A small-bore copper water pipe is located within a concrete wall as shown in figure Q2. After a prolonged spell of mild weather, during which the wall temperature has become uniform at 10 °C, the...

SOLUTIONS
Question No. 1 Section A (11 JAN 2010)
Answer - (a)
The conduction equation for a sphere with uniform volumetric heat generation (H) is
given by
(1
2){(∂/ ∂r) (r2 ∂T/∂r)} + (1
2Sin θ) {(∂/ ∂ θ) (Sine ∂T/∂ θ)} + (1
2Sin2θ) (∂2T/∂φ2) + (H/k) = (1/α)
(∂T/∂t)
As per question there is following conditions.
1. One dimensional
2. steady state or (∂T/∂t) = 0
So for sphere the conduction equation will be
(1
2)(∂/ ∂r) (r2 ∂T/∂r) +H/k = 0
Now we can find out the temperature distribution by using integration
After integration we have
2 ∂T/∂r = - (H/k)(r3/3) + C1
Again Integrating
T = - (Hr2/6k)-(C1
) + C2
At r=0, ∂T/∂r = 0 so C1 = 0
At r= R, T= TS
So C2= TS + HR
2/6k
So putting the value of C1 & C2
T = - (Hr2/6k) + TS + HR
2/6k
T- TS = (H/6k) [R
2-r2]
So temperature distribution can be written as
T- TS = (H/6k) [R
2-r2]
Answer - (b)
As per question we have
K= 16 W/mk
R= 2 cm
H= 1 MW/m3
h= 15W/m2k
As we know that energy balance equation
HV = Ah (TS-Tѳ)
106 x (4/3) ΠR3 = (4ΠR2) x15 (TS-20)
(106/3)(2x10-2) = 15(TS-20)
TS = 464.44
0C
Steady State temperature at the centre of sphere
T- TS = (H/6k) [R
2-r2] at the centre of sphere r= 0
T= TS + (H/6k) R
2
T= 464.44 + (106 x 4 x 10-4)/96
T = 464.44 + 4.167
T = 468.6 0C
Section A- 11 JAN-2010
Question no. 3 Answer
As per given question we have following data.
1. Continuous Belt speed = U
2. Liquid film thickness = h
3. Liquid density =  and viscosity is 
4. Flow is fully developed ,steady ,parallel and laminar with no shear stress
Answer A-
Governing Equation for the flow & boundary condition
Governing equation is as following
a) Navier-stokes equation-
 (DV

Dt)=  g

-

p +  V

2
But as given in question
 Flow is fully developed
 Navier equation for y-direction will be meaningless
Because v=0 and gy=0
so in this case navier stokes equation is as following
 gx +  (d
2u/dy2) = 0
) Shear stress
 XY= YX= [(∂v/∂x) + (∂u/∂y)]
But as per given in question v=0 so  YX = (∂u/∂y)
Boundary condition is as following
a) At Y =0 , u = U
) At y =h ,  YX =  ((∂u/∂y)=0
Because su
ounding atmosphere produces no shear stress at the free surfaces.
Answer B-
Equation to describe the velocity profile.
As we have stated above the Navier stokes equation
 gx +  (d
2u/dy2) = 0
But gx = -g (because acceleration due to gravity in X-direction will be downward as direction
given in question)
So
- g +  (d
2u/dy2) = 0
h
U
 And 
Y
X
d2u / dy2 =  g/
du/dy = ( g/)y + C1
We have boundary condition that, at y =h,  YX =  ((∂u/∂y) =0
C1= - g h/
So after putting value in above equation
du/dy = ( g/)y -  g h/
Again integrate it
u= ( g y2/2) – ( g h/) y +C2
As per boundary condition we have, At Y =0, u = U
So, u = U
u= ( g y2/2) – ( g h/) y +U
Where u is function of y
So Velocity profile equation is
u(y) = ( g y
2/2) – ( g h/) y +U
Answer C-
Volume flow Rate -
Q= ∫u. dA
Q= ∫u b dy, where b is depth
Q= ∫ [( g y2/2) – ( g h/) y +U] b dy
And limit will be y= 0 to y= h
Q= b [( g /2) (y3/3) - ( g h/) (y2/2) +U Y] and limit is y=0 to y=h
After putting the value of limit
Q= b [-( g h3/3) +Uh]
Where b is depth that we have assumed
Eq.No.1
Eq.No.2
So volume flow rate
Q= b [-( g h3/3) +Uh]
Question No. 5 - Section B (11 JAN 2010)
Answer - (a)
Heat and mass transfer analog-
We can understand heat and mass transfer separately or by combining these both.
Analogy means the similarity in the principle between these two process even these two
process are different with each other.
In case of Diffusion and convection –There is similarity between the equation and
principles for heat transfer & mass transfer
But in case of radiation there is no similarity.
Analogy is based on following-
1. Heat transfer takes place due to temperature difference & this temperature
difference produces the driving potential for heat transfer.
2. Mass transfer takes place due to concentration difference or we may say that this
difference produces the driving potential for mass transfer.
Now we can write over here the conductive and convective heat transfer equations
and respective equation for mass transfer also.
A. Constitutive equations –
Heat transfer
Heat transfer is governed by fourier law of heat conduction
Q= - (k/A)dT/dX
q= -k dT/dX
Where q is heat flux (W/m2), A is total area perpendicular to the direction of heat
transfer
k is thermal conductivity and dT/dX is temperature gradient
Mass transfer-
Mass transfer is governed by Fick’s Law of diffusion
J= - D (∂φ/∂X)
Where ∂φ/∂X is concentration gradient
D is Diffusivity or diffusion co-efficient
B. Conductive transfer equation-
Heat transfer-
In this case Heat diffusion equation is used
ϱ CP (∂T/∂X) = Q + ∇ (k. ∇T)
Where ϱ = density, CP = heat capacity, k= thermal conductivity and ∇T = temperature
gradient
Mass transfer –
In this case mass diffusion equation is used
(∂φ/∂T)+v ∇ φ = D ∇2φ
Where φ is concentration, ∇ φ is concentration gradient and v is fluid velocity vector.
C. Convective transfer equation-
Heat transfer-
It is governed by following equation.
q= h A (TS-Tѳ)
where q is heat transfer rate ,A is area of surface , h is convective heat transfer co-
efficient , TS is surface temperature and Tѳ is reference or free stream temperature
Mass transfer –
It is governed by following equation
M = h A (CS-Cѳ)
Where h is mass transfer co-efficient
Answer - (b)
T00 = 52
0C
u= 12 m/s
Pressure = 1 atm
L= 1 m
Let us consider for steady state
Properties = at (50+27)/2 = 39.5 0C
As we do not have table, so we will assume the properties of air. We can find the
properties of air from table provided for 39.5 degree centigrade.
Let the properties of air at 39.5 0C is as following
v= 16.92x10-6 m2/s, Pr= 0.711 and k = 0.0271 W/mk
(RE)L = u L/v = 12x1/ (16.92x10
-6)
(RE)L = 709219.85
NUX = [0.037 Rex
(4/5) - 871] Pr1/3
Putting the value of (RE)L and Pr
NUX = 805.49
As we know that NUX = h L/k
So h L/k = 805.49
Or h = 805.49x k/L
...