Experiment No.7

PIN FIN APPARATUS

AIM:

01. To determine the variation of temperature along the length of the pin fin under forced convection.

02. To determine the value of heat transfer co-efficient under forced condition and to find

a) Theoretical values of temperatures along the length of the fin.

b) Effectiveness and efficiency of the pin-fin for insulated and boundary condition.

Theory:

The heat transfer from a heated surface to the ambient is given by the relation q=hA∆T. In this relation h is the heat transfer coefficient , ∆T is the temperature difference and A is area of heat transfer. To increase q, h may be increased or surface area may be increased. In some cases it is not possible to increase the value of heat transfer coefficient and the temperature difference ∆T and thus the only alternative is to increase the surface area of heat transfer. The surface area is increased by attaching extra material in the form of rod ( circular or rectangular) on the surface where we have to increase the transfer rate. "This extra material attached is called the extended surface or fins".

The fins may be attached on a plane surface, then they are called plane surface fins. If the fins are attached on the cylindrical surface, they are called as circumferential fins. the cross section of fin may be circular, rectangular or parabolic.

TEMPERATURE DISTRIBUTION AND HEAT TRANSFER FROM FINS FROM END INSULATED CONDITION:

Temperature distribution along the length of the fin is

θθ0 = T−T ∞T 0−T ∞= coshm(L−x)

coshmL

Where T= Temperature at any distance x from the fin

T0= Temperature at x=0

T∞= Ambient Temperature

L= Length of the fin

m=√ hpKA hc= Convective heat transfer coefficient

p= perimeter of the fin

A= area of the fin

K= thermal conductivity of the fin

Heat flow q=θ√hPKA tanhmL

EFFECTIVENESS OF FINS:

Effectiveness of a fin is defined as the ratio of the heat transfer with fin to the heat transfer from the surface without fins.

For end insulated condition:

∈= θ0x √ hPKA tanhmLhAθ0

= √ PKhA tanh mL

The efficiency of fin is defined as the ratio of the actual heat transferred by the fin to the maximum heat transferred by the fin area were at base temperatures.

η = θ0√hPKA tanhmLhPL θ0

η = tanhmLmL

EXPERIMENTAL PROCEDURE:

01. Connect the equipment to electric power supply.

02. Keep the thermocouple selector switch to zero position.

03. Turn the dimmerstat knob clockwise and adjust the power input to the heater to the desired value.

04. Switch on the blower.

05. Set the air flow rate to any desired value by adjusting the difference in mercury levels in the manometer.

06. Allow the unit to stabilize.

07. Turn the thermocouple selector switch clockwise and note down the temperatures T1 to T6.

08. Note down the difference in level of manometer.

09. Repeat the experiment for different power input to the heater.

OBESERVATION TABLE:

Sr.No Voltmeter in Volts

Ammeter in Amps

Power in

Manometer in mm of Hg

Temperature reading in deg. c

Watts

h1 h2 H T1 T2 T3 T4 T5 T6OBSERVATION:

Where d0= Diameter of orifice=0.02 m or 10 mm

dp= internal diameter of pipe= 3.81 cm

ρm= Density of manometric fluid= 13600 Kg/m3

ρa= Density of air = 1170 Kg/m3

CALCULATIONS:

1. Velocity of orifice = V0= √2 gh(ρm−ρ a)ρa∗1−β 4

Where β=dodp

2. Va = Velocity of air in the duct= Velocity at orifice∗cross−sectional areaof orifice

Cross sectional areaof duct

Va= Vo∗π

4∗do2

Widthof duct∗breadthof duct=Vo∗π

4∗do2

W∗B

3.Average Surface Temperature of Fin is given by,

Ts=T 1+T 2+T 3+T 4+T 5+T 6

6

4. T∞= Ambient Temperature=

5. Tm= Mean Temperature=Ts+T ∞

2

1) From DRY AIR atmospheric pressure table-1

µ=1.86*10^-5 Kg/s m, Pr= 0.72, K=0.026 W/m k , df=0.012 m

Re=Vadiμ

The relation for Nu is

2) Nu= C*Re^n*Pr^1/3

The value of C, Pr and n from standard table

For Re=0.4 to 4.0, C=0.989, n=0.33

Re=4 to 40, C=0.911, n=0.385

Re=40 to 4000, C=0.683, n=0.466

Re=4000 to 40000, C=0.293, n=0.618

Re=40000 to 400000, C=0.27, n=0.805

3) hc=Nu Kdf

Where K(Tm) of air at 44.16º C is 0.026 W/mºk

Thermal conductivity of fin material

K=110 W/mºk

4)

m=√ hpKA where P= perimeter= πdf

A= Area= π4∗d02

Temperature distribution is given by

T−T ∞T 0−T ∞= coshm(L−x)

coshmL

T= T∞+(T 0−T ∞)coshm(L−x)

coshmL

Where X is the thermocouple distance at different positions.

X=X1, X2, X3, X4, X5, X6

Where T=Temperatures at above points

T=T1, T2, T3, T4, T5, T6

Given thermocouple distance X1=20 mm= 0.02 m

X2= 50 mm= 0.05 m

X3=80 mm= 0.08 m

X4=110 mm= 0.11 m

X5=140 mm = 0.14 m

X6= 170 mm= 0.17 m

T1= T∞+(T 0−T ∞)coshm(L−x)

coshmL

Distance X in mts Temperature from experiment in deg. c

Temperature from calculation in deg. c

Effectiveness of fin= √ hPKA tanhmLhAθ 0

Efficiency of fin = tanhmLmL