EML 4304L Thermal Fluids Lab
Thermal Conduction
Experiment # 3
Mechanical Engineering Department
FAMU/FSU College of Engineering
Outline
•Purpose of the lab
•Fundamental Equations
•Unit 3 and Unit 4 Analysis
•Unit 1 and Unit 2 Analysis
•Error Analysis
PURPOSE
• Conduct a series of thermal conduction
experiments which examines the effects on heat
transfer with varying cross-sectional area and
distance.
– Using this thermal conduction information derive
Fourier’s law of thermal conduction.
• Analyze the temperature variance in a series of
metal rods that are in physical contact.
– From this information determine thermal resistance
and contact resistance.
Q coolant=m Cp dT / t
Q conduction=-k A dT / dx
Heat flow
for units
3 and 4
Heat flow
for units
1 and 2
Fundamental
Equations
q
mwC p T
t
Rate of heat flow at the heat sink
mw = mass of cooling water displaced in time t
Cp = Specific heat of water at constant pressure
T = (Tout - Tin) of cooling water
t = time required to displace a volume Vw of water
(kg)
(kJ/kg °C)
(°C)
(s)
This equation is used to determine the amount of energy that is
being absorbed by the coolant. Once this is determined for each
unit, it is assumed to be the constant rate of conduction through
each material.
Qcond = KA(ΔT/Δx)
Rate of heat conduction
K = thermal conductivity constant
(W/m °C)
A = cross-sectional area
(m2)
T = temperature difference across the material
(°C)
x = distance between temperature readings
(m)
Used to determine rate of heat conduction through
a body based on material properties, area,
temperature difference, and length of material.
Qcond = -KA dT/dx
Fourier’s law of heat conduction
-K = thermal conductivity constant
(W/m °C)
A = cross-sectional area
(m2)
dT = differential element for temperature
(°C)
dx = differential element for distance
(m)
Used to determine rate of heat conduction through a body
based on material properties, area, and temperature/distance
gradient.
• Q=KA ΔT/Δx
Q=-KA dT/dx
• dT/dx = temperature gradient
T
dT/dx
x
Calculations
Conservation of Energy
Qin = Q out
Q conduction = Q coolant
dT
 k  A
 m  C p  T
dx
This equation assumes there is no heat loss
through the system boundary. Though each
unit is insulated, there will still be some heat
loss.
T
Rt ,c 
q
Thermal Contact Resistance
Rt,c = Thermal contact resistance
(ºC/W)
T = Temperature change
(ºC)
q = Heat flux
(W)
Calculates thermal contact resistance for a
given temperature discontinuity and a known
power input.
Heat Conduction
for Units 3 and 4
Thermocouple Placement
for Units 3 and 4
UNIT #3
Q=KA*ΔT/Δx
Diameter is a function of x:
D(x)=D0+mx
D(x)=1”+(x/(11+1/16))
Unit #3
Area can also be written as
a function of x:
A = (p/4) d2
A(x) = (p/4) d(x)2
Q=KA*dT/dx
Unit #3
Q = -k A T/x = -k A(x) dT/dx
Q

Q 


k A ( x) 
1
A ( x)
dx
dT
dx

k  1 d T

Unit #3
1
Q
dx
A( x)
k
T  T0
Once k is solved for, the temperature
can be found for any distance, x.
T( x)

Q
T0
k 

1
A ( x)
dx
UNIT #4
Q=KA*ΔT/Δx
K = coeff. of therm. conductivity
Qin
Qout
x2
x1
NOTE: K is
unknown and
must be
determined
Thermal Contact
Resistance
Determination Units
#1 and #2
Thermal contact resistance(Rc) is a discontinuity in
the temperature gradients between two materials in
contact. The value is determined by projecting the
temperature gradients, calculating the temperature
difference, and dividing the temperature difference
by the power that is transmitted through the
materials.
Factors affecting thermal contact resistance:
1 - Surface Roughness
2 - Type of materials in contact
3 - Temperature materials are at
4 - Pressure applied to materials
5 - Type of fluid trapped at interface
elements are enclosed in the insulating jacket. Figure1 illustrates the schematics of the apparatus. The dimensions of the tapered rod are indicated
in Fig. 2.
Units #1 and #2
Steel
(Mg)
Cu
(Al)
Stainless
Steel
T10
T1
x2
Contact Resistance
x1
Thermocouple placement for
units 1 and 2
Ideal Thermal Conduction
Material 1
T1
Material 2
T3
T4
T2
T2 = T3
Actual Thermal Conduction
Material 1
T1
Material 2
T2
Temperature
profile due to
thermal
contact
resistance
T2 = T3
T4
T3
Material 1
Projected
Slope T2
Material 2
T2
ΔT
Temperature profile due
to thermal contact
resistance
T3
Projected
Slope T3
Temperature vs Distance
Material 1
Material 2
Temperature(ºF)
Material 3
Discontinuities
where ΔT must
be determined
Distance(inches)
Thermal Contact Resistance
Calculation
Rt,c = ΔT/Q
Q = Qwtr
Qwtr=mwCp(ΔT)
ΔT (determined by projection of slope
and measuring difference in
temperatures)
Errors
• Time
Flow rate
Steady State
• Heat Losses
Not perfectly insulated
• Unit #4 Thermocouples #3 and #5
Inconsistent readings
Heat flow
for units
3 and 4