Heat Transportation

Storyboard

Heat transport through a system composed of multiple media can be estimated by analyzing how heat is conducted within each medium and transferred at each interface. The calculation is performed using the specific parameters of each medium and interface, as well as the temperatures at both ends of the system, thereby providing the temperatures at each interface.

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Mechanisms

Iframe

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Code
Concept

Mechanisms

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Heat transport

Concept

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The basic system includes a transfer generated by the temperature difference ($\Delta T$), which consists of the temperature difference at internal interface ($\Delta T_i$), the temperature difference in the conductor ($\Delta T_0$), and the temperature difference at external interface ($\Delta T_e$). Therefore:

$ \Delta T = \Delta T_i + \Delta T_0 + \Delta T_e $



With the heat flow rate ($q$) being responsible for the transfer between the interior and the conductor, using the internal transmission coefficient ($\alpha_i$):

$ q = \alpha_i \Delta T_i $



Conduction involves the thermal conductivity ($\lambda$) and the conductor length ($L$):

$ q = \displaystyle\frac{ \lambda }{ L } \Delta T_0 $



And the transfer from the conductor to the exterior, with the external transmission coefficient ($\alpha_e$), is represented by:

$ q = \alpha_e \Delta T_e $



All this is graphically represented by:

ID:(7723, 0)



Heat transport between two systems via a third medium

Concept

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The heat flow rate ($q$) is calculated from the total transport coefficient (multiple medium, two interfaces) ($k$) and the temperature difference ($\Delta T$) using the following equation:

$ q = k \Delta T $



where the total transport coefficient (multiple medium, two interfaces) ($k$) is derived from the external transmission coefficient ($\alpha_e$), the internal transmission coefficient ($\alpha_i$), the thermal conductivity ($\lambda$), and the conductor length ($L$) through this equation:

$\displaystyle\frac{1}{ k }=\displaystyle\frac{1}{ \alpha_i }+\displaystyle\frac{1}{ \alpha_e }+\displaystyle\frac{ L }{ \lambda }$



This is represented in the image below:

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Temperature Profile

Concept

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Typically, the temperature variation within a conductor follows a linear pattern. However, in the case of gaseous and/or liquid media in contact with the conductor, there is a gradual temperature variation from the center of the medium to the surface, as depicted in the following image:



the outer surface temperature ($T_{es}$) depends on the outdoor Temperature ($T_e$), the coefficient of total transportation ($k$), the external transmission coefficient ($\alpha_e$), and the temperature difference ($\Delta T$):

$ T_{es} = T_e + \displaystyle\frac{ k }{ \alpha_e } \Delta T $



the inner surface temperature ($T_{is}$) is a function of the indoor temperature ($T_i$) and the internal transmission coefficient ($\alpha_i$):

$ T_{is} = T_i - \displaystyle\frac{ k }{ \alpha_i } \Delta T $



and the temperature difference ($\Delta T$):

$ \Delta T = T_i - T_e $

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Total heat flow transportation

Concept

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When the material includes multiple conductors connected in series, the total transport coefficient (multiple medium, two interfaces) ($k$) is calculated from the external transmission coefficient ($\alpha_e$), the internal transmission coefficient ($\alpha_i$), the thermal conductivity element i ($\lambda_i$), and the element length i ($L_i$) using the equation:

$\displaystyle\frac{1}{ k }=\displaystyle\frac{1}{ \alpha_i }+\displaystyle\frac{1}{ \alpha_e }+\sum_i\displaystyle\frac{ L_i }{ \lambda_i }$



This process is illustrated in the following diagram:

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Model

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Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$k$
k
Coefficient of total transportation
W/m K
$L$
L
Conductor length
m
$\alpha_e$
alpha_e
External transmission coefficient
W/m^2K
$\alpha_i$
alpha_i
Internal transmission coefficient
W/m^2K
$\lambda$
lambda
Thermal conductivity
W/m K

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$q$
q
Heat flow rate
W/m^2
$dQ$
dQ
Heat transported
J
$T_i$
T_i
Indoor temperature
K
$T_{is}$
T_is
Inner surface temperature
K
$T_e$
T_e
Outdoor temperature
K
$T_{es}$
T_es
Outer surface temperature
K
$S$
S
Section
m^2
$\Delta T$
DT
Temperature difference
K
$\Delta T_e$
DT_e
Temperature difference at external interface
K
$\Delta T_i$
DT_i
Temperature difference at internal interface
K
$\Delta T_0$
DT_0
Temperature difference in the conductor
K
$dt$
dt
Time variation
s

