EML 4304L Thermal Fluids Lab Thermal Conduction

Experiment # 3

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.

Heat flowfor units 3 and 4

Q conduction=-k A dT / dx

Q coolant=m Cp dT / t

Heat flowfor units 1 and 2

Fundamental

Equations

Rate of heat flow at the heat sinkmw = mass of cooling water displaced in time t (kg)

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

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

t = time required to displace a volume Vw of water (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.

t

TCmq 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 EnergyQin = 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 still be some heat loss.

TCmdx

dTAk p

Thermal Contact Resistance

Rt,c = Thermal contact resistance (ºC/W)

T = Temperature change (ºC)

q = Heat flux (W)

RT

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 = (/4) d2

A(x) = (/4) d(x)2

Q=KA*dT/dx

Unit #3

Q = -k A T/x = -k A(x) T/x

Q k A x( )dT

dx

Q x1

A x( )

d k T1

d

Unit #3

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

T x( ) T 0Q

kx

1

A x( )

d

0

)(1

TT

dxxA

Qk

UNIT #4

x2 x1

QinQout

Q=KA*ΔT/Δx

K = coeff. of therm. conductivity 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 Roughness2 - Type of materials in contact3 - Temperature materials are at4 - Pressure applied to materials5 - 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.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.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

Cu (Al)

Contact Resistance

T1T10

x1x2

Stainless Steel

Steel (Mg)

Thermocouple placement for units 1 and 2

Ideal Thermal Conduction

T1

T2

T3

T4

Material 1 Material 2

T2 = T3

Actual Thermal Conduction

T1T2

T3

T4

Material 1 Material 2

T2 = T3

Temperature profile due to thermal contact resistance

Material 1 Material 2

T2

T3Projected Slope T3

Projected Slope T2

ΔTTemperature 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

Rt,c = ΔT/Q

Qwtr=mwCp(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 #5

Inconsistent readings

Heat flowfor units 3 and 4