Heat transfer Problems

Solution-1
Let T be the temperature of the interface.
Then in equilibrium
Heat flowing from left= heat flowing in the right
then
$K_1 \frac {(T_1-T)}{L_1}=K_2 \frac {(T-T_2)}{L_2}$
or $T= \frac {K_1T_1/L_1+K_2T_2/L_2}{K_1/L_1+K_2/L_2}$

Now
$K_1 \frac {(T_1-T)}{L_1}=K_2 \frac {(T-T_2)}{L_2} =K \frac {(T_1-T_2)}{L_1+L_2}$
or
$K= \frac {L_1+L_2}{L_1/K_1+L_2/K_2}$

Heat current.
$Q=K \frac {(T_1-T_2)}{L_1+L_2}$
$Q= \frac {T_1-T_2}{L_1/K_1+L_2/K_2}$

Heat will be transferred to the gas through the bottom of the cylinder and as the temperature of the gas increased, it will displace piston(Volume increase) since pressure is constant(atmospheric pressure)
Let x be the distance moved the piston at time t
Let T be the temperature of the gas at that moment.

Suppose an amount of heat dQ is transferred in time dt.

Then
$dQ=C_P dT$

Also $\frac {dQ}{dt}=\frac {KA(T_s-T)}{L_1}$

So
$C_P dT=\frac {KA(T_s-T)}{L_1}$ ----1

Also let $V_0$ is the initial volume of the gas.
Then $P_a V_0=RT_0$

or $V_0=\frac {RT_0}{P_a}$

Let V be the volume at time t
then
$ \frac {V}{T}=\frac {V_0}{T_0}$ as Pressure remains constant throughout the process

Substituting V₀ from last expression
then
$V= \frac {RT}{P_a}$
$V_0 + Ax= \frac {RT}{P_a}$
then

$ \frac {RT_0}{P_a}+ Ax= \frac {RT}{P_a}$

$T=T_0 + \frac {P_aAx}{R}$

Differentiating both sides
$Adx=\frac {RdT}{P_a}$

Putting this is 1
$C_PAP_aL_1dx=KAR[T_s-T_0-(\frac {P_a Ax}{R})]dt$
$KRdt = \frac {C_P P_a L_1 dx}{T_s - T_0 - (P_a A x/R)}$
Integrating both sides with right side as upper and lower limit (t,0) and left side as upper and lower limit (x,0)
$\int_{0}^{t} KRdt = \int_{0}^{x} \frac {C_P P_a L_1 dx}{T_s - T_0 - (P_a A x/R)}$

$ \frac {kAt}{L_1C_P}=ln[\frac {(T_s-T_0)}{(T_s-T_0-P_aAx/R)}]$
or $\frac {T_s-T_0}{T_s-T_0-P_aAx/R}= e^{kAt/L_1C_P}$
or $T_s-T_0-(P_aAx/R)=(T_s-T_0)e^{-kAt/L_1C_P}$
or
$x= \frac {R(T_s-T_0)(1-e^{-2kAt/7L_1R})}{P_a A}$

from ----1

$ \frac {2KAdt}{7RL_1}=\frac {dT}{(T_s-T_0)}$

Integrating both sides with right side as upper and lower limit (t,0) and left side as upper and lower limit (T,$T_0$)

$ln \frac {T_s-T_0}{T_s-T}=\frac {2KAt}{7RL_1}$
or $T=T_s-(T_s-T_0)e^{-2KAt/7RL_1}$

We have according to Newton laws of cooling.
$ \frac {dT}{dt}=-k(T-T_a)$
$ \frac {dT}{(T-T_a)}=-kdt$

Integrating both sides with right side as upper and lower limit ($T_1$,$T_0$) and left side as upper and lower limit ($t_1$,0)

$ \int \frac {dT}{(T-T_a)}= \int -kdt$

$ln [\frac {(T_1-T_a)}{(T_0-T_a)}]=-kt_1$
$k = \frac {ln [(T_1-T_a)/(T_0-T_a)]}{t_1}$


The body continues to lose heat till it temperature becomes equal to surrounding.
So maximum heat body can lose

$\Delta Q_m=ms(T_0-T_a)$
Now if the body loses 80% of this heat in time $t_2$ and temperature is $T_{80}$
then
$.80 \Delta Q_m=ms(T_0-T_{80})$
$.8(T_0-T_a)=(T_0-T_{80})$

$T_{80}=.2T_0+.8T_a$.

