Heat Transfer Concept Page - 3

Example

Heat transferred by conduction is a function of temperature

Example:
One face of a copper cube of edge 10 cm is maintained at 1000C${100}^{0}C$ and the opposite face is maintained at 0 0C${}^{0}C$. All other surfaces are covered with an insulating material. Find the amount of heat flowing per second through the cube. Thermal conductivity of copper is 385Wm−1C−1$385W{m}^{-1}{C}^{-1}$.
Solution: 
The heat flows from the hotter face towards the cooler face. The area of cross-section perpendicular to the heat flow is A=(10cm)2$A={(10cm)}^2$
The amount of heat flowing per second is 
△Q△t=KAT1−T2x=385×0.01×(100−0)0.1=3850W${\displaystyle \frac{\mathrm{△}Q}{\mathrm{△}t}}=KA{\displaystyle \frac{T_1-T_2}{x}}=385\times 0.01\times {\displaystyle \frac{(100-0)}{0.1}}=3850W$

Definition

Using conduction in first law of thermodynamics

Example:
100g$100g$ of water is heated from 30oC${30}^{o}C$ to 50oC${50}^{o}C$. Ignoring the slight expansion of the water, find the the change in its internal energy.(specific heat of water is 4184J/Kg/K$4184J/Kg/K$).
Solution:ΔQ=ΔU+ΔW$\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W$
Since ΔW=0$\mathrm{\Delta }W=0$, ΔQ=ΔU$\mathrm{\Delta }Q=\mathrm{\Delta }U$
ΔU=msΔT=0.1×4.184×20=8.368kJ$\mathrm{\Delta }U=ms\mathrm{\Delta }T=0.1\times 4.184\times 20=8.368kJ$  

Example

Application of conduction in phase change

Example:
A cylindrical rod with one end in a steam chamber and the other end is in ice. It is found that 1 gm of ice melts per second. If the rod is replaced by another one of same material double the length and double area of cross section, find the mass of ice that melts per second.Solution:
Thermal conduction in a metal bar is given by
Q=KA(Δθ)tl=miLi$Q={\displaystyle {\displaystyle \frac{KA(\mathrm{\Delta }\theta )t}{l}}=m_i{L}_i}$
mi∝Al$m_i\propto {\displaystyle {\displaystyle \frac{A}{l}}}$
m2m1=A2A1×l1l2=2AA×l2l=1${\displaystyle {\displaystyle \frac{m_2}{m_1}}={\displaystyle \frac{A_2}{A_1}}\times {\displaystyle \frac{l_1}{l_2}}={\displaystyle \frac{2A}{A}}\times {\displaystyle \frac{l}{2l}}=1}$
m2=m1=1gm$m_2=m_1=1gm$

Result

Analogy of current flow with heat flow

Thermal Conduction	Electric Conduction
Heat Flow Rate: Q	Electric Current: I
Temperature Difference: T1−T2$T_1-T_2$	Potential Difference: V1−V2$V_1-V_2$
Thermal Resistance: Rthermal=dxkA$R_{thermal}={\displaystyle \frac{dx}{kA}}$	Electric Resistance: R

Example

Simple Problems Based on Heat

Q. It takes 487.5 J to heat 25 grams of copper from 25 C to 75 C.
What is the specific heat in Joules/gC?
Sol :  Step 1 : Given :  Q=487.5 J,  m=25 grams,  T1=25oC,T2=75oC$Q=487.5J,m=25grams,T_1={25}^{o}C,T_2={75}^{o}C$

         Step 2 : Formula to be used : Q=mcΔT$Q=mc\mathrm{\Delta }T$ ............. (1)
           where Q$Q$ = heat energy
                      m$m$ = mass
                      c$c$ = specific heat
                      ΔT$\mathrm{\Delta }T$ = change in temperature

         Step 3 :  On putting the given data in above equation (1),
           478.5=25×c×(75−25)$478.5=25\times c\times (75-25)$       
           c=487.525×50$c={\displaystyle \frac{487.5}{25\times 50}}$             
           c=0.39Jgm oC$c=0.39{\displaystyle \frac{J}{gm{}^{o}C}}$

