DERIVATION OF THE LAW OF COOLING IN NATURAL CONVECTION.pdf

In this book, the law we derive the law of cooling in natural convection for a cooling liquid based on theory that a cooling liquid exposed to air at the free surface loses heat both by natural convection and by evaporation from the free surface. From experimental results we deduce the nature of the equation of heat loss by evaporation. Using this we solve the general cooling equation to predict temperature against time for a cooling body. We also investigate the quadratic nature of the cooling law and deduce the condition under which Newton’s law of cooling is obeyed. We derive the nature of the heat transfer coefficient using the Stanton number. The law developed is easy to work with compared to the correlations of heat transfer coefficient in natural convection found in literature which make predicting temperature against time for a cooling body difficult.

1. THE RATE OF COOLING IN
NATURAL CONVECTION
EXPLAINED
For cooling liquids
Wasswa Derrick 2/16/25 Mathematical Physics

2. 1

3. 2

PREFACE
A BRIEF NOTE ON THE RATE OF COOLING IN NATURAL CONVECTION
It is known from scientific literature that the rate of cooling in natural
convection is modelled using Newton’s law of cooling but with a fact that the
heat transfer coefficient to be a function of the Grashoff number [1].
The Grashoff number is given as:
??????�=
??????��
3
∆�
??????
2

Where:
∆�=(�−�
�
)=����������� ������
The Grashoff number is evaluated at the mean film temperature of the cooling
body and surrounding medium. This means that all the parameters in the Gr
number for example the viscosity and density of the surrounding medium have
to be evaluated at this mean temperature.
For a cooling body whose temperature is varying with time, this means we have
to evaluate the Gr number at time intervals for the temperatures got and solve
the cooling equation to get the temperature evolution with time. Truly all the
work above can only be done using a complex computer program and this
makes solving the cooling equation very difficult or impossible to predict the
times taken to cool to a particular temperature.
One can go ahead and even check in literature the available correlations for the
heat transfer coefficient in natural convection for example the Churchill and
Chu [2]correlations for the Nusselt number to get the heat transfer coefficient
but again this would be extremely difficult or impossible to use it to predict the
temperature variation with time of a cooling body.
Another issue with the heat transfer coefficient correlations is that the heat
transfer coefficient has to be evaluated for every surface enclosing the liquid of
which each surface has a different correlation of heat transfer coefficient of
Grasshoff number.
The aim of this paper is to develop an empirical law of cooling in natural
convection based on theory and experiment that is analytical for which we can
predict the temperature against time of a cooling body by simply integrating.

4. 3

TABLE OF CONTENTS
PREFACE ............................................................................................................................................ 2
THE RATE OF COOLING IN NATURAL CONVECTION ......................................................... 4
VERIFICATION OF THE EQUATION ABOVE ........................................................................... 6
DERIVATION OF THE COOLING LAW ABOVE ....................................................................... 9
WHEN DOES NEWTON ’S LAW OF COOLING HOLD? ......................................................... 12
WHAT IS THE GENERAL LAW OF HEAT LOSS BY EVAPORATION? ............................ 13
THE NATURE OF THE HEAT TRANSFER COEFFICIENT h AND THE GENERAL
LAW OF COOLING .......................................................................................................................... 15
TO SHOW HOW THE STEFAN BOLTZMANN LAW REDUCES TO NEWTON ’S LAW
OF COOLING ................................................................................................................................ 18
HOW DO WE DEAL WITH CASES WHERE THERE IS A POWER SUPPLY TO THE
FLUID IN THE CONTAINER? ...................................................................................................... 20
REFERENCES .................................................................................................................................. 21

5. 4

THE RATE OF COOLING IN NATURAL CONVECTION
To explain the rate of cooling in natural convection [3], we first of all have to
know the factors that affect the cooling rate of a cooling body:
One of these factors that is crucial, is the nature of the surface on which the
cooling body is resting. For example, a cooling body resting on a cold cement
will cool faster and differently from that resting on a poor conductor like wood.
So, to be exactly uniform in our experiments, we shall only consider cooling
bodies resting on wood (a poor conductor).
NB
All the experimental deductions got in this text is for cooling bodies resting on
wood.
THEORY
To explain the rate of cooling in natural convection, we have to know the
causes of heat loss. For a cooling liquid in a container, the causes of heat loss
are:
 Convection
 Evaporation from the free surface

