Critical Thickness of Insulation MCQ Quiz in বাংলা - Objective Question with Answer for Critical Thickness of Insulation - বিনামূল্যে ডাউনলোড করুন [PDF]
Last updated on Apr 14, 2025
Latest Critical Thickness of Insulation MCQ Objective Questions
Top Critical Thickness of Insulation MCQ Objective Questions
Critical Thickness of Insulation Question 1:
Inner and outer radius of the hollow cylinder and hollow sphere are r, 3r, 2r and 3r respectively. Insulation is applied on both, hollow cylinder and hollow sphere. Both are made up of same material and have same environmental conditions. Find the difference in critical thicknesses of insulation. (Notations have their usual meanings.)
Answer (Detailed Solution Below)
Critical Thickness of Insulation Question 1 Detailed Solution
Explanation:
\({\rm{Critical\;radius\;of\;cylinder}} = \frac{k}{{{h_0}}}\)
Critical radius of sphere \(= \frac{{2k}}{{{h_0}}}\)
Now,
Critical thickness of insulation cylinder \({\rm{}} = \frac{k}{{{h_0}}} - 3r\)
Critical thickness of insulation for sphere \(= \frac{{2k}}{{{h_0}}} - 3r\)
Now,
Difference in critical thicknesses \(= \left( {\frac{{2k}}{{{h_0}}} - 3r} \right) - \left( {\frac{k}{{{h_0}}} - 3r} \right)\)
∴ Difference in critical thicknesses \({\rm{}} = \frac{k}{{{h_0}}}\)
Critical Thickness of Insulation Question 2:
There exists a "critical thickness" beyond which the heat transfer rate
Answer (Detailed Solution Below)
Critical Thickness of Insulation Question 2 Detailed Solution
Concept:
Concept of the critical thickness of insulation:
- In a plane wall, the heat transfer rate depends only on the thickness of insulation. Higher will be the thickness of insulation, lower will be the heat transfer rate. This is due to the fact the outer surface have always the same area.
- But in cylindrical and spherical coordinates, the addition of insulation also increases the outer surface area which decreases the convection resistance at the outer surface.
- At the same time conduction resistance increases due to an increase in insulation thickness. So, to optimize the balance of conductive and convective resistance, a critical radius is decided. At this critical radius, the equivalent conduction and convection resistance will be minimum and heat transfer will be maximum.
- Moreover, in some cases, a decrease in the convection resistance due to the increase in surface area can be more significant than an increase in conduction resistance due to thicker insulation.
Explanation:
As a result, the total resistance may actually decrease resulting in increased heat flow.
- Thickness up to which heat flow increases and beyond which heat flow decreases is termed critical thickness. For cylinders and spheres, it is the critical radius (rc ) so option 1 is correct.
- That can be derived from the critical radius of insulation depending on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h.
For cylinder
\({{\bf{r}}_{\bf{c}}} = \;\frac{k}{h}\)
For sphere
\({{\bf{r}}_{\bf{c}}} = \;\frac{{2k}}{h}\)
Critical Thickness of Insulation Question 3:
An insulating material with a thermal conductivity, k = 0.12 W/mK is used for a pipe carrying steam. The local coefficient of heat transfer (h) to the surroundings is 4 W/m2K. In order to provide effective insulation, the minimum outer diameter of the pipe should be
Answer (Detailed Solution Below)
Critical Thickness of Insulation Question 3 Detailed Solution
Concept:
In order to provide effective insulation, the radius of insulation must be greater than the critical radius of insulation as at a critical radius of insulation maximum heat loss takes place.
The critical radius of insulation for a cylindrical pipe \({r_c} = \frac{k}{h}\)
Where k is the thermal conductivity of the insulating material and h is the heat transfer coefficient of the surrounding medium.
