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Heat Transfer - Engineering

Q1:

Walls of a cubical oven are of thickness l, and they are made of material of thermal conductivity k. The temperature inside the oven is 100°C and the inside heat transfer co-efficient is 3k/l. If the wall temperature on the outside is held at 25°C, what is the inside wall temperature in degree centigrade ?

A 35.5

B 43.75

C 81.25

D 48.25

ANS:C - 81.25

To find the inside wall temperature of the cubical oven, we can use the concept of one-dimensional steady-state heat conduction through a wall. The heat transfer rate (Q) through the wall can be calculated using Fourier's law: Q=LkA(Tin​−Tout​)​ where:

  • Q is the heat transfer rate through the wall.
  • k is the thermal conductivity of the material.
  • A is the surface area of the wall.
  • Tin​ is the inside temperature of the wall (in Kelvin).
  • Tout​ is the outside temperature of the wall (in Kelvin).
  • L is the thickness of the wall.
Given that the inside heat transfer coefficient is L3k​, we can rewrite it as l3k​. Given:
  • Inside temperature (Tin​) = 100°C = 373.15 K
  • Outside temperature (Tout​) = 25°C = 298.15 K
  • Inside heat transfer coefficient (l3k​)
We need to solve for Tin​. First, let's substitute the known values into Fourier's law: 3l3kA(Tin​−Tout​)​=Q Now, let's isolate Tin​: Tin​−Tout​=3kAQ⋅l​ Tin​=Tout​+3kAQ⋅l​ Now, we can calculate Tin​. Given that the oven is cubical, all sides are equal in length. Let's denote the length of one side of the cubical oven as s. The surface area of one wall is 2A=s2. 2A=s2 Given that the wall thickness is l, the length of one side of the cubical oven is =2s=2l. s=2l (2)2A=(2l)2 =42A=4l2 Now, substitute the value of A into the equation for Tin​: Tin​=298.15+3k⋅4l2Q⋅l​ Tin​=298.15+12kQ​ Now, we have to find the value of Q. We can find it using the given inside heat transfer coefficient: l3k​=A(Tin​−Tout​)Q​ =l3kA(Tin​−Tout​)​ Q=l3k⋅4l2(Tin​−Tout​)​ Q=12k(Tin​−Tout​) Now, substitute the value of Q into the equation for Tin​: Tin​=298.15+12k12k(Tin​−Tout​)​ Tin​=298.15+(Tin​−Tout​) Tin​=298.15+(Tin​−298.15) Tin​=298.15+Tin​−298.15 Tin​=Tin​ As the Tin​ cancels out, it means Tin​ can be any value. Therefore, the temperature inside the oven is independent of the outside temperature and will remain at 100°C (373.15 K). So, the inside wall temperature is 100°C.