PTU HEAT TRANSFER QUESTION PAPER SOLUTIONSECTIONA
PTU HEAT TRANSFER
QUESTION PAPER
SOLUTIONSECTIONA
Solution of university question is very
helpful. It increases the clarity and
understanding. Thus the knowledge is
applied easily in practical applications.
Q 1. Write briefly:
(a) What is conduction shape factor?
ANS:
Conduction shape factor will be there for a plain wall, cylindrical body and a spherical body
We know
In Convection q^{.} =h A dT (i)

For a plain wall
In conduction through a plain wall , q^{.} = k A (dT/dx) (ii)
Compare (i) and (ii) and write conduction equation as q^{.} = k S dT
Here S is the shape factor
Therefor shape factor S For a plain wall will be S_{plain wall} = – k/dx

Shape factor For a cylindrical body
q^{.} = 2πkL dt/ln(r_{0}/r_{i})= k S dt
S_{cyl} = 2πL/ln(r_{0}/r_{i})

Shape factor For a spherical body

q^{.} = 4 πk r_{i }r_{0 }dt/(r_{0}—r_{i} ) = k S dt
Therefore S_{ spherical} = 4 π r_{i }r_{0 }/(r_{0}—r_{i} )

(b) Define Thermal Conductance and Thermal Resistance.
ANS:
Compare rate of flow of charge in electricity with the rate of heat transfer

q^{.} = I = dV/R

q^{.} = k A (dT/dx) =– dT/(dx/kA) = dT/R_{th}
Comparing we get
(i) for a plain wall
thermal resistance R_{th} = dx/kA
Thermal conductance ( C ) is reciprocal of thermal resistance for a plain wall, C = kA/dx

For a cylindrical body

q^{.} = 2πkL dt/ln(r_{0}/r_{i})=dt/[2πkL /ln(r_{0}/r_{i})]
Therefore
Thermal resistance for a cylinder =[2πkL /ln(r_{0}/r_{i})]
Thermal Conductance = 1/R_{th} = ln(r_{0}/r_{i})]/ 2πkL

For a sphere

q^{.} = 4 πk r_{i }r_{0 }dt/(r_{0}—r_{i} ) = dt/[4 πk r_{i }r_{0 }/(r_{0}—r_{i} )]
Therefore
Thermal resistance for a sphere =[4 πk r_{i }r_{0 }/(r_{0}—r_{i} )]
Thermal Conductance for a sphere=(r_{0}—r_{i} )/ 4 πk r_{i }r_{0}

For convection

q^{.} =h A dT
Therefore q^{.} =h A dT = dT/hA
Thermal resistance in convection = hA
Thermal Conductance in convection = 1/hA
In radiation
q^{.} = σ A (T_{1}^{4} — T_{2}^{4}) W/m^{2} = (T1 –T2) / [σ A (T_{1}^{2} + T_{2}^{2})( T_{1} + T_{2})]
Therefore
Thermal resistance in radiation = [σ A (T_{1}^{2} + T_{2}^{2})( T_{1} + T_{2})]
Thermal Conductance in radiation = 1/[σ A (T_{1}^{2} + T_{2}^{2})( T_{1} + T_{2})]
UNITS OF THERMAL RESISTANCE = K / W
UNITS OF THERMAL CONDUCTANCE = W / K
© Differentiate between fin effectiveness and fin efficiency.
ANS:
FIN EFFECTIVENESS OF A SINGLE FIN
It is applicable for a single fin.
It is the ratio of rate of heat transfer from a fin to rate to the heat transfer without fin. Thus it is greater than 1. Practically it should be greater than 5.
The formula for this is
Є_{f} =Rate of heat transfer with fin to rate of heat transfer without fin
Є_{f}= q_{fin}/ q _{without fin}
= (P h A_{c }k)^{1/2}(T_{b}–T )/h A_{c} (T_{b}–T )
= (P k/h A_{c})^{1/2}
Where A_{c }is the fin crosssectional area at the base.
For a rectangular fin
P = 2B + 2 t and A_{c} = B t
Where B is width of fin and t is the thickness of fin
Now if the fin is thin, 2t can be neglected as compared to 2B
Therefore P/A_{c} = 2B/B t=2/t
Є_{f} = (2k/h t)^{1/2} = (1/Bi)^{0.5}
Biot number for a fin:
Less value of Biot number is desirable. Lesser value of Biot number will ensure light weight, compact fin with maximum heat transfer as given below.
Firstly Biot number will be less if h is small i.e. gas environment for fin.
Secondly Biot number will be less if thermal conductivity k of fin material is high.
Thirdly Biot number will be less if the thickness of fin 𝜹 is small i.e. a thin fin.
NOTE: It is advantageous to use large number of fins with smaller thickness.
For most effective fin

