Determination of Heat Transfer Coefficient Through Pin - Fin

Aim

To determine the temperature distribution of the PIN - FIN for forced convection and to find the FIN efficiency

Specifications:

Duct width (b) = 150mm

Duct height (w) = 100mm

Orifice Diameter (d_O) = 24mm

Orifice coefficient (C_dd) = 0.6

Fin length (L)= 14.5cm

Fin diameter (D_f) = 12mm (Characteristic length)

Diagram:

Apparatus required:

• Experimental set up of a Pin-Fin inside a horizontal tube with forced convection Environment
• Stabilized power supply
• Stop Clock

Procedure:

1. Switch on the heater and set an input voltage of (60V) Using the voltage regulator.
2. The heater starts to conduct heat through the pin-fin. Allow the heater to heat the Pin- Fin to a sufficient temperature.
3. The air forced through the blower removes the heat from the pin-fin by convection.
4. Observe the temperature readings of thermocouple at different location on the pin-fin and also the forced air temperature at equal intervals of time and tabulate them.
5. Using appropriate formulae calculate the heat transfer coefficient and fin efficiency.
6. Plot the temperature distribution along the length of the fin.

Observations:

Formula Used

Ref:-Heat and Mass Transfer Data book by C.P Kothandaraman & Subramannian , New Age publishers.

Flim Temperature (T_f) = --------------

Volume of air flowing through the duet

Vo = Cd*a₁*a₂*V2gh_a/(a²1- a²2)

Where Cd = Co-efficient of orifice = 0.6

g= Gravitational constant = 9.81 m/sec ²

h = heat of pipe (p_w/p_a)h

a₁= area of the Pipe

a₂ = area of the orifice

Velocity of air in the pipe = V / (W + B)

W = width of duct

B = Breadth of the duct

Reynold's Number = Va x ρ a / μ _a

Where Va - Velacity of the duct

ρ_a = Density of the duct

μ_a= Viscosity of air at T8 ° C

Prandltl Number = ( C p_a * μ_a) / K_a

Where Cp_a = Specific heat of air

μ_a = Viscosity of air

K_a = Thermal conductivity of air

Heat Transfer Coefficient Calculation

Nusselt number (N_nu)

For 40 < NRe < 4000

N_nu = 0.683 (N_Re) * 0.466 (N_pr)^0.331

For 1 < NRe < 4

N_nu = 0.989 (N_Re) 0.33 (N_pr) ^0.33

For 4< NRe < 40

For 4000 < N_Re < 40000 N_Re

N_nu = 0.193 ( N_Re) 0.618 (N_pr) 0.33

For N_Re > 40000

N_nu = 0.0266 ( N_Re) 0.805 (N_pr) 0.333

Heat transfer co - efficient h = N_nu * (K_a / L)

K_a = Thermal conductivity of air

L = Length of Fin

Efficiency of the Pin-fin = [ ( tan h mL) /mL]

H= heat transfer co-efficient

L = Length of the fin

m = V(hp / (K_bx A))

P = Perimeter of the fin = ( μ* Diameter of the fin)

A = cross section area of the fin

K_b= Themal conductivity of brass rod

Temperature distribution = T x = { [(Cosh m(L-X) /Cosh (M )] * (To -Ta)} + T

X= distance between thermocouple and heater

Distance between thermocouples = 20 mm

Evaluation of the heat transfer co -efficient (h)

N_nav = (hD)/K = 1.1 (Gr Pr) ^1/6 for 1/10 < Gr Pr < 104

N_nav = 0.53 (Gr Pr)^1/4 for 104 < Gr Pr < 109

N_nav = 0.13 (Gr Pr) ^1/3 for 109 < Gr Pr < 1012

Where Nnav = average Nusselt Number = (h D)/K

D= Diameter of fin

K = Thermal conductivity of air

G_r = Grashof number - g β Δ;D³/ γ²

β = Coefficient of thermal expansion - 1/ ( T_av + 273) ΔT = (T_av -T _amb)

Pr = Prandtl number = (μCp/K)

Result:

The heat transfer rates of the fin is determined and the values are

Temperature distribution:

Viva-Voce Questions

What is Orifice - meter?

Orifice- meter is used to measure discharge.

What is the function of blower?

Blower is an external mechanical device which is essential for forced convection process

What is Newton's law of cooling?

The rate equations for convective heat transfer between a surface and an adjacent fluid is prescribed by Newton’s law of cooling

What is the range of 'h' for Natural convection in gases & liquids?

Range of 'h'for natural convection in gases is 3-25W/m2 -k & for liquids it is 50- 350W/m2 -k.. Boundary layer is a thin layer at the surface where gradients of both velocity& temperature are large.

What is Nusselt Number?

Nusselt Number represents the enhancement of heat transfer through a fluid layer as a result of convection relative to conduction across a same layer. Larger the Nu, more effective is convection.

What is a boundary layer?

Boundary layer is a thin layer at the surface where gradients of both velocity & temperature are large.