ME2355 FORCED CONVECTION - Computer Programming


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Sunday, March 13, 2011



            To find the heat transfer coefficient under forced convection environment.

Description of apparatus:
The important relationship between Reynolds number, Prandtl number and Nusselt number in heat exchanger design may be investigated in this self contained unit.
The experimental set up (see sketch) consists of a tube through which air is sent in by a blower.  The test section consists of a long electrical surface heater on the tube which serves as a constant heat flux source on the flowing medium.  The inlet and outlet temperatures of the flowing medium are measured by thermocouples and also the temperatures at several locations along the surface heater from which an average temperature can be obtained.  An orifice meter in the tube is used to measure the airflow rate with a ‘U’ tube water manometer.
An ammeter and a voltmeter is provided to measure the power input to the heater.
A power regulator is provided to vary the power input to heater.
A multi point digital temperature indicator is provided to measure the above thermocouples input.
A valve is provided to regulate the flow rate of air.

Inlet temp. of  air
Outlet temp. of air
Temperatures along the duct

  1. Switch on the mains.
  2. Switch on the blower.
  3. Adjust the regulator to any desired power input to heater.
  4. Adjust the position of the valve to any desired flow rate of air.
  5. Wait till steady state temperature is reached.
  6. Note manometer readings h1 and h2.
  7. Note temperatures along the tube.  Note air inlet and outlet temperatures
  8. Note volt meter and ammeter reading.
  9. Adjust the position of the valve and vary the flow rate of air and repeat the experiment.
  10. For various valve openings and for various power inputs and readings may be taken to repeat the experiments.  The readings are tabulated

The heat input Q = h A L M T D = m cp (temp. of tube – temp. of air)
                        M = mass of air.           cp = specific heat of air.
LMTD=(Avg Temp Of tube – outlet air temp) – (Avg. temp of tube – inlet air temp.)
1n x        (Avg. temp of tube – outlet temp. of air)

(Avg. temp of tube – inlet temp. of air)

H= heat transfer co-efficient. A = area of heat transfer = T1d1
From the above the heat transfer co-efficient ‘h’ can be calculated.  These experimentally determined values may be compared with theoretical values.
Calculate the velocity of the air in the tube using orifice meter / water manometer.
The volume of air flowing through the tube (Q) = (cd21222gh0 ) / (a12 – a22 ) m3 / sec.
ho = heat of air causing the flow.
    = (h1 – h2)℮w/ ℮a
h1 and h2 are manometer reading in meters.
a 1= area of the tube.
a2 = area of the orifice.
Hence the velocity of the air in the tube V = Q / a1 m/sec heat transfer rate and flow rates are expressed in dimension less form of Nusselt number and Reynolds number which are defined as
            Nu = h D/k                  Re = Dv/ υ
            D = Dia. Of the pipe
            V = Velocity of air
            K = Thermal conductivity of air.
The heat transfer co-efficient can also be calculated from Dittus-Boelter correlation.
            Nu = 0.023 Re 0.8 Pr 0.4
Where Pr is the Prandtl number for which air can be taken as 0.7.  The Prandtl number represents the fluid properties. The results may be represented as a plot of Nu exp/ Nu corr.  Vs Re which should be a horizontal line.


V         A         T1        T2        T3        T4        T5        T6        h1cm    h2cm
50        1          35        42        45        46        47        38        9          19
Avg. Temp. Of heater = (42 + 45 + 46 + 47) / 4 = 45oC
Avg. Temp. of Air = (35 + 38) / 2 = 36.5oC
Vol. Of air flow Q = (Cda1a22gh) / (a12 – a22 )
                        Cd = 0.6
                        A1 = π/4 (0.04)2 = 1.256 X 10-3
                        A2 = π/4 (0.02)2 = 3.14 X 10-4
                        H = ℮water/℮air  (h1 – h 2) mtrs
                           = 1000/1.16 (0.1) = 86.2 mtrs.
Q = [0.6 X 1.256 X 10-3 X 3.14 X 10-4 (2 X 9.81 X 4)] / (1.256X10-3)2                                                                                                                                   (3.14X10-4)2
    = 9.73 X 10-6 / 1.256 X 10-3 = 8.002 x 10-3
Velocity of air flow = Q / a1 = 6.37 m/sec
Re = D/ r = 15023
R = kinematic viscosity at mean temp.
Using forced convection correlation
Nu = hD /k = 0.023 Re 0.8 Pr 0.4
Pr at mean temp = 0.699
= 0.023(15023)0.8 (0.699)0.4
hD/k = 43.7 k= Thermal conductivity of air at mean temp
h = 43.7 X 28.56 X 10-3 / (0.04)
= 31.2 w/mc.

            The heat transfer coefficient is found to be ---------------- W/m2K

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