In this book we derive the generalized cooling law in natural convection for all temperature excesses from empirical results using the fact that Newton's law of cooling is obeyed for temperature excesses less than 30C. We derive an equation from empirical results that predicts the rate of cooling for a given temperature excess and then integrate and solve this equation to give us a straight line that predicts the temperature at a given time for a cooling body in natural convection for all temperature excesses

1. THE RATE OF COOLING IN NATURAL
CONVECTION EXPLAINED
ABSTRACT
In this book we derive the generalized cooling law in
natural convection for all temperature excesses
from empirical results using the fact that Newton’s
law of cooling is obeyed for temperature excesses
less than 30C
Wasswa Derrick
NATURAL CONVECTION

3. THE EMPIRICAL EQUATION OF THE RATE OF COOLING IN
FREE/NATURAL CONVECTION USING MODIFIED NEWTON’S LAW OF
COOLING.
wasswaderricktimothy7@gmail.com
The equation for the rate of cooling in natural convection is given by:
𝑚𝐶
𝑑𝑇
𝑑𝑡
= −ℎ𝐴𝑠(𝑇 − 𝑇𝑜)[1 + 0.75(
𝑇 − 𝑇𝑜
30
)2
]
Or
𝑚𝐶
𝑑𝑇
𝑑𝑡
= −ℎ𝐴𝑠(𝑇 − 𝑇𝑜)[1 + 𝑐(
𝑇 − 𝑇𝑜
30
)2
]
Or
𝒎𝑪
𝒅𝑻
𝒅𝒕
= −𝒉𝑨𝒔∆𝑻[𝟏 + 𝒄(
∆𝑻
𝟑𝟎
)𝟐
] … … … . 𝒂)
Where:
𝑐 = 0.75
𝑇𝑜 = 𝑟𝑜𝑜𝑚 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒
∆𝑇 = 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑒𝑥𝑐𝑒𝑠𝑠 = (𝑇 − 𝑇𝑜)
PROOF
From literature and experiment, it is known that in natural convection,
Newton’s law of cooling is obeyed when the temperature excess is less than
𝟑𝟎℃.
Looking at the equation a) above, when:
∆𝑇
30
≪ 1 𝑂𝑟 𝑤ℎ𝑒𝑛 ∆𝑇 ≪ 30℃
Then
(
∆𝑻
𝟑𝟎
)𝟐
≈ 𝟎
And we get Newton’s law of cooling i.e.,
𝒎𝑪
𝒅𝑻
𝒅𝒕
= −𝒉𝑨𝒔∆𝑻

4. Provided the temperature excess is less than 𝟑𝟎℃ Newton’s law of cooling is
obeyed.
We can rearrange Newton’s law of cooling and get:
𝒅𝑻
𝒅𝒕
= −
𝒉𝑨𝒔
𝒎𝑪
∆𝑻
Or
𝒅𝑻
𝒅𝒕
= −𝒌∆𝑻 … … 𝒃)
Where:
𝒌 =
𝒉𝑨𝒔
𝒎𝑪
On solving the equation b) above and using experimental results, we can
measure off k.
The solution to equation b) above is:
𝐥𝐧(𝑻 − 𝑻𝒐) = −𝒌𝒕 + 𝐥𝐧(𝑻𝒊 − 𝑻𝒐)
After getting to know k for a given body, we then resort to its cooling rate for
high temperature excesses. The equation obeyed is:
𝒎𝑪
𝒅𝑻
𝒅𝒕
= −𝒉𝑨𝒔∆𝑻 [𝟏 + 𝒄(
∆𝑻
𝟑𝟎
)𝒏
]
OR
𝒅𝑻
𝒅𝒕
= −𝒌∆𝑻 [𝟏 + 𝒄(
∆𝑻
𝟑𝟎
)𝒏
]
HOW DO WE VERIFY THAT n=2?
To verify the above, we consider cooling at high temperature excesses. From
the graph of temperature T against time for the cooling body, we evaluate the
slope or rate of change of temperature given by (a, b) at two high temperature
excesses(∆𝑻𝑎, ∆𝑻𝑏) respectively.
Since we know k, we can rearrange the equation above and have:
𝒅𝑻
𝒅𝒕
⁄
−𝒌∆𝑻
= 𝟏 + 𝒄(
∆𝑻
𝟑𝟎
)𝒏
Then get:

