Heat Transfer by Conduction at Base Calculator

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Basics of HMT

Conduction

Convection

Convective Mass Transfer

✖Thermal Conductivity of Fin is a material property that represents at which rate heat energy is transferred through the fin by conduction.ⓘ Thermal Conductivity [k_fin]			+10% -10%
✖Cross Sectional Area of Fin is the surface area of the Fin which is perpendicular to the direction of Heat Flow.ⓘ Cross Sectional Area of Fin [A_cs]			+10% -10%
✖Perimeter of the Fin refers to the total length of the outer boundary of the fin that is exposed to the surrounding medium.ⓘ Perimeter of the Fin [P]			+10% -10%
✖Convective Heat Transfer Coefficient is a parameter that quantifies the rate of heat transfer between a solid surface and the surrounding fluid due to convection.ⓘ Convective Heat Transfer Coefficient [h]			+10% -10%
✖Base Temperature is the temperature at the base of the fin.ⓘ Base Temperature [t_o]			+10% -10%
✖Ambient Temperature is the temperature of the ambient or surrounding fluid which is in contact with the fin.ⓘ Ambient Temperature [t_a]			+10% -10%

✖Rate of Conductive Heat Transfer is the amount of heat flow through a body per unit time, usually measured in watts (joules per second).ⓘ Heat Transfer by Conduction at Base [Q_fin]

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Formula

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Heat Transfer by Conduction at Base

Formula

`"Q"_{"fin"} = ("k"_{"fin"}*"A"_{"cs"}*"P"*"h")^0.5*("t"_{"o"}-"t"_{"a"})`

Example

`"43.20445W"=("205W/(m*K)"*"0.00009m²"*"0.046m"*"30.17W/m²*K")^0.5*("573K"-"303K")`

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Heat Transfer by Conduction at Base Solution

STEP 0: Pre-Calculation Summary

Formula Used

Rate of Conductive Heat Transfer = (Thermal Conductivity*Cross Sectional Area of Fin*Perimeter of the Fin*Convective Heat Transfer Coefficient)^0.5*(Base Temperature-Ambient Temperature)
Q_fin = (k_fin*A_cs*P*h)^0.5*(t_o-t_a)
This formula uses 7 Variables

Variables Used

Rate of Conductive Heat Transfer - (Measured in Watt) - Rate of Conductive Heat Transfer is the amount of heat flow through a body per unit time, usually measured in watts (joules per second).
Thermal Conductivity - (Measured in Watt per Meter per K) - Thermal Conductivity of Fin is a material property that represents at which rate heat energy is transferred through the fin by conduction.
Cross Sectional Area of Fin - (Measured in Square Meter) - Cross Sectional Area of Fin is the surface area of the Fin which is perpendicular to the direction of Heat Flow.
Perimeter of the Fin - (Measured in Meter) - Perimeter of the Fin refers to the total length of the outer boundary of the fin that is exposed to the surrounding medium.
Convective Heat Transfer Coefficient - (Measured in Watt per Square Meter per Kelvin) - Convective Heat Transfer Coefficient is a parameter that quantifies the rate of heat transfer between a solid surface and the surrounding fluid due to convection.
Base Temperature - (Measured in Kelvin) - Base Temperature is the temperature at the base of the fin.
Ambient Temperature - (Measured in Kelvin) - Ambient Temperature is the temperature of the ambient or surrounding fluid which is in contact with the fin.

STEP 1: Convert Input(s) to Base Unit

Thermal Conductivity: 205 Watt per Meter per K --> 205 Watt per Meter per K No Conversion Required
Cross Sectional Area of Fin: 9E-05 Square Meter --> 9E-05 Square Meter No Conversion Required
Perimeter of the Fin: 0.046 Meter --> 0.046 Meter No Conversion Required
Convective Heat Transfer Coefficient: 30.17 Watt per Square Meter per Kelvin --> 30.17 Watt per Square Meter per Kelvin No Conversion Required
Base Temperature: 573 Kelvin --> 573 Kelvin No Conversion Required
Ambient Temperature: 303 Kelvin --> 303 Kelvin No Conversion Required

STEP 2: Evaluate Formula

Substituting Input Values in Formula

Q_fin = (k_fin*A_cs*P*h)^0.5*(t_o-t_a) --> (205*9E-05*0.046*30.17)^0.5*(573-303)

Evaluating ... ...

