Question

Water at an average temperature of \(180^{\circ} \mathrm{F}\) and an average velocity of \(4 \mathrm{ft} / \mathrm{s}\) flows through a thin-walled \(\frac{3}{4}\)-indiameter tube. The water is cooled by air that flows across the tube with a velocity of \(12 \mathrm{ft} / \mathrm{s}\) at an average temperature of \(80^{\circ} \mathrm{F}\). Determine the overall heat transfer coefficient.

Short answer

Question: Calculate the overall heat transfer coefficient between hot water flowing inside a tube and cooler air flowing outside the tube. Answer: To calculate the overall heat transfer coefficient (U), follow these steps: 1. Find the convection heat transfer coefficients for water (inside the tube) and air (outside the tube) using the Nusselt number and given fluid properties. 2. Determine the inner and outer surface areas of the thin-walled tube. 3. Calculate the logarithmic mean temperature difference (LMTD) using the given average water and air temperatures. 4. Use the heat transfer rate equation to relate the overall heat transfer coefficient (U) to the convection heat transfer coefficients and surface areas: \(U = \frac{1}{\frac{1}{h_{water}} + \frac{1}{h_{air}}}\) Substitute the convection heat transfer coefficients from steps 1 and 2 into this equation to find the overall heat transfer coefficient per unit length (U) for the given conditions.

Step-by-step solution

Find the convection heat transfer coefficient for water (inside the tube)

To calculate the heat transfer coefficient for water, we will use the following equation: \(h_{water} = k \frac{Nu}{D}\) Here, \(Nu\) is the Nusselt number, \(k\) is the fluid's thermal conductivity, and \(D\) is the tube's diameter. To find the Nusselt number, we will use the Sieder-Tate correlation: \(Nu = 0.027 \cdot Re^{0.8} \cdot Pr^{0.33}\) Reynolds Number (Re) and Prandtl Number (Pr) can be found using the given fluid properties (temperature and velocity). Please note that the properties of water may change slightly due to differing temperatures, so be prepared to adjust calculations accordingly. In this case, we assume constant properties for simplicity.

Find the convection heat transfer coefficient for air (outside the tube)

The process is similar to finding the heat transfer coefficient for water. We will use the Colburn analogy to find the Nusselt number for the air-cooled tubes: \(Nu = 0.023 \cdot Re^{0.8} \cdot Pr^{0.3}\) You can then find the Reynolds and Prandtl numbers for air, as we did for water, using the given temperature and velocity data.

Determine the Inner and Outer Surface Areas of the Thin-walled Tube

To find the surface areas, we need the tube's diameter and length. The diameter is given as \(\frac{3}{4}\) inches, which needs converting to feet: \(D_{inner} = \frac{3}{4} \cdot \frac{1 \mathrm{ft}}{12 \mathrm{in}}\) The outer diameter (D_outer) can be determined by assuming negligible wall thickness, so it is the same as D_inner. We will need to know the length (L) of the tube to calculate the surface area, which is not provided in the problem. We will use L as a variable in the next steps and find the heat transfer coefficients per unit length.

Calculate the Logarithmic Mean Temperature Difference

The Logarithmic Mean Temperature Difference (LMTD) is necessary to find the heat transfer rate. In this problem, we are given the average water and air temperatures, so we can use them directly to derive LMTD: \(LMTD = \frac{T_{water} - T_{air}}{\ln{\left(\frac{T_{water}}{T_{air}}\right)}}\)

Determine the Overall Heat Transfer Coefficient via the Heat Transfer Rate Equation

