Thursday, August 20, 2020

H C Verma solutions, HEAT TRANSFER, EXERCISES, Q31-Q40, Chapter-28, Concepts of Physics, Part-II

 Heat Transfer


EXERCISES, Q31 - Q40


   31. Four identical rods AB, CD, CF and DE are joined as shown in figure (28-E8). The length cross-sectional area and thermal conductivity of each rod are l, A and K respectively. The ends A, E and F are maintained at temperatures T₁, T₂ and T₃ respectively. Assuming no loss of heat to the atmosphere, find the temperature at B.    
Figure for Q-31



Answer:  Let the temperature at B = T. Length of AB = l, length of BDE and BCF = l+l/2 =3l/2.

Assume heat flow rate in AB = Q, in BDE =Q' and in BCF = Q". Then,

Q = Q'+Q"

→KA(T₁-T)/l = KA(T-T₂)/(3l/2) +KA(T-T₃)/(3l/2)

→T₁-T = 2(T-T₂ +T-T₃)/3 

→3T₁ -3T = 4T-2T₂-2T₃

→7T = 3T₁+2(T₂+T₃)

→T = {3T₁ +2(T₂+T₃)}/7      




 


    32. Seven rods A, B, C, D, E, F and G are joined as shown in figure (28-E9). All the rods have equal cross-sectional area A and length l. The thermal conductivities of the rods are KA = KC = Ko, KB = KD = 2Ko, KE = 3Ko, KF = 4Ko and KG =5Ko. The rod E is kept at a constant temperature T1 and the rod G is kept at a constant temperature T2 (T2>T1).

(a) Show that the rod F has a uniform temperature T =(T₁+2T₂)/3. 

(b) Find the rate of heat flow from the source which maintains the temperature T₂.  
Figure for Q-32



Answer: (a) Rods A and C are identical and rods B and D are identical. Hence the heat flow rate in steady-state in A+B and C+D will also be identical. It means the junction of A and B, as well as the junction of B and C, will have the same temperature. That is the ends of rod F will have the same temperature resulting in a uniform temperature in rod F = T (say). 

        Due to the uniform temperature in F, there will be no heat flow through it. Hence the heat flow through A = heat flow through B.

→K₀A(T-T₁)/l = 2K₀A(T₂-T)/l

→T-T₁ =2T₂ -2T

→3T = T₁ +2T₂

→T = (T₁+2T₂)/3.  


(b) With the above value of T,

T₂ -T = T₂-(T₁+2T₂)/3

        =(T₂-T₁)/3

The rate of heat flow from the source will be the sum of heat flow in rod B and D. The heat flow rate in both of these two rods are the same. Hence the rate of heat flow from the source

 = 2*2KₒA(T₂-T₁)/3l

 = 4KₒA(T₂-T₁)/3l.   





   33. Find the rate of heat flow through a cross-section of the rod shown in figure (28-E10) (θ₂>θ₁). The thermal conductivity of the material of the rod is K.  
The figure for Q-33



Answer:  Consider a section at a distance x from the left end and across a thickness dx, let the temperature difference be = dθ. Radius y here increases by dy across the thickness dx. See the diagram below. 
Diagram for problem-33

From the similar triangles,

dy/dx = (r₂-r₁)/L

→Ldy = (r₂-r₁)dx 

 Heat flow rate across this section,

Q =K(πy²)dθ/dx

Now y = r₁+x(r₂-r₁)/L

 →Q = πK{r₁+x(r₂-r₁)/L}²dθ/dx

→Qdx/{r₁+x(r₂-r₁)/L}² =πKdθ

Integrating we get,

→-[Q/{r₁+x(r₂-r₁)/L}(r₂-r₁)/L] =πK[θ]

Putting the limits,

{-QL/(r₂-r₁)}*{1/(r₁+r₂-r₁) -1/r₁} =πK(θ₂-θ₁)

→{-QL/(r₂-r₁)}{1/r₂-1/r₁} =πK(θ₂-θ₁)

→{QL/(r₂-r₁)}{(r₂-r₁)/r₁r₂} =πK(θ₂-θ₁)

→ Q = πKr₁r₂(θ₂-θ₁)/L.   






   34. A rod of negligible heat capacity has length 20 cm, area of cross-section 1.0 cm² and thermal conductivity 200 W/m-°C. The temperature of one end is maintained at 0°C and that of the other end is slowly and linearly varied from 0°C to 60°C in 10 minutes. Assuming no loss of heat through the sides, find the total heat transmitted through the rod in these 10 minutes.  



