how to calculate temp gradient of two parts of a rod whose ends are maintained at 0&100C and a point source of 25W is placed at middle

To calculate the temperature gradient of a rod with its ends maintained at 0°C and 100°C, while a point source of heat (25W) is placed in the middle, we can apply the principles of heat conduction and steady-state temperature distribution. Let's break this down step by step.

Understanding the Setup

We have a rod that is 1 meter long, with one end at 0°C and the other at 100°C. The heat source in the middle introduces additional heat into the system. The goal is to find out how the temperature varies along the length of the rod due to these conditions.

Key Concepts

• Steady-State Condition: This means that the temperature distribution does not change over time.
• Heat Conduction: Governed by Fourier's law, which states that the heat transfer rate through a material is proportional to the negative gradient of temperature and the area through which heat is flowing.

Applying Fourier's Law

Fourier's law can be expressed as:

q = -kA(dT/dx)

Where:

• q = heat transfer rate (W)
• k = thermal conductivity of the rod material (W/m·K)
• A = cross-sectional area of the rod (m²)
• dT/dx = temperature gradient (K/m)

Setting Up the Equation

In our case, the total heat input into the rod is the sum of the heat from the ends and the point source. The heat from the ends is constant, while the point source adds a localized increase in temperature. We can denote the temperature at any point along the rod as T(x), where x is the distance from the 0°C end.

Temperature Distribution

To find the temperature distribution, we can set up the differential equation based on the heat balance:

0 = kA(d²T/dx²) + Q

Where Q is the heat added by the point source. For a point source, we can consider it as a Dirac delta function, leading to:

Q = 25W

Solving the Differential Equation

Integrating this equation twice will give us the temperature distribution. The general solution will have the form:

T(x) = Ax + B + (Q/(kA)) * (x - x₀)

Where x₀ is the location of the heat source (0.5 m in this case). The constants A and B can be determined using boundary conditions:

• T(0) = 0°C
• T(1) = 100°C

Finding the Constants

Using the boundary conditions, we can solve for A and B. After substituting the values and solving the equations, we can find the temperature at any point along the rod.

Calculating the Temperature Gradient

Once we have the temperature distribution function T(x), the temperature gradient can be found by taking the derivative:

dT/dx = A + (Q/(kA))

This gradient will vary along the length of the rod, particularly around the point source where the heat is added. The exact values will depend on the thermal conductivity of the rod material and its cross-sectional area.

Conclusion

By following these steps, you can effectively calculate the temperature gradient in a rod with specified boundary conditions and a point heat source. This approach not only illustrates the principles of heat conduction but also emphasizes the importance of boundary conditions in thermal analysis.