6. 5


So, from experimental results, we can postulate the cooling law to be of the
form
���� �� ℎ��� ����=ℎ��� ���� �� �������??????��+ℎ��� ���� �� ��������??????�� ���� ���� �������
�??????
�??????
��
=�??????
�
(??????−??????
�
)+
�
�??????
??????
�
(??????−??????
�)
�

Where:
??????
�=??????�??????�??????�� ����� ������� ���� �� ����??????�� ����
??????=�����−����??????���� ���� �� ���� ������� �� ��������??????��
�=??????�??????�??????�� ���� �� ����??????�� ����
�=����??????�??????� ℎ��� �����??????�� �� ����??????�� ����
�= ����������� ??????� ����??????��
�
�=���� ����������� ??????� ����??????��

7. 6

VERIFICATION OF THE EQUATION ABOVE
Rearranging the equation above we get:
��
��
=
ℎ??????
�
��
(�−�
�
)+
ℎ
1??????
���
�
(�−�
�)
2

When the temperature is decreasing with time, we choose the negative
temperature gradient i.e.,
−
��
��
=
ℎ??????
�
��
(�−�
�
)+
ℎ
1??????
���
�
(�−�
�)
2

Dividing through by the temperature excess we get:
�
(??????−??????
�
)
−�??????
��
=
�
�??????
�????????????
�
(??????−??????
�
)+
�??????
�
�??????

For a beaker containing cooling water with the free surface open to air, we have
??????
�=2��(�+�)
??????=��
2

�=��
2
��
Upon simplifying the equation above, we get;

−�
(??????−??????
�
)
�??????
��
=
�
�
�??????????????????
�
(??????−??????
�
)+
�??????
�
�??????

OR
−�
(??????−??????
�
)
�??????
��
=
�
�
�??????????????????
�
(??????−??????
�
)+
�
�??????
(
??????
�
??????
)
Where:
??????
�
??????
=������� ���� �� ������ ����� �� ������� ���??????
The following experimental results are for cooling water in a plastic beaker of
surface area to volume ratio of 148.052�
−1
�??????�ℎ ����ℎ �=3.5�� as shown in the
diagram above.

8. 7


From experimental results, using Microsoft Excel or a graph, plotting a graph
of
−�
(??????−??????�)
�??????
��
against (??????−??????
�
) gave a straight-line graph with a positive intercept as
stated by the equation above. e.g., for cooling water of surface area to volume
ratio of 148.052�
−1
�??????�ℎ ����ℎ �=3.5�� cooling in a beaker resting on wood the
graph looked as below

From the graph above
ℎ=
8.5101�
�
2
�
��� ℎ
1=
879.942�
�
2
�

To verify that evaporation really affects the rate of cooling through the open
cross-sectional area, it was found that by fixing the surface area to volume
ratio of cooling water in two beakers but varying the heights L of the cooling
y = 2E-05x + 0.0003
R² = 0.9927
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
0.0014
0 10 20 30 40 50 60(
-
1
/(
??????
−
??????
??????
) )*
??????
??????
/
??????
??????
(??????−????????????)
A graph of (-1/(??????−????????????) )*????????????/????????????against (??????−????????????)
Series1
Linear (Series1)

9. 8

water, the gradient of the graph above was as given by the equation above i.e.,
the gradient was inversely proportional to the depth L of the cooling liquid:
ℎ
1
����
�

The value of the heat transfer coefficient can be got from the intercept of the
graph above.
So generally, to get the temperature against time of a cooling liquid in natural
convection, we have to solve the equation stated above and below as:
−
��
��
=
ℎ??????
�
��
(�−�
�
)+
ℎ
1??????
���
�
(�−�
�)
2

