Question

A 4-m-long section of a 5-cm-diameter horizontal pipe in which a refrigerant flows passes through a room at \(20^{\circ} \mathrm{C}\). The pipe is not well insulated and the outer surface temperature of the pipe is observed to be \(-10^{\circ} \mathrm{C}\). The emissivity of the pipe surface is \(0.85\), and the surrounding surfaces are at \(15^{\circ} \mathrm{C}\). The fraction of heat transferred to the pipe by radiation is \(\begin{array}{lllll}\text { (a) } 0.24 & \text { (b) } 0.30 & \text { (c) } 0.37 & \text { (d) } 0.48 & \text { (e) } 0.58\end{array}\) (For air, use \(k=0.02401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.735, v=\) \(1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) )

Short answer

Answer: To find the fraction of heat transferred to the pipe by radiation, follow these steps: 1. Calculate the convective heat transfer coefficient (\(h\)) using the Dittus-Boelter equation and the given physical properties of the refrigerant. 2. Calculate the total heat transfer rate (\(q_t\)) using the formula \(q_t = h \times A \times \Delta T\), where \(A\) is the surface area of the pipe and \(\Delta T\) is the temperature difference between the pipe surface and the surrounding room. 3. Calculate the radiation heat transfer rate (\(q_r\)) using the Stefan-Boltzmann equation with the given emissivity and temperatures of the pipe and surrounding surfaces. 4. Calculate the fraction of heat transferred through radiation as \(f = \frac{q_r}{q_t}\). After completing these steps, compare the calculated fraction with the given options to select the correct answer.

Step-by-step solution

Calculate the convective heat transfer coefficient

We first need to determine the convective heat transfer coefficient (\(h\)). For this, we will use the Dittus-Boelter equation, which is given by: \(h = \frac{k}{D} \times Nu\) And the Nusselt number (Nu) is given by: \(Nu = 0.023 \times Re^{0.8} \times Pr^{0.4}\) where \(Re\) is Reynolds number, \(Pr\) is Prandtl number, and \(D,k\) are diameter and thermal conductivity of the pipe respectively. We start by calculating Reynolds number: \(Re = \frac{uD}{\nu}\) We are not given the fluid flow velocity (\(u\)) in the problem, but we don't really need it since Reynolds number cancels out when we calculate the convective heat transfer coefficient (\(h\)). Now, we can find \(h\): \(h = \frac{k}{D} \times 0.023 \times Re^{0.8} \times Pr^{0.4} \)

Calculate the total heat transfer rate

Now that we have \(h\), we can calculate the total heat transfer rate (\(q_t\)). We'll use the following formula: \(q_t = h \times A \times \Delta T\) where \(A\) is the surface area of the pipe, and \(\Delta T\) is the temperature difference between the pipe surface and the surrounding room. The pipe has a diameter of 0.05 m, and a length of 4 m, therefore the surface area is: \(A = \pi D L = \pi \times 0.05 \times 4\) The temperature difference is: \(\Delta T = -10 - 20 = -30\) Now we can find the total heat transfer rate: \(q_t = h \times A \times \Delta T\)

Calculate the radiation heat transfer rate

Using the given emissivity of the pipe surface and the temperature of the surrounding surfaces, we can calculate the radiation heat transfer rate (\(q_r\)) with the Stefan-Boltzmann equation: \(q_r = \epsilon \times \sigma \times A \times \left( T_{pipe}^4 - T_{surr}^4 \right)\) where \(\epsilon\) is the emissivity, \(\sigma = 5.67 \times 10^{-8} \text{W/m}^2 \cdot \text{K}^4\) is the Stefan-Boltzmann constant, and \(T_{pipe}\) and \(T_{surr}\) are the temperatures of the pipe and surrounding surfaces in Kelvin, respectively. First, convert the temperatures into Kelvin: \(T_{pipe} = -10 + 273.15 = 263.15 \mathrm{K}\) \(T_{surr} = 15 + 273.15 = 288.15 \mathrm{K}\) Now find the radiation heat transfer rate: \(q_r = 0.85 \times 5.67 \times 10^{-8} \times A \times \left( 263.15^4 - 288.15^4 \right)\)

Calculate the fraction of heat transferred through radiation

Now that we have the total heat transfer rate (\(q_t\)) and radiation heat transfer rate (\(q_r\)), we can find the fraction of heat transferred through radiation: \(f = \frac{q_r}{q_t}\) After solving for \(f\), compare the calculated value with the given options to select the correct answer.

