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1. Name Haleemullah
Roll no D-19-EE-04
Department Energy Environment
Subject Heat And Mass Transfer
Submitted Sir Aamir Raza

2. Assignment
Heat & Mass Transfer
Q no 1
Draw a boiling curve and identify the different boiling regimes and explain the different
characteristics of each regime?
Boiling is the term which explains the rise in temperature of liquid in contact with a solid
surface. It occurs at solid to liquid intersection, when the solid surface’s temperature is
maintained slightly above the saturation point of liquid.
Boiling curve represents different the different regions through which boiling process is carried
out. The different regimes in a boiling curve are:
1. Natural Convection,
2. Nucleate Boiling,
3. Transition Boiling,
4. Film Boiling.

3. Natural Convection regime
When the liquid (in this case water) is heated above it’s saturation point (2 – 7 degrees for
water) bubbles begin to appear from the bottom of liquid where it is in contact with solid
surface. Natural convection starts to process and convection currents are formed in the fluid. It
uses the heat in the solid surface, and by convection, transfers the heat through-out the fluid.
Bubbles form and then disintegrate within the liquid without rising to the top and slowly the
temperature of fluid rises.
Nucleate Boiling regime
As the temperature of liquid rises, curve moves upward towards the nucleate boiling region
where bubbles are forming and popping at a quick rate. Heat flux is very large in this region due
to agitation in the liquid caused by trapping of vapors and evaporation. Thus, liquid reaches the
maximum heat flux point with increase in temperature of liquid.
Transition Boiling regime
Beyond the nucleate boiling region lies transition boiling. At this point heat flux is at it’s
maximum but as soon as it reaches the critical point it stars to drop gradually in transition
boiling region. This phenomenon occurs because of the presence of vapors. These vapors act as
a screen or a film, taking up large heating surface and slowing down the heat transfer rate. In
this region both the nucleate boiling and next boiling region, which is filmboiling, exists.
Film Boiling regime
At this point the temperature of liquid is somewhere between 100 – 120 degree Celsius. Heat
flux drops to it’s minimum because of the availability of vapors between heating surface and
liquid. After this drop in heat flux, heat flux begins to rise slowly as the excess temperature
increases. This excess temperature is responsible for increase in heat transfer rate above 120
degree Celsius.
Q no 2
Explain the filmcondensation on a vertical plate?
Film condensation on a vertical plate
Condensation is a phenomenon which happens when the temperature of a vapor is dropped
below its saturation temperature. In Film Condensation, vapors cling to a solid surface and then
form a layer of liquid on the surface, wetting the solid surface. This liquid layer forms a thin
screen or a film which is actually resistant to heat transfer. The temperature of this filmis now
lower than the vapors thus the heat is released and transferred across the film to the solid
surface. When this filmcondensation acts on a vertical plate, film is under the influence of
gravity which is pulling the filmdownwards.

4. Due to this downward motion, three prime flow regimes are established. At the top is Laminar
flow with Reynold number 0, then is Wavy Laminar with Reynold number between 30 and
1800 and finally Turbulent at the bottom of vertical plate with Reynold number above 1800. It
is noted that velocity of vapor is negligible and so it doesn’t drags the liquid filmand so the
acceleration of filmis also negligible. So the heat transfer within the filmis considered to be
linear.
Q no 3
A counter-flow double-pipe heat exchanger is to heat water from 20°C to 80°C at a rate of 1.2
kg/s. The heating is to be accomplished by geothermal water available at160°C at a mass flow
rate of 2 kg/s. The inner tube is thin-walled and has a diameter of 1.5 cm. If the overall heat
transfer coefficient of the heat exchanger is 640 W/m2 · °C, determine the length of the heat
exchanger required to achieve the desired heating.
Data
m (water)= 1.2 kg/s
m (geo)= 2 kg/s
D= 1.5 cm
= 0.015 m
We know that,
Cp (water)= 4.18 kJ/kg.C
Cp (geo)= 4.31 kJ/kg.C
Solution,
Using water,
Q = [mCp (Tout – T in)]
=(1.2)(4.18)(60)
=301 kW
Now for geothermal,
Q=[mCp(T out – T in)]
T out= T in – Q/mCp
=160 – 301/(4.31)(2)
=125° C

5. As we know the inlet and outlet temperatues,
dT1= T hin – T cout
=160 -80
=80 degree
dT2= T hout – Tcin
= 125-20
= 105° C
dTlm= dT1 – dT2/ ln (dT1/dT2)
= 80 – 105/ ln (80/105)
=92° C
Surface Area of exchanger can be determined by
Q= U A dTlm
A=Q /(U)(dTlm)
=301,000/(640)(92)
= 5.11 m2
We can find length by formula of area
A= π d l
5.11/(3.14)(0.015) =L
L= 108 m