A copper pipe carrying refrigerant at T°C is covered by cylindrical insulation of thermal conductivity k W/mK. The heat transfer coefficient over the insulation surface is h W/m2K. The critical radius of insulation would be.
k/h
Understanding the Critical Radius of Insulation
When we add insulation to a cylindrical pipe, like the copper pipe carrying refrigerant mentioned in the question, we affect heat transfer in two main ways:
- Conduction Resistance: The insulation material adds resistance to heat flow by conduction through the material itself. As the insulation thickness increases, the conduction resistance increases.
- Convection Resistance: The outer surface area of the insulation increases as more insulation is added. A larger surface area enhances heat transfer by convection from the surface to the surrounding environment. This effectively decreases the convection resistance.
The total thermal resistance is the sum of the conduction resistance of the insulation and the convection resistance at the outer surface. Heat transfer is inversely proportional to the total thermal resistance.
For a cylindrical pipe, adding insulation initially increases the heat transfer up to a certain thickness, and beyond that thickness, it starts to decrease the heat transfer. This happens because initially, the decrease in convection resistance due to the increased surface area is more significant than the increase in conduction resistance. The radius at which the heat transfer is maximum (or total thermal resistance is minimum) is called the critical radius of insulation.
Calculating the Critical Radius for a Cylinder
The formula for the critical radius of insulation ($r_c$) for a cylinder is derived by finding the minimum of the total thermal resistance with respect to the outer radius of the insulation. The total thermal resistance per unit length of a cylindrical pipe with insulation is given by:
\( R_{total}' = \frac{\ln(r_o/r_i)}{2 \pi k} + \frac{1}{2 \pi r_o h} \)
where:
- \( r_i \) is the inner radius (radius of the pipe)
- \( r_o \) is the outer radius of the insulation
- \( k \) is the thermal conductivity of the insulation
- \( h \) is the convection heat transfer coefficient at the outer surface
To find the critical radius ($r_c$), we differentiate the total resistance with respect to \( r_o \) and set the derivative to zero:
\( \frac{d R_{total}'}{d r_o} = \frac{1}{2 \pi k r_o} - \frac{1}{2 \pi r_o^2 h} = 0 \)
Solving for \( r_o \), which is the critical radius \( r_c \):
\( \frac{1}{2 \pi k r_c} = \frac{1}{2 \pi r_c^2 h} \)
\( \frac{r_c^2}{r_c} = \frac{2 \pi k}{2 \pi h} \)
\( r_c = \frac{k}{h} \)
So, the critical radius of insulation depends only on the thermal conductivity of the insulation material ($k$) and the convection heat transfer coefficient ($h$) at the outer surface, not on the pipe's radius or the refrigerant temperature.
Applying the Formula to the Question Parameters
The question provides:
- Thermal conductivity of insulation = \( k \) W/mK
- Heat transfer coefficient over the insulation surface = \( h \) W/m²K
Using the formula derived above, the critical radius of insulation is \( r_c = k/h \).
This matches one of the given options.
Important Questions from Conduction
In M - L - t - T system, the dimension of thermal diffusivity is -
The transfer of heat through the molecules of matter in any body is called ________.
Unit of thermal diffusivity is
When heat is transferred from one particle of hot body to another by actual motion of the heated particles, it is referred to as heat transfer by:
Which of the following is a case of steady state heat transfer?