Calculations


First, select the equation: to , then, select the variable: to

Calculations

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

Variable Given Calculate Target : Equation To be used




Equations

#
Equation

$\displaystyle\frac{1}{ k }=\displaystyle\frac{1}{ \alpha_i }+\displaystyle\frac{1}{ \alpha_e }+\displaystyle\frac{ L }{ \lambda }$

1/ k =1/ alpha_i + 1/ alpha_e + L / lambda


$ \Delta T = \Delta T_i + \Delta T_0 + \Delta T_e $

DT = DT_i + DT_0 + DT_e


$ \Delta T = T_i - T_e $

DT = T_i - T_e


$ \Delta T_0 = T_{is} - T_{es} $

DT_0 = T_is - T_es


$ \Delta T_e = T_{es} - T_e $

DT_e = T_es - T_e


$ \Delta T_i = T_i - T_{is} $

DT_i = T_i - T_is


$ q = \alpha_e \Delta T_e $

q = alpha_e * DT_e


$ q = \alpha_i \Delta T_i $

q = alpha_i * DT_i


$ q \equiv \displaystyle\frac{1}{ S }\displaystyle\frac{ dQ }{ dt }$

q = dQ /( S * dt )


$ q = k \Delta T $

q = k * DT


$ q = \displaystyle\frac{ \lambda }{ L } \Delta T_0 $

q = lambda * DT_0 / L


$ T_{es} = T_e + \displaystyle\frac{ k }{ \alpha_e } \Delta T $

T_es = T_e + k * DT / alpha_e


$ T_{is} = T_i - \displaystyle\frac{ k }{ \alpha_i } \Delta T $

T_is = T_i - k * DT / alpha_i

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Temperature difference

Equation

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The temperature difference ($\Delta T$) is calculated by subtracting the outdoor temperature ($T_e$) and the indoor temperature ($T_i$), which is expressed as:

$ \Delta T = T_i - T_e $

$T_i$
Indoor temperature
$K$
5208
$T_e$
Outdoor temperature
$K$
5207
$\Delta T$
Temperature difference
$K$
10161

ID:(15116, 0)



Temperature difference conductor to medium

Equation

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The temperature difference at external interface ($\Delta T_e$) is calculated by subtracting the outer surface temperature ($T_{es}$) from the outdoor temperature ($T_e$):

$ \Delta T_e = T_{es} - T_e $

$T_e$
Outdoor temperature
$K$
5207
$T_{es}$
Outer surface temperature
$K$
5214
$\Delta T_e$
Temperature difference at external interface
$K$
10167

ID:(15118, 0)



Medium to conductor temperature difference

Equation

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The temperature difference at internal interface ($\Delta T_i$) is calculated by subtracting the inner surface temperature ($T_{is}$) from the indoor temperature ($T_i$):

$ \Delta T_i = T_i - T_{is} $

$T_i$
Indoor temperature
$K$
5208
$T_{is}$
Inner surface temperature
$K$
5212
$\Delta T_i$
Temperature difference at internal interface
$K$
10166

ID:(15117, 0)



Surface temperature difference

Equation

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In the case of a solid, and similarly for a liquid, we can describe the system as a structure of atoms held together by something that behaves like a spring. When both ends have temperatures of ($$), with the inner surface temperature ($T_{is}$) and the outer surface temperature ($T_{es}$):

$ \Delta T_0 = T_{is} - T_{es} $

$T_{is}$
Inner surface temperature
$K$
5212
$T_{es}$
Outer surface temperature
$K$
5214
$\Delta T_0$
Temperature difference in the conductor
$K$
10165

ID:(15120, 0)



Total temperature variation

Equation

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In the process of heat transfer, the temperature gradually decreases from the system with the highest temperature (internal) to the one with the lowest temperature (external). In this process, it first decreases from the internal average temperature to the temperature difference at internal interface ($\Delta T_i$), then to the temperature difference in the conductor ($\Delta T_0$), and finally to the temperature difference at external interface ($\Delta T_e$). The sum of these three variations equals the total drop, that is, the temperature difference ($\Delta T$), as shown below:

$ \Delta T = \Delta T_i + \Delta T_0 + \Delta T_e $

$\Delta T$
Temperature difference
$K$
10161
$\Delta T_e$
Temperature difference at external interface
$K$
10167
$\Delta T_i$
Temperature difference at internal interface
$K$
10166
$\Delta T_0$
Temperature difference in the conductor
$K$
10165

ID:(15115, 0)



