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Inner surface temperature of pipe with eccentric lagging Formula

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Eccentric Lagging Inner Surface Temperature is the temperature at the inner surface of the wall either plane wall or cylindrical wall or spherical wall, etc. Check FAQs

T ie = (Q e \cdot ((\frac{1}{2 \cdot π \cdot k e \cdot L e}) \cdot (\ln (\frac{\sqrt{({(r 2 + r 1)}^{2}) - e^{2}} + \sqrt{({(r 2 - r 1)}^{2}) - e^{2}}}{\sqrt{({(r 2 + r 1)}^{2}) - e^{2}} - \sqrt{({(r 2 - r 1)}^{2}) - e^{2}}})))) + T oe

T_ie - Eccentric Lagging Inner Surface Temperature?Q_e - Eccentric Lagging Heat Flow Rate?k_e - Eccentric Lagging Thermal Conductivity?L_e - Eccentric Lagging Length?r₂ - Radius 2?r₁ - Radius 1?e - Distance Between Centers of Eccentric Circles?T_oe - Eccentric Lagging Outer Surface Temperature?π - Archimedes' constant?

Inner surface temperature of pipe with eccentric lagging Example

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Here is how the Inner surface temperature of pipe with eccentric lagging equation looks like with Values.

Here is how the Inner surface temperature of pipe with eccentric lagging equation looks like with Units.

Here is how the Inner surface temperature of pipe with eccentric lagging equation looks like.

25 = (3021.485 \cdot ((\frac{1}{2 \cdot 3.1416 \cdot 15 \cdot 7}) \cdot (\ln (\frac{\sqrt{({(12.1 + 4)}^{2}) - {1.4}^{2}} + \sqrt{({(12.1 - 4)}^{2}) - {1.4}^{2}}}{\sqrt{({(12.1 + 4)}^{2}) - {1.4}^{2}} - \sqrt{({(12.1 - 4)}^{2}) - {1.4}^{2}}})))) + 20

25K = (3021.485W \cdot ((\frac{1}{2 \cdot 3.1416 \cdot 15W/(m*K) \cdot 7m}) \cdot (\ln (\frac{\sqrt{({(12.1m + 4m)}^{2}) - {1.4m}^{2}} + \sqrt{({(12.1m - 4m)}^{2}) - {1.4m}^{2}}}{\sqrt{({(12.1m + 4m)}^{2}) - {1.4m}^{2}} - \sqrt{({(12.1m - 4m)}^{2}) - {1.4m}^{2}}})))) + 20K

25 = (3021.485 \cdot ((\frac{1}{2 \cdot 3.1416 \cdot 15 \cdot 7}) \cdot (\ln (\frac{\sqrt{({(12.1 + 4)}^{2}) - {1.4}^{2}} + \sqrt{({(12.1 - 4)}^{2}) - {1.4}^{2}}}{\sqrt{({(12.1 + 4)}^{2}) - {1.4}^{2}} - \sqrt{({(12.1 - 4)}^{2}) - {1.4}^{2}}})))) + 20

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Inner surface temperature of pipe with eccentric lagging Solution

Follow our step by step solution on how to calculate Inner surface temperature of pipe with eccentric lagging?

FIRST Step Consider the formula

T ie = (Q e \cdot ((\frac{1}{2 \cdot π \cdot k e \cdot L e}) \cdot (\ln (\frac{\sqrt{({(r 2 + r 1)}^{2}) - e^{2}} + \sqrt{({(r 2 - r 1)}^{2}) - e^{2}}}{\sqrt{({(r 2 + r 1)}^{2}) - e^{2}} - \sqrt{({(r 2 - r 1)}^{2}) - e^{2}}})))) + T oe

Next Step Substitute values of Variables

∴

T ie = (3021.485W \cdot ((\frac{1}{2 \cdot π \cdot 15W/(m*K) \cdot 7m}) \cdot (\ln (\frac{\sqrt{({(12.1m + 4m)}^{2}) - {1.4m}^{2}} + \sqrt{({(12.1m - 4m)}^{2}) - {1.4m}^{2}}}{\sqrt{({(12.1m + 4m)}^{2}) - {1.4m}^{2}} - \sqrt{({(12.1m - 4m)}^{2}) - {1.4m}^{2}}})))) + 20K

Next Step Substitute values of Constants

∴

T ie = (3021.485W \cdot ((\frac{1}{2 \cdot 3.1416 \cdot 15W/(m*K) \cdot 7m}) \cdot (\ln (\frac{\sqrt{({(12.1m + 4m)}^{2}) - {1.4m}^{2}} + \sqrt{({(12.1m - 4m)}^{2}) - {1.4m}^{2}}}{\sqrt{({(12.1m + 4m)}^{2}) - {1.4m}^{2}} - \sqrt{({(12.1m - 4m)}^{2}) - {1.4m}^{2}}})))) + 20K

Next Step Prepare to Evaluate

∴

T ie = (3021.485 \cdot ((\frac{1}{2 \cdot 3.1416 \cdot 15 \cdot 7}) \cdot (\ln (\frac{\sqrt{({(12.1 + 4)}^{2}) - {1.4}^{2}} + \sqrt{({(12.1 - 4)}^{2}) - {1.4}^{2}}}{\sqrt{({(12.1 + 4)}^{2}) - {1.4}^{2}} - \sqrt{({(12.1 - 4)}^{2}) - {1.4}^{2}}})))) + 20

Next Step Evaluate

∴

T ie = 25.0000002141715K

LAST Step Rounding Answer

∴

T ie = 25K

Inner surface temperature of pipe with eccentric lagging Formula Elements

Variables

Constants

Functions

Eccentric Lagging Inner Surface Temperature

Eccentric Lagging Inner Surface Temperature is the temperature at the inner surface of the wall either plane wall or cylindrical wall or spherical wall, etc.

Symbol: T_ie

Measurement: TemperatureUnit: K

Note: Value can be positive or negative.

Formulas to find Eccentric Lagging Inner Surface Temperature

Eccentric Lagging Heat Flow Rate

Eccentric lagging Heat Flow Rate is the amount of heat that is transferred per unit of time in some material. Heat is the flow of thermal energy driven by thermal non-equilibrium.

Symbol: Q_e

Measurement: PowerUnit: W

Note: Value can be positive or negative.

Formulas to find Eccentric Lagging Heat Flow Rate

Eccentric Lagging Thermal Conductivity

Eccentric Lagging Thermal Conductivity is expressed as amount of heat flows per unit time through a unit area with a temperature gradient of one degree per unit distance.

Symbol: k_e

Measurement: Thermal ConductivityUnit: W/(m*K)