Unit 42: Heat Transfer and
Combustion
Lesson 6: Conduction-Convection
Systems
Aim
NDGTA
• LO1: Understanding Heat Transfer Rates for
Composite Systems.
Conduction-Convection
Systems
NDGTA
• Heat that is conducted through a body must
frequently be removed (or delivered) by some
convection process.
• For example, the heat lost by conduction through
a furnace wall must be dissipated to the
surroundings through convection.
• In heat exchanger applications a finned-tube
arrangement might be used to remove heat from
a hot liquid – the heat is conducted through the
material and finally dissipated to the
surroundings by convection.
Conduction-Convection
Systems
NDGTA
• Consider a one-dimensional fin exposed to a
surrounding fluid at a temperature T∞.
dqconv= hPdx(T-T∞)
t
A
qx
d
x
Base
L
X
qx + dx
Z
Conduction-Convection
Systems
NDGTA
• The temperature at the base of the fin is T0.
Energy on left face = energy out right face + energy lost in convection
As previous…
q = hA(Tw – T∞)
Where the Area A in this equation is the surface area
for convection.
Let the cross sectional area be A and the perimeter be
P.
Conduction-Convection
Systems
Energy on left face = qx = - kAdT/dx
Energy out right face = qx+dx = - kAdT/dx|x+dx
= - kA dT + d2T dx
dx dx2
Energy lost in convection = hPdx(T – T∞)
NDGTA
Conduction-Convection
Systems
NDGTA
Energy on left face = energy out right face + energy lost in convection
- kAdT = - kA dT + d2T dx + hPdx(T – T∞)
dx
dx dx2
Thus…
d2T dx - hP(T – T∞) = 0
dx2
kA
now let (T – T∞) = θ
Conduction-Convection
Systems
NDGTA
Thus…
d2θ dx - hPθ = 0
dx2
kA
One boundary condition occurs when θ = θ0 = (T0 – T∞) at x = 0
The other boundary conditions depend on the physical situation…
CASE 1: the fin is very long and the temperature at the end of the fin is
essentially that of the surrounding fluid
CASE 2: the fin is of finite length and loses heat by convection from its
end
CASE 3: the end of the fin is insulated so that dT/dx = 0 at x = L
Conduction-Convection
Systems
NDGTA
If we let m2 = hP/kA then we can derive a general solution
for the equation…
θ = C1e-mx + C2emx
For case 1 the boundary conditions are θ = θ0 at x = 0 and
θ = 0 at x = ∞
Thus… θ = T – T∞ = e-mx
i.e.
θ = θ0e-mx
θ0 T0 - T∞
Conduction-Convection
Systems
NDGTA
For case 3 the boundary conditions are θ = θ0 at x = 0 and
dθ/dx = 0 at x = L
Thus… θ0 = C1 + C2
i.e.
0 = m(- C1e-mL + C2emL)
Solving for constant C1 and C2 gives…
θ = e-mx +
e-mx = cosh[m(L-x)]/cosh(mL)
θ0 1 + e-2mL
1 + e2mL
Conduction-Convection
Systems
NDGTA
For case 2 the result is…
T – T ∞ = cosh[m(L-x)] + (h/mK)sinh[m(L-x)]
T0 – T ∞
cosh(mL) + (h/mK)sinh(mL)
Note all of the heat lost by the fin must be conducted into
the base at x = 0. Using the equations for temperature
distribution we can compute the heat loss…
Conduction-Convection
Systems
NDGTA
Thus…
q = - kA dT
dx
x=0
An alternative method of integrating convection heat loss
could be used…
q=
0
L
L
hP(T - T∞)dx =
hPθdx
0
Conduction-Convection
Systems
NDGTA
Thus for CASE 1…
q = -kA(-mθe-mθ) = √(hPkAθ0)
And for CASE 3…
q = -kAθ0m 1
1
= √(hPkA)θ0 tanh(mL)
1 + e-2mL 1+ e2mL
For CASE 2…
q = √(hPkA)(T0 - T∞) sinh(mL) + (h/mk)cosh(mL)
cosh(mL) + (h/mk)sinh(mL)
Conduction-Convection
Systems
NDGTA
• For this is has been assumed that…
1. that there is a substantial temperature gradient
only in the x-direction which is only true if the fin
is sufficiently thin
2. The convection coefficient h is seldom uniform
over the entire surface of the fin. To deal with
this in practical terms finite difference
techniques must be employed.