Definition

Thermal resistance

Thermal resistance is a heat property and a measurement of a temperature difference by which an object or material resists a heat flow. Thermal resistance is the reciprocal of thermal conductance.
Thermal resistance =L(kA)$={\displaystyle \frac{L}{(kA)}}$, measured in KW−1$K{W}^{-1}$
L$L$ is the length of the material (measured on a path parallel to the heat flow) (m)
k$k$ is the thermal conductivity of the material (W/(Km))
A$A$ is the cross-sectional area (perpendicular to the path of heat flow) (m2${}^2$)

Example

Thermal resistance of a spherical conductor

A system has two concentric spheres of radii r1$r_1$ and r2$r_2$, kept at temperatures T1$T_1$ and T2$T_2$ respectively. The radial rate of flow of heat between the two concentric spheres is proportional to:Consider an elemental spherical shell of thickness dx at radius x.
Thermal resistance of this shell is given by
dR=dxK(4πx2){FromR=LKA}∫dR=R=1K∫r2r1dx4πx2=14πK[1r1−1r2]=(r2−r1)4πK(r1r2)$$\begin{gathered} {\displaystyle dR={\displaystyle \frac{dx}{K(4\pi x^2)}}\{FromR={\displaystyle \frac{L}{KA}}\}} \\ {\displaystyle \int dR=R={\displaystyle \frac{1}{K}}{\int }_{r_1}^{r_2}{\displaystyle \frac{dx}{4\pi x^2}}} \\ ={\displaystyle \frac{1}{4\pi K}}[{\displaystyle \frac{1}{r_1}}-{\displaystyle \frac{1}{r_2}}] \\ ={\displaystyle \frac{(r_2-r_1)}{4\pi K\left(r_1{r}_2\right)}} \end{gathered}$$

Therefore, rate of heat flow is
H=KAT1−T2R=KA(T1−T2)(4πK(r1r2)r2−r1)H∝(r1r2)r2−r1$$\begin{gathered} H=KA{\displaystyle \frac{T_1-T_2}{R}}=KA(T_1-T_2)\left({\displaystyle \frac{4\pi K\left(r_1{r}_2\right)}{r_2-r_1}}\right) \\ H\propto {\displaystyle \frac{\left(r_1{r}_2\right)}{r_2-r_1}} \end{gathered}$$

Example

Thermal resistance of a cylindrical conductor

A cylinder of radius R made of a material of thermal conductivity k1$k_1$ is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of material of thermal conductivity k2$k_2$. The ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state.Find the effective thermal conductivity.We know that for a cylinder, rate of heat flow  Q=KA(T1−T2)L$Q={\displaystyle \frac{KA(T_1-T_2)}{L}}$ , where A is he area.
now for the radius of internal cylinder is R$R$ and  thermal conductivity will be k1$k_1$Then Q1=k1A1(T1−T2)L$Q_1={\displaystyle \frac{k_1{A}_1(T_1-T_2)}{L}}$
and for radius of out cylinder 2R$2R$ thermal conductivity will be k2$k_2$  Q2=k2A2(T1−T2)L$Q_2={\displaystyle \frac{k_2{A}_2(T_1-T_2)}{L}}$
Let effect thermal conductivity is K$K$ thenQ=Q1+Q2$Q=Q_1+Q_2$⇒KA(T1−T2)L$\Rightarrow {\displaystyle \frac{KA(T_1-T_2)}{L}}$=k1A1(T1−T2)L$={\displaystyle \frac{k_1{A}_1(T_1-T_2)}{L}}$+k2A2(T1−T2)L$+{\displaystyle \frac{k_2{A}_2(T_1-T_2)}{L}}$so A1=πR2$A_1=\pi R^2$
A2=π(4R2−R2)$A_2=\pi ({4R}^2-R^2)$
so equating these,we get
K(4R2)=k1R2+k2(4R2−R2)$K(4R^2)=k_1{R}^2+k_2(4R^2-R^2)$
K=k1+3k2$K=k_1+3k_2$
K=k1+3k24

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