Answer: Length, L = 20 cm =0.20 m,

Area of cross section, A =1 cm² =0.0001 m².

K =200 W/m-°C.       

T =0°C and T' varires from 0°C to 60°C in 10 min i.e. in 600 s, 

so T' =(t/600)*60°C =t/10 °C

Suppose dQ heat flowsin dt time, then   

dQ/dt =KA(T'-T)/L

→dQ = {KA(t/10 -0)/L}*dt

→dQ = (KA/10L)tdt, integrating,

→Q = (KA/10L)[t²/2], putting limit t=0 to 600 s we get

→Q =KA(600)²/20L

  =200*0.0001*360000/(20*0.2)

  =1800 J.  





   35. A hollow metallic sphere of radius 20 cm surrounds a concentric metallic sphere of radius 5 cm. The space between the two spheres is filled with nonmetallic material. the inner and outer spheres are maintained at 50°C and 10°C respectively and it is found that 100 J of heat passes from the inner sphere to the outer sphere per second. Find the thermal conductivity of the material between the spheres.   



Answer: 
Figure for Q-34
   Consider a hollow sphere of nonmetallic material between the two spheres of radius r and thickness dr. The temperature difference across the thickness = dθ. If the heat flow rate = H, then

H =K(4πr²)dθ/dr

→Hdr/r² =4πKdθ

Integrating,

-H[1/r] = 4πK[θ]

Putting the value of r from 0.05 m to 0.20 m and θ from 50°C to 10°C. 

-H{1/0.20 -1/0.05} = 4πK(50-10)

→-H(5 -20) =4πK*40

→15H =160πK

→K = 15H/160π

→K = 15*100/160π

      = 2.98 ≈ 3.0W/m-°C.




 


   36. Figure (28-E11) shows two adiabatic vessels, each containing a mass m of water at different temperatures. The ends of a metal rod of length L, area of cross-section A and thermal conductivity K, are inserted in the water as shown in the figure. Find the time taken for the difference between the temperatures in the vessels to become half of the original value. The specific heat capacity of water is s. Neglect the heat capacity of the rod and the container and any loss of heat to the atmosphere.  
Figure for Q-36


 

Answer: Let the temperature of the left box =T' and right box =T''. Difference T =T"-T'. 

Initially the rate of heat flow in a small interval of time dt,

dQ/dt =KA(T"-T')/L

→dQ ={KA(T"-T')/L}dt

If the temperature of left box in this time dt is H' then,

ms(H'-T') ={KA(T"-T')/L}dt

→H' = T' + {KA(T"-T')/Lms}dt

So the temperature of the left box increases by {KA(T"-T')/Lms}dt and the temperature of the right box decreases by this amount because the mass and specific heat of the water on both sides is the same. Thus the temperature difference will be double this amount.

→dT =-2{KA(T"-T')/Lms}dt

{Negative sign is for with the increase in time the temperature difference decreases} 

→dT =-{2KAT/Lms)dt

→dt = -(Lms/2KA)(dT/T)

Integrating we get.

t =-(Lms/2KA)*[lnT]

Since T is the initial temperature difference and T/2 the final difference, the limits of T on the right side will be from T to T/2. Putting the limits,

t = -(Lms/2KA){ln(T/2)-lnT}

  =(Lms/2KA){lnT -ln(T/2)}

  =(Lms/2KA)*ln2.    

 

 



   37. Two bodies of masses m₁ and m₂ and specific heat capacities s₁ and s₂ are connected by a rod of length l, cross-sectional area A, thermal conductivity K and negligible heat capacity. The whole system is thermally insulated. At time t = 0, the temperature of the first body is T₁ and the temperature of the second body is T₂(T₂>T₁). Find the temperature difference between the two bodies at time t.  



Answer: At time t = 0, the temperature difference T = T₂-T₁. Suppose initially in a very small time interval dt the amount of heat transferred is dQ. So,

dQ/dt =KAT/l 

→dQ = KATdt/l

If the temperature of the first body after time dt is T', then 

m₁s₁(T'-T₁) =KATdt/l

→T' =T₁+(KAT/lm₁s₁)dt

Similarly, the temperature of the second body after time dt will be

T" = T₂-(KAT/lm₂s₂)dt

Now the difference in the temperature =T"-T'.

