Heat Transfer Per Unit Length For Annular Space Between Concentric Cylinders Calculator

Formula Used:

\[ q' = \frac{2\pi k_{eff}}{\ln\left(\frac{D_o}{D_i}\right)} (T_i - T_o) \]

Effective Thermal Conductivity:

W/m·K

Outside Diameter:

Inside Diameter:

Inside Temperature:

Outside Temperature:

Heat Transfer Per Unit Length:

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1. What Is Heat Transfer Per Unit Length For Annular Space Between Concentric Cylinders?

This calculation determines the rate of heat transfer through the annular space between two concentric cylinders per unit length. It's essential in thermal engineering applications involving pipes, heat exchangers, and insulated systems.

2. How Does The Calculator Work?

The calculator uses the formula:

\[ q' = \frac{2\pi k_{eff}}{\ln\left(\frac{D_o}{D_i}\right)} (T_i - T_o) \]

Where:

\( q' \) — Heat transfer per unit length (W/m)
\( k_{eff} \) — Effective thermal conductivity (W/m·K)
\( D_o \) — Outside diameter (m)
\( D_i \) — Inside diameter (m)
\( T_i \) — Inside temperature (K)
\( T_o \) — Outside temperature (K)
\( \pi \) — Mathematical constant pi (≈3.14159)

Explanation: The formula calculates conductive heat transfer through cylindrical geometry, accounting for the logarithmic mean area between the two surfaces.

3. Importance Of Heat Transfer Calculation

Details: Accurate heat transfer calculation is crucial for designing thermal systems, determining insulation requirements, optimizing energy efficiency, and preventing thermal damage in engineering applications.

4. Using The Calculator

Tips: Enter all values in consistent SI units. Ensure outside diameter is greater than inside diameter. Temperature difference should be positive for heat flow from inside to outside.

5. Frequently Asked Questions (FAQ)

Q1: What is the significance of the logarithmic term?
A: The logarithmic term accounts for the varying surface area between concentric cylinders, providing the correct mean area for heat transfer calculation.

Q2: When is this formula applicable?
A: This formula applies to steady-state, radial heat conduction through homogeneous materials between concentric cylindrical surfaces with constant temperatures.

Q3: What if the cylinders are not concentric?
A: The formula assumes perfect concentricity. For eccentric cylinders, more complex correction factors are needed.

Q4: How does effective thermal conductivity differ from regular conductivity?
A: Effective thermal conductivity may account for composite materials or porous media, representing the overall conductive property of the annular space.

Q5: Can this be used for transient heat transfer?
A: No, this formula is for steady-state conditions only. Transient heat transfer requires more complex time-dependent solutions.