Applications of the Energy Equation
Solids with a Uniform Temperature
Suppose a metal sphere of uniform temperature, T.
Heat is transferred by convection with a heat transfer
coefficient, h. The temperature of the surroundings is
Ta
Therefore the energy balance is
ρC pV dT = – hA T – Ta
dt
(Please note the difference between this equation and 11.1.1 in the text. This is the
correct form)
If the sphere is initially at T0, how does one describe the cooling of the
sphere?
dT
hA dt
The equation is separable so that T – T = –
ρC pV
a
T–T
a
hA t
It follows that the solution is ln T – T = –
ρC pV
0
a
Or more explicitly
ChE 333- Lecture 6
Spring 2000
hA
T – Ta
= e – ρC V t
T0 – Ta
p
1
Adimensionalization
If I had defined the following:
ρC pV
T – Ta
θ =
and τ = hA
T0 – Ta
The solution has a simple expression ... Θ = e-t/τ
Measurement of a Convective Heat Transfer Coefficient
Suppose a sphere of radius R in a stagnant gas of infinite extent. The
heat flux from the sphere through the gas is given by Fourier’s law
q r = – k g dT
dr
The heat flow through a spherical shell is constant so
r 2q r = – r 2k g dT = C
dr
The boundary conditions are T = Ts at r = R and T -> Ta at r -> ∞
The solution becomes
R
T – Ta
= r
TR – Ta
ChE 333- Lecture 6
Spring 2000
2
We can calculate the heat flux at the surface as
qr
r = R
= – k g dT
dr
= kg
r =R
TR – Ta
R
We can define a heat transfer coefficient as
kg
qr
h ≡
=
TR – Ta
R
The corresponding Nusselt Number for heat transfer in a stagnant gas is
Nu = hD = 2 This represents a lower bound for convective heat
transfer, given that there is no gas flow. We expect
kg
the heat transfer coefficient to be larger.
ChE 333- Lecture 6
Spring 2000
3
The experiment
The thermal conductivity of air is 0.014 Btu/ft-°F. If the sphere was 1
cm. in diameter, then h = 0.014(1/2.54(12)) = 4.2 Btu/hr-ft2-°F
Note that h = = 4.2 Btu/hr-ft2-°F = 0.00117 Btu/sec-ft2-°F
If ρ = 436 lb/ft3 and Cp = 0.12 Btu/lb.-°F, then
τ=
0.00117 6
436 0.12 1 / 30
= 0.00403 sec
If the heat transfer coefficient were 5 times larger (Nu = 10) then τ =
0.02 sec,
If the heat transfer coefficient were 10 times larger (Nu = 100) then τ =
0.2 sec,
It should be clear that we cannot make a reasonable verification of a
uniform temperature until we solve for the temperature field in the
sphere.
ChE 333- Lecture 6
Spring 2000
4