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What is the Expression for Thermal Conductivity in Series and Parallel?
1. Series combination: Two rectangular slabs, S1 and S2, of different materials are in contact [Fig.]. Cross sectional area of each slab is A; their breadths are x1 and x2 and the coefficients of thermal conductivity are k1 and k2 respectively. The composite slab is heated from the left of Sx and heat is conducted through the junction to S2. Heat will be released from the right surface of S2.
Let us assume that at steady state, the temperatures on the left hand surface and on the right hand surface of the composite slab are θ2 and θ1 respectively. The temperature at the junction is θ such that θ2 > θ > θ1.
Hence, heat conducted through S1 in time t,
In steady state, Q1 = Q2 = Q (say)
Now, let us consider a single slab of thickness (x1 + x2) and area of cross section A, such that if a temperature difference of (θ2 – θ1) is maintained between its two opposite surfaces,
then the same amount of heat will be conducted through the slab per second. In this case the coefficient of thermal con-ductivity is called the equivalent thermal conductivity of the composite slab.
Let the equivalent thermal conductivity of the composite slab be k. Then,
\(\frac{Q}{t}\) = \(\frac{k A\left(\theta_2-\theta_1\right)}{x_1+x_2}\) = \(\frac{A\left(\theta_2-\theta_1\right)}{\frac{x_1+x_2}{k}}\) …… (2)
Comparing equations (1) and (2) we have,
\(\frac{x_1+x_2}{k}\) = \(\frac{x_1}{k_1}+\frac{x_2}{k_2}\) …… (3)
If the composite slab is made up of n number of slabs with thickness x1, x2, ….., xn and coefficient of thermal conductivity k1, k2, ……, kn respectively, then
\(\frac{x_1+x_2+\cdots+x_n}{k}\) = \(\frac{x_1}{k_1}\) + \(\frac{x_2}{k_2}\) + ……. + \(\frac{x_n}{k_n}\) …… (4)
Temperature of the junction:
From equation (1),
2. Parallel Combination: In this case, two rectangular slabs S’1 and S’2 of same length x but of different cross sectional area A1 and A2 respectively are joined at their ends so that the left ends of both slabs are kept at a temperature θ2 and the right ends are kept at temperature θ1 [Fig.]. Let k1 and k2 be the coefficients of thermal conductivity of S’1 and S’2 respectively, then heat conducted through slab S’1 in
and heat conducted through slab S’2 in time t,
Q2 = \(\frac{k_2 A_2\left(\theta_2-\theta_1\right) t}{x}\)
∴ The total heat passing through the composite slab in time t,
Q = Q1 + Q2 = \(\frac{\left(\theta_2-\theta_1\right) t}{x}\)(k1A1 + k2A2)
∴ \(\frac{Q}{t}\) = \(\frac{\theta_2-\theta_1}{x}\)(k1A1 + k2A2) …… (6)
Now, let us consider a single slab of length x, cross sectional area A such that if a temperature difference (θ2 – θ1) is maintained between its two opposite surfaces, the same amount of heat will be conducted through the slab per second.
Let, k be the equivalent thermal conductivity’ of the composite slab. Then,
\(\frac{Q}{t}\) = \(\frac{k A\left(\theta_2-\theta_1\right)}{x}\) …… (7)
Comparing equations (6) and (7) we get,
k1A1 + k2A2 = kA = k(A1 + A2) [∵ A = A1 + A2]
∴ k = \(\frac{k_1 A_1+k_2 A_2}{\left(A_1+A_2\right)}\) …. (8)
If the composite slab is made up of n number of slabs with cross sectional areas A1, A2, ….,An and coefficients of thermal conductivity k1, k2, ……, kn respectively, then,
k1A1 + k2A2 + ….. + knAn = k(A1 + A2 + …… + An)
∴ k = \(\frac{k_1 A_1+k_2 A_2+\cdots+k_n A_n}{\left(A_1+A_2+\cdots+A_n\right)}\) …. (9)
Numerical Examples
Example 1.
Four metal pieces of the same surface area and of thickness 1 cm, 2 cm, 3 cm and 4 cm respectively are connected serially with each other. Coefficients of thermal conductivity of the respective pieces are 0.2 CGS, 0.3 CGS, 0.1 CGS and 0.4 CGS units. Find the equivalent conductivity of the system.
Solution:
It is known that
Example 2.
A thick composite plate is formed by two plates of equal thickness kept one over the other. If the conductivities of the materials of the constituent plates are k1 and k2, show that the equivalent conductivity of the thick plate, k = \(\frac{2 k_1 k_2}{k_1+k_2}\).