Q_fin = 43.2044539266497

STEP 3: Convert Result to Output's Unit

43.2044539266497 Watt --> No Conversion Required

FINAL ANSWER

43.2044539266497 ≈ 43.20445 Watt <-- Rate of Conductive Heat Transfer

(Calculation completed in 00.004 seconds)

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Created by Rushi Shah

K J Somaiya College of Engineering (K J Somaiya), Mumbai

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Indian Institute of Information Technology (IIIT), Guwahati

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< 13 Heat and Mass Transfer Calculators

Heat Transfer by Conduction at Base

Go Rate of Conductive Heat Transfer = (Thermal Conductivity*Cross Sectional Area of Fin*Perimeter of the Fin*Convective Heat Transfer Coefficient)^0.5*(Base Temperature-Ambient Temperature)

Heat Exchange by Radiation due to Geometric Arrangement

Go Heat Transfer = Emissivity*Area*[Stefan-BoltZ]*Shape Factor*(Temperature of Surface 1^(4)-Temperature of Surface 2^(4))

Black Bodies Heat Exchange by Radiation

Go Heat Transfer = Emissivity*[Stefan-BoltZ]*Area*(Temperature of Surface 1^(4)-Temperature of Surface 2^(4))

Heat Transfer According to Fourier's Law

Go Heat Flow Through a Body = -(Thermal Conductivity of Material*Surface Area of Heat Flow*Temperature Difference/Thickness)

One Dimensional Heat Flux

Go Heat Flux = -Thermal Conductivity of Fin/Wall Thickness*(Temperature of Wall 2-Temperature of Wall 1)

Newton's Law of Cooling

Go Heat Flux = Heat Transfer Coefficient*(Surface Temperature-Temperature of Characteristic Fluid)

Non Ideal Body Surface Emittance

Go Real Surface Radiant Surface Emittance = Emissivity*[Stefan-BoltZ]*Surface Temperature^(4)

Convective Processes Heat Transfer Coefficient

Go Heat Flux = Heat Transfer Coefficient*(Surface Temperature-Recovery temperature)

Thermal Conductivity given Critical Thickness of Insulation for Cylinder

Go Thermal Conductivity of Fin = Critical Thickness of Insulation*Heat Transfer Coefficient at Outer Surface

Diameter of Rod Circular Fin given Area of Cross-Section

Go Diameter of Circular Rod = sqrt((Cross-sectional area*4)/pi)

Critical Thickness of Insulation for Cylinder

Go Critical Thickness of Insulation = Thermal Conductivity of Fin/Heat Transfer Coefficient

Thermal Resistance in Convection Heat Transfer

Go Thermal Resistance = 1/(Exposed Surface Area*Co-efficient of Convective Heat Transfer)

Heat Transfer

Go Heat Flow Rate = Thermal Potential Difference/Thermal Resistance

< 13 Conduction, Convection and Radiation Calculators

Heat Transfer by Conduction at Base

Go Rate of Conductive Heat Transfer = (Thermal Conductivity*Cross Sectional Area of Fin*Perimeter of the Fin*Convective Heat Transfer Coefficient)^0.5*(Base Temperature-Ambient Temperature)

Heat Exchange by Radiation due to Geometric Arrangement

Go Heat Transfer = Emissivity*Area*[Stefan-BoltZ]*Shape Factor*(Temperature of Surface 1^(4)-Temperature of Surface 2^(4))

Black Bodies Heat Exchange by Radiation

Go Heat Transfer = Emissivity*[Stefan-BoltZ]*Area*(Temperature of Surface 1^(4)-Temperature of Surface 2^(4))

Heat Transfer According to Fourier's Law

Go Heat Flow Through a Body = -(Thermal Conductivity of Material*Surface Area of Heat Flow*Temperature Difference/Thickness)

One Dimensional Heat Flux

Go Heat Flux = -Thermal Conductivity of Fin/Wall Thickness*(Temperature of Wall 2-Temperature of Wall 1)

Newton's Law of Cooling

Go Heat Flux = Heat Transfer Coefficient*(Surface Temperature-Temperature of Characteristic Fluid)

Non Ideal Body Surface Emittance

Go Real Surface Radiant Surface Emittance = Emissivity*[Stefan-BoltZ]*Surface Temperature^(4)

