Heat Transfer Test 1

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Heat Transfer Test 1

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Weekly Quiz Competition

Question 1
1 / -0

The inner surface of a nickel-steel cylinder (k = 19 W/mK) of inner diameter = 1 cm and outer diameter = 7 cm is maintained at a constant temperature of 110°C, while the outer surface is held at 50°C. Calculate the heat lost per meter length of the pipe (in kW).

A
3.5-4.2

B
3.54-4.21

C
3.57-4.26

D
3.59-4.22

Solution

Concept:
Heat transfer through a hollow pipe:
\({R_{th}} = \frac{{\ln \left( {\frac{{{r_o}}}{{{r_i}}}} \right)}}{{2\pi kL}}\)
\(Q = \frac{{{\rm{\Delta }}T}}{{{R_{th}}}} = \frac{{2\pi kL\left( {{T_i} - {T_o}} \right)}}{{\ln \left( {\frac{{{r_o}}}{{{r_i}}}} \right)}} = \frac{{2\pi kL\left( {{T_i} - {T_o}} \right)}}{{\ln \left( {\frac{{{D_o}}}{{{D_i}}}} \right)}}\)
Given:
D_i = 0.01 m, D_o = 0.07 m, k = 19 W/mK
T_i = 110°C, T_o = 50°C
Calculation:
\(Q = \frac{{2\pi \times 19 \times \left( {110 - 50} \right)}}{{\ln \left( {\frac{{0.07}}{{0.01}}} \right)}} = 3680.97\;W = 3.68\;kW\)
Question 2
1 / -0

A pipe of 25 mm outer diameter carries steam. The heat transfer coefficient between the cylinder and the surrounding is 5 W/m²K. It is proposed to reduce the heat lost from the surrounding pipe by adding insulation having a thermal conductivity of 0.05 W/mK. Which of the following statements is true?

A
The outer radius if the pipe = the critical radius

B
The outer radius of the pipe is less than the critical radius

C
Adding the insulation will reduce the heat lost

D
Adding the insulation will increase the heat lost

Solution

Concept:
In order to provide effective insulation, the radius of insulation must be greater than the critical radius of insulation as at a critical radius of insulation maximum heat loss takes place.
Calculation:
Given:
r₂ = 0.125 m = 12,5 mm, k_i = 0.05 W/mK
h_o = 5 W/m²K

\(r_c = \frac{{{k_i}}}{{{h_o}}} = 10mm\)
\(r_c < r_2,\)the addition of insulation will reduce heat loss
Question 3
1 / -0

A long slender bar has one of its end in a furnace and has reached a steady-state condition. The temperature at 25 cm from the furnace was found to be T. If now the heat transfer coefficient is increased to ‘n’ times such that the same temperature is now found at a location of 15 cm, then calculate the value of nⁿ.

A
36.32

B
25

C
17.08

D
15.49

Solution

Concept:
The slender bar could be considered as a long fin.
For long fin the temperature distribution follows
\(\frac{{T - \;{T_\infty }}}{{{T_0} - \;{T_\infty }\;}} = {e^{ - mx}}\)
Where, m = fin parameter = \(\sqrt {\frac{{hP}}{{kA}}} \)
T₀ = Furnace temperature, T_∞= surrounding temperature
For same temperature at two locations we get,
mx = constant.
Solve for h by using formula for m
Calculation:
\(\frac{{T - {T_\infty }}}{{{T_0} - {T_\infty }}} = {e^{ - mx}}\)
mx = constant (for same Temperature and x₁ and x₂)
m₁x₁ = m₂x₂
\({\left( {\sqrt {\frac{{hP}}{{kA}}} \;} \right)_1}{x_1} = {\left( {\sqrt {\frac{{hP}}{{kA}}} \;} \right)_2}\;{x_2}\)
Squaring on both the sides
∴ h₁× x₁² = h₂× x₂²(since all other parameters are same)
∴ h₁ (0.25)² = h₂ (0.15)²
∴ h₂ = 2.77 h₁
∴ n = 2.777.
∴ nⁿ = 17.080
Question 4
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A rod of length 25 cm and diameter 20 mm is connected between two walls which are maintained at a temperature of 353 K and air at 298 K flows over the rod. If the thermal conductivity of the rod is 250 W/mK and the convective heat transfer coefficient is 40 W/m²K, then calculate the temperature at the center of the rod (℃).

A
68-69

B
684-691

C
687-696

D
689-692

Solution

Concept:
For this type of fin the temperature at mid-point (T) is given by,
\(\frac{{T - \;{T_\infty }}}{{{T_0} - \;{T_\infty }}} = \frac{1}{{\cosh \left( {\frac{{mL}}{2}} \right)}}\)
\({\rm{Where}},{\rm{\;m\;}} = {\rm{\;fin\;parameter\;}} = \sqrt {\frac{{hP}}{{kA}}} = \sqrt {\frac{{4h}}{{kD}}} \)
D = diameter of rod, L = Length of fin, A = cross-sectional area, P = perimeter
k = thermal conductivity (W/mK), h = heat transfer coefficient (W/m²K)
T₀ = wall temp, T_∞ = surrounding temp.
Calculation:
Given:
D = 20 mm, L = 0.25 m, h = 40 W/m²K, k = 250 W/mK
\(m = \sqrt {\frac{{4h}}{{kD}}} = \sqrt {\frac{{4\; \times \;40\; \times \;1000}}{{250\; \times \;20}}} \)
∴ m = 5.6568
\(T = {T_\infty } + \frac{{\left( {{T_0} - {T_\infty }} \right)}}{{\cos h\;\left( {m\frac{L}{2}} \right)}}\)
\({\rm{T}} = 25 + \frac{{\left( {353 - 298} \right)}}{{1.26}}\)
∴ T = 68.63°C