Question

A hollow pipe of 10 mm outer diameter is to be insulated by thick cylindrical insulation having thermal conductivity of 1 W/m K. The surface heat transfer coefficient on the insulation surface is 5 W/m$ ^2 $ K. What is the minimum effective thickness of insulation for causing the reduction in heat leakage from the insulated pipe?

• A. \( 195 \text{ mm} \)
• B. \( 200 \text{ mm} \)
• C. \( 205 \text{ mm} \)
• D. \( 210 \text{ mm} \)

Hint:

For cylindrical insulation, calculate the critical radius \( r_c = k/h \). If the pipe's outer radius is less than \( r_c \), heat transfer increases with insulation up to \( r_c \). To reduce heat leakage, the total insulation thickness must be such that the outer radius of the insulation exceeds \( r_c \). The minimum effective thickness for reduction is \( r_c - r_{pipe,outer} \).

Answer 1

The Correct Option is A

Step 1: Identify the concept of Critical Radius of Insulation.
For a cylindrical or spherical object, adding insulation does not always reduce heat transfer. Sometimes, up to a certain thickness, adding insulation can actually increase heat transfer. This is due to the combined effect of decreasing conductive resistance and increasing convective surface area. The insulation thickness at which heat transfer is maximum (or minimum effective thickness for reduction in heat leakage) is called the critical radius of insulation. Beyond this critical radius, adding more insulation will reduce heat transfer.
Step 2: Recall the formula for Critical Radius of Insulation for a cylinder.
For a cylindrical object, the critical radius of insulation \( r_c \) is given by: \[ r_c = \frac{k}{h} \] where:
\( k \) = thermal conductivity of the insulation material
\( h \) = surface heat transfer coefficient on the outer surface of the insulation
Step 3: Calculate the critical radius of insulation.
Given:
Thermal conductivity of insulation \( k = 1 \text{ W/m K} \).
Surface heat transfer coefficient \( h = 5 \text{ W/m}^2 \text{ K} \).
\[ r_c = \frac{1 \text{ W/m K}}{5 \text{ W/m}^2 \text{ K}} = 0.2 \text{ m} \] Convert to mm: \[ r_c = 0.2 \times 1000 \text{ mm} = 200 \text{ mm} \]
Step 4: Determine the minimum effective thickness for heat leakage reduction.
The critical radius \( r_c \) represents the radius of the insulation where heat transfer is maximum. If the outer radius of the pipe is less than \( r_c \), adding insulation will initially increase heat transfer until the outer radius of the insulation reaches \( r_c \). Beyond \( r_c \), adding more insulation will decrease heat transfer.
The question asks for the "minimum effective thickness of insulation for causing the reduction in heat leakage". This means we need the total outer radius of the insulation to be greater than the critical radius. The given outer diameter of the hollow pipe is 10 mm. So, the outer radius of the pipe is \( r_o = \frac{10}{2} = 5 \text{ mm} \). Since \( r_o = 5 \text{ mm}<r_c = 200 \text{ mm} \), the pipe's bare surface is below the critical radius. This means heat transfer will initially increase as insulation is added, reaching a maximum at \( r_c = 200 \text{ mm} \). To reduce heat leakage, the outer radius of the insulation must exceed this critical radius. The insulation thickness \( t_{ins} \) is the difference between the outer radius of the insulation and the outer radius of the pipe: \( t_{ins} = r_{ins,outer} - r_o \) For reduction in heat leakage, \( r_{ins,outer} \) must be greater than \( r_c \). The minimum value for \( r_{ins,outer} \) that causes reduction is just infinitesimally greater than \( r_c \). So, the minimum effective thickness would be when \( r_{ins,outer} \approx r_c \). Therefore, the minimum effective thickness of insulation is: \[ t_{ins,min} = r_c - r_o \] \[ t_{ins,min} = 200 \text{ mm} - 5 \text{ mm} = 195 \text{ mm} \] Adding a thickness of 195 mm means the outer radius of the insulation will be \( 5 + 195 = 200 \text{ mm} \), which is exactly the critical radius. Any thickness greater than 195 mm will result in the outer radius of insulation being greater than the critical radius, thereby reducing heat leakage. Thus, 195 mm is the minimum thickness to start causing a reduction in heat leakage (beyond the maximum). The final answer is $\boxed{\text{1}}$.