10. 9

DERIVATION OF THE COOLING LAW ABOVE
The general law of cooling in natural convection is given by Newton’s law of
cooling assuming the mass �
?????? and surface area ??????
�?????? are constant and no heat
loss by evaporation:
�
�??????
�??????
��
=�??????
��
(??????−??????
�
)
It’s known that evaporation causes a loss of mass ∆�(�) which is a function of
time and so the effective mass in the cooling law above becomes (�
??????−∆�(�))
and effective surface area becomes (??????
�??????−∆??????
�(�)). The law of cooling becomes:
(�
??????−∆�(�))�
��
��
=ℎ(??????
�??????−∆??????
�(�))(�−�
�
)
But for a cylindrical beaker
∆??????
�=2��∆�
So, the heat loss ℎ∆??????
�
(�−�
�
) means that there is heat loss normal to this
surface area but the actual loss of heat by evaporation occurs vertically not
normal to this surface area and so ℎ∆??????
�
(�−�
�
)=0
The cooling law then becomes:
(�
�−∆�(�))??????
�??????
��
=�??????
��
(??????−??????
�
)−−−�)
The rate of heat loss by evaporation is given as:
��
��
??????(??????−??????
�
)=
��
�
??????
�
??????(??????−??????
�)
�
−−−−�)
Equation 1) gives the rate of change of temperature with the inclusion of
change in mass due to loss to evaporation ∆� . where:
∆�(�)=∫��
Equation 2) gives the rate of change of mass due to evaporation. The
superscript i refers to the initial state of the cooling body when reading begin to
be taken or when the stop watch starts to record time.
To get the rate of change of temperature we first have to get ∆� and we get it
from equation 2.
Considering a cylindrical beaker of cross-sectional area, A,
From 2) we have:

11. 10

��
��
�(�−�
�
)=
2ℎ
1
�
�
??????(�−�
�)
2

��=
2ℎ
1
�
��
??????(�−�
�)��
∆�=∫��
∆�=∫
2ℎ
1
�
��
??????(�−�
�)��
Upon substituting in equation 1) above we have:
(�
??????−∆�)�
��
��
=ℎ??????
�??????
(�−�
�
)
�
??????�
��
��
−∆��
��
��
=ℎ??????
�??????
(�−�
�
)
�
??????�
��
��
−∫
2ℎ
1
�
��
??????(�−�
�)���
��
��
=ℎ??????
�??????
(�−�
�
)
Which simplifies to:

�
??????�
��
��
−∫
2ℎ
1
�
�
??????(�−�
�)��=ℎ??????
�??????
(�−�
�
)
�
??????�
��
��
−
2ℎ
1
�
�
??????∫(�−�
�)��=ℎ??????
�??????
(�−�
�
)
Upon integrating, we get:
�
??????�
��
��
−
2ℎ
1
�
�
??????[
(�−�
�)
2
2
]+�=ℎ??????
�??????
(�−�
�
)
Where: �=??????�������??????�� ��������
To get D we know that at temperature �=�
�,
�??????
��
=0 and so we get �=0
So, the cooling law becomes:
�
??????�
��
��
−
ℎ
1
�
�
??????(�−�
�)
2
=ℎ??????
�??????
(�−�
�
)
We finally get:
�
�??????
�??????
��
=�??????
��
(??????−??????
�
)+
�
�
??????
�
??????(??????−??????
�)
�

12. 11

Which is the cooling law we observed. Or
�
�??????
�??????
��
=�??????
��
(??????−??????
�
)+
�
�
??????
�
??????(??????−??????
�)
�

Where:
�
??????=??????�??????�??????�� ���� �� ����??????�� ����
??????
�??????=??????�??????�??????�� ������� ���� �� ����??????�� ����
??????= �����−����??????�� ���� �� ��������??????��
�=����??????�??????� ℎ��� �����??????�� �� ����??????�� �??????��??????�
For a cylindrical beaker, we notice from the above that the rate of change of
temperature only depends on the initial mass and initial surface area and
cross-sectional area of the beaker.

13. 12

WHEN DOES NEWTON’S LAW OF COOLING HOLD?

