PTUHEAT TRANSFER PAPER SOLUTIONSECTIONA
PTUHEAT TRANSFER PAPER SOLUTION
SECTIONA
Solution of University question paper
increases clarity. It increases the deep
understanding. Thus these concepts are
used in real practical applications.
Q1. Write briefly :
a) Explain Thermal conductivity. How it is different from apparent thermal conductivity?
b) What is meant by critical thickness of insulation? How it is calculated in case of a cylinder?
c) State Stefan Boltzmann’s law.
d) Why thin fins are preferred over a thick fin?
e) What is the limitation of LMTD method?
f) Define shape factor of radiation. Calculate all the shape factors of an isosceles
vertical triangle.
g) State Grashoff’s. What are the forces associated with it?
h) Define effectiveness and NTU of a heat exchanger.
i) Discuss hydrodynamic boundary layer. Which non dimensional number governs the relative magnitude of hydrodynamic and thermal boundary layers?
j) State Buckingham pi theorem. Explain repeating variables and their method of selection.
SOLUTION
(a) Definition of k, Thermal conductivity
Fourier law gives q^{.} = – k A ∂t/∂x
k becomes equal to q^{.} when A=1, ∂t=1 and ∂x =1
Hence k is defined as rate of heat transfer
through a wall thickness of 1 meter,
in a surface of area of 1 m^{2},
due to a temperature difference of 1^{0}C
i.e. k=q^{.}
Hence Fourier law gives the definition of thermal conductivity k. It is the ability of a material to conduct heat.
How it is different from apparent thermal conductivity?
Apparent thermal conductivity is variable with respect to temperature, composition and moisture content. Theoretical thermal conductivity is assumed to be constant.
(b) What is meant by critical thickness of insulation? How it is calculated in case of cylinder?
CRITICAL THICKNESS OF INSULATION
It is that thickness of insulation at which there is a least thermal resistance. The rate of heat transfer is maximum. There is no critical radius of insulation for a plane wall. However, it is applicable only to cylindrical and spherical bodies. Adding insulation in a cylinder/sphere increases conductive resistance. At the same time it decreases the convective resistance by increasing the surface area. The heat transfer may increase or decrease depending upon which effect is more.
For a cylindrical surface r_{cr} = k/h_{o }
For a spherical surface r_{cr }=2k/h_{o}
CALCULATION OF CRITICAL THICKNESS FOR A CYLINDRICAL SURFACE
Inside radius =r_{i}, Insulation thickness ‘ x ‘,Outer radius with insulation r_{o}
Inside hot temperature =t_{h}
Outside cold air temperature = t_{c}
_{q}^{.}_{ (at any radius ‘r’ will be) =}(t_{h} – t_{c}) / ((ln(r_{o}/r)/2𝝅kL +1/2𝝅h_{o}rL))
HT will be maximum if dq^{.}/dr = 0
We get, for a cylinder, critical radius r_{cr} = k/h_{o}
Critical radius is outer radius with insulation.
r_{cr} = r_{o}= outer radius with insulation
© Define Stefan Boltzmann’s law.

This law gives the emissive power (heat flux) of a black body. It is as proportional to the FOURTH power of the ABSOLUTE TEMPERATURE.

E_{b} ∝ T^{4}.