5. (
𝒅𝑻
𝒅𝒕
⁄
−𝒌∆𝑻
− 𝟏) = 𝒄(
∆𝑻
𝟑𝟎
)𝒏
For temperature excess ∆𝑻𝑎 we have
𝑑𝑇
𝑑𝑡
⁄ = 𝒂
For temperature excess ∆𝑻𝑏 we have
𝑑𝑇
𝑑𝑡
⁄ = 𝒃
Upon substitution, we have:
(
𝒂
−𝒌∆𝑻𝑎
− 𝟏) = 𝒄(
∆𝑻𝑎
𝟑𝟎
)𝒏
… . . 𝟏)
Similarly for temperature excess , ∆𝑻𝑏 we have
(
𝒃
−𝒌∆𝑻𝑏
− 𝟏) = 𝒄(
∆𝑻𝑏
𝟑𝟎
)𝒏
… . . 𝟐)
Dividing equation 1 and 2 above we get
(
𝒂
−𝒌∆𝑻𝑎
− 𝟏)
(
𝒃
−𝒌∆𝑻𝑏
− 𝟏)
= (
∆𝑻𝑎
∆𝑻𝑏
)𝒏
and taking natural logarithms we get
𝒍𝒏 (
(
𝒂
−𝒌∆𝑻𝑎
− 𝟏)
(
𝒃
−𝒌∆𝑻𝑏
− 𝟏)
) = 𝒏 × 𝒍𝒏(
∆𝑻𝑎
∆𝑻𝑏
)
From experiment, n was got equal to 2 i.e.,
𝒏 = 𝟐
Using either equation 1 or 2, we can measure off an approximate value of c.
We call it an approximate value because from experimental value we don’t get n
as exactly 2 but close to 2.
The value of c is chosen so that when we integrate the equation of cooling
below
𝒎𝑪
𝒅𝑻
𝒅𝒕
= −𝒉𝑨𝒔∆𝑻 [𝟏 + 𝒄(
∆𝑻
𝟑𝟎
)𝟐
]

6. And get the solution for all temperature and time, all the points will lie as close
as possible to the straight-line graph. The value of c that fits the above criteria
is 𝑐 = 0.75
So, the rate of cooling in natural convection is given by:
𝒎𝑪
𝒅𝑻
𝒅𝒕
= −𝒉𝑨𝒔∆𝑻 [𝟏 + 𝟎. 𝟕𝟓(
∆𝑻
𝟑𝟎
)𝟐
]
EXPERIMENTAL VERIFICATION
Using temperature-time data for water in a glass beaker of surface area to
volume ratio
𝐴
𝑉
= 156.1842𝑚−1
(𝑟𝑎𝑑𝑖𝑢𝑠 = 2.375𝑐𝑚) below:
Room temperature 𝑇0 = 24℃
For temperature excesses less than 30℃ ,Newton’s law of cooling is obeyed as
shown below using experimental results.
𝑚𝐶
𝑑𝑇
𝑑𝑡
= −ℎ𝐴𝑠(𝑇 − 𝑇𝑜)
The solution is:

7. 𝐥𝐧(𝑻 − 𝑻𝒐) = −
𝒉𝑨𝒔
𝒎𝑪
𝒕 + 𝐥𝐧(𝑻𝒊 − 𝑻𝒐)
𝑘 =
ℎ𝐴𝑠
𝑚𝐶
𝐥𝐧(𝑻 − 𝑻𝒐) = −𝒌𝒕 + 𝐥𝐧(𝑻𝒊 − 𝑻𝒐)
𝒌 =
𝒉𝑨𝒔
𝒎𝑪
= 𝟎. 𝟎𝟎𝟎𝟔
TO VERIFY THAT n=2
Recalling
𝒍𝒏 (
(
𝒂
−𝒌∆𝑻𝑎
− 𝟏)
(
𝒃
−𝒌∆𝑻𝑏
− 𝟏)
) = 𝒏 × 𝒍𝒏(
∆𝑻𝑎
∆𝑻𝑏
)
From the table above, choosing the temperature excesses below:
∆𝑻𝑎 = 61℃ 𝑎𝑛𝑑 ∆𝑻𝑏 = 51℃
And their respective cooling rates
𝒂 = −0.1073242 (
℃
𝑠
) 𝑎𝑛𝑑 𝒃 = −0.07072159 (
℃
𝑠
)
Inserting in the formula above, we get:
y = -0.0006x + 3.2581
0
0.5
1
1.5
2
2.5
3
3.5
0 1000 2000 3000 4000 5000
𝐥𝐧(𝑻−𝑻𝒐
)
time t
A graph of 𝐥𝐧(𝑻−𝑻𝒐 ) against time t

