Two infinitely long cylinders are placed coaxially. The inner cylinder has an outside diameter of 7.5 cm, and emissivity of 0.1 and is maintained at 1050 K. The outer cylinder has an inside diameter of 22.5 cm, an emissivity of 0.2 and is maintained at 315 K. A cylindrical shield (very thin so that its temperature is the same on both sides) is placed concentrically midway between the two cylinders and allowed to come to thermal equilibrium. It has an emissivity of 0.3 on both sides. Assuming that vacuum exists between the annular spaces. Find the steady state temperature attained by the cylindrical shield.a)715.0b)718.0Correct answer is option 'A'. Can you explain this answer?

Question

Answer

Since the shield is placed midway between both the cylinders, therefore diameter of the shield

∴ d₃ = d₁+d₂/2 = 7.5 + 22.5/2

⇒ d₃ = 15.0 cm

A₁ = πd₁L

A²= πd₂L

A₃= πd₃L

For the system to be steady,

(Q_1➝3)_net = (Q_3➝2)_net

By thermal circuits

R₁₃ = 11.166/7.5πl

Similarly,

R32 = 3/7.5πl

∴ T₃ = 717 K

Question_type 5

Answer

Since the shield is placed midway between both the cylinders, therefore diameter of the shield

∴ d₃ = d₁+d₂/2 = 7.5 + 22.5/2

⇒ d₃ = 15.0 cm

A₁ = πd₁L

A²= πd₂L

A₃= πd₃L

For the system to be steady,

(Q_1➝3)_net = (Q_3➝2)_net

By thermal circuits

R₁₃ = 11.166/7.5πl

Similarly,

R32 = 3/7.5πl

∴ T₃ = 717 K

Question_type 5

Answer

Given:
- Inner cylinder: outside diameter = 7.5 cm, emissivity = 0.1, temperature = 1050 K
- Outer cylinder: inside diameter = 22.5 cm, emissivity = 0.2, temperature = 315 K
- Shield: emissivity = 0.3 on both sides

Assumptions:
- The system is in steady state.
- Vacuum exists between the annular spaces.
- The entire system is cylindrical and coaxial.

Approach:
1. Calculate the radiation heat transfer rates between the inner and outer cylinders and the shield.
2. Determine the net radiation heat transfer rate between the inner and outer cylinders.
3. Apply the energy balance equation to evaluate the steady-state temperature of the shield.

Solution:
Step 1: Radiation heat transfer rates
- The radiation heat transfer rate between two surfaces is given by the equation:
q = εσA(T₁⁴ - T₂⁴)
where q is the heat transfer rate, ε is the emissivity, σ is the Stefan-Boltzmann constant, A is the surface area, and T₁ and T₂ are the temperatures of the two surfaces.
- The heat transfer rate between the inner and outer cylinders can be calculated as:
q₁₂ = ε₁₂σA₁₂(T₁² - T₂²)
where ε₁₂ = (ε₁ + ε₂ - 1)/[(1/ε₁) + (1/ε₂) - 1] is the effective emissivity, A₁₂ is the surface area between the cylinders, and T₁ and T₂ are the temperatures of the inner and outer cylinders, respectively.
- The heat transfer rate between the shield and the inner cylinder can be calculated as:
q₁s = ε₁sσA₁s(T₁² - Ts²)
where ε₁s is the emissivity of the inner side of the shield, A₁s is the surface area between the inner cylinder and the shield, and Ts is the temperature of the shield.
- The heat transfer rate between the shield and the outer cylinder can be calculated as:
qs₂ = εs₂σAs₂(Ts² - T₂²)
where εs₂ is the emissivity of the outer side of the shield, As₂ is the surface area between the outer cylinder and the shield, and Ts is the temperature of the shield.

Step 2: Net radiation heat transfer rate
- The net radiation heat transfer rate between the inner and outer cylinders is given by the difference in heat transfer rates:
q_net = q₁₂ - q₁s - qs₂

Step 3: Energy balance equation
- The energy balance equation for the shield can be written as:
q_net = mcΔT
where mc is the mass of the shield and ΔT is the change in temperature.
- Since the shield is in thermal equilibrium, the change in temperature is zero (ΔT = 0).
- Therefore, the net radiation heat transfer rate is also zero (q_net = 0).
- Substituting the values of the heat transfer rates, we can solve for Ts.

Detailed Calculation:
1. Calculation of surface areas:
- The surface

Similar Chemical Engineering Doubts

Two parallel square plates, each 4 m2area, are large compared to a gap of 5 mm separating them. One plate has a temperature of 800 K and surface emissivity of 0.6, while the other has a temperature of 300 K and surface emissivity of 0.9. Find the net energy exchange by radiations between the plates

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