Its symbol is E. Mathematically
E_{b}=emissive power (heat flux) = σ_{b} T^{4} W/m^{2}
Where σ_{b} is Boltzmann’s constant, σ_{b}=5.67 x 10^{8} W/m^{2}K^{4}
T is the absolute temperature in K = 273+ ^{0}C
iv. Black body emissive power is correlated for a grey body.
grey body, E_{g}=emissive power (heat flux) = ε σ_{b} T^{4} W/m^{2}
Where ε= emissivity = (E_{g}/ E_{b}) _{T=Constant}
d) Why thin fins are preferred over a thick fin?
Thin fins are more effective and also have more rate of heat transfer.
Effectiveness of a fin = Є_{f} = (kP/hA_{c})^{1/2} and A_{c}=t x b
Thus Є_{f} is inversely proportional to the thickness
e) What is the limitation of LMTD method?
The limitation is that the inlet as well as outlet temperatures of both hot and cold fluids should be known.
LMTD = (θ_{max}—θ_{min})/ ln (θ_{max}/θ_{min})
For a parallel flow heat exchanger
θ_{max} =t_{hi} –t_{ci}
θ_{min} = t_{ho} –t_{co}
f) Define shape factor of radiation. Calculate all the shape factors of an isosceles
vertical triangle.
In general, its symbol is F_{ij } and has no units.
It is the ratio of energy received by body J out of energy emitted by body I
OR
F_{12} = fraction of radiation leaving surface 1 intercepted by surface 2
Say for example 140 units of energy emitted by body A. 35 units of energy are received by a body B. Then F_{AB}= 35/140=0.25
Number of shape factors
For n surfaces, n^{2} shape factors are required.
Example for three bodies, 9 shape factors are required.
Shape factors for a vertical isosceles triangle
Consider a three sided enclosure in the form of a vertical isosceles triangle. One angle will be 90^{0}, while the other two angles will be 45^{0} each.
There will be 9 shape factors for a 3 sided enclosure.
Let the vertical side is named as 2.
Therefore F21 + F22 + F23 =1
Firstly F22 =0
Secondly F21 + F23 =1
Thirdly F21 = F23
Therefore F21 =F23 = 0.5
(F22 =, F21 = F23 0.5)
A1 F12 = A2 F21 (By reciprocity theorem)
F12 = (A2/A1) F21 = √2 F21 = 1.414 x 0.5 = 0.7070
F13 = f12 =0.7070
(F11 = 0, F13 = f12 =0.7070)
Similarly (F33 = 0, F31 = f32 = 0.7070)
( g) Define Grashoff’s Number. What are the forces associated with it?
GRASHOFF’S NUMBER, Gr
It is used in analyzing free convection.
Gr= Buoyant force x inertia force/ (viscous force)^{2}
Gr = L^{3}g β ΔT/ν^{2}
Where
L is the length of the plate
g is acceleration due to gravity
β = 1/T_{av}
ΔT is the temperature difference
ν is the kinematic viscosity
Forces associated are Buoyant force, inertia force and viscous force.
h) Define effectiveness and NTU of a heat exchanger.
Effectiveness Є= q^{.}_{actual }/q^{.}_{max}
q^{.}_{max} = C_{min }(t_{max} – t_{min})=C_{min}(t_{hot }in – t_{cold in})
(i) Parallel flow heat exchanger, Effectiveness
Є = (1 –e^{—NTU(1+C)})/(1+C)
(ii) counter flow heat exchanger, effectiveness
Є = (1 –e^{—NTU(1–C)})/(1–Ce^{—NTU(1—C)})
C = C_{min}/C_{max}
^{NTU—It is one of the methods of analyzing heat exchangers.}
Method NTU is used when LMTD cannot be calculated or the four temperatures are not given.
Where NTU = Number of transfer units (dimensionless) which represents the size of heat exchanger. NTU measures the effectiveness of the heat exchanger. It is a fictitious term.
NTU= U A/ C_{min}
Where U is the overall heat transfer coefficient
A is the area of the heat exchanger
C_{min} is the minimum heat capacity among the hot and cold fluids in the heat exchanger
i) Define hydrodynamic boundary layer. Which non dimensional number governs the relative magnitude of hydrodynamic and thermal boundary layers?
Boundary layer is 3 dimensional region (space) over a plate (or inside a pipe) where velocity gradient is present. It is due to viscous effect of the fluid. In this velocity is zero at the solid boundary. There is free velocity of the fluid at the outer surface of the boundary layer. Thus it is also called a velocity boundary layer. Wall shear stress is the most important parameter of velocity boundary layer.
Comparison of hydrodynamic and thermal boundary layers

Hydrodynamic boundary layer 
Thermal boundary layer 
1. 
Velocity variation 
Temperature variation 
2. 
Used in fluid flow as well as in heat transfer 
Used in heat transfer 
3. 
δ (B.L.THICKNESS) 
δ_{th}(T.B.L. thickness) 
4. 
δ = δ_{th}Pr^{1/3} 
δ_{th} = δ Pr^{ –1/3} 
5. 
Reynolds number governs the hydrodynamic layer 
Nusselt and Prandtl numbers govern the thermal boundary layers. 
j) State Buckingham pi theorem. Explain repeating variables and method of their selection.
Buckingham π theorem is the most commonly used method of dimensional analysis.
Buckingham theorem states
(i) An equation has total n physical variables
(ii) m is the fundamental quantities (M,L,T, angle, Temperature)
(iii) Then there will be π terms (dimensionless groups) = n – m.
Solve simple and complex problems in fluid flow and heat transfer by this method. It is an improvement over Rayleigh’s Method. Rayleigh’s method fails to specify the number of dimensionless groups formed in the equation.
Repeating variables: The repeating variables are a set of variables which among themselves cannot form a dimensionless group. Length, diameter, displacement, velocity, acceleration and density if arranged in any manner cannot cancel the dimensions. Hence cannot form dimensionless groups. Hence these are repeating variables.
(k) How to select repeating variables:

First identify the variables involved in the process under consideration i.e. heat transfer or fluid flow or pressure drop. Now mark few of these as repeating variables.

For example, forced convection heat transfer coefficient depends upon

(i) d (pipe diameter),

(ii) V (the fluid velocity),

(ii) ρ(density of the fluid),

(iv) μ(viscosity of the fluid),

(v) c_{p}( Specific heat)

(vi) k (thermal conductivity.

Now Repeating Variables can be d, V and ρ. This selection comes by experience or by practice. Normally consider linear dimension, velocity and density to be the Repeating Variables. After some experience looking at the variables, one can easily select the repeating variables.
https://mesubjects.net/wpadmin/post.php?post=6434&action=edit PTU HT PAPER SOL A
https://mesubjects.net/wpadmin/post.php?post=6732&action=edit PTU HT PAPER SOL B
https://www.mesubjects.net/wpadmin/post.php?post=675&action=edit Q.A HT