Free Convection And Radiation Effects In Heat Transfer Engineering Essay

Published: Last Edited:

This essay has been submitted by a student. This is not an example of the work written by our professional essay writers.


The experiment we carried out to find the free convection heat transfer coefficient 'h' for the two bars made of brass (oxidised & polished). The values of 'h' for both the bars were found and it was observed that the convective heat transfer coefficient is higher for dull surfaces than polished surfaces. It is because of the turbulent flow of air over the dull surface of the oxidised bar, which is responsible for more heat transfer and the contact surface area of the dull rod is more than that of the polished rod which is responsible for the rapid heat transfer due to convection and radiation. We then calculated the heat lost by radiation (by the dull surface) to know how much that plays a part in heat loss. We found the value of 'h' for brass through formulas, compared it with our experimental values and came to the conclusion that the experiment works.


In experiment we used two brass bars one was highly polished and the other was highly oxidised (dull). The aims were to heat the bars up to a certain temperature i.e. 120oC (To) after which they were allowed to be cooled down by free convection in air. The experimental h is found by this procedure for both the bars and then compared to know which bar looses heat more by convection. Also a theoretical h value is found by formulas. We found that our experimental values are close to these theoretical values, it means our experimental work could be used for other materials too.


Convection: Heat transfer due to convection occurs by the bulk motion of fluid. There are two types of convection; forced and free.

Forced convection: In which the fluid is forced by some means ( fans, pumps etc)

Free convection: Natural or free convection is a type of heat transfer in which the fluid motion is not forced by any external source but only by density differences in the fluid occurring due to temperature gradients.

The part heat transfer occurred due to the radiation also,

Radiation: The electro magnetic radiation emitted due to the temperature of the material or matter.

Hot metalwork from a blacksmith. The yellow-orange glow is

the visible part of the thermal radiation emitted due to the

high temperature. Everything else in the picture is glowing

With thermal radiation as well, but less brightly and at longer

wavelengths than the human eye can see. An infrared camera

will show this radiation.(Wikipedia)

Surface can also affect the heat transfer; a shinny or polished metal surface which has low emissivity can be used to reduce heat transfer.

In this experiment, we will cool down the two brass bars (polished and oxidised) by free convection.

Newton's Law of Cooling could better explain this:

Qconv = h.A. (Ts - T∞).

Qconv = rate of heat transfer by convection

A = surface area

Ts = temperature of the surface of the body

T∞ = ambient temperature (room temperature)

h = constant known as the convection heat transfer coefficient.

The convection heat transfer coefficient 'h' is defined as "the rate of heat transfer between a solid surface and a fluid per unit surface area per unit temperature difference."


In experiment we got two brass bars we have to find out the convective heat transfer coefficient for these bars which are enclosed in the wooden boxes. One box contained the 'polished rod' while the other contained the 'oxidised rod' or 'Dull rod' with their sides attached to the side walls of the box. The front wall was made up of glass for our observation with an opening at the top and bottom, for passage of air into and out of the box. The boxes were designed such a way that they could reduce unwanted air movement (draught). Thermocouples were attached to both the rods in order to display their current temperature to us on a digital display. We suppose the temperature variation with time is uniform. There were two electric strip heaters that could be used to heat the rods inside the boxes. The gloves were there to protect our hands from burning while dealing with the strip heaters. A thermometer was fixed in a corner of the lab from which we got the ambient temperature. It was fixed on a distance from the heat source (strip heaters) to get the real readings of the ambient temperature.

Oxidised bar polished bar

crs2.JPG crse1.JPG Thermometer


We took the ambient temperature of the lab using a thermometer fixed at one of the corners of the lab. The reading came out to be

Ta = 21oC.


We have been divided into two groups one group took the readings on the box containing the polished bar and another took the readings on the box containing the oxidised bar. We placed a strip heater on top of the polished rod. We then, let the rod heat up until the digital display showed its temperature to be around 120oC. We turned off the strip heaters and using gloves carefully removed them from the bars and put them aside. Then we let the bars cool down slowly and note down the readings from digital display carefully. We started taking readings on stop watch when digital meter was showing 100oC. We note down the time after every 5degree drop in temperature of the rod and kept on taking reading until the temperature of the rod dropped to 500C. The other group did the same procedure for the box containing the oxidized bar.

We collected the following data, tabulated and plotted graph on the base of this data.

