Saturday, August 15, 2020

H C Verma solutions, HEAT TRANSFER, EXERCISES, Q21-Q30, Chapter-28, Concepts of Physics, Part-II

 Heat Transfer


EXERCISES, Q21 - Q30


   21. A hollow tube has a length l, inner radius R₁ and outer radius R₂. The material has a thermal conductivity K. Find the heat flowing through the walls of the tube if (a) the flat ends are maintained at temperatures T₁ and T₂ (T₂ > T₁) (b) the inside of the tube is maintained at temperature T₁ and the outside is maintained at T₂.  



Answer: (a) Temperature difference, ΔT = T₂-T₁, Cross-sectional area, A =π(R₂²-R₁²), Thickness, x = l.

Hence heat flowing through the walls of the tube 

=KA(ΔT)/x

=Kπ(R₂²-R₁²)(T₂-T₁)/l


(b) Here thickness, x = R₂-R₁, 

Consider a very thin tube of thickness dR at distance R from the center. Its area, A =2πRl, Temperature difference between is walls =dT.

Heat flow-rate through the walls of this tube, q = K*2πRl*dT/dR

→dT =(q/2πlK)*dR/R

Integrating both sides between the limits,

→∫dT = (q/2πlK)∫dR/R

→[T] =(q/2πlK)*[ln R]

→T₂-T₁ ={q*ln (R₂/R₁)}/(2πlK)

→q =2πlK(T₂-T₁)/ln (R₂/R₁)

 




   22. A composite slab is prepared by pasting two plates of thickness L₁ and L₂ and thermal conductivity K₁ and K₂. The slabs have equal cross-sectional areas. Find the equivalent conductivity of the slab.



Answer: The plates are in series. Hence the equivalent thermal resistance will be the sum of individual thermal resistances.

i.e. R = R₁+R₂

→L/KA =L₁/K₁A + L₂/K₂A, where A is the area of cross-section, K equivalent conductivity and L is total thickness.

→(L₁+L₂)/K =L₁/K₁ +L₂/K₂

→(L₁+L₂)/K =(L₁K₂+L₂K₁)/K₁K₂

→K/(L₁+L₂) =K₁K₂/(L₁K₂+L₂K₁)

→K =K₁K₂(L₁+L₂)/(L₁K₂+L₂K₁) 



 


   23. Figure (28-E2) shows a copper rod joined to a steel rod. The rods have equal length and equal cross-sectional area. The free end of the copper rod is kept at 0°C and that of the steel rod is kept 100°C. Find the temperature at the junction of the rods. The conductivity of copper = 390 W/m-°C and that of the steel = 46 W/m-°C.  
Figure for Q-23






Answer: Assume the length of each rod = L, area of cross-section = A and the temperature of the junction =T. Given that,

The conductivity of copper, K =390 W/m-°C, the conductivity of steel, K' = 46 W/m-°C.

   Since the rods are connected in series, the same amount of heat is flowing per second through both of them. Hence,

KA(T-0)/L = K'A(100-T)/L

→KT =100K'-K'T

→T(K+K') =100K'

→T =100K'/(K+K')

→T =100*46/(46+390)

→T =10.6°C.  




 

   24. An aluminum rod and a copper rod of equal length of 1.0 m and cross-sectional area 1 cm² are welded together as shown in figure (28-E3). One end is kept at a temperature of 20°C and the other at 60°C. Calculate the amount of heat taken out per second from the hot end. Thermal conductivity of aluminum = 200 W/m-°C and of copper =390 W/m-°C.  
Figure for Q-24



Answer: The length of each rod, L =1 m, Area of cross-section, A =1 cm² =0.0001 m², Temperature of one end, T = 20°C, the temperature of another end, T' = 60°C. Thermal conductivity of aluminum, K =200 w/m-°C, the thermal conductivity of copper, K' =390 W/m-°C. Let the thermal resistance of aluminum = R and that of copper =R'. Since the rods are connected in parallel, the equivalent thermal resistance (R") of the rods is given as,

1/R" = 1/R +1/R', 

→1/(L/K"A') =1/(L/KA)+1/(L/K'A)

{where A' is total area =2A and K" is the equivalent conductivity of the rods}

 →K"A' =KA +K'A

→K"*2A =KA+K'A

→K" = (K+K')/2 =(200+390)/2

→K" =295 W/m-°C.