8. 𝑙𝑛 (
(
0.1073242
−0.0006 × 61
− 1)
(
0.07072159
−0.0006 × 51
− 1)
) = 𝑛 × 𝑙𝑛(
61
51
)
𝑛 = 2.1662
From what we have got n=2
Using the general equation
𝑚𝐶
𝑑𝑇
𝑑𝑡
= −ℎ𝐴𝑠∆𝑇 [1 + 0.75(
∆𝑇
30
)2
]
𝑑𝑇
𝑑𝑡
= −(
ℎ𝐴𝑠
𝑚𝐶
)∆𝑇 [1 + 0.75(
∆𝑇
30
)2
]
It is also shown below that a graph of
𝒅𝑻
𝒅𝒕
against (𝑻 − 𝑻𝒐)[𝟏 + 𝟎. 𝟕𝟓(
𝑻−𝑻𝒐
𝟑𝟎
)𝟐
] is a
straight-line graph through the origin for all temperature excesses in natural
convection. (For water in a glass beaker of (
𝐴
𝑉
= 156.1842 𝑚−1
)
Let us call
(𝑻 − 𝑻𝒐)[𝟏 + 𝟎. 𝟕𝟓(
𝑻 − 𝑻𝒐
𝟑𝟎
)𝟐
] = 𝒁
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0 50 100 150 200 250 300 350
-(dT/dt)
Z
A graph of -(dT/dt) against Z

9. The equation above can be integrated to find the temperature profile with time
for a cooling body for all time.
Let us solve the governing differential equation below:
𝒅𝑻
𝒅𝒕
= −
𝒉𝑨𝒔
𝒎𝑪
(𝑻 − 𝑻𝒐)[𝟏 + 𝟎. 𝟕𝟓(
𝑻 − 𝑻𝒐
𝟑𝟎
)𝟐
]
Let 𝑦 = (𝑇 − 𝑇𝑜)
𝑑𝑦
𝑑𝑡
= −𝑑𝑦[1 + 𝑏𝑦2
]
Where:
𝑑 =
ℎ𝐴𝑠
𝑚𝐶
𝑏 =
0.75
302
𝒅𝒚
𝒅𝒕
= −(𝒅𝒃)𝒚[𝒚𝟐
+
𝟏
𝒃
]
∫
𝑑𝑦
𝑦[𝑦2 +
1
𝑏
]
= −(𝑑𝑏)∫ 𝑑𝑡 … . 𝑖)
From literature it is known that the integral below is equal to:
∫
𝒅𝒙
𝒙[𝒙𝟐 + 𝒂𝟐]
=
𝟏
𝟐𝒂𝟐
𝐥𝐧 (
𝒙𝟐
𝒙𝟐 + 𝒂𝟐
)
By analogy the solution to equation i) above is:
∫
𝒅𝒚
𝒚[𝒚𝟐 +
𝟏
𝒃
]
=
𝒃
𝟐
𝐥𝐧 (
𝒚𝟐
𝒚𝟐 +
𝟏
𝒃
)
Upon substitution in equation i), we get
𝑏
2
ln (
𝑦2
𝑦2 +
1
𝑏
) = −(𝑑𝑏) ∫ 𝑑𝑡
𝑡
0
1
2
ln (
𝑦2
𝑦2 +
1
𝑏
) = −(𝑑)∫ 𝑑𝑡
𝑡
0
Upon substitution of the values, we get

