The aim of this experiment is to find the value of convection heat transfer coefficient h.Convection in this case is free, i.e.,the fluid motion(air) is not influenced by any external source,for example,a fan or pump.The air was allowed to pass freely(without external source)from the surfaces of two brass bars,one is oxidized(dull surface) and the other is polished.A number of equations(Nusselt number,Grashof number,Prandtl number) were used to find the theoretical value of 'h' and then it was compared with the experimentally calculated value of 'h'.During the experiment ,it is observed that the rise or fall of temperature for the oxidized bar is quicker than the polished bar.The reason behind this is that the oxidized bar having dull surface is not only transferring heat by convection but also by radiation.
This experiment is about the measurement of free convection heat transfer coefficient 'h',experimentally and theoretically.These values were then compared to know the difference between experimentally and theoretically calculated value of 'h'.For this purpose,two brass bars with different surface finishing is used.One bar is oxidized having dull or rough surface and the other bar has polished surface.Both bars have same composition,means both are made of brass.The bars were heated to 120 oC and then allowed to cool at room temperature(21 oC).The values of temperature and time were noted and tabulated to evaluate our experimental and theoretical work.
Radiation also took part in the heat transfer of oxidized bar as it has more surface area than the polished bar.It is observed that the experimental values of 'h' for both bars(oxidized & polished) were different than the theoretical values of 'h'.It is because of certain irregularities that occur in performing the experiment.These irregularities will be discussed in the section"conclusion"of the report
EXPERIMENTAL PROCEDURE:
A few terminologies must be defined before proceeding further.
Brass: It is an alloy of copper and zinc.The proportions of zinc and copper can be varied to create a range of brasses with varying properties.It has a muted yellow colour ,similar to gold.
Convection: It is one of the three modes of heat transfer through the motion of hot fluid from one place to another.It occurs in liquids and gases but not in solids and can be defined as vertical circulation that results from differences in density ultimately brought about by differences in temperature,resulting in the transfer of heat.
Warm air
Cool air farooq pic.jpg
Convection has two types:
Free convection: Free or natural convection is caused by fluid motion due to density differences.In free convection,the fluid would be at rest in the absence of temperature variations .
Example: hot air rising off the surface of a radiator.
Forced Convection: Forced or assisted convection happens when motion of the fluid is imposed externally (such as by a fun or pump).Example: a fan-powered heater, where a fan blows cool air past a heating element, heating the air. When a person blows on their food to cool it, he is using forced convection.
Convection heat transfer coefficient 'h': The transfer of heat per unit surface through convection was first described by Newton and the relation is known as the Newton's Law of Cooling.
The equation for convection can be expressed as:
q = k A dT
where
q = heat transferred per unit time (W)
A = heat transfer area of the surface (m2)
k = convective heat transfer coefficient of the process (W/m2K or W/m2oC)
dT = temperature difference between the surface and the bulk fluid (K or oC)
The convection heat transfer coefficient - k - is dependent on the type of media, gas or liquid, the flow properties such as velocity, viscosity and other flow and temperature dependent properties.
Heat Transfer Coefficients - Units
1 W/m2K = 0.85984 kcal/h m2 oC = 0.1761 Btu/ ft2 h oF
1 Btu/ft2 h oF = 5.678 W/m2 K = 4.882 kcal/h m2 oC
1 kcal/h m2 oC = 1.163 W/m2K = 0.205 Btu/ ft2 h oF
In general the convective heat transfer coefficient for some common fluids is within the ranges:
Free Convection - Air : 5 - 25 (W/m2K)
Free Convection - Water: 20 - 100 (W/m2K)
Forced Convection - Air: 10 - 200 (W/m2K)
Forced Convection - Water: 50 - 10.000 (W/m2K)
Radiation: It is due to the transfer of heat in the form of waves through space.No medium is required for this type of heat transfer.
350px-Heattransfer.jpg
EXPERIMENTAL SETUP:
Two rectangular shape wooden boxes with a front side made of transparent glass are used in this experiment.Both boxes are of the same size and at a distance from eachother.The boxes are open from the top.The front side of the boxes are fixed above the bottom.This is because we want the air to move freely from the front bottom to the top.This sort of arrangement is suitable to minimize the amount of unwanted air from the surrounding(humans,fans etc). One of the box contains a polished bar and the other one contains an oxidized bar.Thermocouples(a sensor used to measure temperature) are attached to both the bars in order to display their temperature on a digital display.Both bars are heated inside the wooden box with the help of strip heaters.The strip heaters used electrical energy to heat the brass bars.
