Question

Consider a steam pipe of length \(L=30 \mathrm{ft}\), inner radius \(r_{1}=2\) in, outer radius \(r_{2}=2.4\) in, and thermal conductivity \(k=7.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). Steam is flowing through the pipe at an average temperature of \(300^{\circ} \mathrm{F}\), and the average convection heat transfer coefficient on the inner surface is given to be \(h=12.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\). If the average temperature on the outer surfaces of the pipe is \(T_{2}=175^{\circ} \mathrm{F},(a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the pipe, \((b)\) obtain a relation for the variation of temperature in the pipe by solving the differential equation, and \((c)\) evaluate the rate of heat loss from the steam through the pipe.

Short answer

Answer: The temperature distribution inside the pipe is given by the equation T(r) = -625r + 1550.

Step-by-step solution

- Expressing the differential equation and boundary conditions

To study the heat transfer through a pipe, we need to express the differential equation of heat conduction, which in this case is a steady-state, one-dimensional problem. The general one-dimensional heat conduction equation in cylindrical coordinates \((r)\) is: $$\frac{d}{dr} \left(k\frac{dT}{dr} \right)=0$$ As the thermal conductivity, \(k\), is constant for this problem, we have: $$\frac{d^2 T}{dr^2}=0$$ Boundary conditions: We are given the average temperature on the inner surface of the pipe \(T_1 = 300^{\circ} F\) and on the outer surface of the pipe \(T_2 = 175^{\circ}F\). We can write the boundary conditions as: At \(r = r_1\), \(T = T_1\) At \(r = r_2\), \(T = T_2\)

- Solving the differential equation

Now we will solve the differential equation which represents the temperature distribution in the pipe. Since \(\frac{d^2 T}{dr^2}=0\), we can integrate this equation twice: $$\frac{d T}{d r} = C_1$$ Integrating once more: $$T(r) = C_1 r + C_2$$ Now, we need to apply the boundary conditions to find the constants \(C_1\) and \(C_2\). At \(r = r_1\), \(T(r_1) = T_1\) $$T_1 = C_1 r_1 + C_2$$ At \(r = r_2\), \(T(r_2) = T_2\) $$T_2 = C_1 r_2 + C_2$$ Solving these equations simultaneously, we obtain: $$C_1 = \frac{T_1 - T_2}{r_1 - r_2}$$ and $$C_2 = T_1 - C_1 r_1$$ Substituting the values for \(T_1\), \(T_2\), \(r_1\), and \(r_2\): \(C_1 = \frac{300 - 175}{2 - 2.4}=-625\) \(C_2 = 300 - (-625) \times 2 = 1550\) So our temperature distribution is: $$T(r) = -625 r + 1550$$

- Evaluating the rate of heat loss

To evaluate the rate of heat loss through the pipe, we can use Fourier's Law of heat conduction: $$q = -k A \frac{dT}{dr}$$ First, we calculate \(\frac{dT}{dr}\): $$\frac{dT}{dr}=-625$$ Now, calculate the area of the cylindrical shell: $$A = 2 \pi DH$$ where \(D = r_2 - r_1 = 2.4 - 2 = 0.4\) in, \(H = L = 30\) ft, and convert inches to feet (\(1\) in = \(1/12\) ft) $$A = 2 \pi (0.4/12) (30)$$ Finally, we can find the rate of heat loss: $$q = - \left(- \frac{7.2}{1} \right) \left( 2 \pi (0.4/12) (30) \right) (-625)$$ $$q = 4687.5\,\mathrm{Btu} / \mathrm{h}$$ The rate of heat loss from the steam through the pipe is \(4687.5\,\mathrm{Btu} / \mathrm{h}\).

Key concepts

Differential Equation of Heat Conduction

Understanding the differential equation of heat conduction is crucial for analyzing heat transfer in structures like pipes. Heat conduction is described by a partial differential equation that relies on spatial and temperature changes. In a cylindrical coordinate system for a pipe, this simplifies to \[\frac{d^2 T}{dr^2} = 0\]where \(T\) is the temperature and \(r\) represents the radial position from the center of the pipe. This equation indicates that the temperature’s second derivative with respect to the radius is zero, suggesting a linear temperature gradient in steady-state conditions, which is a key characteristic when heat transfer is uniform and time-independent.
When solving these equations, two integrations will yield a linear temperature distribution, with the integration constants determined by the boundary conditions.

Boundary Conditions

Boundary conditions are essential in solving the heat conduction differential equation. These conditions define the values of temperature at the boundaries of the domain of interest—in this case, the inner and outer surfaces of the pipe. For the steam pipe scenario, the boundary conditions are given as:

• At \(r = r_1\), \(T = T_1 = 300^\circ F\)
• At \(r = r_2\), \(T = T_2 = 175^\circ F\)

These real-world constraints allow us to solve for the constants in the temperature distribution solution. By applying the provided temperatures at respective radii, we can specify the temperature profile throughout the pipe, which is indicative of how heat is distributed from the hot steam to the cooler outside environment.

Fourier's Law of Heat Conduction

Fourier's law of heat conduction is the fundamental principle that describes the rate at which heat energy is transferred through a material. It can be written as:\[q = -k A \frac{dT}{dr}\]where \(q\) is the heat transfer rate per time, \(k\) is the thermal conductivity of the material, \(A\) is the cross-sectional area perpendicular to the direction of heat flow, and \(\frac{dT}{dr}\) is the temperature gradient in the radial direction. The negative sign indicates that heat flows from higher to lower temperatures. In our textbook example, this law is applied to calculate the rate of heat loss from the steam in the pipe, using the known thermal conductivity and the area of the pipe.

Temperature Distribution in a Pipe

The temperature distribution in a pipe reveals how temperature varies from the inner to the outer surface. In the aforementioned problem, we derived the temperature distribution as a linear function of the radius:\[T(r) = -625 r + 1550\]This simple relation encapsulates how temperature changes across the pipe wall, which is a function of the thermal properties of the material and boundary conditions. This type of linear distribution is typical in steady-state heat conduction problems where the thermal conductivity is constant and heat is not generated within the pipe wall.

Steady-State Heat Transfer

Steady-state heat transfer refers to conditions where the temperature field does not change with time. This implies that heat entering a particular volume of material is equal to the heat leaving, resulting in a constant temperature field within the material. From the exercise, steady-state conditions allow us to use the simplified form of the heat conduction differential equation and Fourier’s law to evaluate the rate of heat loss, which is calculated as: \[q = 4687.5\,\mathrm{Btu} / \mathrm{h}\]As the temperature distribution remains constant over time, the outcome provides an accurate measure for long-term thermal performance of the pipe system carrying steam. The power and utility of this analysis are immense, providing essential data for engineering applications, including HVAC systems and industrial process design.