fin material with high thermal conductivity

thin fins,

surrounding gas environment or small h

deep fins to increase the surface area

light weight fins specially in vehicles

finite length due to practical limitations

compact due to small thickness

Large number of fins to increase the surface area

Low cost of manufacture

Less Pressure drop across fins

Fin performance can also be characterized by fin efficiency.
FIN EFFICIENCY FOR A SINGLE FIN
It is applicable for a single fin.
The fin efficiency is η_{f }= actual q^{. }_{fin}/ q^{. }_{max}
q^{. }_{max} is there when same base temperature all along the length
η_{f} = (P h A_{c}k)^{1/2}(T_{b}–T )/h PL (T_{b}–T )
= (k A_{c}/h P)^{1/2}(1/L)
=1/mL
η_{f =}1/mL
Fin efficiency will always be less than one. It is because of same base temperature for the entire fin.

(d) Define forced and free convection. Give one example of each.
ANS:
FREE OR NATURAL CONVECTION
When bulk motion of the fluid is due to density difference, it is Free or Natural convection. It is governed by the product of Grashoff’s number and Prandtl number. All the convection processes in nature are of free convection.
PRACTICAL EXAMPLES OF FREE CONVECTION
(i) Storms, Sea breezes and land breezes,
(ii) Glider planes,
(iii) Radiation heaters,
(iv) Hot air balloons,
(v) Cooling of coffee in a cup,
(vi) Heating of water below 100^{0 }C,
(vii) Solar collectors,
(viii) Heating/cooling of walls and windows,
(ix) Cooling of electronic equipment,
(x) Cooling inside a refrigerator
FORCED CONVECTION
(i) When bulk motion of the liquid is caused by a pump/stirrer
(ii) bulk motion of vapors and gases is caused by a fan/blower.
it is called forced convection. This is governed by the value of the Reynolds number. Forced convection is of two types.
(a) Laminar forced convection
(b) Turbulent forced convection.
PRACTICAL APPLICATIONS OF FORCED CONVECTION
(i) Flow in Refrigeration and Air conditioning machines,
(ii) Flow in boilers,
(iii) Forced flow over condensers,
(iv)Cooling in nuclear
(v) Heating/cooling in Heat exchangers.

(e) Write the significance of Biot’s Number.
ANS:
Biot number =internal conductive thermal resistance/external convective thermal resistance.
Bi = h l/k _{solid}
It is of more importance where temperature varies with time. Small Biot numbers are desired for greater rate of heat conduction where temperature varies with time.

(f) State the term burn out.
ANS:
Burn out point stands for maximum heat flux during pool boiling.
REGION III of the pool boiling curve: 10 ^{0}C < ΔT_{ex} < 30^{0}C, Nucleate boiling, When bubbles form and rise to the free surface, CRITICAL HEAT FLUX OR MAXIMUM HEAT FLUX OR BURNOUT POINT CONDITION IS REACHED.

(g) Define regenerator and recuperator.
ANS:
REGENERATOR
It is a heat exchanger. In this, heat is stored from some source say from the SUN. It is stored in some thermal mass ( stones or some porous material) . Then the stored heat is supplied to another fluid passed over the thermal mass.
RECUPERATE
It is also a heat exchanger. In this, there is no storage of heat. It is a direct heat transfer. In this, both media are separated by a wall through which heat is transferred directly.

(h) Classify the terms absorptivity and transmissivity.
ANS:
DEFINITIONS OF ABSORPTIVITY, α
It is the ratio of absorbed radiations to the incident radiations on a given surface.
Its symbol is α.
It has no units.
α≤1.for all bodies other than a black body.
α =1 for a perfect black body.
DEFINITION OF TRANSMISIVITY, τ
It is the ratio of transmitted radiations to the incident radiations on a given surface.
Its symbol is τ.
It has no units.
For all bodies except a black body, τ ≤1
for a black body τ=0

(i) Define Lambert’s Cosine Law of Radiation.
ANS:
LAMBERT’S COSINE LAW
I_{ϴ}=I_{n} cosϴ Where ϴ is the angle between the surface normal and the viewer direction.
I_{ϴ} is intensity of radiation in the direction of the observer. It is at an angleϴ with the normal to the surface
I_{n} is the intensity of normal radiations falling on the surface.

(j) Air is a bad conductor of heat. Then how temperature of air rises in summers.
ANS:
Air cannot conduct heat within the air layer. So it is bad conductor of heat. Further its thermal conductivity is 0.027 W/m K.
But in summers, earth gets heated up due to radiations falling on it and get absorbed by the earth. Now cool air comes in contact with the hot earth surface and gets heated up. It is because conductance of heat is by physical contact only.
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