ . Similarly, consider the inner surface of the plywood siding as $C$ and the outer surface as $D$ . Draw the diagram showing the different junction $A$ , $B$ , $C$ , $D$ and the direction of flow of heat.

Here, ${h_i}$ , ${h_o}$ are the convective heat transfer coefficient for the inside surface of the house and the outside air respectively. Also, ${T_{\infty ,i}}$ and ${T_{\infty ,o}}$ are the temperature inside the house and the outside of the house respectively. Similarly, ${L_p}$ , ${L_b}$ , and ${L_s}$ are the length of the plasterboard surface, glass fiber blanket surface, and plywood siding surface respectively.

Consider the temperature of each junction as ${T_A}$ , ${T_B}$ , ${T_C}$ and ${T_D}$ . Now, take the value of thermal conductivities of different materials from the table “Thermophysical properties of common materials”. Take the value of conductive heat transfer coefficient for glass fiber blanket.

${k_b} = 0.038{\rm{ }}\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}$

Take the value of conductive heat transfer coefficient for plywood siding.

${k_s} = 0.12{\rm{ }}\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}$

Take the value of conductive heat transfer coefficient for plasterboard.

${k_p} = 0.17{\rm{ }}\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}$

From the Figure (a), it is clear that the heat transfer from the house to the inner surface $A$ occurs through convection. Similarly, in the case of $A$ to $B$ , $B$ to $C$ , and $C$ to $D$ , it occurs through conduction. Finally, heat flows through the surface $D$ to the outside air by convection. Draw the diagram showing the thermal resistance.

Here, ${R_1}$ is the thermal resistance due to convection between the house and the inner surface $A$ , ${R_2}$ , ${R_3}$ , ${R_4}$ are the thermal resistance due to conduction between the surface $A$ and $B$ , $B$ and $C$ , $C$ and $D$ respectively.

As shown in Figure (b), all the resistance is in series. Thus, write the expression for the total resistance by using concept of conduction and convection resistance.

$\begin{array}{c}\\{R_{th}} = {R_1} + {R_2} + {R_3} + {R_4} + {R_5}\\\\ = \frac{1}{{{h_i}A}} + \frac{{{L_p}}}{{{k_p}A}} + \frac{{{L_b}}}{{{k_b}A}} + \frac{{{L_S}}}{{{k_S}A}} + \frac{1}{{{h_o}A}}\\\end{array}$ …… (1)

Calculate the value of total thermal resistance.

Substitute $30{\rm{ W/}}{{\rm{m}}^2} \cdot {\rm{K}}$ for ${h_i}$ , $10{\rm{ mm}}$ for ${L_p}$ , $0.17\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}$ for ${k_p}$ , $100{\rm{ mm}}$ for ${L_b}$ , $0.038\frac{{\rm{W}}}{{{{\rm{m}}^{\rm{2}}}}}$ for ${k_b}$ , $20{\rm{ mm}}$ for ${L_s}$ , $0.12\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}$ for ${k_S}$ , $350{\rm{ }}{{\rm{m}}^2}$ for $A$ , and $60{\rm{ W/}}{{\rm{m}}^2} \cdot {\rm{K}}$ for ${h_o}$ in the equation (1).

${R_{th}} = \left[ \begin{array}{l}\\\frac{1}{{\left( {30{\rm{ W/}}{{\rm{m}}^2} \cdot {\rm{K}}} \right)\left( {350{\rm{ }}{{\rm{m}}^2}} \right)}} + \frac{{\left( {10{\rm{ mm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{1000{\rm{ mm}}}}} \right)}}{{\left( {0.17\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}} \right)\left( {350{\rm{ }}{{\rm{m}}^2}} \right)}}\\\\ + \frac{{\left( {100{\rm{ mm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{1000{\rm{ mm}}}}} \right)}}{{\left( {0.038\frac{{\rm{W}}}{{{{\rm{m}}^{\rm{2}}}}}} \right)\left( {350{\rm{ }}{{\rm{m}}^2}} \right)}} + \frac{{\left( {20{\rm{ mm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{1000{\rm{ mm}}}}} \right)}}{{\left( {0.12\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}} \right)\left( {350{\rm{ }}{{\rm{m}}^2}} \right)}} + \frac{1}{{\left( {60{\rm{ W/}}{{\rm{m}}^2} \cdot {\rm{K}}} \right)\left( {350{\rm{ }}{{\rm{m}}^2}} \right)}}\\\end{array} \right]$

Further, solve,

$\begin{array}{c}\\{R_{th}} = \left( {9.52 + 16.81 + 751.88 + 47.6 + 4.76} \right) \times {10^{ - 5}}\frac{{\rm{K}}}{{\rm{W}}}\\\\ = 8.3 \times {10^{ - 3}}\frac{{\rm{K}}}{{\rm{W}}}\\\end{array}$

The temperature inside the room is given as $20^\circ {\rm{ C}}$ . Convert the unit.