Performing the experiment:
First we noted the room temperature with the help of a thermometer which was placed in a corner of the laboratory.The reading shown by the thermometer was 21 oC.The strip heaters were placed on the top of the brass bars inside each wooden box.Each rod(oxidized and Polished) was heated upto 120 oC.After reaching 120 oC,we switched off the strip heaters and placed them outside the box in a small separate container.The bars were then allowed to cool down slowly without any external source.We were observing the temperature drop for both the bars on the digital display.When the temperature of the bar reached 100oC,we started our stop watch.We noted time(sec) for every degree temperature(oC) drop of the bar.We kept on noting the time(sec) until the temperature of the bar reached 50oC.The time corresponding 100oC was taken as 0 seconds
Note: There were five members in our group,I and Lee performed experiment on the polished bar while the others on the oxidized bar.
The temperature,time and the derived results were tabulated in the following table.
Derived Results:
The temperature ratio,
θo / θ = To -Ta / T- Ta
Where,
To = Temperature up to which each bar was heated(120oC).
T = Temperature drop from 100 oC to 50oC.
T = Room temperature(oC).
Polished Bar
Next we plotted a graph with Ln(θo/θ) plotted on the Y-axis and time plotted on the X-axis. As shown below:
The curves we obtained were not linear so we added in a 'best fit curve' and from the equation of that curve we got to know their slopes. The slopes were:
mpolished = 0.0021 moxidized = 0.0024
Next we have the relation: m= hA/ÏCV
The bars we had were 9.5mm in diameter and 305mm long. And the density and specific heat of brass are 8500 kg/m3 and 370 J/kgK respectively, we calculated:
hpol = [ (mpol)( Ï).(C).(V) ] / A hoxd= [ (moxd)( Ï).(C).(V) ] / A
hpol = 15.68 W/m2K hoxd = 17.926 W/m2K
The above values of h we got from experimental procedures.
THE VALUE OF 'h' BY FORMULAS:
We have the relation, Nusselt number:
Nu = 0.47 (Gr Pr)1/4 (1.1)
After finding the value of Nu we can to find the value of h using the relation:
Nu = hD/k (1.2)
Where ,k is the thermal conductivity of air = 2.78 x 10-2 W/mK
But before that, we need to calculate the Grashof number: Gr = gβθD3/ v2
Where g is the acceleration due to gravity and β is the coefficient of cubical expansion which for air is the reciprocal of ambient temperature (1/Ta). The kinematic viscosity v for air is 1.76 x 10-5 m2/s.
Here θ varies so we take:
θaverage = (100oC - 50oC) / 2 = 55oC.
Hence,
Gr = 30396.299
Prandtl number: Pr = µC/k, whose value for air is 0.7
Putting the above values in equation (1.1) we get:
Nu = 5.67
And hence by equation (1.2) we get:
h = 16.6 W/m2K
To know the percentage difference between the correlation value and experimental value we calculated:
% difference = [ ( h - hpol ) / h ] x 100
% difference = 5.4%
THE AMOUNT OF HEAT LOST BY RADIATION:
The difference between the experimentally found values of h for the polished and oxidised bar gives the heat lost by radiation. i.e.:
hraditaion = hoxd - hpol
hraditaion = 0.9
Next the radiant heat flux was calculated:
q = hraditaion x θaverage
q = 295.2 W/m2
By Stefan Boltzman law; (σ is the Stefan Boltzman constant = 56.7 x 10-9 W/m2K4)
q = ε σ (T4 - Ta4)
ε = 0.713
In this experiment we supposed that temperature is only a function of time and not space (meaning we suppose the temperature distribution over the bars is uniform throughout the time). This was also validated when we calculated the Biot modulus which should have been less than 0.1 for steady state systems.
Bi = h[L] / k
Where, [L] = V/A of the cylindrical rod.
Bi = 0.00036 W/mK
Precautions:
Stand at a distance from the boxes so as to allow a natural and an undisturbed draft of air to flow in and out of the boxes.
Handle the strip heater carefully, in order to avoid any burns/accidents. Use gloves if provided.
Place the thermometer measuring the ambient temperature of the room at a distance from the boxes so that the heat of the strip heaters does not affect its reading. Plus, make sure the thermometer is placed in such a condition that it does measure an average temperature of the room (meaning avoid placing the thermometer in the hot or cold spots of the room). Preferably measure the temperature near the boxes before the experiment begins.
Try to measure the time with the help of the stop watch as accurately as humanly possible.
