What is net radiation heat exchange per square meter per unit time for two very large plates at temperature values of 800 K and 500 K? Emissivities of hot and cold plates are 0.8 and 0.6 respectively. Stefan-Boltzmann constant is - 5.67x10-8W/m2K4a)1026 kW/m2b)10.26 kW/m2c)102.6 kW/m2d)1.026 kW/m2Correct answer is option 'B'. Can you explain this answer?

Question

Answer

For two very large parallel plates, the net radiation heat exchange (using thermal circuit) is given by,

= 10.26 kW/m²

Answer

For two very large parallel plates, the net radiation heat exchange (using thermal circuit) is given by,

= 10.26 kW/m²

Answer

To calculate the net radiation heat exchange per square meter per unit time between two large plates at different temperatures, we can use the Stefan-Boltzmann law, which states that the rate of heat transfer by radiation between two objects is proportional to the fourth power of the absolute temperature difference between them.

The formula for net radiation heat exchange is given by:

Q = εσA(T₁⁴ - T₂⁴)

Where:
Q = Net radiation heat exchange per square meter per unit time (W/m²)
ε = Emissivity of the hot plate
σ = Stefan-Boltzmann constant (5.67x10⁻⁸ W/m²K⁴)
A = Surface area of the plates (assumed to be the same for both plates)
T₁ = Temperature of the hot plate (in Kelvin)
T₂ = Temperature of the cold plate (in Kelvin)

In this case, the temperature of the hot plate (T₁) is given as 800 K, and the temperature of the cold plate (T₂) is given as 500 K. The emissivity of the hot plate (ε₁) is given as 0.8, and the emissivity of the cold plate (ε₂) is given as 0.6.

Substituting the given values into the formula, we get:

Q = 0.8 * 5.67x10⁻⁸ * A * (800⁴ - 500⁴)

To simplify the calculation, we can use the fact that the difference between the fourth powers of two numbers can be expressed as the product of their squares multiplied by the sum and difference of the numbers:

(800⁴ - 500⁴) = (800² + 500²) * (800² - 500²)

Now, let's calculate the net radiation heat exchange:

Q = 0.8 * 5.67x10⁻⁸ * A * (800² + 500²) * (800² - 500²)

Q = 0.8 * 5.67x10⁻⁸ * A * (640000 + 250000) * (640000 - 250000)

Q = 0.8 * 5.67x10⁻⁸ * A * 890000 * 390000

Q ≈ 10.26 W/m²

Therefore, the net radiation heat exchange per square meter per unit time between the two large plates is approximately 10.26 kW/m².

Top Courses for Chemical Engineering

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams