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Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions Formula

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Temperature 1 is the degree or intensity of heat present in a substance or object. Check FAQs

t 1 = - \frac{q G \cdot b^{2}}{2 \cdot k} \cdot (\frac{x}{b} - {(\frac{x}{b})}^{2}) + T 1

t₁ - Temperature 1?q_G - Internal Heat Generation?b - Wall Thickness?k - Thermal Conductivity?x - Thickness?T₁ - Surface Temperature?

Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions Example

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Here is how the Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions equation looks like with Values.

Here is how the Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions equation looks like with Units.

Here is how the Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions equation looks like.

130.3241 = - \frac{100 \cdot {12.6019}^{2}}{2 \cdot 10.18} \cdot (\frac{4.2667}{12.6019} - {(\frac{4.2667}{12.6019})}^{2}) + 305

130.3241K = - \frac{100W/m³ \cdot {12.6019m}^{2}}{2 \cdot 10.18W/(m*K)} \cdot (\frac{4.2667m}{12.6019m} - {(\frac{4.2667m}{12.6019m})}^{2}) + 305K

130.3241 = - \frac{100 \cdot {12.6019}^{2}}{2 \cdot 10.18} \cdot (\frac{4.2667}{12.6019} - {(\frac{4.2667}{12.6019})}^{2}) + 305

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Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions Solution

Follow our step by step solution on how to calculate Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions?

FIRST Step Consider the formula

t 1 = - \frac{q G \cdot b^{2}}{2 \cdot k} \cdot (\frac{x}{b} - {(\frac{x}{b})}^{2}) + T 1

Next Step Substitute values of Variables

∴

t 1 = - \frac{100W/m³ \cdot {12.6019m}^{2}}{2 \cdot 10.18W/(m*K)} \cdot (\frac{4.2667m}{12.6019m} - {(\frac{4.2667m}{12.6019m})}^{2}) + 305K

Next Step Prepare to Evaluate

∴

t 1 = - \frac{100 \cdot {12.6019}^{2}}{2 \cdot 10.18} \cdot (\frac{4.2667}{12.6019} - {(\frac{4.2667}{12.6019})}^{2}) + 305

Next Step Evaluate

∴

t 1 = 130.324094010629K

LAST Step Rounding Answer

∴

t 1 = 130.3241K

Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions Formula Elements

Variables

Temperature 1

Temperature 1 is the degree or intensity of heat present in a substance or object.

Symbol: t₁

Measurement: TemperatureUnit: K

Note: Value can be positive or negative.

Formulas to find Temperature 1

Internal Heat Generation

Internal Heat Generation is defined as the conversion of electrical, chemical, or nuclear energy into heat (or thermal) energy which leads to a rise in temperature throughout the medium.

Symbol: q_G

Measurement: Power DensityUnit: W/m³

Note: Value should be greater than 0.

Formulas to find Internal Heat Generation

Wall Thickness

Wall Thickness is simply the width of the wall that we are taking under consideration.

Symbol: b

Measurement: LengthUnit: m

Note: Value should be greater than 0.

Formulas to find Wall Thickness

Thermal Conductivity

Thermal Conductivity is rate of heat passes through specified material, 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

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

Note: Value should be greater than 0.

Formulas to find Thermal Conductivity

Thickness

Thickness is the distance from one end to the desired end of the body or object.

Symbol: x

Measurement: LengthUnit: m

Note: Value should be greater than 0.

Formulas to find Thickness

Surface Temperature

Surface Temperature is the temperature at or near a surface. Specifically, it may refer to as Surface air temperature, the temperature of the air near the surface of the earth.

Symbol: T₁

Measurement: TemperatureUnit: K

Note: Value can be positive or negative.

Formulas to find Surface Temperature

Other formulas in Steady State Heat Conduction with Heat Generation category

Go Maximum Temperature in Plane Wall with Symmetrical Boundary Conditions

T max = T 1 + \frac{q G \cdot b^{2}}{8 \cdot k}

Go Location of Maximum Temperature in Plane Wall with Symmetrical Boundary Conditions

X = \frac{b}{2}

Go Maximum Temperature in Solid Cylinder

T max = T w + \frac{q G \cdot {R cy}^{2}}{4 \cdot k}

Go Maximum Temperature in Solid Sphere

T max = T w + \frac{q G \cdot {R s}^{2}}{6 \cdot k}

How to Evaluate Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions?

Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions evaluator uses Temperature 1 = -(Internal Heat Generation*Wall Thickness^2)/(2*Thermal Conductivity)*(Thickness/Wall Thickness-(Thickness/Wall Thickness)^2)+Surface Temperature to evaluate the Temperature 1, The Temperature inside plane wall at given thickness x with symmetrical boundary conditions formula gives the value of temperature at desired thickness inside the plane wall provided with an internal heat generation source. Temperature 1 is denoted by t₁ symbol.

How to evaluate Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions using this online evaluator? To use this online evaluator for Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions, enter Internal Heat Generation (q_G), Wall Thickness (b), Thermal Conductivity (k), Thickness (x) & Surface Temperature (T₁) and hit the calculate button.

FAQs on Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions

What is the formula to find Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions?

The formula of Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions is expressed as Temperature 1 = -(Internal Heat Generation*Wall Thickness^2)/(2*Thermal Conductivity)*(Thickness/Wall Thickness-(Thickness/Wall Thickness)^2)+Surface Temperature. Here is an example- 130.3241 = -(100*12.601905^2)/(2*10.18)*(4.266748/12.601905-(4.266748/12.601905)^2)+305.

How to calculate Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions?

With Internal Heat Generation (q_G), Wall Thickness (b), Thermal Conductivity (k), Thickness (x) & Surface Temperature (T₁) we can find Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions using the formula - Temperature 1 = -(Internal Heat Generation*Wall Thickness^2)/(2*Thermal Conductivity)*(Thickness/Wall Thickness-(Thickness/Wall Thickness)^2)+Surface Temperature.

Can the Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions be negative?

Yes, the Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions, measured in Temperature can be negative.

Which unit is used to measure Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions?

Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions is usually measured using the Kelvin[K] for Temperature. Celsius[K], Fahrenheit[K], Rankine[K] are the few other units in which Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions can be measured.

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Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions Formula

Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions Example

Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions Solution

Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions Formula Elements

Other formulas in Steady State Heat Conduction with Heat Generation category

How to Evaluate Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions?

FAQs on Temperature Inside Plane Wall at given Thickness x with Symmetrical Boundary Conditions

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