Solved Examples on Heat Transfer:-

Question 1:-An idealized representation of the air temperature as a function of distance from a single-pane window on a calm, winter day is shown in below figure. The window dimensions are 60 cm × 60 cm × 0.50 cm.

(a)At what rate does heat flow out through the window? (Hint: The temperature drop across the glass is very small.)(b)Estimate the difference in temperature between the inner and outer glass surfaces.

Concept:-The rate

Hat which heat is transferred through the slab is,(a) directly proportional to the area (

A) available.(b) inversely proportional to the thickness of the slab Δ

x.(c) directly proportional to the temperature difference Δ

T.So,

H=kAΔT/ ΔxHere,

kis the proportionality constant and is called thermal conductivity of the material.From the equation

H=kAΔT/ Δx, the temperature difference ΔTwill be,Δ

T=HΔx/kAThe temperature gradient is defined as,

Temperature gradient = Δ

T/ Δx

Solution:-(a)To find out the rate of heat flows out through the window, first we have to find out the temperature gradient both inside and outside the window.

Inside, the temperature gradient Δ

T/ Δxwill be,Δ

T/ Δx= (20^{°}C- 5^{°}C)/8 cm (Since, Δx= 8 cm)= ((20 +273)K- (5+273) K)/(8 cm×10

^{-2}m/ 1 cm)= 293 K-278 K/0.08 m

= 187.5 K/m

Rounding off to two significant figures the temperature gradient will be 190 K/m.

Similar result will be occur for also outside.

Area

Afrom which heat will flow from the window will be,

A= (60 cm ×60 cm)^{2}= [(60 cm×10

^{-2}m/ 1 cm) (60 cm×10^{-2}m/ 1 cm)]^{2}= (0.6 m)

^{2}To find out the heat flow

Hthrough the air, substitute 0.026 W/m. K for thermal conductivityk, (0.6 m)^{2}forAand 190^{°}C/m for ΔT/ Δxin the equationH=kAΔT/ Δx,

H=kAΔT/ Δx= (0.026 W/m. K) (0.6 m)

^{2}(190 K/m)= 1.8 W

The value that we arrived at is the rate that heat flows through the air across an area the size of the window on either side of the window. Therefore the rate of heat flows out through the window would be 1.8 W.

(b) To find out the difference in temperature Δ

Tbetween the inner and outer glass surfaces, substitute 1.8 W forH, 0.50 cm for Δx, 1.0 W/m. K forkand (0.6 m)^{2}forAin the equation ΔT=HΔx/kA,Δ

T=HΔx/kA= (1.8 W) (0.50 cm)/( 1.0 W/m. K)( (0.6 m)

^{2})= (1.8 W) (0.50 cm×10

^{-2}m/1 cm )/( 1.0 W/m. K)( (0.6 m)^{2})= 0.025 K

From the above observation we conclude that, the difference in temperature Δ

Tbetween the inner and outer glass surfaces would be 0.025 K.________________________________________________________________________________________

Question 2:-Two identical rectangular rods of metal are welded end to end as shown in below figure

(a), and 10 J of heat flows through the rods in 2.0 min. How long would it take for 30 J to flow through the rods if they are welded as shown in below figure(b).

Concept:-The rate

Hat which heat is transferred through the slab is,(a) directly proportional to the area (

A) available.(b) inversely proportional to the thickness of the slab Δ

x.(c) directly proportional to the temperature difference Δ

T.So,

H=kAΔT/ ΔxHere

kis the proportionality constant and is called thermal conductivity of the material.

Solution:-As the temperature difference in this case Δ

T=T_{hot}–T_{cold}is same for both the case, thus the thermal conductivity will be same for both the cases.To obtain the rate of heat transfer

H_{o}for the rectangular rod which are connected in series (configuration (a)), substitute 2Lfor Δxin the equationH_{o}=kAΔT/ Δx,

H_{o}=kAΔT/ Δx=

kAΔT/2LTo obtain the rate of heat transfer (

H_{o})^{'}for the rectangular rod which are connected in parallel (configuration (b)), substitute 2AforAin the equationH_{o}=kAΔT/ Δx,(

H_{o})^{'}=kAΔT/ Δx=

k(2A) ΔT/LSo, (

H_{o})^{'}/H_{o}= (k(2A) ΔT/L)/ (kAΔT/2L)= 4

So, (

H_{o})^{'}= 4H_{o}Again also rate of heat transfer

His defined as,

H=W/tHere

Wis the energy andtis the time.So, time

twill be,

t =W/HTo find the rate of heat transfer

H_{o}for the rectangular rod which are connected in series (configuration (a)), substitute 10 J forWand 2 min fortin the equationt =W/H,