Solution:
It is known, \(\frac{x_1+x_2}{k}\) = \(\frac{x_1}{k_1}\) + \(\frac{x_2}{k_2}\)
Here, x1 = x2 = x (say)
∴ \(\frac{x+x}{k}\) = \(\frac{x}{k_1}\) + \(\frac{x}{k_2}\) or, \(\frac{2}{k}\) = \(\frac{1}{k_1}\) + \(\frac{1}{k_2}\) or, k = \(\frac{2 k_1 k_2}{k_1+k_2}\)
Example 3.
A 75 cm long copper rod and a 125 cm long steel rod are joined face to face. Each rod is of a circular cross section of diameter 2 cm. Temperatures at the two ends of the composite rod are 100 °C and 0 °C and the outer surface of the rod is insulated. Find the temperature at the junction of the two rods. What is the rate of conduction of heat through the junction?
k for copper = 9.2 × 10-2 kcal ᐧ m-1 ᐧ °C-1 ᐧ s-1 and k for steel = 1.1 × 10-2 kcal ᐧ m-1 ᐧ °C-1 ᐧ s-1.
Solution:
Let the temperature of the junction = θ [Fig.]
Conductivity of copper,
k1 = \(\frac{9.2 \times 10^{-2} \times 10^3}{10^2}\) = 0.92 CGS unit
and that of steel,
Area of cross section of each rod, A = π(1)2 = π cm2
∴ Rate of flow of heat through the junction,
\(\frac{Q}{t}\) = \(\frac{k_1 A\left(\theta_2-\theta\right)}{l_1}\) = \(\frac{k_2 A\left(\theta-\theta_1\right)}{l_2}\)
or, k1l2(100 – θ) = 0.11 × 75 × θ or, θ = 93.3°C
∴ Rate of conduction,
\(\frac{Q}{t}\) = \(\frac{k_2 A\left(\theta-\theta_1\right)}{l_2}\) = \(\frac{0.11 \times \pi \times 93.3}{125}\) = 0.258 cal ᐧ s-1
Example 4.
The thickness of each metal in a composite bar is 0.01 m and the temperature of the two external surfaces are 100 °C and 30 °C. If conductivities of the metals be 0.2 CGS unit and 0.3 CGS unit respectively, find the temperature on the interface. [HS 12]
Solution:
Let the temperature on the interface = θ. At steady state, rate of flow of heat will be the same through both the plates.
∴ \(\frac{Q}{t}\) = \(\frac{k_1 A(100-\theta)}{x_1}\) = \(\frac{k_2 A(\theta-30)}{x_2}\)
or, \(\frac{0.2(100-\theta)}{1}\) = \(\frac{0.3(\theta-30)}{1}\) or, 200 – 2θ = 3θ – 90
or, 5θ = 290 or θ = 58°C
Example 5.
Three metal rods of the same length and area of cross section are attached in series. Conductivities of the three metals are k, 2k and 3k. Free end of the first rod is kept at 200°C while the other end of the combination is kept of 100 °C. Find the temperatures of the two junctions at steady state. Assume that no heat loss occurs due to radiation. [WBJEE 2000]
Solution:
Let the required temperatures be θ1 and θ2.
At steady state,
\(\frac{Q}{t}\) = \(\frac{k A\left(200-\theta_1\right)}{l}\) = \(\frac{2 k A\left(\theta_1-\theta_2\right)}{l}\) = \(\frac{3 k A\left(\theta_2-100\right)}{l}\)
∴ 200 – θ1 = 2(θ1 – θ2) = 3(θ2 – 100)
As 200 – θ1 = 2(θ1 – θ2)
∴ 3θ1 – 2θ2 = 200 …… (1)
Also, 200 – θ1 = 3(θ2 – 100)
or, θ1 + 3θ2 = 500 ….. (2)
Solving (1) and (2)
θ1 = 145.45°C and θ2 = 118.18°C
Example 6.
A composite block is constructed with three plates of equal thickness and of equal cross sectional area. The coefficients of conductivity of the three plates are k1, k2 and k3 respectively. If the coefficient of conductivity of the composite block is k, then prove that k = \(\frac{3}{\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}}\)
Solution:
Let the thickness of each plate be d and cross sec-tional area be A. Q amount of heat flows through each plate in time t [Fig.]. The temperatures of the two ends of the com-posite slab are T1 and T2 respectively. Also, the temperatures of the consecutive junctions are T1’ and T2’.