We can now use the heat transfer coefficients, surface areas, and LMTD to find the overall heat transfer coefficient (U) using the following heat transfer rate equation: \(Q = U \cdot A \cdot LMTD\) Where Q is the heat transfer rate, A is the surface area, and LMTD is the logarithmic mean temperature difference. However, we will not find Q explicitly because it is not required in this problem. Instead, we will express U for per unit length of the tube to solve the problem. The overall heat transfer coefficient (U) can be related to the convection heat transfer coefficients for water (inside) and air (outside) as: \(\frac{1}{U \cdot A} = \frac{1}{h_{water} \cdot A_{inner}} + \frac{1}{h_{air} \cdot A_{outer}}\) Assuming that the inner and outer surface areas are equal (since the tube is thin-walled): \(U = \frac{1}{\frac{1}{h_{water}} + \frac{1}{h_{air}}}\) By substituting the convection heat transfer coefficients from steps 1 and 2 into this equation, we can find the overall heat transfer coefficient per unit length (U) for the given conditions.

Key concepts

Convection Heat Transfer Coefficient

The convection heat transfer coefficient quantifies the ease with which heat is transferred between a fluid and a solid surface. Higher values indicate more effective heat transfer. In practical terms, this coefficient plays a critical role in designing systems where heat exchange between a fluid and solid is vital, such as in radiators, condensers, or heating elements.

For instance, in calculating the heat transfer for water inside a tube, the coefficient depends on the fluid's velocity, its properties at the given temperature, and the physical dimensions of the heat exchange surface. Due to the fluid's motion, convection can either be forced by external means, such as a pump or fan, or natural, driven by density differences within the fluid due to temperature gradients.

Nusselt Number

The Nusselt number (Nu) is a dimensionless value crucial for determining the convection heat transfer coefficient. It represents the ratio of convective to conductive heat transfer across a fluid boundary layer.

For a given set of conditions, it's influenced by the flow's character, categorized by the Reynolds and Prandtl numbers. The formulas, like the Sieder-Tate correlation, provide a method to calculate the Nusselt number for more complex situations where theoretical values become hard to predict. The Nusselt number is not a fixed value but changes with fluid properties, temperature, and flow conditions.

Logarithmic Mean Temperature Difference (LMTD)

LMTD is a measure used in heat exchanger design and analysis, assessing the average temperature driving force between the hot and cold streams across the heat exchanger. It accounts for the temperature difference variation along the heat exchanger, yielding a more accurate assessment of the overall temperature change than a simple arithmetic mean would.

Using LMTD allows for solving complex heat transfer problems where the temperatures of fluid streams can vary considerably from one end of the exchanger to the other. Recognizing how to calculate the LMTD is vital for students aiming to solve real-world engineering problems involving thermal systems.

Reynolds Number

The Reynolds number (Re) is a dimensionless quantity expressing the ratio of inertial forces to viscous forces within a fluid flow. It is employed to predict the flow pattern in different fluid flow situations, like whether the flow will be laminar or turbulent.

In heat transfer problems, the Reynolds number is used to determine the nature of the convection – whether heat transfer will be more effective due to turbulent mixing or less effective due to a smooth, laminar flow. Knowledge of the specific Reynolds number for a flow helps engineers optimize systems for the most efficient heat transfer.

Prandtl Number

The Prandtl number (Pr) is another dimensionless number in fluid dynamics and heat transfer. It represents the ratio of momentum diffusivity (viscosity) to thermal diffusivity. In layman's terms, it helps predict how a fluid's temperature field will evolve in response to its flow field.

This is particularly relevant when the fluid velocity and temperature are changing, such as around heating or cooling surfaces. For students, understanding the Prandtl number helps grasp how heat conduction and fluid motion interact with each other - crucial for engineering applications such as the design of heat exchangers and assessments of convective heat transfer performance.

Heat Transfer Rate Equation

The heat transfer rate equation, often symbolized by Q, serves as a fundamental bridge linking thermal energy movement to the prevailing physical conditions. It's articulated as the product of the heat transfer coefficient (U), the area over which the transfer occurs (A), and the LMTD.

This equation underlies most heat transfer calculations and offers a basis for comparing theoretical and experimental data. Understanding how to manipulate and apply this equation is critical for students facing real-world problems in thermal management, be it in designing cooling systems or optimizing industrial processes for better energy efficiency.