Thermal Conduction4

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EML 4304L Thermal Fluids Lab Thermal Conduction

Experiment # 3

Mechanical Engineering DepartmentFAMU/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 conductionexperiments which examines the effects on heattransfer with varying cross-sectional area anddistance.

– Using this thermal conduction information derive Fourier’s law of thermal conduction.

• Analyze the temperature variance in a series ofmetal rods that are in physical contact.

– From this information determine thermal resistanceand contact resistance.



Heat flow

for units

3 and 4

Q conduction=-k A dT / dx

Q coolant=m Cp dT / t

H fl



Heat flow

for units

1 and 2



Fundamental

Equations



Rate of heat flow at the heat sink

mw = mass of cooling water displaced in time t (kg)

C p = Specific heat of water at constant pressure (kJ/kg °C)

T = (Tout - T in ) of cooling water (°C)

t = time required to displace a volume V w of water (s)

This equation is used to determine the amount of energy that isbeing absorbed by the coolant. Once this is determined for each

uni t, it is assumed to be the constant rate of conduction through

each mater ial.

t

T C m

q

pw



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.



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)

Qcond = -KA dT/dx

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

x

T

dT/dx



Conservation of EnergyQ in = Q out

Q conduction = Q coolant

Calculations

This equation assumes there is no heat loss

through the system boundary. Though each

unit is insulated, there will sti l l be some heat

loss.

T C mdx

dT Ak p



Thermal Contact Resistance

R t,c

= Thermal contact resistance (ºC/W)

T = Temperature change (ºC)

q = Heat flux (W)

R T

qt c,

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 #3Q=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 k A x( )dT

dx

Q x1

A x( )

d k 1

d



Unit #3

Once k is solved for, the temperaturecan be found for any distance, x.

T x( ) T 0Q

k

1

A x( )

d

0

)(

1

T T

dx x AQk



UNIT #4

x2 x1

QinQout

Q=KA*ΔT/Δx

K = coeff. of therm. conductivity NOTE: K is

unknown andmust be

determined



Thermal ContactResistance

Determination Units#1 and #2



Thermal contact resistance(R c) is a discontinuity in

the temperature gradients between two materials in

contact. The value is determined by projecting thetemperature 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 at4 - 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.


in Fig. 2.


in Fig. 2.



Units #1 and #2

Cu

(Al)

Contact Resistance

T1T10

x1x2

Stainless

Steel

Steel

(Mg)





Ideal Thermal Conduction

T1

T2

T3

T4

Material 1 Material 2

T2 = T3



Actual Thermal Conduction

T1T2

T3

T4


T2 = T3

Temperature

profile due to

thermalcontact

resistance




T2

T3

Projected

Slope T3

Projected

Slope T2

ΔT

Temperature profile due

to thermal contact

resistance



Temperature vs Distance

Temperature(ºF)

Distance(inches)

Material 1

Material 2

Material 3

Discontinuities

where ΔT must be determined



Thermal Contact Resistance

Calculation

R t,c = ΔT/Q

Qwtr =mwC p(ΔT)

Q = Qwtr

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 #5Inconsistent readings



Heat flow



Heat flowfor units

3 and 4

Documents

Thermal Conduction4