${T_{\infty ,}}_i = \left( {t^\circ {\rm{ C}} + 273} \right){\rm{ K}}$ …… (2)

Substitute $20^\circ {\rm{ C}}$ for $t^\circ {\rm{ C}}$ in the equation (2).

$\begin{array}{c}\\{T_{\infty ,}}_i = \left( {20^\circ {\rm{ C}} + 273.15} \right){\rm{ K}}\\\\ = 293.15{\rm{ K}}\\\end{array}$

The outer temperature is given as $- 15^\circ {\rm{ C}}$ . Convert the unit.

Substitute $- 15^\circ {\rm{ C}}$ for $t^\circ {\rm{ C}}$ in the equation (2).

$\begin{array}{c}\\{T_{\infty ,}}_o = \left( { - 15^\circ {\rm{ C}} + 273.15} \right){\rm{ K}}\\\\ = {\rm{258}}{\rm{.15 K}}\\\end{array}$ .

Calculate the rate of heat transfer rate through the wall.

$q = \frac{{\Delta T}}{{{R_{th}}}}$

Substitute $293.15{\rm{ K}}$ for ${T_{\infty ,}}_i$ , ${\rm{258}}{\rm{.15 K}}$ for ${T_{\infty ,}}_o$ , and $8.3 \times {10^{ - 3}}\frac{{\rm{K}}}{{\rm{W}}}$ for ${R_{th}}$ .

$\begin{array}{c}\\q = \frac{{293.15{\rm{ K}} - {\rm{258}}{\rm{.15 K}}}}{{8.3 \times {{10}^{ - 3}}\frac{{\rm{K}}}{{\rm{W}}}}}\\\\ = 4216.86{\rm{ W}}\\\end{array}$

(B)

The wind raises the convective coefficient of the outer surface of the house as $300{\rm{ W/}}{{\rm{m}}^{\rm{2}}} \cdot {\rm{K}}$ . Calculate the value of total thermal resistance.

Substitute $30{\rm{ W/}}{{\rm{m}}^2} \cdot {\rm{K}}$ for ${h_i}$ , $10{\rm{ mm}}$ for ${L_p}$ , $0.17\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}$ for ${k_p}$ , $100{\rm{ mm}}$ for ${L_b}$ , $0.038\frac{{\rm{W}}}{{{{\rm{m}}^{\rm{2}}}}}$ for ${k_b}$ , $20{\rm{ mm}}$ for ${L_s}$ , $0.12\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}$ for ${k_S}$ , $350{\rm{ }}{{\rm{m}}^2}$ for $A$ and $300{\rm{ W/}}{{\rm{m}}^2} \cdot {\rm{K}}$ for ${h_o}$ in the equation (1).

${R_{th,{\rm{ New}}}} = \left[ \begin{array}{l}\\\frac{1}{{\left( {30{\rm{ W/}}{{\rm{m}}^2} \cdot {\rm{K}}} \right)\left( {350{\rm{ }}{{\rm{m}}^2}} \right)}} + \frac{{\left( {10{\rm{ mm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{1000{\rm{ mm}}}}} \right)}}{{\left( {0.17\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}} \right)\left( {350{\rm{ }}{{\rm{m}}^2}} \right)}}\\\\ + \frac{{\left( {100{\rm{ mm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{1000{\rm{ mm}}}}} \right)}}{{\left( {0.038\frac{{\rm{W}}}{{{{\rm{m}}^{\rm{2}}}}}} \right)\left( {350{\rm{ }}{{\rm{m}}^2}} \right)}} + \frac{{\left( {20{\rm{ mm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{1000{\rm{ mm}}}}} \right)}}{{\left( {0.12\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}} \right)\left( {350{\rm{ }}{{\rm{m}}^2}} \right)}} + \frac{1}{{\left( {300{\rm{ W/}}{{\rm{m}}^2} \cdot {\rm{K}}} \right)\left( {350{\rm{ }}{{\rm{m}}^2}} \right)}}\\\end{array} \right]$

Further, solve,

$\begin{array}{c}\\{R_{th,{\rm{ New}}}} = \left( {9.52 + 16.81 + 751.88 + 47.6 + 0.952} \right) \times {10^{ - 5}}\frac{{\rm{K}}}{{\rm{W}}}\\\\ = 8.267 \times {10^{ - 3}}\frac{{\rm{K}}}{{\rm{W}}}\\\end{array}$

Calculate the new rate of heat transfer through the walls.