Thermal Resistance in Conduction

Go Thermal Resistance = (Thickness)/(Thermal Conductivity of Fin*Cross Sectional Area)

Convective Processes Heat Transfer Coefficient

Go Heat Flux = Heat Transfer Coefficient*(Surface Temperature-Recovery temperature)

Thermal Conductivity given Critical Thickness of Insulation for Cylinder

Go Thermal Conductivity of Fin = Critical Thickness of Insulation*Heat Transfer Coefficient at Outer Surface

Critical Thickness of Insulation for Cylinder

Go Critical Thickness of Insulation = Thermal Conductivity of Fin/Heat Transfer Coefficient

Thermal Resistance in Convection Heat Transfer

Go Thermal Resistance = 1/(Exposed Surface Area*Co-efficient of Convective Heat Transfer)

Heat Transfer

Go Heat Flow Rate = Thermal Potential Difference/Thermal Resistance

Heat Transfer by Conduction at Base Formula

Heat Transfer in fins

Fins are the extended surface protruding from a surface or body and they are meant for increasing the heat transfer rate between the surface and the surrounding fluid by
increasing heat transfer area.

How to Calculate Heat Transfer by Conduction at Base?

Heat Transfer by Conduction at Base calculator uses Rate of Conductive Heat Transfer = (Thermal Conductivity*Cross Sectional Area of Fin*Perimeter of the Fin*Convective Heat Transfer Coefficient)^0.5*(Base Temperature-Ambient Temperature) to calculate the Rate of Conductive Heat Transfer, Heat Transfer by conduction at Base is defined as the process in which the molecules are moved from the region of higher temperature to lower temperature. Rate of Conductive Heat Transfer is denoted by Q_fin symbol.

How to calculate Heat Transfer by Conduction at Base using this online calculator? To use this online calculator for Heat Transfer by Conduction at Base, enter Thermal Conductivity (k_fin), Cross Sectional Area of Fin (A_cs), Perimeter of the Fin (P), Convective Heat Transfer Coefficient (h), Base Temperature (t_o) & Ambient Temperature (t_a) and hit the calculate button. Here is how the Heat Transfer by Conduction at Base calculation can be explained with given input values -> 43.20445 = (205*9E-05*0.046*30.17)^0.5*(573-303).

FAQ

What is Heat Transfer by Conduction at Base?

Heat Transfer by conduction at Base is defined as the process in which the molecules are moved from the region of higher temperature to lower temperature and is represented as Q_fin = (k_fin*A_cs*P*h)^0.5*(t_o-t_a) or Rate of Conductive Heat Transfer = (Thermal Conductivity*Cross Sectional Area of Fin*Perimeter of the Fin*Convective Heat Transfer Coefficient)^0.5*(Base Temperature-Ambient Temperature). Thermal Conductivity of Fin is a material property that represents at which rate heat energy is transferred through the fin by conduction, Cross Sectional Area of Fin is the surface area of the Fin which is perpendicular to the direction of Heat Flow, Perimeter of the Fin refers to the total length of the outer boundary of the fin that is exposed to the surrounding medium, Convective Heat Transfer Coefficient is a parameter that quantifies the rate of heat transfer between a solid surface and the surrounding fluid due to convection, Base Temperature is the temperature at the base of the fin & Ambient Temperature is the temperature of the ambient or surrounding fluid which is in contact with the fin.

How to calculate Heat Transfer by Conduction at Base?

Heat Transfer by conduction at Base is defined as the process in which the molecules are moved from the region of higher temperature to lower temperature is calculated using Rate of Conductive Heat Transfer = (Thermal Conductivity*Cross Sectional Area of Fin*Perimeter of the Fin*Convective Heat Transfer Coefficient)^0.5*(Base Temperature-Ambient Temperature). To calculate Heat Transfer by Conduction at Base, you need Thermal Conductivity (k_fin), Cross Sectional Area of Fin (A_cs), Perimeter of the Fin (P), Convective Heat Transfer Coefficient (h), Base Temperature (t_o) & Ambient Temperature (t_a). With our tool, you need to enter the respective value for Thermal Conductivity, Cross Sectional Area of Fin, Perimeter of the Fin, Convective Heat Transfer Coefficient, Base Temperature & Ambient Temperature and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.

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