Heat Transfer Test 5

Result

Heat Transfer Test 5

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Weekly Quiz Competition

Question 1
1 / -0

Bodies A and B have thermal emissivity of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are same. The two bodies emit total radiant power at the same rate. Find the temperature of B if the temperature of A is 5802 K.

A
645 K

B
1734 K

C
1934 K

D
None of these

Solution

Concept:
The radiation energy emitted by a body per unit time is given by:
E_b = ϵAσT⁴
Where ϵ is emissivity of the body.
σ = The Stefan – Boltzmann constant = 5.67 × 10^-8W m^-2K^-4
Calculation:
ϵ_A = 0.01, ϵ_B = 0.81, A_A = A_B = A, E_A = E_B = E, T_A = 5802 K
E_A = E_B
ϵ_AA_AσT_A⁴ = ϵ_BA_BσT_B⁴
ϵ_AT_A⁴ = ϵ_BT_B⁴
\(\frac{{{T_B}}}{{{T_A}}} = {\left( {\frac{{{\epsilon_A}}}{{{\epsilon_B}}}} \right)^{\frac{1}{4}}} = {\left( {\frac{{0.01}}{{0.81}}} \right)^{\frac{1}{4}}} = {\left( {\frac{1}{{81}}} \right)^{\frac{1}{4}}} = {\left( {\frac{1}{{{3^4}}}} \right)^{\frac{1}{4}}} = \frac{1}{3}\)
\({T_B} = \frac{1}{3}{T_A} = \frac{1}{3} \times 5802 = 1934\;K\)
Question 2
1 / -0

It is observed that intensity of radiation is maximum in case of solar radiation at a wavelength of 0.49 microns. Assuming the sun as a black body, its emissive power is ___________ × 10⁷ W/m²
Wein displacement constant = 0.289 × 10^-2 mK.
Stefan – Boltzmann constant, σ = 5.67 × 10^-8 W/m²K⁴

A
6.5-7.0

B
6.54-7.01

C
6.57-7.06

D
6.59-7.02

Solution

Concept:
Wein’s displacement law:
λ_max.T = 2898 μmK
It states that the “product of absolute temperature and wavelength at which emissive power of a black body is a maximum, is constant”.
Calculation:
Given: λ_max = 0.49 microns = 0.49 × 10^-6
λ_max.T = 0.289 × 10^-2
0.49 × 10^-6 × T = 0.289 × 10^-2
T = 5898 K
Emissive power, E = σT⁴ = 5.67 × 10^-8 × (5898)⁴
E = 6.86 × 10⁷ W/m²
Question 3
1 / -0

Two infinite parallel plates with 7 radiation shields are held close together and are maintained at 1200 K and 800 K respectively. The plates and shields have an emissivity of 0.66. If 5 more shields are now inserted between them, then calculate the percentage change in heat transfer _______

A
37-39

B
374-391

C
377-396

D
379-392

Solution

Concept:
Heat transfer rate per unit area,
\(\frac qA = \frac{{\sigma \left( {T_1^4 - \;T_2^{4\;}} \right)}}{{\left( {\frac{2}{\epsilon} - 1\;} \right)\;\left( {n + 1} \right)}}\)
Calculation:
\(\frac{q}{A} = \frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{\left( {\frac{2}{e} - 1} \right)\left( {n + 1} \right)}}\)
Initially, n = 7
\({\left( {\frac{q}{A}} \right)_i} = \frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{\left( {\frac{2}{E} - 1} \right)\left( 8 \right)}}\)
Finally, n = 7 + 5 = 12
\({\left( {\frac{q}{A}} \right)_f} = \frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{\left( {\frac{2}{E} - 1} \right)\left( {13} \right)}}\)
\(\% \;reduction = \left( {1 - \frac{{{{\left( {\frac{q}{A}} \right)}_f}}}{{{{\left( {\frac{q}{A}} \right)}_i}}}} \right) \times 100\)
\(\% \;reduction = \left( {1 - \frac{8}{{13}}} \right) \times 100\)
∴ % reduction = 38.46%