[by componendo and dividendo process]
Again, since thickness of the composite slab is 3 d,
Q = \(\frac{k A\left(T_1-T_2\right) t}{3 d}\) or, \(\frac{Q d}{A t}\) = \(\frac{T_1-T_2}{\frac{3}{k}}\) …. (2)
From (1) and (2) we get,
\(\frac{1}{k_1}\) + \(\frac{1}{k_2}\) + \(\frac{1}{k_3}\) = \(\frac{3}{k}\) or, k = \(\frac{3}{\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}}\)
Example 7.
The temperature of a room is kept fixed at 20 °C with the help of an electric heater of resistance 2011. The heater is connected to a 200 V main. Temperature is same everywhere in the room. The area of the window through which heat is conducted outside from the room is 1 m2 and the thickness of the glass is 0.2 cm. Then calculate the temperature outside. [Given, coefficient of thermal conductivity of glass = 0.002 cal ᐧ m-1 ᐧ ° C-1 ᐧ s-1 and J = 4.2 J/s
Solution:
Heat generated by the heater in 1 s,
Example 8.
The temperature of a room is kept fixed at 20 °C. With the help of an electric heater when the temperature outside is -10°C. The total area of the walls in the room is 137 m2. The walls have three layers—the innermost layer is made of wood and 2.5 cm thick, the middle layer is made of cement and 1 cm thick, the outermost layer is made of bricks and 25 cm thick. Then calculate the power of the heater. [Given, coefficients of thermal conductivity of wood, cement and brick are 0.125 W ᐧ m-1 ᐧ °C-1, 1.5 W ᐧ m-1 ᐧ °C-1 and 1.0 W ᐧ m-1 ᐧ °C-1 respectively]
Solution:
If k is the equivalent thermal conductivity of the wall made of three layers of different materials then,
Heat conducted outside through the walls per second
= \(\frac{k A\left(\theta_2-\theta_1\right)}{d}\) = \(\frac{0.624 \times 137 \times\{20-(-10)\}}{28.5 \times 10^{-2}}\)
= 8998.7 W ≈ 8999 J ᐧ s-1
To keep temperature of the room fixed the heater must generate 8999 J or 8.999 kJ heat per second.
Hence, power of the heater = 9 kW.
Example 9.
The temperature gradient at earth’s surface is 320 C / km and average thermal conductivity of earth is 0.008 CGS unit. Find the loss of heat per day from the earth’s surface taking its radius to be 6000 km.
Solution:
We know, Q = k ᐧ A\(\frac{d \theta}{d x}\)t [\(\frac{d \theta}{d x}\) = \(\frac{\theta_2-\theta_1}{d}\)]
Here, k = 0.008 CGS unit, A = 4π(6000 × 105)2 cm2
\(\frac{d \theta}{d x}\) = 32°C/km = \(\frac{32}{10^5}\)°C/cm,
t = 1 day = 24 × 60 × 60 s
∴ Q = \(\frac{0.008 \times 4 \pi \times 36 \times 10^{16} \times 32 \times 24 \times 60 \times 60}{10^5}\)
= 1.0006 × 1018 cal
Example 10.
Three rods made of material X and three rods of material Y are connected as shown in Fig 9.9. Each rod has equal length and equal cross section. If the temperatures of the end A and the junction E are 60°C and 10°C respectively, find the temperatures of the junctions B. C and D. Coefficients of thermal conductivity of the materials X and Y are 0.92 CGS units and 0.46 CGS unit respectively.
Solution:
Let the temperatures of the junctions B, C and D be θ1, θ2 and θ3 respectively.
Now, heat conducted from A to B
= heat conducted from B to C + heat conducted from B to D
or, 0.46(60 – θ1) = 0.92(θ1 – θ2) + 0.46(θ1 – θ3)
or 60 – θ1 = 2(θ1 – θ2) + (θ1 – θ3)
or, 4θ1 – 2θ1 – θ3 = 60 ……. (1)
Again, heat conducted from B to D = heat conducted from D to C + heat conducted from D to E
or, 0.46(θ1 – θ3) = 0.92(θ3 – θ2) + 0.46(θ3 – 10)
or, θ1 – θ3 =2(θ3 – θ1) + (θ3 – 10)
or, -θ1 – 2θ2 + 4θ3 = 10 ………… (2)
Similarly, heat conducted from B to C + heat conducted from D to C = heat conducted from C to E
or, 0.92(θ1 – θ2) + 0.92(θ3 – θ2) = 0.92(θ2 – 10)
or, θ1 – θ2 + θ3 – θ2 = θ2 – 10
or, θ1 – 3θ2 + θ3 = -10 …….. (3)
Solving (1), (2) and (3), we get,
θ1 = 30°C; θ2 = θ3 = 20°C