Now again from Newton law of cooling
$ \frac {dT}{(T-T_a)}=-kdt$

Integrating both sides with right side as upper and lower limit (T₈₀,T₀) and left side as upper and lower limit ($t_2$,0)

$ \int \frac {dT}{(T-T_a)}=\int (-k)dt$
$ln \frac {(T_{80}-T_a)}{(T_0-T_a)}=-kt_2$

Substituting the value of k and T₈₀
$t_2=ln( \frac {5}{k})$

Now
$T- T_a=(T_1-T_a)e^{-kt}$
Taking logarithm on both sides
$ln(T- T_a)=ln(T_1-T_a) -kT$

which is similarly to
y=mx+c


So graph will be a straight line.

We have

Q=-αAT^1/2dT/dx
Minus sign is there Temperature decreases with distance from the end $T_1$
Qdx=-αT^1/2dT---1
Integrating both sides with right side as upper and lower limit (l,0) and left side as upper and lower limit ($T_2$,$T_1$)
∫Qdx=-∫αAT^1/2dT
$Q=\frac {2 \alpha A(T_1^{3/2}-T_2^{3/2})}{3l}$

Now again Integrating equation 1 with right side as upper and lower limit (x,0) and left side as upper and lower limit (T,$T_1$)

∫Qdx=-∫αAT^1/2dT
$Qx= \frac {2 \alpha A(T_1^{3/2}-T^{3/2})}{3}$

Substituting the value of Q from last expression
$T^{3/2}=T_1^{3/2} +( \frac {x}{l})(T_2^{3/2}-T_1^{3/2})$

or $T=T_1[1+(x/l)[(\frac {T_2}{T_1})^{3/2} -1]]^{2/3}]$

Thermal Resistance of Rod AB=(L/KA)
Thermal Resistance of Rod BC=(2L/KA)
Thermal Resistance of Rod BD=(2L/KA)
Thermal Resistance of Rod CE=(L/KA)
Thermal Resistance of Rod DE=(L/KA)
Thermal Resistance of Rod EF=(L/KA)

Now Rod BC and CE are in series So Total Thermal resistance
R_BCE=R_BC+R_CE
=(3L/KA)

Similarly BD and DE are in series
then
R_BDE=R_BD+R_DE
=(3L/KA)

Now Rod BCE and BDE are in parallel, so Net thermal resistance between ends B and E
(1/R)=(KA/3L)+(KA/3L)
R=(3L/2KA)


Now AB ,BE and EF are in series ,so net thermal resistance of the system is
R=(L/KA) + (3L/2KA) + (L/KA)
R=(7L/2KA)

So Net heat flow through the system
Q=(T_1-T_2)/R
Q=(2KA/7L)(T_1-T_2)

Now heat flow in Rod AB
Q=KA(T_B -T_1)/L
Substituting the value of Q from equation 1

T_B=(9T_1-2T_2)/7

Similarly for the heat flowing rod EF

Q=KA(T_E -T_2)/L
Substituting the value of Q from equation 1
T_E=(2T_1+5T_2)/7

Now Heat flow through rod BC =Heat flow through rod CE
(KA/2L)(T_B-T_C)=(KA/L)(T_C-T_E)
Substituting the values of T_B and T_E
we get
T_C=(13T_1+8T_2)/21

Similarly the temperature at D
T_D=(13T_1+8T_2)/21

Ratio of heat transfer =1:1

Let us draw two spherical shell with radii r and r+dr concentric with given system and Let T and T +dT be the temperature at them the heat flow through it

Q=-K(4πr²)(dT/dr)
-4πKdT=Q(dr/r²)

Integrating both sides with right side as lower and Upper limit (T_a,T_b) and left side as Lower and upper limit (a,b)

4πk(T_a-T_b)=Q[(1/a)-(1/b)]
$Q= \frac {4\pi kab(T_a-T_b)}{b-a}$

Considering a small portion of rod at distance x of thickness dx from the end whose temperature is T₁

Heat flow through the portion of the Rod is

Q=-K₂(d²x/l)(dT/dx) -K₁d(d -dx/l)(dT/dx)
as the edge at the portion at x for K₂ is given by dx/l
-Q=dT/dx[K₂(d²x/l) +K₁d² -K₁(d²x/l)]
or
$-dT = \frac {Qdx}{K_2^2d^2x/l +K_1d^2 -K_1d^2x/l}


Integrating both sides with right side as lower and Upper limit (T₁,T₂) and left side as lower and upper limit (0,l)
 
$Q = \frac {d^2(T_1-T_2)(K_2-K_1)}{l[ln (K_2/K_1)]}$

Now Q is also given by

Q=Kd²(T₁-T₂)/l

where K is the equivalent thermal conductivity of the system

So comparing
K=(K₂-K₁)/ln (K₂/K₁)
 

Amount of the radiation by the block
$=e \sigma AT_b^4$
where T_b is body temperature.

