Heat Transfer between Two Long Concentric Cylinder given Temp, Emissivity and Area of Both Surfaces Calculator

✖The Surface Area of Body 1 is the area of body 1 through which the radiation takes place.ⓘ Surface Area of Body 1 [A₁]			+10% -10%
✖Temperature of Surface 1 is the temperature of the 1st surface.ⓘ Temperature of Surface 1 [T₁]			+10% -10%
✖Temperature of Surface 2 is the temperature of the 2nd surface.ⓘ Temperature of Surface 2 [T₂]			+10% -10%
✖The Emissivity of Body 1 is the ratio of the energy radiated from a body's surface to that radiated from a perfect emitter.ⓘ Emissivity of Body 1 [ε₁]			+10% -10%
✖The Surface Area of Body 2 is the area of body 2 upon which the radiation takes place.ⓘ Surface Area of Body 2 [A₂]			+10% -10%
✖The Emissivity of Body 2 is the ratio of the energy radiated from a body's surface to that radiated from a perfect emitter.ⓘ Emissivity of Body 2 [ε₂]			+10% -10%

✖Heat Transfer is the amount of heat that is transferred per unit of time in some material, usually measured in watts (joules per second).ⓘ Heat Transfer between Two Long Concentric Cylinder given Temp, Emissivity and Area of Both Surfaces [q]

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Heat Transfer between Two Long Concentric Cylinder given Temp, Emissivity and Area of Both Surfaces Solution

STEP 0: Pre-Calculation Summary

Formula Used

Heat Transfer = (([Stefan-BoltZ]*Surface Area of Body 1*((Temperature of Surface 1^4)-(Temperature of Surface 2^4))))/((1/Emissivity of Body 1)+((Surface Area of Body 1/Surface Area of Body 2)*((1/Emissivity of Body 2)-1)))
q = (([Stefan-BoltZ]*A₁*((T₁^4)-(T₂^4))))/((1/ε₁)+((A₁/A₂)*((1/ε₂)-1)))
This formula uses 1 Constants, 7 Variables

Constants Used

[Stefan-BoltZ] - Stefan-Boltzmann Constant Value Taken As 5.670367E-8

Variables Used

Heat Transfer - (Measured in Watt) - Heat Transfer is the amount of heat that is transferred per unit of time in some material, usually measured in watts (joules per second).
Surface Area of Body 1 - (Measured in Square Meter) - The Surface Area of Body 1 is the area of body 1 through which the radiation takes place.
Temperature of Surface 1 - (Measured in Kelvin) - Temperature of Surface 1 is the temperature of the 1st surface.
Temperature of Surface 2 - (Measured in Kelvin) - Temperature of Surface 2 is the temperature of the 2nd surface.
Emissivity of Body 1 - The Emissivity of Body 1 is the ratio of the energy radiated from a body's surface to that radiated from a perfect emitter.
Surface Area of Body 2 - (Measured in Square Meter) - The Surface Area of Body 2 is the area of body 2 upon which the radiation takes place.
Emissivity of Body 2 - The Emissivity of Body 2 is the ratio of the energy radiated from a body's surface to that radiated from a perfect emitter.

STEP 1: Convert Input(s) to Base Unit

Surface Area of Body 1: 34.74 Square Meter --> 34.74 Square Meter No Conversion Required
Temperature of Surface 1: 202 Kelvin --> 202 Kelvin No Conversion Required
Temperature of Surface 2: 151 Kelvin --> 151 Kelvin No Conversion Required
Emissivity of Body 1: 0.4 --> No Conversion Required
Surface Area of Body 2: 50 Square Meter --> 50 Square Meter No Conversion Required
Emissivity of Body 2: 0.3 --> No Conversion Required

STEP 2: Evaluate Formula

Substituting Input Values in Formula

q = (([Stefan-BoltZ]*A₁*((T₁^4)-(T₂^4))))/((1/ε₁)+((A₁/A₂)*((1/ε₂)-1))) --> (([Stefan-BoltZ]*34.74*((202^4)-(151^4))))/((1/0.4)+((34.74/50)*((1/0.3)-1)))

