HEAT TRANSFER LAB

PRESENTATIONSUBJECT : FORCED CONVECTION

Presented By : RAHUL MONDAL -08

&

UPENDRA MANDAL-09

3rd YearPOWER ENGG.

Objective:

1.To determine experimentally Forced Convection Heat Transfer Co-efficient & Nu

2.To compare the theoretical Nu with the Experimental Nu

Assumptions:

1.Steady State condition2.Heat transfer only through forced convection

Photo of the Set up:

Experimental Set up

Schematic Diagram of the Experimental Set up: L=test length

Dia of the test part(D)Valve

Orifice

Heater

Pump

T1 T2 T3

TinTout

Heat flux

Direction of air flow

Pipe line

Convection

Heat transfer rate q” = hA( Ts - T ) W [ Newton’s cooling law ]

h=heat transfer coefficient (W /m2K)

(h is not a property. It depends on geometry ,nature of flow, thermodynamics properties etc.)

Convection is the mode of heat transfer due to the bulky motion of the medium.

Forced Convection: Here the motion of the medium is governed by some external agent , as Blower or Fan.

Forced Convection takes place due to the internal flow &External flow of the medium

Convection rate equation

Main purpose of convective heat transfer analysis is to determine:

• flow field

• temperature field in fluid

• heat transfer coefficient, h

q’’=heat flux = h(Ts - T)

q’’ = -k(T/ y)y=0

Hence, h = [-k(T/ y)y=0] / (Ts - T)

The expression shows that in order to determine h, we must first determine the temperature distribution in the thin fluid layer that coats the wall.

• Nusselt No. Nu = hx / k = (convection heat transfer strength)/ (conduction heat transfer strength)

• Prandtl No. Pr = / = (momentum diffusivity)/ (thermal diffusivity)

• Reynolds No. Re = U x / = (inertia force)/(viscous force)

Viscous force provides the dampening effect for disturbances in the fluid. If dampening is strong enough laminar flow

Otherwise, instability turbulent flow critical Reynolds number

Forced convection: Non-dimensional groupings

Empirical Correlations

Observation Table:

Sl no. V(volt) I(A) T1 T2 T3 Tin Tout h(m) 1. 49 0.26 40.5 40.4 40.3 44.6 40.7 0.104

2. 68 0.36 42 41.9 41.8 49 42.2 0.111

3. 92 0.53 44.3 44 43.9 60.2 44.6 0.114

Calculation

:For Experimental Nu & h:

Where d1= diameter of the delivery pipe A1= Cross sectional area of the pipe =1.256×10-3 M2

where d2= diameter of the orifice & A2=area of the orifice =3.14×10-4 M2

Step:1

Surface area of the testing pipe = πdl = 0.02355 M2

Where d= diameter of the pipe. l= length of the test pipe

Step:2

= Volume flow rate of the air through the pipe

= co-efficient of discharge of the orifice

= cross sectional area of the pipe

=area of the orifice

g =gravitational acceleration

=density of the water used in manometer at ambient temperature

ρ=density of the test fluid(air) at ambient temperature

=8.6116×10-3 M3/s

Step:3

= density of the air at ambient temperature

= mass flow rate or the test fluid (air) in the pipe

=volume flow rate of the test fluid (air) in the pipe

=9.5716×10-3 Kg/s

Step:4

= rate of heat transfer through the test part of the pipe= specific heat of the air at constant pressure at film temperature= Avg. temp of the air= mean temperature of the air at the inlet of the test section= mean temperature of the air at the outlet of the test section= mass flow rate of the air through the test part of the pipe

Where T f =(Tin + Tout )/2

cp

=37.6095 W

Step:5

=heat transfer rate through the test part of the pipe

= surface area of the test section

= average mean temperature of the air in the test part

= average surface temperature of the pipe in the test part

=Experimental heat transfer co-efficient of the test fluid (air)

=727.446 W/m2 k

Step:6

=Experimental Nusselt Number

=Experimental Heat transfer co-efficient

= Diameter of the test part of the pipe

=Thermal conductivity of the air at film temperature

=666.16

Step:7

Calculation of Theoretical Nu & h

Where V=Velocity of the air through the pipe =Volume flow rate of the air = Cross sectional area of the test section

=17.552 m/s

Step:8

= Reynolds Number of the air flow = density of the air at film temperature

=diameter of the test section

=co-efficient of viscosity of the air at film temperature

= velocity of the air through the pipe

=25.306×103

Step:9

From Dittus-Boelter Equation:

Where the range of validity: 0.7≤Pr≤160; Re≥10,000 ; L/D≥10

=Theoretical value of the Nusselt Number = Reynolds number of the air in the pipe = Prandtl Number of the air at film temp

=637.20

Step:10

=(Nuth.Ka)/D

= theoretical heat transfer co-efficient

= theoretical Nusselt Number

= thermal conductivity of the air at film temperature

=Diameter or the test section

=695.82 W/ k

Step:11

Final table:

• There after we have to plot a graph in log-log paper

The final table contains the Nu exp & Nu th

Sl. No. Nu exp Nu th

1. 666.16 637.16

2. 698.404 670.18

3. 700.44 666.305

The Final Graph:

104 105 106

Experimental Curve

Gradient(m)= 0.75General Correlation:

We get the value of “m” & “c” from the graph

The Experimental Correlation:

The Theoretical Correlation:

Nuexp=0.368 Re0.75 Pr0.3

Application:

1. Among many ways to dissipate heat in electronic components, forced convection cooling is the most effective.2. Forced convection is used in cooling towers in Power Plants & in other industries ( Forced Draft Fan. )3. It is also applicable for drying purpose(ie, in washing machine).4. Forced convection can also be used for heating purpose etc.

We are really very much obliged to all of them without whom this project can not be possible. Special thanks to –

Dr. Apurba kumar Santra

Prof. Amitava Datta

Mr. Bireswar Paul

Mr. Atish Nandi

Mr. Prakash Ghosh

THANK YOU