Now the amount of heat taken out per second from the hot end,

=K"A'(T'-T)/L

=295*2*0.0001*40/1 J

=2.36 J.       




 

   25. Figure (28-E4) shows an aluminum rod joined to a copper rod. Each of the rods has a length of 20 cm and an area of cross-section 0.20 cm². The junction is maintained at a constant temperature of 40°C and the two ends are maintained at 80°C. Calculate the amount of heat taken out from the cold junction in one minute after the steady-state is reached. The conductivities are KAl = 200 W/m-°C and KCu = 400 W/m-°C. 
Figure for Q-25



Answer: Here length, L =20 cm =0.20 m, cross-sectional area, A =0.2 cm² =2x10⁻⁵ m².

The amount of heat taken out per minute from the cold junction,

=60{KAl*A(80-40)/L +KCu*A(80-40)/L}

=60*A*40(200+400)/L

=2400*2x10⁻⁵*600/0.20

=144 J. 




 

   26. Consider the situation shown in figure (28-E5). The frame is made of the same material and has a uniform cross-sectional area everywhere. Calculate the amount of heat flowing per second through a cross-section of the bent part if the total heat taken out per second from the end at 100°C is 130 J. 
Figure for Q-26


Answer: Let us assume that the amount of heat flowing per second in the bent part is Q' and in the straight part is Q". Also that the temperature of the junction at the left end = T' and of the right end =T". 

Heat flow through bent section,

Q' =KA(T"-T')/L', and through the straight section,

Q" =KA(T"-T')/L".

Hence, Q'/Q" = L"/L' =60/(60+2*5)

→Q'/Q" =60/70 =6/7

→Q" =7Q'/6.  

     Given that the total heat taken out per second from the end at 100°C = Q =130 J. It will also be the total heat flowing out per second through bent and straight sections taken together. Hence, 

Q =Q'+Q"

→130 =Q'+7Q'/6

→13Q'/6 = 130

→Q' =6*130/13 =60 J. 




 

   27. Suppose the bent part of the frame of the previous problem has a thermal conductivity of 780 J/m-s-° whereas it is at 390 J/m-s-°C for the straight part. Calculate the ratio of the rate of heat flow through the bent part to the rate of heat flow through the straight part. 



Answer: The conductivity of the bent portion, K = 780 J/m-s-°C, for the straight part, K' =390 J/m-s-°C. 

Length of the bent part, L = 70 cm =0.70 m

Length of the straight part, L' = 60 cm =0.60 m

If the uniform area of cross-section = A, then the rate of heat flow through the bent part 

I =Q'/t = KA(T"-T')/L

The rate of heat flow through the straight part,

I' =Q"/t = K'A(T"-T')/L'

Hence the ratio,

I/I' ={KA(T"-T')/L}/{K'A(T"-T')/L'}

     = KL'/K'L

     = 780*0.60/(390*0.70)

     = 2*6/7

     = 12/7

     = 12:7. 




 

   28. A room has a window fitted with a single 1.0 m x 2.0 m glass of thickness 2 mm. (a) Calculate the rate of heat flow through the closed window when the temperature inside the room is 32°C and that outside is 40°C. (b) The glass is now replaced by two glass panes, each having a thickness of 1 mm and separated by a distance of 1 mm. Calculate the rate of heat flow under the same conditions of temperature. Thermal conductivity of window glass = 1.0 J/m-s-°C and that of air = 0.025 J/m-s-°C.