To = Each bar is heated up to 120 oC

Ta = Ambient Temperature of the Lab = 20 oC

T = Temperature observed during the heat transfer.

θo/θ = ( To - Ta ) / ( T - Ta )

Ln θo/θ = (To - Ta) / ( T - Ta )

Oxidized Bar

T (°C)

time (sec)

θ o/θ














































Plotted graph with Ln(θo/θ) plotted on the Y-axis and Time plotted on the X-axis:

Polished Bar

T (°C)

time (sec)

θ o/θ














































The curves are not linear so we added a 'best fit line' and from the equation of curves we got their slopes.

mpolished= 0.0018

moxidised= 0.0022

As m= hA/ρCV

A = Cylinder surface area = 2Ï€r2L + 2Ï€rL

V = volume = πr2L

The oxidised and polished bars we had got diameter D = 9.5mm, length L= 305mm. The density and specific heat of brass are ρ= 8500 kg/m3 and C=370 J/kgK respectively so we can calculate the value of " h ",

hpol = (mpol) ρCV / A

hpol = 13.53 W/m2K

hoxd = (moxd) ρCV ] / A

hoxd = 16.12 W/m2K

These are the experimental values of 'h'

3.5 The Correlated value of 'h'

Nusselt number:

Nu = 0.47 (Gr Pr) 1/4 (A)

we can to find the value of h using the relation:

Nu = hD/k (B)

k is the thermal conductivity of air = 2.78 x 10-2 W/mK

we have to calculate the Grashof number: Gr = gβθD3/ v2

Where g is the acceleration due to gravity and the kinematic viscosity v for air is 1.76 x 10-5 m2/s. The coefficient of cubical expansion β; which for air is the reciprocal of ambient temperature (1/Ta). Here θ varies so we take average:

θaverage = (T - Ta) / total readings

θaverage = (100+95+90+85+80+75+70+65+60+55+50)oC / 11 = 75-21 = 54 oC


Gr = 30169.286

Prandtl number: Pr = µC/k, whose value for air is 0.7

Putting the above values in equation (A) we get:

Nu = 5.66

And hence by equation (B) we get:

h = 16.58

The percentage difference between the correlation (theoretical) and experimental value of 'h' calculated:

% difference = (h - hpol) / h x 100

% difference = 18.39


Difference between the experimental values of h for the oxidised and polished bar is equal to the heat lost by the radiation.

hraditaion = hoxd - hpol

hraditaion = 2.59

Radiant heat flux:

q = hraditaion x θaverage

q = 827.31 W/m2

Stefan Boltzman law:

q = ε σ (T4 - Ta4)

σ is the Stefan Boltzmann constant = 56.7 x 10-9 W/m2K4

ε = 0.890

We suppose the temperature distribution over the bars is uniform throughout the time. Its mean the temperature is only a function of time and not space. This is also validated when we calculated the Biot modulus, for steady state system it must be less than 0.1

Bi = h L / k

L = V/A for the cylindrical bar

Bi = 0.00026 W/mK


From the slope m for the polished bar is 0.0018 and for the oxidized bar is 0.0022 we can predict that the rate of cooling and heating of oxidized bar is higher than the polished bar.

This is further validated the values of convection coefficient of heat transfer as hpol = 13.53 W/m2K and hoxd = 16.12 W/m2K. The heat transfer by convection in the oxidized bar is 1.29 times more than in the polished bar.

e ofThe valu h calculated by the correlation came out to be 16.58 W/m2K, while that by experimental method came out to be 13.53 W/m2K and 16.12W/m2K with the error 18.39%. It validates that this experiment works successfully in finding out the convection heat transfer coefficient and can be used to find the value of h for any material. The greater error in case of polished bar validate that we have done some error during taking the readings.

We considered that there is uniform temperature distribution throughout the bar. We ignore the internal resistance of the body to conduction of heat which could make the experiment more complex and would require reading of temperature at different points on the bar. Thus such an assumption is known as a lumped system and the Biot number; which is the ratio of the internal resistance of a body to heat conduction to its external resistance to heat convection.

We also see that the oxidized bar is losing more heat with higher rate than the polished bar because radiation is playing a vital role in loss of heat from the oxidized bar which we found to be 3.67 W/m2K.

The emissivity 'ε' value showed us that the polished bar has an emissivity of 0.89 which means it absorbs about 89% of the radiations falling on it.