H_{o}=W/t= 10 J/ 2 min

= 5 J/min

To find out (

H_{o})^{'}, substitute 5 J/min forH_{o}in the equation (H_{o})^{'}= 4H_{o},(

H_{o})^{'}= 4H_{o}= 4(5 J/min)

= 20 J/min

Using equation

t =W/H, the timetfor 30 J to flow through the rods if they were welded in the figure parallel configuration will be,

t=W/(H_{o})^{'}To obtain the time

tfor 30 J to flow through the rods if they were welded in the figure parallel configuration, substitute 30 J forWand 20 J/min for (H_{o})^{'}in the equationt=W/(H_{o})^{'},

t=W/(H_{o})^{'}=(30 J)/(20 J/min)

= 1.5 min

From the above observation we conclude that, the time

tfor 30 J to flow through the rods if they were welded in the figure parallel configuration would be 1.5 min._______________________________________________________________________________________________

Question 3:-Ice has formed on a shallow pond and a steady state has been reached with the air above the ice at -5.2ºC and the bottom of the pond at 3.98ºC. If the total depth of ice + water is 1.42 m, how thick is the ice? (Assume that the thermal conductivities of ice and water are 1.67 and 0.502 W/m.K, respectively.)

Concept:-The temperature

T_{x}at the interface of a compound slab is equal to (T_{1}R_{2}+T_{2}R_{1}) /(R_{1}+R_{2}).So,

T_{x}= (T_{1}R_{2}+T_{2}R_{1}) /(R_{1}+R_{2})Here,

T_{1},T_{2}are the temperature of two surfaces (with T_{2}> T_{1}) andR_{1},R_{2}are thermal resistance of the two materials.

Solution:-At the interface between ice and water

T_{x}= 0^{°}CSubstitute 0

^{°}C forT_{x}in the equationT_{x}= (T_{1}R_{2}+T_{2}R_{1}) /(R_{1}+R_{2}),(

T_{1}R_{2}+T_{2}R_{1}) /(R_{1}+R_{2}) =T_{x}(

T_{1}R_{2}+T_{2}R_{1}) /(R_{1}+R_{2}) = 0So,

T_{1}R_{2}+T_{2}R_{1}= 0Or,

k_{1}T_{1}/L_{1}+k_{2}T_{2}/L_{2}= 0(

k_{1}T_{1}L_{2}+k_{2}T_{2}L_{1}) /L_{1}L_{2}= 0

k_{1}T_{1}L_{2}+k_{2}T_{2}L_{1}= 0

k_{1}T_{1}L_{2}+k_{2}T_{2}(L-L_{2}) = 0

k_{1}T_{1}L_{2}+k_{2}T_{2}L-k_{2}T_{2}L_{2}= 0

k_{2}T_{2}L=k_{2}T_{2}L_{2}-k_{1}T_{1}L_{2}= (

k_{2}T_{2}-k_{1}T_{1})L_{2}Or,

L_{2}=k_{2}T_{2}L/(k_{2}T_{2}-k_{1}T_{1})Here

Lis the depth of ice + water andL_{2}is the depth of ice.To find out the thickness

L_{2}of ice (depth of ice), substitute 1.67 W/m. K for thermal conductivity of icek_{2}, -5.20^{°}C forT_{2}, 1.4 m forL, 0.502 W/m . K for thermal conductivity of waterk_{1}and 3.98^{°}C forT_{1}in the equationL_{2}=k_{2}T_{2}L/(k_{2}T_{2}-k_{1}T_{1}),

L_{2}=k_{2}T_{2}L/(k_{2}T_{2}-k_{1}T_{1})=(1.67 W/m. K) (-5.20

^{°}C) (1.4 m) / ((1.67 W/m. K) (-5.20^{°}C) –(0.502 W/m . K) (3.98^{°}C))=(1.67 W/m. K) (-5.20 +273) K (1.4 m) / ((1.67 W/m. K) (-5.20 + 273) K–(0.502 W/m . K) (3.98 +273) K)

= 635.06 W/(447.23 -139.04) W/m

= 2.06 m

Rounding off to one significant figure, the thickness of the ice will be 2 m.