10. 1
2
ln(
(𝑇 − 𝑇𝑜)2
(𝑇 − 𝑇𝑜)2 + (
1
𝑏
)2
) = −
ℎ𝐴𝑠
𝑚𝐶
𝑡 + 𝐷
Using the initial condition, we can eliminate D.
𝑇 = 𝑇𝑠 𝑎𝑡 𝑡 = 0
1
2
ln (
(𝑇𝑠 − 𝑇𝑜)2
(𝑇𝑠 − 𝑇𝑜)2 + (
1
𝑏
)2
) = 𝐷
1
2
ln (
(𝑇𝑠 − 𝑇𝑜)2
(𝑇𝑠 − 𝑇𝑜)2 + (
1
𝑏
)2
) −
1
2
ln(
(𝑇 − 𝑇𝑜)2
(𝑇 − 𝑇𝑜)2 + (
1
𝑏
)2
) =
ℎ𝐴𝑠
𝑚𝐶
𝑡
ln (
(𝑇𝑠 − 𝑇𝑜)2
(𝑇𝑠 − 𝑇𝑜)2 + (
1
𝑏
)2
) − ln (
(𝑇 − 𝑇𝑜)2
(𝑇 − 𝑇𝑜)2 + (
1
𝑏
)2
) = 2
ℎ𝐴𝑠
𝑚𝐶
𝑡
𝐥𝐧
(
(𝑻𝒔 − 𝑻𝒐)𝟐
(𝑻𝒔 − 𝑻𝒐)𝟐 + (
𝟏
𝒃
)𝟐
(𝑻 − 𝑻𝒐)𝟐
(𝑻 − 𝑻𝒐)𝟐 + (
𝟏
𝒃
)𝟐
)
= 𝟐
𝒉𝑨𝒔
𝒎𝑪
𝒕
Where:
𝑏 =
0.75
302
The above is the solution for the cooling curve.
Plotting a graph of 𝐥𝐧 (
(𝑻𝒔−𝑻𝒐)𝟐
(𝑻𝒔−𝑻𝒐)𝟐+(
𝟏
𝒃
)𝟐
(𝑻−𝑻𝒐)𝟐
(𝑻−𝑻𝒐)𝟐+(
𝟏
𝒃
)𝟐
) against time t gives a straight line graph
through the origin for all temperatures and time as shown for the graph below
of cooling water in a beaker of
𝐴
𝑉
= 156.1842𝑚−1
Let us call 𝐥𝐧 (
(𝑻𝒔−𝑻𝒐)𝟐
(𝑻𝒔−𝑻𝒐)𝟐+(
𝟏
𝒃
)𝟐
(𝑻−𝑻𝒐)𝟐
(𝑻−𝑻𝒐)𝟐+(
𝟏
𝒃
)𝟐
) = 𝑷

11. It can be shown that the solution above works for other surface areas to
volume ratios of water cooling in a beaker.
For example, when the surface area to volume ratio
𝐴
𝑉
= 76.1905𝑚−1
, the graph
of p against time looked as below for all temperature excesses and time as a
straight-line graph.
y = 0.001x
-1
0
1
2
3
4
5
6
7
8
0 1000 2000 3000 4000 5000 6000 7000 8000
P
time t(seconds)
A Graph of P against time
y = 0.0005x
-1
0
1
2
3
4
5
6
0 2000 4000 6000 8000 10000 12000 14000
P
time t(seconds)
A Graph of P against time

12. Procedure for determining c.
From the graph of P against time above, we can vary the value of c until a value
is reached for which a straight-line graph is got through most of the points and
the origin. Using the above procedure, the value of c was found to be
𝒄 = 𝟎. 𝟕𝟓
𝐍𝐁:
When using the equations above, it should be noted that the size of the body or
the surface area to volume ratio of the body should remain constant as stated
by the equations above. If the size of the body changes significantly for example
a cooling body with a lot of evaporation happening so that the fraction of the
mass lost to evaporation compared to the original mass of the liquid is great,
then the equation doesn’t hold.
Otherwise for large bodies for example cooling water in a beaker, the fraction of
water lost due to evaporation is not significant compared to the water in the
beaker and so the cooling equation can be taken to hold.

13. 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.,
𝑃 = 𝑉𝐼
To solve for how the temperature varies with time, we solve the equation below:
𝒎𝑪
𝒅𝑻
𝒅𝒕
= 𝑷 − 𝒉𝑨𝒔(𝑻 − 𝑻𝒐)[𝟏 + 𝟎. 𝟕𝟓(
𝑻 − 𝑻𝒐
𝟑𝟎
)𝟐
]
To get the temperature evolution with time.
In steady state,
𝑑𝑇
𝑑𝑡
= 0
𝑷 = 𝒉𝑨𝒔(𝑻 − 𝑻𝒐)[𝟏 + 𝟎. 𝟕𝟓(
𝑻 − 𝑻𝒐
𝟑𝟎
)𝟐
]