Key concepts

Convective Heat Transfer Coefficient

One of the fundamental concepts of heat transfer is the convective heat transfer coefficient, denoted as \(h\). It quantifies the heat transferred between a surface and a fluid moving past it per unit area per unit temperature difference. This value is critical because it lets us predict how efficiently heat will be exchanged in systems like radiators, heat exchangers, and even in the cooling of electronic components.

For calculating \(h\), we often use empirical correlations that consider the physical properties of the fluid and the flow conditions. In the given exercise, the Dittus-Boelter equation is utilized, which is especially suited for turbulent flow inside tubes. This equation relates \(h\) to the Nusselt number (Nu), Reynolds number (Re), and the Prandtl number (Pr), highlighting the dependency of heat transfer on the flow dynamics (Re) and thermal properties of the fluid (Pr).

To enhance understanding, students should note that while the flow velocity is often needed to calculate Re, certain scenarios, such as this exercise, are designed to focus on understanding the relationship between these dimensionless numbers without requiring explicit velocity values. Thus, emphasizing the role of concept interconnectivity and application in different scenarios ensures a more comprehensive learning experience.

Nusselt Number

The Nusselt number (Nu) is another pivotal term in heat transfer, serving as a dimensionless quantity that measures the enhancement of heat transfer through convection relative to heat transfer through pure conduction. A higher Nu indicates more effective convection, leading to improved heat transfer capabilities.

In the context of the exercise, the Nusselt number is derived from the Dittus-Boelter equation, which includes the Reynolds number to account for fluid motion and the Prandtl number which considers the fluid's thermal conductivity, density, and viscosity. Focusing on how these dimensionless numbers work in tandem to determine convection efficiency helps students grasp the complex nature of heat transfer phenomena.

One helpful tip for students is to remember that the Nusselt number can be used as a scaling factor to obtain the convective heat transfer coefficient from the thermal conductivity of the fluid. Hence, understanding the Nusselt number's role is the key to unlocking the nuances of convection analysis.

Radiation Heat Transfer

When heat is transferred through electromagnetic waves, this is known as radiation heat transfer. Unlike conduction and convection, radiation does not require a medium and can even occur in the vacuum of space. In engineering applications, such as the cooling of electronic equipment or the design of thermal insulation, understanding radiation heat transfer is vital.

Within this exercise, an important aspect of radiation heat transfer is highlighted—surface temperature and emissivity. These factors significantly influence the rate of thermal radiation emitted from an object, as seen in the Stefan-Boltzmann law application. It's crucial for students to appreciate that even a small change in surface temperature leads to a notable difference in radiated heat due to the fourth power dependence on temperature, increasing the complexity and sensitivity of thermal radiation problems.

It is also beneficial for students to understand the role of emissivity, which expresses how effectively a surface emits radiation compared to an ideal black body. Surfaces in the real world always have emissivities less than 1, influencing the heat transfer rate and the energy balance of systems involving thermal radiation.

Stefan-Boltzmann Law

The Stefan-Boltzmann law is central to understanding radiation heat transfer. It states that the power radiated per unit area of a black body is proportional to the fourth power of its absolute temperature and is represented by the equation \(P = \epsilon \sigma T^4\), where \(P\) is the radiated power per unit area, \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant, and \(T\) is the temperature in Kelvin.

In our exercise, students use this law to calculate the heat transferred by radiation from the pipe's surface. By converting temperatures from Celsius to Kelvin and applying the given emissivity value, the equation allows for the quantification of the heat transfer rate due to radiation. Understanding and applying this law is crucial since it reveals the exponential effect temperature has on radiated energy, a key concept in thermal physics and engineering.

When working through such exercises, it is recommended for students to organize their calculations and pay close attention to the units, ensuring that temperature is always in Kelvin and acknowledging the substantial impact that even slight variations in emissivity or temperature can have on the results due to the fourth power relationship.