RESULTS:
Firstly we saw that the slope m for the polished bar is 0.0021 and for the oxidized bar is 0.0024 which shows that the oxidized bar has been cooling at a faster pace than the polished bar.
This was further backed up the values of convection coefficient of heat transfer as hpol = 15.68 W/m2K and hoxd = 17.926 W/m2K. This means that heat by convection is lost more by the oxidized bar than the polished bar. The heat lost by convection in the oxidized bar was 1.143 times more than in the polished bar.
The value of h calculated by the correlation came out to be 16.6 W/m2K, while that by experimental method came out to be 15.68 = 15.7 W/m2K, with a very small error of 5.4% which 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. This error can further be decreased if some errors are corrected.
Further we see that the oxidized bar is losing more heat than the polished bar because radiation is playing a vital role in loss of heat from the oxidized bar which we found to be 0.9 W/m2K.
Next the ε value showed us that the polished bar has an emissivity of 0.7 which means it absorbs about 70% of the radiations falling on it.
In our experiment we have considered that there is uniform temperature distribution throughout the bar. We did not take in account of the internal resistance of the body to conduction of heat otherwise our experiment could have become 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) should have a value less than 0.1 which is true in our case.
UNCERTAINTIES:
From the slopes calculated we approximate we have an error of ±0.00005 hence we can state that there is an uncertainty of ±0.00005 in the slopes calculated.
Next as h= m Ï C V / A Thus,
(Δh/h) = [(Δm/m)2]
(Δh/15.68) = [(0.00005/0.0021)2]0.5
Δh = ±0.373
Hence, we can calculate that there is an error of ±0.373 existing in the value of h as all other factors are constant such as Ï C V A (all the data provided in the lab paper handout).
Taking now the value of Gr = gβθD3/ v2 here we see that g, D and v are given in the handouts.
We have β= 1/Ta. The thermometer reading the ambient room temperature has the smallest division of 1oC thus we can say the ambient temperature has an error of:
ΔTa = ±0.5oC. Thus,
(Δβ/β) = [ΔTa/ Ta]
(Δβ/0.05) = [0.5/20]
Δβ = ±0.00125
θaverage = Taverage -Ta, so:
(Δθaverage/ θaverage) = [ (Δ Taverage/ Taverage)2 + (Δ Ta/ Ta)2 ]0.5
(Δθaverage/55) = [ (0.05/75)2 + (0.5/ 20)2]
Δθaverage = ±1.37
(ΔGr/Gr) = [(Δβ/β)2 + (Δθaverage/ θaverage)2]0.5
(ΔGr/30396.3) = [(0.00125/0.05)2 + (1.37/ 55)2]0.5
ΔGr = ±1072.7
The value of Pr is given in data hence no error calculations for that.
Nu = 0.47 (Gr Pr)1/4
Here only Gr has an error so:
(ΔNu/Nu) = 1/4(ΔGr/Gr)
(ΔNu/5.676) = 1/4(1072.7/30396.3)
ΔNu = ±0.05
Now as h = Nu k / D, as k a D are given in data hence;
(Δh/h) = (ΔNu/Nu)
(Δh/16.6) = (0.05/5.676)
Δh = ±0.14
If we compare both the experimental and calculated h values we see that:
Δh = ±0.14 (calculated) Δh = ±0.373 (experimental)
There is more chance of an error in the experimental value than the one calculated from formulas which is quite obvious and expected.#
1.6 CONCLUSIONS:
Clearly from the values of hpol = 15.68 W/m2K & hoxd = 17.926 W/m2K we conclude that the convection coefficient of heat transfer is higher for dull surfaces than for highly polished surfaces. This statement holds true for materials other than brass too. What happens actually is that dull surfaces consist of bumpy or irregular surfaces hence they increase the surface area per unit length in contact with the air flowing around it. This increase the heat lost by convection. Also with increase in surface area comes increase in heat lost hue to radiation.
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The aims of our little experiment have been achieved as we were successful in finding the values of h both by experimental methods and by formulas with a very small error of 5.4%. And even this error was accounted for when we calculated the uncertainties in our measurements. And after comparing the experimental and theoretical values, we in a way validated that this experiment works in finding the convection heat transfer coefficient for any other material too.
We do have some uncertainties in our experimental procedures about ±0.3 to more exact but still the experiment gives us a pretty good picture of the value of h for any given material.
The irregularities seen were only in the straight line graph between Ln(θo/θ) and time. The line obtained was not very straight whose obvious reason is the human error induced in the readings due to the inaccurate timing of the stop watch by the student.