Critical radius for sphere is given by \({r_c} = \frac{{2k}}{h}\)
Calculation:
Given:
k = 0.12 W/mK, h = 4 W/m2K,
Critical radius = \({r_c} = \frac{k}{h} = \frac{{0.12}}{4} = 0.03\;m = 30\;mm\)
Minimum diameter = 2 × radius = 60 mmCritical Thickness of Insulation Question 4:
The critical radius of insulation for a cylindrical object, like a pipe, is the radius at which:
Answer (Detailed Solution Below)
Critical Thickness of Insulation Question 4 Detailed Solution
Explanation:
Critical thickness is the thickness up to which heat flow increases and after which heat flow decreases.
- The insulation radius at which resistance to heat flow is minimum and consequently heat flow rate is maximum is called “critical radius”.
- Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h.
The critical radius of insulation for a cylindrical body:
\({r_{cr,cylinder}} = \frac{k}{h}\)
The critical radius of insulation for a spherical shell:
\({r_{cr,sphere}} = \frac{{2k}}{h}\)
Important Points
- If the radius of insulation is less than the critical Radius then the rate of heat transfer from the cylinder increases with the addition of insulation.
- If the radius of insulation is equal to the critical radius of insulation rate of heat transfer from the cylinder reaches a maximum when the radius of insulation is equal to the critical radius of insulation.
- If the radius of insulation is greater than the critical radius of insulation then the rate of heat transfer from the cylinder decreases with the addition of insulation.
Critical Thickness of Insulation Question 5:
The figure below shows a cross section through an insulated heating pipe which is made from steel (k = 45 W/mK) with an inner radius of 150 mm and outer radius of 155 mm. The pipe is coated with 100 mm thickness of insulation having thermal conductivity of k = 0.06 W/mK. Air at inlet temperature Ti = 60°C flow through the pipe has a value of convective coefficient hi = 35 W/m2K. The outside surface of the pipe is surrounded by air which is at 15°C and convective heat transfer coefficient on this surface has a value of ho = 10 W/m2K. Find the critical insulation radius?
Answer (Detailed Solution Below)
Critical Thickness of Insulation Question 5 Detailed Solution
Concept:
The insulation radius at which resistance to heat flow is minimum and consequently heat flow rate is maximum is called “critical radius”.
The critical radius of insulation for a cylindrical body:
\({r_{cr,cylinder}} = \frac{k}{h}\)
where, k = conductivity of insulation, h = convective heat transfer coefficient of outer body
Calculation:
Given:
k = 0.06 W/mK, h = 10 W/m2K
Critical radius for cylinder
\(\begin{array}{l} {r_c} = \frac{{{k_{insulation}}}}{{{h_{outside}}}}\\ = \frac{{0.06}}{{10}} = 0.006\;m = 6~mm \end{array}\)
Critical Thickness of Insulation Question 6:
A metal rod of 2 cm diameter has a conductivity of 40 W/mK, which is to be insulated with an insulating material of conductivity of 0.1 W/mK. If the convective heat transfer coefficient with the ambient atmosphere is 5 W/m2K, The critical thickness of insulation will be
Answer (Detailed Solution Below)
Critical Thickness of Insulation Question 6 Detailed Solution
Concept:
The value of Router for which the heat transfer rate is maximum is called the critical radius of insulation.
The thickness up to which heat flow increases and after which heat flow decreases is termed as Critical thickness.
The insulation radius at which resistance to heat flow is minimum and consequently heat flow rate is maximum is called “critical radius”.
Critical radius of insulation for a cylindrical body:
\({r_{cr,cylinder}} = \frac{k}{h}\)
Critical radius of insulation for a spherical shell:
\({r_{cr,sphere}} = \frac{{2k}}{h}\)
Critical thickness = rcr - r
Calculation:
Given:
Thermal conductivity of insulating material, k = 0.1 W/mK, h = 5 W/m2K, r = 1 cm
Critical radius of insulation for an cylindrical body
\({r_c} = \frac{k}{{{h}}} = \frac{{0.1}}{5} = 2cm\)
Critical thickness of insulation
= 2 – 1 = 1 cm
Critical Thickness of Insulation Question 7:
If a foam insulation is added to a 4cm outer diameter pipe as shown in the figure, the critical radius of insulation (in cm) is _____________
Answer (Detailed Solution Below) 4.9 - 5.1
Critical Thickness of Insulation Question 7 Detailed Solution
Concept:
The critical radius of insulation rc is the radius upto which the addition of material increases the rate of heat loss and after critical radius addition of heat results into decrease in heat loss.
for cylinder \({r_c} = \frac{k}{h}\)
for sphere \({r_c} = \frac{2k}{h}\)
where, k is thermal conductivity of insulating material in W/m-K, and h is heat transfer coefficient of surrounding fluid in W/m2-K.