Evaluating ... ...

q = 547.335263755058

STEP 3: Convert Result to Output's Unit

547.335263755058 Watt --> No Conversion Required

FINAL ANSWER

547.335263755058 ≈ 547.3353 Watt <-- Heat Transfer

(Calculation completed in 00.020 seconds)

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Created by Ayush gupta

University School of Chemical Technology-USCT (GGSIPU), New Delhi

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University of Hawaiʻi at Mānoa (UH Manoa), Hawaii, USA

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< Radiation Heat Transfer Calculators

Net Heat Exchange given Area 1 and Shape Factor 12

LaTeX Go Net Heat Transfer = Surface Area of Body 1*Radiation Shape Factor 12*(Emissive Power of 1st Blackbody-Emissive Power of 2nd Blackbody)

Net Heat Exchange given Area 2 and Shape Factor 21

LaTeX Go Net Heat Transfer = Surface Area of Body 2*Radiation Shape Factor 21*(Emissive Power of 1st Blackbody-Emissive Power of 2nd Blackbody)

Net Heat Exchange between Two Surfaces given Radiosity for Both Surface

LaTeX Go Radiation Heat Transfer = (Radiosity of 1st Body-Radiosity of 2nd Body)/(1/(Surface Area of Body 1*Radiation Shape Factor 12))

Net Heat Transfer from Surface given Emissivity, Radiosity and Emissive Power

LaTeX Go Heat Transfer = (((Emissivity*Area)*(Emissive Power of Blackbody-Radiosity))/(1-Emissivity))

< Important Formulas in Radiation Heat Transfer Calculators

Area of Surface 1 given Area 2 and Radiation Shape Factor for Both Surfaces

LaTeX Go Surface Area of Body 1 = Surface Area of Body 2*(Radiation Shape Factor 21/Radiation Shape Factor 12)

Area of Surface 2 given Area 1 and Radiation Shape Factor for Both Surfaces

LaTeX Go Surface Area of Body 2 = Surface Area of Body 1*(Radiation Shape Factor 12/Radiation Shape Factor 21)

Emissive Power of Blackbody

LaTeX Go Emissive Power of Blackbody = [Stefan-BoltZ]*(Temperature of Blackbody^4)

Absorptivity given Reflectivity and Transmissivity

LaTeX Go Absorptivity = 1-Reflectivity-Transmissivity

Heat Transfer between Two Long Concentric Cylinder given Temp, Emissivity and Area of Both Surfaces Formula

LaTeX Go

What is Radiation?

Radiation is energy that comes from a source and travels through space at the speed of light. This energy has an electric field and a magnetic field associated with it, and has wave-like properties. You could also call radiation “electromagnetic waves”.

What is Emissivity?

Emissivity is defined as the ratio of the energy radiated from a material's surface to that radiated from a perfect emitter, known as a blackbody, at the same temperature and wavelength and under the same viewing conditions. It is a dimensionless number between 0 (for a perfect reflector) and 1 (for a perfect emitter).

How to Calculate Heat Transfer between Two Long Concentric Cylinder given Temp, Emissivity and Area of Both Surfaces?

Heat Transfer between Two Long Concentric Cylinder given Temp, Emissivity and Area of Both Surfaces calculator uses Heat Transfer = (([Stefan-BoltZ]*Surface Area of Body 1*((Temperature of Surface 1^4)-(Temperature of Surface 2^4))))/((1/Emissivity of Body 1)+((Surface Area of Body 1/Surface Area of Body 2)*((1/Emissivity of Body 2)-1))) to calculate the Heat Transfer, The Heat Transfer between Two Long Concentric Cylinder given Temp, Emissivity and Area of Both Surfaces formula is a function of emissivity and temperature of both surface, area of heat transfer. Heat Transfer is denoted by q symbol.

