Thermal transfer Characteristics of TCM
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THERMAL TRANSFER CHARACTERISTICS ON HEATER Jorge Hernandez PhD. Stundent DIMES -UNICAL
Transcript of Thermal transfer Characteristics of TCM
THERMAL TRANSFER CHARACTERISTICS ON HEATER
Jorge Hernandez PhD. Stundent DIMES -UNICAL
Heat Transfer
β’ Temperature conduction is performed from a hot body to a cold body.
β’ Thermal Equilibrium happens in two bodies have the same temperature.
Thermal Transfer Variables
Difference Temperature
β’ Contact surface of two bodies experiences a difference in temperature.
β’ While thermal equilibrium come to happen, the difference go to zero.
Material thermal characteristic
β’ In previous example, heat was transferred from water through the metal to water.
β’ The rate of heat transfer depends on material through heat is transferred.
β’ Heat transfer coefficient (k) express thermal conduction characteristic of a material.
β’ Large value of K means thermal conductor and a lower value means thermal isulators.
Material thermal characteristic
β’ k is determined experimentally and its unit is π
π. Β°πΆ
Source: http://www.roymech.co.uk/Related/Thermos/Thermos_HeatTransfer.html
Material k Material k
Aluminum 237 Porcelain 1.05
Brass 110 Wood 0.13
Copper 398 Water 0.58
Gold 315 Air 0.024
Material body characteristics
β’ Area (A) β Wider areas conduct more heat and this means
greater heat dissipation. Not good for our project.
β’ Thickness (d) β Thin walls conduct heat in faster way. This affect
the rate thermal transfer.
Mathematical Equation
Mathematical Equation β’ Variables:
β Temperature difference: (T1 β T2) β Thermal Conductivity Coefficient of the Material: k β Area: A β Thickness: d
(1) πππππ = ππ
= πππ‘π‘ (ππ β ππ) [π½π½π½π½ππ½π½
= π]
(2) πππ‘π‘ = π .π΄π
[πΒ°πΆ
] Average Thermal Conductance (3) π ππ‘π‘ = 1
ππ‘π‘π‘ [Β°πΆ
π] Average Thermal Resistance
Information extracted from reference [1], assuming that areas are flat and regular.
Application Problem Case
Heater V1
Practical Considerations β Contact area of SafeFET
(heat source size) is smaller than the opposite side area and even the DUT contact area.
β Location of heat source is important to note because of thermal distribution on sink.
β Also, thermal resistance of electrical isulator must be considered.
Mathematical Resolution
β’ Calculation of Total Thermal Resistance based on reference [1] and [2]. 1. Calculate π ππ‘π‘. 2. Calculate π π π π 3. Calculate π π π π .
Calculation π ππ‘π‘ according to [1] π΄ = 0.0π8 π₯ 0.0ππ [π2]
π΄ = 3.96π₯π0β4 [π2]
π = 0.003 [π]
π = ππ0 [π
π. Β°πΆ]
From (π) and (3):
π ππ‘π‘ = π
π. π΄=
0.003(ππ0)(3.96π₯π0β4)
Β°πΆπ
πΉπππ = ππ. πππππβπ[Β°πͺπΎ
]
Equations for π ππ‘π‘ β’ In [2], Spreading Thermal Resistance (π ππ‘π‘) is stablished
in function of π ππ‘π‘ and the difference between size of the heating areas in each side of the sink.
β’ π ππ‘π‘ is an additional quantity that is needed for determining the maximum heat sink temperature.
β’ Also, [2] considers the location of heating source on the sink surface. πͺπ [ π
π]
β’ Ideally, this should be in the center of the same. πͺπ = π
Equations for π ππ‘π‘ and π ππ‘π
(4) π ππ‘π‘ = πΆπ π₯ π΄π. π΄π½
π . π. π΄π. π΄π½π₯
Ξ». π. π΄π. π ππ‘π‘ + tanh (Ξ». π)π + Ξ». π. π΄π. π ππ‘π‘. tanh (Ξ». π)
[Β°πΆπ
]
Where: Ξ»=
π32οΏ½
π΄π+
ππ΄π½
β’ π΄π: Footprint area of the heat sink base-plate β’ π΄π½: Contact area of the heat source
(5) π ππ‘π= π ππ‘π‘ + π ππ‘π‘
Calculation of π ππ‘π‘ according to [2]
π΄π = 0.0π8 π₯ 0.0ππ = 3.96π₯π0β4[π2] π΄π½ = 0.0π04 π₯ 0.0π0π = π05.04π₯π0β6[π2]
Ξ» = π3
2οΏ½
π΄π+
ππ΄π½
= 377.39 [ππ
]
Calculation of π ππ‘π‘ according to [2] β’ From (4) πππ assuming that heat source is placed in the center of
the base-plate:
π ππ‘π‘ = πΆπ π₯ π΄π. π΄π½
π . π. π΄π. π΄π½π₯
Ξ». π. π΄π. π ππ‘π‘ + tanh (Ξ». π)π + Ξ». π. π΄π. π ππ‘π‘. tanh (Ξ». π)
[Β°πΆπ
]
tanh 377.39 π₯ 0.003 = 0.8ππ76 Ξ». π. π΄π. π ππ‘π‘ = 377.39 π₯ ππ0 π₯ 3.96π₯π0β4 π₯ 68.87π₯π0β3 = π.π3ππ
π ππ‘π‘ = π π₯ 3.96π₯π0β4. π05.04π₯π0β6
ππ0 . π(3.96π₯π0β4)(π05.04π₯π0β6)π₯
π.944π.9π9
[Β°πΆπ
]
πΉπππ = π. ππππβπ [Β°πͺπΎ
]
Calculation of π ππ‘π according to [2] β’ From (5):
π ππ‘π = π ππ‘π‘ + π ππ‘π‘
π ππ‘π = 68.87π₯π0β3 + 5.ππ₯π0β3 [Β°πΆπ
]
πΉπππ = ππ. πππππβπ [Β°πͺπΎ
]
Note.- Line graph shows the Cf variations in function of the distance from the center of the heat sink to the heat source placed along the center line at y=0 and -37.5 < x < 37.5 [mm]. Cf is case dependent.
