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INSTRUCTION MANUAL

FOR

PIN FIN APPARATUS

Manufactured by:

ROORKEE EQUIPMENT & MODELS PVT LTD

Factory : C-18 Ram Nagar Industrial Area, Ram Nagar
Roorkee Distt-Haridwar,
Roorkee-247 667.
Email: rempvtltd@yahoo.in , remtender007@gmail.com
Website: www.rempvtltd.com
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PIN FIN APPARATUS

AIM

1. To draw the variation of temperature along the length of pin fin under forced
convection.

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

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

(b) Effectiveness and efficiency of the pin-fin for insulated and boundary
condition.
Heat Transfer from a Pin-Fin (By Natural & forced convection)
 Fin: Pin type
 Material: Aluminum
 Size: Dia. 20 mm, length-170 mm approx.
 Air Duct: made of MS
 Fan : standards make
 Heater: Band type Nichrome wire
Control Panel Equipped with
 Digital Temperature controller: PID controller 0-199.9⁰C
 Energy meter: Digital Indicator: 0-199.9⁰C with multi-channel switch.
 Temperature sensor: RTD PT-100 type -8 No’s
 Standard make on/off switch, mains indicator etc.

THEORY

The heat transfer from a heated surfaces to the ambient is given by the relation q = h.A
T. In this relation h is the heat transfer coefficient. T is the temperature difference and
A is the 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 heat transfer rate. “This extra material attached is called the extended
surface or fins”. The fins may be attached on a plane surface, and then they are called
plane surface fins. If the fins are attached on the cylindrical surface, they are called
circumferential fins. The cross-section of the fin may be circular, rectangular, triangular
or parabolic.
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TEMPERATURE DISTIBUTION AND HEAT TRANSFER FROM FINS FROM END

CONDITION.
Temperature distribution along the length

 T - T Cosh m (L – x)
of the fin is ------ = ------------- = ---------------------
o To - T Cosh mL

Where T = Temperature at any distance x on the fin

To = Temperature at x = 0

T = Ambient temperature

L = Length of the fin

_____

/ hcP hc = convection heat transfer coefficient

m = / ------ p = perimeter of the fin

 KA A = area of the fin

K = thermal conductivity of the fi

_______

Heat flow q = o  hc PKA tan h mL

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

______

/ hc PKA tan h (mL)

 = o / -----------------------
 hc A o
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_____

/ PK

 = / --------- tanh( mL)

 hcA

The efficiency of a fin is defined as the ratio of the actual heat transferred by the fin to
the maximum heat transferred by the fin if the entire fin area were at base
temperature.
______

o ( hc PKA) tan h mL

 = -------------------------------
hc PL o

tan h (mL)
 = -------------
mL

SPECIFICATIONS

Length of the fin = 170 mm = .17 m

Diameter of the fin = 20 mm = .02 m

Parameter of fin (π.d) = 0.0628 m

Thermal conductivity of fin material (Aluminum) = 204 w/mK

Diameter of the orifice meter do = 0.015 m

Duct Length = 0.9 m

Width of the duct = 0.200 m

Height of duct = 0.150 m

Area of the duct = 0.63 m2

Coefficient of discharge of the orifice meter = 0.61

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Density of manometer fluid (Water) = 872.5 kg/m3

Type Thermocouple = Copper/Constantan wire

Heater to Distance Thermocouple T1: 20 mm

Heater to Distance Thermocouple T2: 55 mm

Heater to Distance Thermocouple T3: 90 mm

Heater to Distance Thermocouple T4: 125 mm

Heater to Distance Thermocouple T5: 160 mm

Ambiment Thermocouple T6

Schematic Diagram
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PROCEDURE

1. Connect the equipment to electric power supply.

2. Keep the thermocouple selector switch to zero position.
3. Turn the dimmer stat knob clockwise and adjust the power input to the
heater to the desired value.
4. Switch on the blower.
5. Set the air flow rate to any desired value by adjusting the difference in
mercury levels in the manometer.
6. Allow the unit to stabilize
7. Turn the thermocouple selector switch clockwise and note down the
temperature T1 to T6.
8. Note down the difference in level of the manometer
9. Repeat the experiment for different power input to the heater.
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SPECIMEN CALCULATIONS
____________________

/ 2 gh (m - a) 1

Velocity at orifice = Vo = Cd / --------------------- x ---------

 a  1-4

m = density of manometer fluid = 872.5 kg/m3

dp = Area of Duct

a = density of air = 1.17 kg/m3

do 0.015

 = ----- = --------------- = .024

dp 0.63

_________________________ ____

/ 2 x 9.81 x 0.038 (872.5 – 1.17) 1

Vo = 0.61 x / ------------------------------------------ -----------

 1.17 1- 0.0244

= 16.24 m/s.

Velocity of air in the duct Va

Velocity at orifice x Cross-sectional area of orifice meter

Va = ------------------------------------------------------------------
Cross-sectional area of duct

Va = Vo x ---- do2
4


16.24 x ---- x (0.015)2
4
Va = -----------------------------
0.63

Va = 0.005 m/s.
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Average surface temperature of fin is given by

T1 + T2 + T3 + T4 + T5
Ts = ------------------------------
5

68 + 62 + 58+ 57 + 55
= ---------------------------------- = 60 oC
5

T = Ambient temperature = 32 oC

60 + 32
Tm = Mean temperature = ------------------- = 46 oC
2

Properties of air at 46 oC

U = 16.96 x 10-6 m2/s Pr = 0.698, K = 0.0237 wmk

Va df
Re = ---------
U
Df = diameter of the fin = 0.02 m

0.005 x 0.02
Re = --------------------- -= 5.10
19.96 x 10-6

The relationship for Nu is

Nu = C Ren Pr1/3

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

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

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

= 4000 to 40,000 C = 0.293 and n = 0.618

= 40,000 to 400,000 C = 0.27 and n = 0.805

For Re = 5.39 C = 0.911 and n = 0.385

Nu = 0.911 (5.39)0.385 (0.698)1/3

= 1.5
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Nuk 1.5 x .0237

hc = ------ = ----------------
d 0.02

= 1.78 w/m2 K

Thermal conductivity of the fin material = 205 w/ m2 K

____ _____________
/ hcP / 1.78 x πx 0.02
M = / -------- = /------------------------ = 1.3
 KA  0.022
205 x  x ----------
4

Temperature distribution is given by

T - T Cosh m (L – x)
--------- = ---------------------
To - T Cosh mL

Cosh m (L-x)
T = T + (To - T) -------------------
Cosh mL

Cosh m (L-x)
T = T + (To - T ) --------------------
Cosh mL

Put value of x and find T

X1 = 0.02 T1 = 68 oC

X2 = 0.055 T2 = 62 oC

X3 = 0.09 T3 = 58 oC

X4 = 0.125 T4 = 57 oC

X5 = 0.16 T4 = 55oC
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_____
/ PK
Effectiveness of fin = / -------- tan h mL
 hcA

parameter of fin p= 3.14x.020= 0.0628

Area of fins = 3.14x.(.020)2/4 = .000314

_____
/ 0.0628x 205
= / --------------------- tan h( 1.3 x.170)
 1.78 x.000314

Effectiveness of fin = 33.26

tan h mL
Efficiency of fin = -------------
mL

= tanh(1.3x.170)/(1.3x.170)

= .98 = 98%

Observation Table

S.N V I Q=V Temp Duct Manometer Effec Efficie

o .I Tem Difference tiven ncy
p ess
of fin
T1 T2 T3 T4 T5 Tҿ h1 h2 H=
h2-
h1

1
Natura
l

2
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