The R-square value from graph is actually the correlation coefficient. It gives us the degree of reliability of the linear relationship between x and y. The value of R = 1 gives an exact linear relation ship between x and y.; R value close to 1 indicate excellent linear reliability and very little uncertainty. Using the equation R2 = 0.993 we can calculate the uncertainty.

R2 = 1-0.993

R = (0.007)2

= ±0.000049 ≈ ±0.00005

Hence we can state that there is an uncertainty of ±0.00005 in the slopes calculated.

as h= m ρ C V / A so,

(Δh/h) 2= (Δm/m)2

(Δh/13.53) = [(0.00005/0.0022)2]0.5

Δh = ±0.375

We can calculate that there is an error of ±0.366 existing in the value of 'h' as all other factors are constant such as ρ C V A, there values are givin.

Now the Grashof number, Gr = gβθD3/ v2, we see that the values of g, D and v are given.

We got β= 1/Ta. The thermometer reading the ambient room temperature has the smallest division of 1oC, we assume that the ambient temperature has an error of:

ΔTa = ±0.5oC. Thus,

(Δβ/β) = [(ΔTa/ Ta)2]0.5

(Δβ/0.047) = [0.5/21]

Δβ = ±0.0010

θaverage = Taverage -Ta, so:

(Δθaverage/ θaverage) = [ (Δ Taverage/ Taverage)2 + (Δ Ta/ Ta)2 ]0.5

(Δθaverage/54) = [ (0.05/75)2 + (0.45/ 21)2]

Δθaverage = ±1.15

(ΔGr/Gr) = [(Δβ/β)2 + (Δθaverage/ θaverage)2]0.5

(ΔGr/30169.2) = [(0.0010/0.047)2 + (1.15/ 54)2]0.5

ΔGr = ±908.20

Nu = 0.47 (Gr Pr)1/4

Only Gr has an error so:

(ΔNu/Nu) = 1/4(ΔGr/Gr)

(ΔNu/5.66) = 1/4(908.2/30169.2)

ΔNu = ±0.04

As the value or Pr = 0.7 is given and it is constant so no error can be calculated.

h = Nu k / D, as k a D are given in data hence;

(Δh/h) = [(ΔNu/Nu)2]1/2

(Δh/16.56) = (0.04/5.66)

Δh = ±0.087

If we compare both the experimental and calculated h values we see that:

Δh = ±0.087 (calculated) Δh = ±0.375 (experimental)

There is uncertainty in both the experimental and correlated value but there is more chance of an error in the experimental value than calculated from formulas which is obvious and expected.

To reduce the uncertainty we should take the following steps:

Keep the distance from the boxes so the natural air can flow in and out of the boxes.

Try to measure the time with the help of the stop watch as accurately as humanly possible.

The thermometer measuring the ambient temperature should place at a distance from the boxes so the heat does not affect its reading.

Place the thermometer in such a place that it could measure the average temperature of the room, not in hot or cold spots of the room.

Use the gloves to handle the strip heater and Handle it carefully, in order to avoid any burns.


We have achieved our aims to find out both the values of 'h' experimental and by formulas and got the error of 18..53%. This error was also accounted for when we calculated the uncertainties in our measurements. After comparing both the experimental and theoretical values we have found that this experiment is also useful in finding the convection heat transfer coefficient 'h' for any other materials.

From the values of convection heat transfer coefficient 'h' that is;

hpol = 13.53 W/m2K & hoxd = 16.45 W/m2K. We conclude that the convection coefficient of heat transfer is higher for dull surfaces than for highly polished surfaces. This statement also validated for materials other than brass too. Actually the dull or oxidise bar hold irregular surface which increase the contact surface area with the air passing through that surface, in a result the heat transfer by convection also increase. Also the air passing through the irregular surfaces become turbulent which also increase the heat lost by convection, this technique is adopt in the manufacturing of the fins of the radiator to insure the maximum heat transfer by making the fins wavy and make the air flow turbulent which pass through fins. The radiation heat transfer depends upon the metal surface the shinny and polish surface which has low emissivity would have less radiant heat loss than the dull surface which is validated by our experiment results.

Although we have some uncertainties in our experiment but still this experiment help us to understand more about the value of h for any material.

The uncertainties seen in the experimental values and in the graph that was not straight line because of the human error that we made during the reading we took due to the inaccurate timing of the watch.