THERMAL TRANSFER SAFEFET - DUT
How does the heat flow?
β’ Considerations: β Contact temperature
between case SafeFET and right side of sink is ππΆπΆ.
β Contact temperature between case DUT and left side of sink is ππΆπΆ.
β Thermal resistence of the electrical isulator must be considered.
Case study extracted from Project β’ Both SafeFET and DUT are heat
sources, but in practice SafeFET deliveres more heat than DUT.
β’ All considerations are referenced to regime work. β SafeFET (DUT 2)
β’ πππ½ = π56.8π Β°πΆ β’ ππΆπΆ = 6.04 π
β DUT 2 β’ ππΆπΆ = 0.π4 ππ β’ ππΆπΆ = π50 Β°πΆ (measured with LM35
and It will be demonstrated mathematically)
Analog focus for resolution
Case Temperature on SafeFET Adapting (π) for this situation:
πππ½ β ππΆπΆ = π ππ‘ππ‘π½ β ππΆ
Where, from datasheet SafeFET : π ππ‘ππ‘π½ = 0.6[Β°πΆ
π] : Thermal resistance junction-case of SafeFET
πππ½ = π56.8π Β°πΆ It is important to note that because both devices are heat sources due to their power consumption:
ππΆ = ππΆπΆ + ππΆπΆ ππΆ = 6.040π4 [π]
Therefore: π56.8π Β°πΆ β ππΆπΆ = 0.6
Β°πΆπ
β 6.040π4 [π]
π»πͺπͺ = πππ. π Β°πͺ
Transfer Temperature between heat sink sides
Adapting (π) for modified value of thermal resistance of the heat sink: ππΆ = πππ‘π (ππΆπΆ β ππΆπ) [π]
Where: πππ‘π = 1π π‘π‘π‘
= π3.5 πΒ°πΆ
ππΆπΆ = π53.π Β°πΆ ππΆπ β‘ Temperature in side contact between heat sink and insulator ππΆ = ππΆπΆ + ππΆπΆ = 6.04π + 0.π4ππ = 6.040π4π
ππΆπ = ππΆπΆ β πππππ
πππ‘π
π»πͺπΊ = πππ. ππ Β°πͺ
Case Temperature on DUT transferred through insulator
Adapting (π) for this situation: ππ½ = πππ‘π (ππΆπ β ππΆπΆ) [π]
Where: ππΆπ = π5π.75 Β°πΆ ππΆπΆ β‘ Case temperature of DUT π·πͺ = π·π«πͺ + π·π«π« = π. ππππππΎ Whit:
πππ‘π = ππ π₯ π΄π
π π= π.6π
πΒ°πΆ
From the vendor site: ππ = 3.5 [ π
π.Β°πΆ]
π΄π = 0.0π5 π₯ 0.0π9 = π85π₯π0β6 [π2] π π = 38ππ₯π0β6 [π] Therefore:
π»πͺπ« = π»πͺπΊ βπ·πͺ
πππ‘π= πππ. ππ Β°πͺ
Junction Temperature on DUT Adapting (π) for this situation, and considering that heat source is the SafeFET:
ππΆπΆ β πππΆ = π ππ‘ππ‘π½ β ππΆ Where: ππΆπΆ = π50.44 Β°πΆ πππΆ β‘ Junction temperature of DUT π·πͺ = π·π«πͺ + π·π«π« = π. ππππππΎ π ππ‘ππ‘πΆ = 0.5 Β°πΆ
Β°π
Therefore:
π»ππ« = π»πͺπ« β π·π. πΉπππππ« = πππ. ππ Β°πͺ
Conclusions β’ Difference of junction temperatures for this case study analyzed.
βπππΆπΆ = πππΆ β πππΆ = π56.8π Β°πΆ β π47.4π Β°πΆ βπ»ππͺπ« = π. π Β°πͺ
β’ Error between temperature setpoint and πππΆ
%πΈπΈπΈπΈπΈ =π50Β°πΆ β π47.4πΒ°πΆ
π50Β°πΆβ π00 = π. ππ %
β’ Thermal Resistance of Heat Sink
πΉπππ = ππ. πππππβπ[Β°πͺπΎ
]
β’ Total Thermal Resistance of Heat Sink according location of heat source criteria.
πΉπππ = ππ. πππππβπ [Β°πͺπΎ
]
β’ Temperature aquisition of the system was adjusted with an LM35, which
performs a sensing of its case temperature, in this way this mathematic process shows this right adjustment with the case temperature value on DUT:
π»πͺπ«π΄ = πππ. ππ Β°πͺ (Mathematical value) π»πͺπ«π¬ = πππ. π Β°πͺ (Experimental value)
References
β’ The Physics Classroom http://www.physicsclassroom.com/class/thermalP/u18l1f.cfm
β’ Calculating spreading resistance in heat sinks http://www.electronics-cooling.com/1998/01/calculating-spreading-resistance-in-heat-sinks/
ΒΏQUESTIONS?