$q = \frac{{\Delta T}}{{{R_{th}}}}$

Substitute $293.15{\rm{ K}}$ for ${T_{\infty ,}}_i$ , ${\rm{258}}{\rm{.15 K}}$ for ${T_{\infty ,}}_o$ , and $8.267 \times {10^{ - 3}}\frac{{\rm{K}}}{{\rm{W}}}$ for ${R_{th,{\rm{ New}}}}$ .

$\begin{array}{c}\\{q_{{\rm{New}}}} = \frac{{293.15{\rm{ K}} - {\rm{258}}{\rm{.15 K}}}}{{8.267 \times {{10}^{ - 3}}\frac{{\rm{K}}}{{\rm{W}}}}}\\\\ = 4233.7{\rm{ W}}\\\end{array}$

Calculate the percentage increase in the rate of heat transfer.

$\% {\rm{ increase}} = \frac{{{Q_n} - Q}}{Q} \times 100$

Substitute $4233.7{\rm{ W}}$ for ${q_{{\rm{New}}}}$ , and $4216.86{\rm{ W}}$ for $q$ .

$\begin{array}{c}\\\% {\rm{ increase}} = \frac{{4233.7{\rm{ W}} - 4216.86{\rm{ W}}}}{{4216.86{\rm{ W}}}} \times 100\\\\ = 0.4\% \\\end{array}$

(c)

Calculate the thermal resistance of plasterboard.

${R_{{\rm{plaster board}}}} = \frac{{{L_p}}}{{A{k_p}}}$

Substitute $10{\rm{ mm}}$ for ${L_p}$ , $350{\rm{ }}{{\rm{m}}^2}$ for $A$ and $0.17\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}$ for ${k_p}$ .

$\begin{array}{c}\\{R_{{\rm{plaster board}}}} = \frac{{\left( {10{\rm{ mm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{1000{\rm{ mm}}}}} \right)}}{{\left( {350{\rm{ }}{{\rm{m}}^2}} \right)\left( {0.17\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}} \right)}}\\\\ = 16.81 \times {10^{ - 5}}\\\end{array}$

Similarly, calculate the thermal resistance of glass fiber blanket.

${R_{{\rm{glass fiber}}}} = \frac{{{L_b}}}{{A{k_b}}}$

Substitute $100{\rm{ mm}}$ for ${L_b}$ , $350{\rm{ }}{{\rm{m}}^2}$ for $A$ and $0.038{\rm{ }}\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}$ for ${k_p}$ .

$\begin{array}{c}\\{R_{{\rm{glass fiber}}}} = \frac{{\left( {100{\rm{ mm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{1000{\rm{ mm}}}}} \right)}}{{\left( {350{\rm{ }}{{\rm{m}}^2}} \right)\left( {0.038{\rm{ }}\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}} \right)}}\\\\ = 751.88 \times {10^{ - 5}}{\rm{ }}\frac{{\rm{K}}}{{\rm{W}}}\\\end{array}$

Similarly, calculate the thermal resistance of glass fiber blanket.

${R_{{\rm{plywood}}}} = \frac{{{L_s}}}{{A{k_s}}}$

Substitute $20{\rm{ mm}}$ for ${L_s}$ , $350{\rm{ }}{{\rm{m}}^2}$ for $A$ and $0.12\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}$ for ${k_S}$ .

$\begin{array}{c}\\{R_{{\rm{plywood}}}} = \frac{{\left( {20{\rm{ mm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{1000{\rm{ mm}}}}} \right)}}{{\left( {350{\rm{ }}{{\rm{m}}^2}} \right)\left( {0.12\frac{{\rm{W}}}{{{\rm{m}} \cdot {\rm{K}}}}} \right)}}\\\\ = 47.6 \times {10^{ - 5}}{\rm{ }}\frac{{\rm{K}}}{{\rm{W}}}\\\end{array}$

Since, the value of ${R_{{\rm{glass fiber}}}}$ is the greatest, therefore, calculate the percentage of thermal resistance of glass fiber blanket to the total thermal resistance.

$\begin{array}{c}\\\% \left( {\frac{{{R_{{\rm{plaster board}}}}}}{{\;{R_{th}}}}} \right) = \frac{{751.88 \times {{10}^{ - 5}}{\rm{ }}\frac{{\rm{K}}}{{\rm{W}}}}}{{8.3 \times {{10}^{ - 3}}\frac{{\rm{K}}}{{\rm{W}}}}} \times 100\\\\ = 90.5\% \\\end{array}$

Therefore, the glass fiber is the controlling resistance.

Ans: Part A

The heat transfer rate through walls is $4216.86{\rm{ W}}$ .

Part B

Percentage increase in the heat transfer rate is $0.4\%$ .

Part C

The controlling resistance is the glass fiber blanket.