Question

Steam at \(450^{\circ} \mathrm{F}\) is flowing through a steel pipe \(\left(k=8.7 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) whose inner and outer diameters are \(3.5\) in and \(4.0\) in, respectively, in an environment at \(55^{\circ} \mathrm{F}\). The pipe is insulated with 2 -in-thick fiberglass insulation \((k=\) \(\left.0.020 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\). If the heat transfer coefficients on the inside and the outside of the pipe are 30 and \(5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\), respectively, determine the rate of heat loss from the steam per foot length of the pipe. What is the error involved in neglecting the thermal resistance of the steel pipe in calculations?

Short answer

Answer: The rate of heat loss from the steam per foot length of the pipe is 1920.52 Btu/h, and the error involved in neglecting the steel pipe's thermal resistance is 0.65%.

Step-by-step solution

Convert units to consistent values

To ensure consistency in our calculations, let's first convert inches to feet: Inner diameter: \(3.5 \,\text{in} = 3.5/12 \,\text{ft} = 0.29167 \,\text{ft}\) Outer diameter: \(4.0 \,\text{in} = 4/12 \,\text{ft} = 0.33333 \,\text{ft}\) Insulation thickness: \(2 \,\text{in} = 2/12 \,\text{ft} = 0.16667 \,\text{ft}\)

Calculate thermal resistances and the overall heat transfer coefficient (U)

There are three types of thermal resistances: 1. Convection inside the pipe (R1): \(\frac{1}{h_1 \cdot A_1} = \frac{1}{30 \cdot (2\pi\cdot0.29167/2) } = 0.01131 \,\text{h} \cdot \text{ft\)^2\(} \cdot{ }^{\circ}\text{F} \cdot \text{Btu}^{-1}\) 2. Conduction through the steel pipe (R2): \(\frac{\ln(D_2/D_1)}{2\pi k_1 L} = \frac{\ln(0.33333/0.29167)}{2\pi \cdot 8.7} = 0.00149 \,\text{h} \cdot \text{ft\)^2\(} \cdot{ }^{\circ}\text{F} \cdot \text{Btu}^{-1}\) 3. Conduction through the insulation (R3): \(\frac{\ln(D_3/D_2)}{2\pi k_2 L} = \frac{\ln((0.33333+0.16667)/0.33333)}{2\pi \cdot 0.020} = 0.25671 \,\text{h} \cdot \text{ft\)^2\(} \cdot{ }^{\circ}\text{F} \cdot \text{Btu}^{-1}\) 4. Convection outside the pipe (R4): \(\frac{1}{h_2 \cdot A_2} = \frac{1}{5 \cdot (2\pi\cdot0.5) } = 0.05305 \,\text{h} \cdot \text{ft\)^2\(} \cdot{ }^{\circ}\text{F} \cdot \text{Btu}^{-1}\) Total thermal resistance (R_total) = R1 + R2 + R3 + R4 = 0.01131 + 0.00149 + 0.25671 + 0.05305 = 0.32257 \,\text{h} \cdot \text{ft\(^2\)} \cdot{ }^{\circ}\text{F} \cdot \text{Btu}^{-1}$ Overall heat transfer coefficient (U) = \(\frac{1}{R_\text{total}} = \frac{1}{0.32257} = 3.099 \,\text{Btu} \cdot \text{h}^{-1} \cdot \text{ft\)^{-2}\(} \cdot { }^{\circ}\text{F}^{-1}\)

Calculate heat loss per foot length of the pipe (Q)

To find the heat loss, we can use the formula: \(Q = U \cdot A \cdot \Delta T\) Area of the outer surface (A): \(2\pi(\frac{D_3}{2})\cdot L = 2\pi(\frac{0.33333+0.16667}{2}) = 1.5708 \,\text{ft\)^2\(}\) Temperature difference (\(\Delta T\)): \(450 \,^{\circ}\text{F} - 55 \,^{\circ}\text{F} = 395 \,^{\circ}\text{F}\) Heat loss (Q): \(3.099 \cdot 1.5708 \cdot 395 = 1920.52 \,\text{Btu/h}\)

