Critical Thickness of Insulation MCQ Quiz in मराठी - Objective Question with Answer for Critical Thickness of Insulation - मोफत PDF डाउनलोड करा

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 \({\rm{}} = \frac{k}{{{h_0}}}\)

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

\({{\bf{r}}_{\bf{c}}} = \;\frac{{2k}}{h}\)

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 = \({r_c} = \frac{k}{h} = \frac{{0.12}}{4} = 0.03\;m = 30\;mm\)

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

1. 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. 
2. 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.
3. 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. 

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

\(\begin{array}{l} {r_c} = \frac{{{k_{insulation}}}}{{{h_{outside}}}}\\ = \frac{{0.06}}{{10}} = 0.006\;m = 6~mm \end{array}\)    

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 = r_cr - 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

Concept:

The critical radius of insulation r_c 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 h_o = 2 W/m²-K

For cylinder,

Critical radius \({r_c} = \frac{k}{{{h_0}}} = \frac{{0.1}}{2} = 0.05m\)

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

k_i = 0.05 W/mK, h_o = 5 W/m²K


\(r_c = \frac{{{k_i}}}{{{h_o}}} = 10mm\)
\(r_c < r_2,\) the addition of insulation will reduce heat loss

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:

\({r_{cr,sphere}} = \frac{{2k}}{h}\)

1. 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. 
2. 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.
3. 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. 

Concept:

The radial thicknss of each material is same.

Theefore, the radii of pipe be r, radii of first material is 2r, radii of second material is 3r, 

Case 1:

Resistance to heat flow is,

\({R_1} = \frac{{\ln \left( {\frac{{r_2}}{r_1}} \right)}}{{2\pi KL}}= \frac{{\ln \left( {\frac{{2r}}{r}} \right)}}{{2\pi KL}}\)

\(∴ {R_1} = \frac{{\ln \left( 2 \right)}}{{2\pi KL}}\)

\({R_2} = \frac{{\ln \left( {\frac{{r_3}}{r_2}} \right)}}{{2\pi (2K)L}}= \frac{{\ln \left( {\frac{{3r}}{2r}} \right)}}{{4\pi KL}}\)

\(∴ {R_2} = \frac{{\ln \left( {1.5} \right)}}{{4\pi K L}}\)

\(∴ {Q_1} = \frac{{{\rm{\Delta }}T\left( {2\pi KL} \right)}}{{\ln \left( 2 \right) + \frac{{{\rm{ln}}\left( {1.5} \right)}}{2}}}\)

Case 2:

\({R_3} = \frac{{\ln \left( {\frac{{r_2}}{r_1}} \right)}}{{2\pi (2K)L}}= \frac{{\ln \left( {\frac{{2r}}{r}} \right)}}{{4\pi KL}}\)

\({R_3} = \frac{{\ln \left( 2 \right)}}{{4\pi KL}}\)

\({R_4} = \frac{{\ln \left( {\frac{{r_3}}{r_2}} \right)}}{{2\pi KL}}= \frac{{\ln \left( {\frac{{3r}}{2r}} \right)}}{{2\pi KL}}\)

\({R_4} = \frac{{\ln \left( {1.5} \right)}}{{2\pi kL}}\)

\(∴ {Q_2} = \frac{{{\rm{\Delta }}T\left( {2\pi kL} \right)}}{{\ln \left( {1.5} \right) + \frac{{{\rm{ln}}\left( 2 \right)}}{2}}}\)

Now when you compare the values of Q1 and Q2

We can see that Q1 < Q­2

∴ We can say that when a material with lower thermal conductivity is used for the inner layer and one with higher thermal conductivity for the outer layer then the heat transfer is less compared to the reverse case.