Answer: Area of glass, A = 1*2 =2 m². The thickness of glass, L = 2 mm =0.002 m, Concuctivity of glass, K = 1 J/m-s-°C, Temperature difference, T'-T = 40-32 = 8°C.


(a) The rate of heat flow through the closed window = KA(T'-T)/L

   = 1*2*8/0.002 J/s

   = 16000/2 J/s

   = 8000 J/s.


 (b) When the glass pane is replaced with two glasses with air between them, then these three are aligned in series. In this case, the equivalent heat resistance (R) is equal to the sum of individual resistances. Hence,

R = R'+R"+R', where R' and R" are the heat resistance of glass and air respectively.

→R = L/KA +L/K'A + L/KA

  = 2*(0.001)/(1*2) + 0.001/(0.025*2)

   =0.001 +0.02

   =0.021 s-°C/J

Hence the heat flowing through the glass = (T'-T)/R

= 8/0.021 J/s

 = 381 J/s.    




 

   29. The two rods shown in figure (28-E6) have identical geometrical dimensions. They are in contact with two heat baths at temperature 100°C and 0°. The temperature of the junction is 70°. Find the temperature of the junction if the rods are interchanged.  

Figure for Q-29


Answer: The rate of heat flow in both the rods will be the same in each case.   

Diagram for Q - 29

Let the length and cross-sectional area of rods be L and A. Then equating the rate of heat flow in both rods,

Case - I, 

K₁A(100-70)/L =K₂A(70-0)/L
→30K₁ = 70K₂
→3K₁ =  7K₂
→K₂/K₁ =3/7 

Case - II,

Consider the temperature of the junction now = T°C,
K₂A(100-T)/L = K₁A(T-0)/L
→K₂(100-T) = K₁T  
→K₂/K₁ = T/(100-T)
→3/7 = T/(100-T)
→300-3T = 7T
→10T = 300
→T = 30°C.
 
 

 


 

   30. The three rods shown in figure (28-E7) have identical geometrical dimensions. Heat flows from the hot end at a rate of 40 W in the arrangement (a). Find the rates of heat flow when the rods are joined as in arrangement (b) and in (c). Thermal conductivities of aluminum and copper are 200 W/m-°C and 400 W/m-°C respectively.  
Figure for Q-30

Answer:  Let the heat resistance of aluminum and copper be Rₐ and R₍ and the equivalent heat resistance = R. The heat flow-rate in the first case,

Q/t = (100-0)/R = 40

→ R =100/40 =2.5

But 2Rₐ+R₍ = R =2.5

→2L/KₐA +L/K₍A =2.5

→2L/200A +L/400A = 2.5

→5L/400A =2.5

→L/A =2.5*400/5 =200


In the arrangement (b),

   Let the equivalent heat resistance of two rods in parallel = R'.

1/R' = 1/Rₐ+1/R₍

→R' = RₐR₍/(Rₐ+R₍)

      =(L/KₐA)(L/K₍A)/{(L/KₐA+L/K₍A)

     =(L/A)(1/KₐK₍)/(1/Kₐ+1/K₍)

     ={200/(200*400)}/(1/200+1/400)

     =(1/400)/{(2+1)/400}

     =1/3  

Hence the equivalent heat resistance of the arrangement, 

R = Rₐ +R' = L/KₐA +1/3

   = 200/200 +1/3

   = 1 +1/3

   = 4/3

Hence the rate of heat flow,

= (100-0)/R 

= 100/(4/3) W

= 300/4 W

= 75 W


In the arrangement (c)

All the rods are in parallel combination. The equivalent heat resistance R is given as

1/R = 1/Rₐ+1/R₍+1/Rₐ

     =2/Rₐ+1/R₍

     =2KₐA/L + K₍A/L

     =2*200A/L +400A/L

     =(400+400)A/L

     =800/(L/A)

     =800/200 

     = 4

Hence the heat flow per second from the arrangement,

=(100-0)/R

=100*(1/R)

=100*4 W

=400 W.        

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CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

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