Calculate heat loss without the thermal resistance of the steel pipe (Q_without_R2)

To calculate this, we use the same formula to find the heat transfer rate without considering R2: R_total_without_R2 = R1 + R3 + R4 = 0.01131 + 0.25671 + 0.05305 = 0.32108 \,\text{h} \cdot \text{ft\(^2\)} \cdot{ }^{\circ}\text{F} \cdot \text{Btu}^{-1}$ Overall heat transfer coefficient (U_without_R2) = \(\frac{1}{R_\text{total without R2}} = \frac{1}{0.32108} = 3.114 \,\text{Btu} \cdot \text{h}^{-1} \cdot \text{ft\)^{-2}\(} \cdot { }^{\circ}\text{F}^{-1}\) Heat loss without considering the steel pipe's resistance (Q_without_R2): \(3.114 \cdot 1.5708 \cdot 395 = 1932.99 \,\text{Btu/h}\)

Calculate the error involved due to neglecting R2

To find the error, we can use the formula: Error = \(\frac{Q_\text{without R2} - Q}{Q} \times 100\%\) Error = \(\frac{1932.99 - 1920.52}{1920.52} \times 100\% = 0.65\%\) The error involved in neglecting the thermal resistance of the steel pipe is 0.65%.

Key concepts

Thermal Resistance

Thermal resistance is a concept in thermodynamics that measures the opposition to heat flow through a material. It's analogous to electrical resistance, which measures how much a material resists the flow of electricity. In our context, thermal resistance is crucial for calculating how well a material insulates, or in other words, slows down the transfer of heat.

For instance, in the step-by-step solution provided, we see individual thermal resistances for various layers around the steam pipe: convection inside the pipe, conduction through the steel and the fiberglass insulation, and finally convection outside the pipe. By calculating the thermal resistance of each layer and summing them up, we can determine the total resistance to heat loss.

In practice, the lower the thermal resistance, the greater the heat flow. Insulation materials, such as fiberglass, are designed with high thermal resistance to minimize heat loss in thermal applications. Conversely, materials with low thermal resistance, like metals, are excellent heat conductors. Knowing the thermal resistance values for different materials is essential for engineers and architects to design energy-efficient systems and structures, such as the steam pipe in the exercise.

Overall Heat Transfer Coefficient

The overall heat transfer coefficient, commonly represented by the symbol U, is a measure that combines the thermal resistances of different layers and interfaces to quantify how easily heat can pass through a system. It's defined as the amount of heat that passes per unit area, per unit temperature difference, per unit time.

In the solution, the overall heat transfer coefficient (U) takes into account all the resistances - including those caused by conduction through the steel pipe and insulation, as well as convection inside and outside the pipe. It's calculated by taking the inverse of the total thermal resistance of the system. The U-factor is particularly important when designing environmental control systems, as it helps to predict the heat loss or gain, which in turn affects energy consumption and efficiency.

Note the step involving the calculation of U and its importance: the lower the U-value, the better the composite material is at insulating. This step is critical to properly assess the performance of the insulation in preventing heat loss, thereby guiding decisions regarding material selection and system design.

Conduction and Convection

Conduction and convection are two primary ways that heat transfers through materials and fluids. Conduction is the process of heat transfer through a material without any movement of the material itself. It occurs at the microscopic level as vibrating, heat-carrying particles bump into their neighbors, transferring energy. This principle is seen at work in the exercise when heat conducts through the steel pipe wall.

Convection, on the other hand, is the heat transfer due to the bulk movement of molecules within fluids (gases and liquids), carrying heat along with them. This is especially pertinent in systems where a fluid, like steam or water, is involved. In the provided exercise, convection is at play both inside and outside the steam pipe, where it carries heat away from the pipe's surface, contributing to the system's thermal resistance.

Understanding both conduction and convection is necessary not only to calculate heat loss effectively but also to design systems with optimal temperature control. These principles help us determine the efficiency of heaters, coolers, and insulators in real-world applications, which is directly linked to energy conservation and operational cost.