Substituting all the values
Amount of the radiation by the block=28.12 W

Amount of radiation absorb from the surrounding
$=e \sigma AT_s^4$
where T_s is surrounding temperature.
Substituting all the values
Amount of the radiation absorb from the surrounding =3.64 W

Net heat flow outside the block=Amount of the radiation by the block - Amount of radiation absorb from the surrounding
=24.38

Now $ms \frac {dT}{dt}=24.38$
$ \frac {dT}{dt}=\frac {24.38}{ms}$
=.06 °C/sec

If the chamber is at the same temperature as the block.
then Heat absorbed=28.12 W
Heat radiated=28.12 W

Let k be the Thermal conductivity of the Rod.
Then

Heat flow through conduction in the Rod=Heat radiation from the Rod
$ \frac {KA (T_1-T_2)}{L}=e \sigma A(T_2^4-T_s^4)$
So
$K= \frac {e \sigma L(T_2^4-T_s^4)}{(T_1-T_2)}$
br /> Also

$Q=e \sigma A(T_2^4-T_s^4)$

Q=KAK^'

$P=e \sigma AT^4$
$T=(\frac {P}{e \sigma A})^{1/4}$

If the body behaves like black-body.
$P_1= \sigma AT^4$

Substituting the value of T from last expression
$P_1= \frac {P}{e}$

% increase in Power
$ =\frac {P/e -P}{P} \times 100 = 150$ %

For the same volume, Surface area of circular plate is maximum and Sphere is minimum.
Now rate of cooling is
$=e \sigma AT_2^4$

So rate of cooling will be maximum for circular plate then the cube and it will be minimum for sphere.

$Q=-K \times 2 \pi r d \times (\frac {dT}{dr})$

$\frac {Qdr}{r}=-2 \pi k d dT$
Integrating both sides with right side as lower and Upper limit ($R_1$,$R_2$) and left side as lower and Upper limit ($T_1$,$T_2$)


$\int \frac {Qdr}{r}= \int (-2 \pi kd)dT$
$Q ln(\frac {R_2}{R_1})=2 \pi kd (T_1-T_2)$

So
$Q= \frac {2 \pi kd(T_1-T_2)}{ln(\frac {R_2}{R_1})}$

Now Integrating (1) both sides with right side as lower and Upper limit ($R_1$,R) and left side as lower and upper limit ($T_1$,T)


$\int \frac {Qdr}{r}= \int (-2 \pi kd)dT$
$Q ln (\frac {R}{R_1})=2\ pi kd(T_1-T)$

Substituting the value of Q

$T=T_1 -(T_1 -T_2) \frac {ln (R/R_1)}{ln (R_2/R_1)}$

Rate of heat loss
$ms \frac {dT}{dt}=4 \pi r^2 \sigma(T^4 - T_s^4)$
Then
Initial rate of cooling
$\frac {dT}{dt}= \frac {4 \pi r^2 \sigma(T^4 - T_s^4)}{ms}$

So $\frac {dT}{dt}$ is proportional to $ \frac {r^2}{m}$

For sphere X
$(\frac {dT}{dt})_X= \frac {kr_1^2}{m_1}$

For sphere Y
$(\frac {dT}{dt})_Y= \frac {kr_2^2}{m_2}$

Now let D be the density of the material

$m_1=\frac {4}{3} \pi r_1^3 D$

$r_1=(\frac {3 \pi D m_1}{4})^{1/3}$

Similarly
$r_2=(\frac {3 \pi Dm_2}{4})^{1/3}$

Ratio of initial rate of cooling
$R= \frac {r_1^2 m_2}{r_2^2 m_1}$ $R=\frac {m_1^{2/3} m_2}{m_2^{2/3} m_1}$ $R= =(\frac {m_2}{m_1})^{1/3}$
$=(\frac {1}{n})^{1/3}$

Comparing from the given expression
n=4

Ratio of heat transfer AXB and AB
$=\frac { \frac {KA(T_a -T_b)}{\pi r}}{ \frac {KA(T_a-T_b)}{2r}}$
$= \frac {2}{\pi} $

Similarly for BYD and BD
$=\frac {2}{\pi}$