Newton’s law of cooling states that:
�??????
�??????
��
=�??????
�
(??????−??????
�
)
OR
��
��
=
ℎ??????
�
��
(�−�
0
)
To arrive at the Newton’s law of cooling, we look at the governing equation for a
cooling liquid i.e.,
�
??????�
��
��
=ℎ??????
�??????
(�−�
�
)+
ℎ
1
�
�
??????(�−�
�)
2

Upon factoring, we get:
�
??????�
��
��
=ℎ??????
�??????
(�−�
�
)[1+
ℎ
1
ℎ�
�
??????
??????
�??????
(�−�
�
)]
When
ℎ
1
ℎ�
�
??????
??????
�??????
(�−�
�
)≪1
We approximate and ignore
ℎ1
ℎ????????????
??????
??????
????????????
(�−�
�
) and arrive at Newton’s law of cooling
�
�??????
�??????
��
=�??????
��
(??????−??????
�
)
So, the condition to arrive at Newton’s law of cooling is:
ℎ
1
ℎ�
�
??????
??????
�??????
(�−�
�
)≪1

14. 13

WHAT IS THE GENERAL LAW OF HEAT LOSS BY
EVAPORATION?
Previously, we derived the heat loss equation by evaporation as:
��
��
�(�−�
�
)=
ℎ
1
�
�
??????(�−�
�)
2

Where:
��=���� �� �??????��??????� ���� �� ��������??????��
�=����??????�??????� ℎ��� �����??????�� �� �??????��??????� ��� ������
But this is not the general law of heat loss by evaporation as we are going to
see:
From experimental results involving a cooling liquid in a container enclosed by
a lid so that evaporation to the surroundings is prohibited, the nature of the
heat loss by evaporation was found to be:


��
��
??????(??????−??????
�
)=
�
�
??????
�
(��??????
�
�
)(??????−??????
�)
�

Where:
�
�=�ℎ�������??????��??????� �����ℎ ??????� ���??????�� 1 ��������
And it was found that:

15. 14

�
�=
1
(
??????
�
)
|1

Where:
(
??????
�
)
|1=������� ���� �� ������ ���??????� ??????� ���??????�� 1
(
??????
�
)
|1=2(
1
�
+
1
�
)
When the lid is removed so that evaporation freely occurs �→∞ so that
(
??????
�
)
|1=2(
1
�
)
And we arrive back to the cooling law:
��
��
�(�−�
�
)=
ℎ
1
�
�
??????(�−�
�)
2

So generally, the evaporation heat loss equation is
��
��
??????(??????−??????
�
)=
�
�
??????
�
(
��
((
??????
??????
)
|�)
�
)(??????−??????
�)
�

In cylindrical enclosures it is given by:
��
��
??????(??????−??????
�
)=
�
�
??????
�
(
��
(�(
�
�
+
�
�
))
�
)(??????−??????
�)
�

As shown in the diagram above.

16. 15

THE NATURE OF THE HEAT TRANSFER COEFFICIENT
h AND THE GENERAL LAW OF COOLING
The total heat transfer coefficient h is the sum of the heat transfer coefficient ℎ
2
due to convection and the radiative heat transfer coefficient ℎ
3 i.e.,
ℎ=ℎ
2+ℎ
3
To derive ℎ
2 for convection, we can use the Stanton number (St) [6] to get an
expression for the heat transfer coefficient h by equating it to a constant a i.e.,
��=
ℎ
2
�
??????��
�
=�
Where;
�
??????=����??????�� �� �??????� �� ���� �����������
�=�����??????�� �� �??????�
�
�=����??????�??????� ℎ��� �����??????�� �� �??????� �� �������� ��������
ℎ
2=��
??????��
�
where:
�=�����??????������ ��������
ℎ
2=��
??????�
��
At room temperature
��������,�=
1
2
�
??????�
2

So
�=√
2�
�
??????

we get:
ℎ
2=��
??????�
�√
2�
�
??????