Calculation:
Given, k = 0.1 W/m-K and ho = 2 W/m2-K
For cylinder,
Critical radius \({r_c} = \frac{k}{{{h_0}}} = \frac{{0.1}}{2} = 0.05m\)
= 5 cmCritical Thickness of Insulation Question 8:
A pipe of 25 mm outer diameter carries steam. The heat transfer coefficient between the cylinder and the surrounding is 5 W/m2K. It is proposed to reduce the heat lost from the surrounding pipe by adding insulation having a thermal conductivity of 0.05 W/mK. Which of the following statements is true?
Answer (Detailed Solution Below)
Critical Thickness of Insulation Question 8 Detailed Solution
Concept:
In order to provide effective insulation, the radius of insulation must be greater than the critical radius of insulation as at a critical radius of insulation maximum heat loss takes place.
Calculation:
Given:
d2= 25 mm Hence, r2 = 0.125 m = 12.5 mm
ki = 0.05 W/mK, ho = 5 W/m2K
\(r_c = \frac{{{k_i}}}{{{h_o}}} = 10mm\)
\(r_c < r_2,\) the addition of insulation will reduce heat loss
Critical Thickness of Insulation Question 9:
An electric cable 1.4 cm in diameter has to be insulated with rubber (thermal conductivity \(0.174\frac{W}{{mK}}\)). The cable is located in air where heat transfer coefficient is \(8.7\frac{W}{{{m^2}K}}\). The critical thickness of insulation is
Answer (Detailed Solution Below)
Critical Thickness of Insulation Question 9 Detailed Solution
Concept:
Critical Thickness of Insulation: The thickness of insulation over a thermal conductivity body at which max rate of heat transfer takes place or minimum resistance occurs is called critical thickness of insulation.
For Cylinder:
Critical Thickness of Insulation = \(\frac{k}{h_o} - r\)
For Sphere:
Critical Thickness of Insulation = \(\frac{2k}{h_o} - r\)
where k is the thermal conductivity of the insulating material, ho is the heat transfer coefficient of the environment, and r is the radius of the thermal conductivity body.
Calculation:
Given:
k = 0.174 W/mK, ho = 8.7 W/m2K, r = 0.7 cm = 0.007 m
The critical thickness of insulation for electrical cables is
t = \((\frac{0.174}{8.7}) - 0.007 \)
t = 0.013 m
t = 1.3 cm
Critical Thickness of Insulation Question 10:
The critical radius of insulation for a spherical shell is
Answer (Detailed Solution Below)
Critical Thickness of Insulation Question 10 Detailed Solution
Explanation:
Critical thickness is the thickness up to which heat flow increases and after which heat flow decreases.
- The insulation radius at which resistance to heat flow is minimum and consequently heat flow rate is maximum is called “critical radius”.
- Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h.
The critical radius of insulation for a cylindrical body:
\({r_{cr,cylinder}} = \frac{k}{h}\)
The critical radius of insulation for a spherical shell:
\({r_{cr,sphere}} = \frac{{2k}}{h}\)
- If the radius of insulation is less than the critical Radius then the rate of heat transfer from the cylinder increases with the addition of insulation.
- If the radius of insulation is equal to the critical radius of insulation rate of heat transfer from the cylinder reaches a maximum when the radius of insulation is equal to the critical radius of insulation.
- If the radius of insulation is greater than the critical radius of insulation then the rate of heat transfer from the cylinder decreases with the addition of insulation.