=(T₂-T₁)+(KAT/l)(-1/m₂s₂ -1/m₁s₁)dt

=(T₂-T₁)-{KA(m₁s₁+m₂s₂)/lm₁s₁m₂s₂}T/dt

=(T₂-T₁)-λT/dt 

Where λ ={KA(m₁s₁-m₂s₂)/lm₁s₁m₂s₂}

→(T"-T')-(T₂-T₁)= λT/dt

L.H.S. is the small change in the temperature difference =-dT. The sign of dT is negative because the first term is smaller than the second term on the L.H.S. 

→-dT/T =λdt 

Integrating,

→ln(T) = -λt

The limit of integration is,

 at t = 0, T=T₂-T₁, 

and at time t = t, T =T, so

ln{T/(T₂-T₁)} =-λt

By definition of log

→T/(T₂-T₁) = e-λt

→T = (T₂-T₁)e-λt

   



  

   38. An amount n (in moles) of a monoatomic gas at an initial temperature T₀ is enclosed in a cylindrical vessel fitted with a light piston. The surrounding air has a temperature Tₛ (>Tₒ) and the atmospheric pressure is pₐ. Heat may be conducted between the surrounding and the gas through the bottom of the cylinder. The bottom has a surface area A, thickness x and thermal conductivity K. Assuming all changes to be slow, find the distance moved by the piston in time t. 



Answer: The process is at constant pressure. Suppose dQ heat is transferred in small time interval dt, then the heat flow rate,

dQ/dt = KA(Tₛ-T)/x 

If the temperature of the gas after time dt is T' then,

nCₚ(T'-T) = KA(Tₛ-T)dt/x ---(i)

For a monoatomic gas, Cₚ =5R/2, where R is the universal gas constant.

The small change in temperature in time dt, dT =T'-T

Now (i) becomes

5nRdT/2 = KA(Tₛ-T)dt/x

→dT/(Tₛ-T) =(2KA/5Rnx)dt   

Integrating we get,

[-ln(Tₛ-T)] = (2KA/5Rnx)[t]

Now the limit of integration is at t = 0, T = Tₒ and at t = t, T = T. Putting the limits,

ln(Tₛ-T)/(Tₛ-Tₒ) =-(2KA/5Rnx)t

→Tₛ-T =(Tₛ-Tₒ)e-2KAt/5Rnx

→T =Tₛ-(Tₛ-Tₒ)e-2KAt/5Rnx 

Subtracting Tₒ from both sides,

T-Tₒ =(Tₛ-Tₒ)-(Tₛ-Tₒ)e-2KAt/5Rnx
→ΔT = (Tₛ-Tₒ)(1-e-2KAt/5Rnx)
But pₐV =nRΔT, and V =A*L, where L is the distance moved by the cylinder, so
pₐAL/nR =(Tₛ-Tₒ)(1-e-2KAt/5Rnx)
→L =(nR/pₐA)(Tₛ-Tₒ)(1-e-2KAt/5Rnx). 


 


   39. Assume that the total surface area of a human body is 1.6 m² and that it radiates like an ideal radiator. Calculate the amount of energy radiated per second by the body if the body temperature is 37°C. Stefan constant 𝜎 is 6.0x10⁻⁸܁ W/m²-K⁴.  



Answer: Area of the body, A =1.6 m². Body temperature, T = 37°C =273+37 K =310 K. Given, Stefan constant, 𝜎 =6.0x10⁻⁸ W/m²-K⁴.

For an ideal radiator, thermal radiation emitted per unit time is given as,

u = 𝜎AT⁴

   =6.0x10⁻⁸*1.6*310⁴ J/s

   = 887 J/s



     

 

   40. Calculate the amount of heat radiated per second by a body of surface area 12 cm² kept in thermal equilibrium in a room at temperature 20°C. The emissivity of the surface =0.80 and 𝜎 =6.0x10⁻⁸ W/m²-K⁴.  



Answer: Surface area of the body, A =12 cm² =0.0012 m². The temperature of the body, T =20+273 K = 293 K. Emissivity of the body, e =0.8 and 𝜎 =6.0x10⁻⁸ W/m²-K⁴. 

The rate of heat radiation is given as, 

u =e𝜎AT⁴

   =0.8*6.0x10⁻⁸*0.0012*293⁴ J/s

  = 0.42 J/s.

  

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CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


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CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




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CHAPTER- 3 - Kinematics - Rest and Motion

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