Calculation of heat conduction

Equation

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The heat flux ($q$) is a function of the thermal conductivity ($\lambda$), the conductor length ($L$) and the temperature difference in the conductor ($\Delta T_0$):

$ q = \displaystyle\frac{ \lambda }{ L } \Delta T_0 $

$L$
Conductor length
$m$
5206
$q$
Heat flow rate
$W/m^2$
10178
$\Delta T_0$
Temperature difference in the conductor
$K$
10165
$\lambda$
Thermal conductivity
$J/m s K$
5204

ID:(7712, 0)



Calculation of the total heat transport by a conductor

Equation

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In this way, we establish a relationship that allows us to calculate the heat flow rate ($q$) as a function of the total transport coefficient (multiple medium, two interfaces) ($k$), and the temperature difference ($\Delta T$):

$ q = k \Delta T $

$k$
Coefficient of total transportation
$W/m^2K$
5174
$q$
Heat flow rate
$W/m^2$
10178
$\Delta T$
Temperature difference
$K$
10161

With the temperature difference at internal interface ($\Delta T_i$), the temperature difference in the conductor ($\Delta T_0$), the temperature difference at external interface ($\Delta T_e$), and the temperature difference ($\Delta T$), we obtain

$ \Delta T = \Delta T_i + \Delta T_0 + \Delta T_e $



which can be rewritten with the heat transported ($dQ$), the time variation ($dt$), the section ($S$)

$ q = \alpha_i \Delta T_i $



$ q = \alpha_e \Delta T_e $



and with the thermal conductivity ($\lambda$) and the conductor length ($L$)

$ q = \displaystyle\frac{ \lambda }{ L } \Delta T_0 $



and

$\displaystyle\frac{1}{ k }=\displaystyle\frac{1}{ \alpha_i }+\displaystyle\frac{1}{ \alpha_e }+\displaystyle\frac{ L }{ \lambda }$



as

$\Delta T = \Delta T_i + \Delta T_0 + \Delta T_e = \displaystyle\frac{1}{S} \frac{dQ}{dt} \left(\displaystyle\frac{1}{\alpha_i} + \displaystyle\frac{1}{\alpha_e} + \displaystyle\frac{L}{\lambda}\right) = \displaystyle\frac{1}{Sk} \displaystyle\frac{dQ}{dt}$



resulting in

$ q = k \Delta T $

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Calculation of heat transmission to the conductor

Equation

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In this way, we establish a relationship that allows us to calculate the heat flow rate ($q$) based on the temperature difference at internal interface ($\Delta T_i$), and the internal transmission coefficient ($\alpha_i$):

$ q = \alpha_i \Delta T_i $

$q$
Heat flow rate
$W/m^2$
10178
$\alpha_i$
Internal transmission coefficient
$W/m^2K$
10163
$\Delta T_i$
Temperature difference at internal interface
$K$
10166

ID:(15113, 0)



Calculation of heat transfer from the conductor

Equation

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In this manner, we establish a relationship that enables us to calculate the heat flow rate ($q$) based on the temperature difference at external interface ($\Delta T_e$), and the external transmission coefficient ($\alpha_e$):

$ q = \alpha_e \Delta T_e $

$\alpha_e$
External transmission coefficient
$W/m^2K$
10162
$q$
Heat flow rate
$W/m^2$
10178
$\Delta T_e$
Temperature difference at external interface
$K$
10167

ID:(15114, 0)



Temperature on the external surface of the conductor

Equation

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The outer surface temperature ($T_{es}$) is not equal to the temperature of the medium, which is the outdoor temperature ($T_e$). This temperature can be calculated from the temperature difference ($\Delta T$), the total transport coefficient (multiple medium, two interfaces) ($k$), and the external transmission coefficient ($\alpha_e$) using the following formula:

$ T_{es} = T_e + \displaystyle\frac{ k }{ \alpha_e } \Delta T $

$k$
Coefficient of total transportation
$W/m^2K$
5174
$\alpha_e$
External transmission coefficient
$W/m^2K$
10162
$T_e$
Outdoor temperature
$K$
5207
$T_{es}$
Outer surface temperature
$K$
5214
$\Delta T$
Temperature difference
$K$
10161

With the heat transported ($dQ$), the time variation ($dt$), the section ($S$), the temperature difference ($\Delta T$), and the total transport coefficient (multiple medium, two interfaces) ($k$), we obtain

$ q = k \Delta T $



which, with the external transmission coefficient ($\alpha_e$) and the temperature difference at external interface ($\Delta T_e$)