ℎ
2=��
�√2��
??????
The radiation heat transfer coefficient is given by
ℎ
3=4????????????�
0
3

17. 16

Where:
??????=������
′
���������=5.67×10
−8
�/(�
2
�
4
)
�
0=���� ����������� ??????� ����??????��
??????=���??????�??????�??????��
So, the heat transfer coefficient h in natural convection is given by:
�=�??????????????????
�
�
+�??????
�√���
�
So, the general law of cooling in natural convection can be stated as:
�
�??????
�??????
��
=??????????????????
��
(??????
�
−??????
�
�
)+�
�??????
��
(??????−??????
�
)+
�
�
??????
�
??????(??????−??????
�)
�

Where:
�
�=�
�??????
�??????
�
�
??????=�����??????�� �� �ℎ� ������
�
??????=����??????�??????� ℎ��� �����??????�� �� ������ �� ���� �����������
�
??????=����??????�� �� ������ �� ���� �����������
When the temperature excess is less than ??????
� �� �������, (we shall show the
derivation later) then Stefan’s law of cooling [7]
�=??????????????????
��
(??????
�
−??????
�
�
)
reduces to
�=4????????????�
0
3
??????
�??????
(�−�
�
)
And the cooling law becomes:
�
??????�
��
��
=4????????????�
0
3
??????
�??????
(�−�
�
)+ℎ
1??????
�??????
(�−�
�
)+
ℎ
1
�
�
??????(�−�
�)
2

OR
�
??????�
��
��
=(4????????????�
0
3
+ℎ
2)??????
�??????
(�−�
�
)+
ℎ
1
�
�
??????(�−�
�)
2

OR
�
??????�
��
��
=ℎ??????
�??????
(�−�
�
)+
ℎ
1
�
�
??????(�−�
�)
2

Where:
ℎ=(4????????????�
0
3
+ℎ
2)

18. 17

ℎ
2=��
�√2��
??????
So, we finally get
�
�??????
�??????
��
=�??????
��
(??????−??????
�
)+
�
�
??????
�
??????(??????−??????
�)
�

Which is what we observed.
In a vacuum Stefan’s law will be the law of cooling since ℎ
2=0 assuming no
evaporation is allowed from the liquid surface.
In case of forced convection, we have:
ℎ=4????????????�
0
3
+��
�√2��
??????+��
??????�
��
for large velocities, h becomes:
ℎ=��
??????�
��
where:
�=�����??????�� �� �ℎ� ��������??????�� ���??????��
In forced convection, Newton’s law of cooling will be obeyed with h given as
above because h will have such a high value that the rate of loss by
evaporation is negligible compared to that lost by forced convection

19. 18

TO SHOW HOW THE STEFAN BOLTZMANN LAW REDUCES TO
NEWTON’S LAW OF COOLING
From
�̇=??????????????????
�(??????
�
−??????
�
�
)
Where: temperature T is in Kelvins
�=(�
0+(�−�
0
))
Upon substitution in the Stefan law, we get:
�̇=??????????????????
�((�
0+(�−�
0
))
4
−�
0
4
)
Factoring out �
0 we get:
�̇=??????????????????
�(�
0
4
(1+
(�−�
0
)
�
0
)
4
−�
0
4
)
Using the Taylor expansion for small x
(1+�)
�
≈1+�� ��� �≪1
We have that for
(�−�
0
)
�
0
≪1
OR
When (�−�
0
)≪�
0
Where �
0=���� ����������� ??????� ����??????��
Truly �
0 is a big value for example when the room temperature is 24℃ then �
0=
297� which means that when the temperature excess is less than 297K or
297℃ we arrive at Newton’s law of cooling as below:

We get:
(1+
(�−�
0
)
�
0
)
4
≈1+4
(�−�
0
)
�
0

And we get upon substitution in the Stefan law:
�̇=??????????????????
�(�
0
4
(1+
(�−�
0
)
�
0
)
4
−�
0
4
)≈??????????????????
�(�
0
4
(1+4
(�−�
0
)
�
0
)−�
0
4
)
Upon simplification we get:

20. 19

�̇=�??????????????????
�
�
??????
�
(??????−??????
�
)

21. 20

HOW DO WE DEAL WITH CASES WHERE THERE IS A
POWER SUPPLY TO THE FLUID IN THE CONTAINER?
Theory:
Let us call the power given by the power supply be given by P. This P could be
electrical power i.e.,
�=�??????
To solve for how the temperature varies with time, we solve the equation below:
�
�??????
�??????
��
=�−(�??????
��
(??????−??????
�
)+
�
�
??????
�
??????(??????−??????
�)
�
)

In steady state,
��
��
=0
�=�??????
��
(??????−??????
�
)+
�
�
??????
�
??????(??????−??????
�)
�

22. 21

REFERENCES

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