How to calculate Heat Transfer between Two Long Concentric Cylinder given Temp, Emissivity and Area of Both Surfaces using this online calculator? To use this online calculator for Heat Transfer between Two Long Concentric Cylinder given Temp, Emissivity and Area of Both Surfaces, enter Surface Area of Body 1 (A₁), Temperature of Surface 1 (T₁), Temperature of Surface 2 (T₂), Emissivity of Body 1 (ε₁), Surface Area of Body 2 (A₂) & Emissivity of Body 2 (ε₂) and hit the calculate button. Here is how the Heat Transfer between Two Long Concentric Cylinder given Temp, Emissivity and Area of Both Surfaces calculation can be explained with given input values -> 547.3353 = (([Stefan-BoltZ]*34.74*((202^4)-(151^4))))/((1/0.4)+((34.74/50)*((1/0.3)-1))).

FAQ

What is Heat Transfer between Two Long Concentric Cylinder given Temp, Emissivity and Area of Both Surfaces?

The Heat Transfer between Two Long Concentric Cylinder given Temp, Emissivity and Area of Both Surfaces formula is a function of emissivity and temperature of both surface, area of heat transfer and is represented as q = (([Stefan-BoltZ]*A₁*((T₁^4)-(T₂^4))))/((1/ε₁)+((A₁/A₂)*((1/ε₂)-1))) or Heat Transfer = (([Stefan-BoltZ]*Surface Area of Body 1*((Temperature of Surface 1^4)-(Temperature of Surface 2^4))))/((1/Emissivity of Body 1)+((Surface Area of Body 1/Surface Area of Body 2)*((1/Emissivity of Body 2)-1))). The Surface Area of Body 1 is the area of body 1 through which the radiation takes place, Temperature of Surface 1 is the temperature of the 1st surface, Temperature of Surface 2 is the temperature of the 2nd surface, The Emissivity of Body 1 is the ratio of the energy radiated from a body's surface to that radiated from a perfect emitter, The Surface Area of Body 2 is the area of body 2 upon which the radiation takes place & The Emissivity of Body 2 is the ratio of the energy radiated from a body's surface to that radiated from a perfect emitter.

How to calculate Heat Transfer between Two Long Concentric Cylinder given Temp, Emissivity and Area of Both Surfaces?

The Heat Transfer between Two Long Concentric Cylinder given Temp, Emissivity and Area of Both Surfaces formula is a function of emissivity and temperature of both surface, area of heat transfer is calculated using Heat Transfer = (([Stefan-BoltZ]*Surface Area of Body 1*((Temperature of Surface 1^4)-(Temperature of Surface 2^4))))/((1/Emissivity of Body 1)+((Surface Area of Body 1/Surface Area of Body 2)*((1/Emissivity of Body 2)-1))). To calculate Heat Transfer between Two Long Concentric Cylinder given Temp, Emissivity and Area of Both Surfaces, you need Surface Area of Body 1 (A₁), Temperature of Surface 1 (T₁), Temperature of Surface 2 (T₂), Emissivity of Body 1 (ε₁), Surface Area of Body 2 (A₂) & Emissivity of Body 2 (ε₂). With our tool, you need to enter the respective value for Surface Area of Body 1, Temperature of Surface 1, Temperature of Surface 2, Emissivity of Body 1, Surface Area of Body 2 & Emissivity of Body 2 and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.

How many ways are there to calculate Heat Transfer?

In this formula, Heat Transfer uses Surface Area of Body 1, Temperature of Surface 1, Temperature of Surface 2, Emissivity of Body 1, Surface Area of Body 2 & Emissivity of Body 2. We can use 3 other way(s) to calculate the same, which is/are as follows -

Heat Transfer = (((Emissivity*Area)*(Emissive Power of Blackbody-Radiosity))/(1-Emissivity))
Heat Transfer = (Area*[Stefan-BoltZ]*((Temperature of Surface 1^4)-(Temperature of Surface 2^4)))/((1/Emissivity of Body 1)+(1/Emissivity of Body 2)-1)
Heat Transfer = Surface Area of Body 1*Emissivity of Body 1*[Stefan-BoltZ]*((Temperature of Surface 1^4)-(Temperature of Surface 2^4))

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