$ q = \alpha_e \Delta T_e $



results in

$k\Delta T = \alpha_e \Delta T_e$



and with the outdoor temperature ($T_e$) and the outer surface temperature ($T_{es}$) and

$ \Delta T_e = T_{es} - T_e $



results in

$ T_{es} = T_e + \displaystyle\frac{ k }{ \alpha_e } \Delta T $

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Temperature on the inner surface of the conductor

Equation

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The inner surface temperature ($T_{is}$) is not equal to the temperature of the medium itself, which is the indoor temperature ($T_i$). This temperature can be calculated from the temperature difference ($\Delta T$), the total transport coefficient (multiple medium, two interfaces) ($k$), and the internal transmission coefficient ($\alpha_i$) using the following formula:

$ T_{is} = T_i - \displaystyle\frac{ k }{ \alpha_i } \Delta T $

$k$
Coefficient of total transportation
$W/m^2K$
5174
$T_i$
Indoor temperature
$K$
5208
$T_{is}$
Inner surface temperature
$K$
5212
$\alpha_i$
Internal transmission coefficient
$W/m^2K$
10163
$\Delta T$
Temperature difference
$K$
10161

With the heat transported ($dQ$), the time variation ($dt$), the section ($S$), the temperature difference ($\Delta T$), and the total transport coefficient (multiple medium, two interfaces) ($k$), we have

$ q = k \Delta T $



which, with the internal transmission coefficient ($\alpha_i$) and the temperature difference at internal interface ($\Delta T_i$)

$ q = \alpha_i \Delta T_i $



results in

$k\Delta T = \alpha_i \Delta T_i$



and with the indoor temperature ($T_i$) and the inner surface temperature ($T_{is}$) and

$ \Delta T_i = T_i - T_{is} $



results in

$ T_{is} = T_i - \displaystyle\frac{ k }{ \alpha_i } \Delta T $

ID:(15121, 0)



Total transportation coefficient (one medium, two interfaces)

Equation

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The value of the coefficient of total transportation ($k$) in the transport equation is determined using the external transmission coefficient ($\alpha_e$), the internal transmission coefficient ($\alpha_i$), the thermal conductivity ($\lambda$), and the conductor length ($L$) as follows:

$\displaystyle\frac{1}{ k }=\displaystyle\frac{1}{ \alpha_i }+\displaystyle\frac{1}{ \alpha_e }+\displaystyle\frac{ L }{ \lambda }$

$k$
Coefficient of total transportation
$W/m^2K$
5174
$L$
Conductor length
$m$
5206
$\alpha_e$
External transmission coefficient
$W/m^2K$
10162
$\alpha_i$
Internal transmission coefficient
$W/m^2K$
10163
$\lambda$
Thermal conductivity
$J/m s K$
5204

With the temperature difference at internal interface ($\Delta T_i$), the temperature difference in the conductor ($\Delta T_0$), the temperature difference at external interface ($\Delta T_e$), and the temperature difference ($\Delta T$), we obtain

$ \Delta T = \Delta T_i + \Delta T_0 + \Delta T_e $



which can be rewritten with the heat transported ($dQ$), the time variation ($dt$), the section ($S$)

$ q = \alpha_i \Delta T_i $



$ q = \alpha_e \Delta T_e $



and with the thermal conductivity ($\lambda$) and the conductor length ($L$)

$ q = \displaystyle\frac{ \lambda }{ L } \Delta T_0 $



as

$\Delta T_i + \Delta T_0 + \Delta T_e = \displaystyle\frac{1}{S} \displaystyle\frac{dQ}{dt} \left(\displaystyle\frac{1}{\alpha_i} + \displaystyle\frac{1}{\alpha_e} + \displaystyle\frac{L}{\lambda}\right)$



so we can define a combined coefficient as

$\displaystyle\frac{1}{ k }=\displaystyle\frac{1}{ \alpha_i }+\displaystyle\frac{1}{ \alpha_e }+\displaystyle\frac{ L }{ \lambda }$

ID:(3486, 0)



Heat flux density

Equation

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The heat flow rate ($q$) is defined in terms of the heat transported ($dQ$), the time variation ($dt$), and the section ($S$) as follows:

$ q \equiv \displaystyle\frac{1}{ S }\displaystyle\frac{ dQ }{ dt }$

$q$
Heat flow rate
$W/m^2$
10178
$dQ$
Heat transported
$J$
10159
$S$
Section
$m^2$
5205
$dt$
Time variation
$s$
10160

ID:(15133, 0)