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Wize University Physics Textbook (Master) > Heat and Temperature

Heat Transfer: Conductors in Series and Parallel

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Conductor In Series

  • For conductors in series:
ΔTt=ΔTA+ΔTB=TH−TC\Delta T_t=\Delta T_A+\Delta T_B=T_H-T_C
  • The heat transfer rate is the same everywhere:
dQAdt=dQBdt=dQtdt→kaAΔTALA=kBAΔTBLB=ktAΔTtLA+LB\displaystyle \begin{array}{l} \displaystyle \frac{dQ_A}{dt}=\frac{dQ_B}{dt}=\frac{dQ_t}{dt}\\ \\\to\displaystyle \frac{k_aA\Delta T_A}{L_A}=\frac{k_BA\Delta T_B}{L_B}=\frac{k_tA\Delta T_t}{L_A+L_B}\end{array}
Where ktk_t is the total conductivity of the system and could be found by:
LA+LBkt=LAkA+LBkB\frac{L_A+L_B}{k_t}=\frac{L_A}{k_A}+\frac{L_B}{k_B}
  • Lk\frac{L}{k}is usually shown by RR and is called the thermal resistance
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Example: Conductors In Series

A compound bar is made as shown below. It is insulated perfectly on its sides. If the crosssection of the bar is 5 cm25\ cm^2, what is the temperature at the junction of two metals and what is the total rate of heat flow?
(kst=50.2 wm⋅K, kBr=109 wm⋅K)\left(k_{st}=50.2\ \frac{w}{m\cdot K},\ k_{Br}=109\ \frac{w}{m\cdot K}\right)


Conductor in Series:
(dQdt)t=(dQdt)st=(dQdt)Br\left(\frac{dQ}{dt}\right)_t=\left(\frac{dQ}{dt}\right)_{st}=\left(\frac{dQ}{dt}\right)_{Br}→kstLstAst(TH−T)=kBrABrLBr(T−Tc)\to\frac{k_{st}}{L_{st}}\cancel{A_{st}}(T_H-T)=\frac{k_{Br}\cancel{A_{Br}}}{L_{Br}}(T-T_c)

→kstLst(75−T)=kBrTLBr\to \frac{k_{st}}{L_{st}}(75-T)=\frac{k_{Br}T}{L_{Br}}→T(kBrLBr+kstLst)=75 kstLst\to T\left(\frac{k_{Br}}{L_{Br}}+\frac{k_{st}}{L_{st}}\right)=75\ \frac{k_{st}}{L_{st}}

→T=75 kst/Lst(kBrLBr+kstLst)=25100770.7=32.57°\to T=\frac{75\ k_{st}/L_{st}}{\left(\frac{k_{Br}}{L_{Br}}+\frac{k_{st}}{L_{st}}\right)}=\frac{25100}{770.7}=32.57\degree


(dQdt)t=(dQdt)st=kstAstLst(TH−T)=50.2×5×10−40.15(75−32.57)\left(\frac{dQ}{dt}\right)_t=\left(\frac{dQ}{dt}\right)_{st}=\frac{k_{st}A_{st}}{L_{st}}\left(T_H-T\right)=\frac{50.2\times5\times10^{-4}}{0.15}\left(75-32.57\right)

→(dQdt)t=7.1J\to\left(\frac{dQ}{dt}\right)_t=7.1J

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Conductors In Parallel

For parallel conductors:
dQtdt=dQAdt+dQBdt and ΔTt=ΔTA=ΔTB→ΔTAALkA+ΔTABLkB=ktAtΔTL\begin{array}{l}\dfrac{dQ_t}{dt}=\dfrac{dQ_A}{dt}+\dfrac{dQ_B}{dt} \text{ and } \Delta T_t=\Delta T_A=\Delta T_B\\ \\\to\dfrac{\Delta TA_A}{L}k_A+\dfrac{\Delta TA_B}{L}k_B=\dfrac{k_tA_t\Delta T}{L}\end{array}
Where ktAt=kAAA+kBABk_tA_t=k_AA_A+k_BA_B is the relation for total conductivity in parallel conductors.


Extra Practice
Consider a double-pane window (4mm glass layers, 20mm gap between them). When it is 20 degrees Celsius inside and -15 degrees Celsius outside, and if the area of the windows is 1 m2,
a) What is the energy loss if the gap is filled with air?
b) What if the gap is filled with Argon instead?
Single-pane windows consist of a single layer of glass. They are light, but double-pane windows are much more energy-efficient; they have a layer of air between two thin layers of glass. In very cold places such as Yellowknife, you may also come across triple-pane windows.
Consider these windows when it is 20 degrees Celsius inside and -40 degrees Celsius outside. The area of the windows is 1 m2.
a) What is the energy loss if a single-pane window was used? (2.2 mm thick glass)
Using the following system, find the temperature at the copper/steel and steel/aluminum interfaces. What is the heat flow through each material? You may assume only conduction occurs, and that the fixed ends are held at constant temperature.
Givens:
ksteel=43[WmK], kCu=401[WmK], kal=205[WmK]k_{steel}=43\left[\frac{W}{mK}\right],\ k_{Cu}=401\left[\frac{W}{mK}\right],\ k_{al}=205\left[\frac{W}{mK}\right]
Why are double pane windows more effective at preventing heat loss than single pane windows of the same thickness? Draw a diagram to demonstrate this property.
How much more effective are double pane windows are if kglass=1.05[WmK]k_{glass}=1.05\left[\frac{W}{mK}\right], kair=0.025[WmK]k_{air}=0.025\left[\frac{W}{mK}\right]. The single pane is 7 mm7\ mmthick, and the double panes are 3 mm3\ mmthick (air accounts for the remaining 1 mm1\ mm).
If we could vacuum pump the double pane window, what could happen to the heat flow through the window?
A home has 4 walls, three of the walls have no window and are made of a layer 1m thick wood and a layer of 1m thick brick. Each wall is 2 meter by 2 meters and the window on one of the walls is 1 meter by 1 meter. The window is a double panned window which means its a 0.5 m layer of glass then 1 m layer of air then a 0.5 m layer of glass. Each wall is also 2 meters thick. If the inside temperature is 20°C20\degree C and the outside temperature is 0°C0\degree C. (Do not consider the roof or floor).
What is the power level that a generator need to provide such that the inside temperature stays at 20°C20\degree C, if the generator is only 70% efficient.
K wood: 0.12 wm⋅k\frac{w}{m\cdot k}
New Practice Question
A compound bar made up of 15cm of Cu with ACu=25cm2A_{Cu}=25cm^{2} and 20 cm of steel with ASt=10cm2A_{St}=10cm^2𝑎𝑡 𝑇 = 293𝐾 is fixed at both ends as shown in the diagram below. How much is the stress in each metal if the system is heated uniformly by a 700 watt heater for 5 minutes?
CCu=390JKg⋅KCSt=470JKg⋅KYCu=11×1010Nm2YSt=20×1010Nm2aCu=1.7×10−51KaSt=1.2×10−51KρCu=9gcm3ρSt=7.85gcm3\begin{array}{cc}C_{Cu}=390 \frac{J}{Kg\cdot K}&C_{St}=470\frac{J}{Kg\cdot K}\\Y_{Cu}=11\times 10^{10} \frac{N}{m^2}&Y_{St}=20\times 10^{10} \frac{N}{m^2}\\a_{Cu}=1.7\times 10^{-5}\frac{1}{K}&a_{St}=1.2\times 10^{-5}\frac{1}{K}\\\rho_{Cu}=9\frac{g}{cm^3}&\rho_{St}=7.85\frac{g}{cm^3}\end{array}