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Chapter 7: Problem 144

Jakob (1949) suggests the following correlation be used for square tubes in a liquid cross-flow situation: $$ \mathrm{Nu}=0.102 \mathrm{Re}^{0.675} \mathrm{Pr}^{1 / 3} $$ Water \((k=0.61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=6)\) at \(50^{\circ} \mathrm{C}\) flows across a \(1-\mathrm{cm}\) square tube with a Reynolds number of 10,000 and surface temperature of \(75^{\circ} \mathrm{C}\). If the tube is \(2 \mathrm{~m}\) long, the rate of heat transfer between the tube and water is (a) \(6.0 \mathrm{~kW}\) (b) \(8.2 \mathrm{~kW}\) (c) \(11.3 \mathrm{~kW}\) (d) \(15.7 \mathrm{~kW}\) (e) \(18.1 \mathrm{~kW}\)

Short Answer

Expert verified
Question: A 2-meter long, 1-cm side square tube is placed inside water with a Reynolds number of 10,000 and a Prandtl number of 6. The thermal conductivity of the water is given as 0.6 W/mK. Calculate the rate of heat transfer between the tube and the water using the following correlation from Jakob (1949) for square tubes in liquid cross-flow situations: Nu = 0.102 × Re^0.675 × Pr^(1/3). Answer: Follow the steps below to find the rate of heat transfer between the tube and water: 1. Calculate the Nusselt number (Nu) using the given correlation: Nu = 0.102 × (10,000)^0.675 × (6)^(1/3) Nu ≈ 394.16 2. Calculate the convective heat transfer coefficient (h) using the Nu, thermal conductivity (k), and characteristic length (L): h = (394.16 × 0.6) / 0.01 h ≈ 23649.6 W/m²K 3. Calculate the surface area (A) of the square tube: A = 4 × (0.01) × (2) A = 0.08 m² 4. Calculate the heat transfer rate (Q) using the convective heat transfer coefficient (h), surface area (A), and the temperature difference (ΔT) between the tube surface and the water (assuming a constant ΔT): Q = 23649.6 × 0.08 × ΔT Q = 1891.97 × ΔT W The heat transfer rate between the tube and water depends on the temperature difference (ΔT) between the tube surface and the water, and is given by Q = 1891.97 × ΔT W.

Step by step solution

01

Calculate the Nusselt number (Nu)

Using the given correlation and the known values for the Reynolds and Prandtl numbers, calculate the Nusselt number (Nu): $$ \mathrm{Nu} = 0.102 \times \mathrm{Re}^{0.675} \times \mathrm{Pr}^{1/3} $$ Substituting the known values: $$ \mathrm{Nu} = 0.102 \times (10,000)^{0.675} \times (6)^{1/3} $$
02

Calculate the convective heat transfer coefficient (h)

Using the Nusselt number (Nu), the thermal conductivity of the fluid (k), and the characteristic length (L): $$ h = \frac{\mathrm{Nu} \times k}{L} $$ The characteristic length (L) is equal to the side length of the square tube, which is given as 1 cm (0.01 m). Using the calculated Nusselt number and the given thermal conductivity, calculate the convective heat transfer coefficient (h).
03

Calculate the heat transfer rate (Q)

Using the convective heat transfer coefficient (h), the surface area (A), and the temperature difference (ΔT) between the tube surface and the water, calculate the heat transfer rate (Q): $$ Q = h \times A \times \Delta T $$ The surface area (A) of the square tube can be calculated as: $$ A = P \times L = 4 \times \text{(side length)} \times \text{(tube length)} $$ Substitute the known values and solve for the heat transfer rate (Q), then compare the result with the given options to choose the correct answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nusselt Number
The Nusselt number (Nu) is a dimensionless parameter of paramount importance in the study of convective heat transfer. It is essentially a comparative measure of convective to conductive heat transfer at a boundary in a fluid flow. The higher the Nusselt number, the more efficient the convective heat transfer relative to conduction. The Nusselt number is calculated using certain key physical properties and flow characteristics of the fluid, such as the Reynolds number and the Prandtl number.

For our example exercise, Jakob's correlation formula provides a specific way to calculate the Nusselt number for water flowing over a square tube. By inserting the given Reynolds number (Re) for the flow velocity and dimensions, and the Prandtl number (Pr) representing the physical properties of water at the specified temperature, we can determine the Nusselt number, which is crucial to finding the heat transfer coefficient.
Reynolds Number
The Reynolds number (Re) is another dimensionless number used to predict flow patterns in different fluid flow situations. It expresses the ratio of inertial forces to viscous forces and is key in determining the regime of the flow, whether it's laminar or turbulent. Laminar flow leads to a lower heat transfer rate compared to turbulent flow, which significantly enhances the mixing and therefore, the rate of convective heat transfer.

In the given problem, we have a Reynolds number of 10,000, which suggests that the flow is turbulent. Turbulent flow tends to have higher Nusselt numbers, implying enhanced convection. Knowing the right Reynolds number for the fluid crossing over the tube is crucial for applying the correlation formula accurately to calculate the Nusselt number.
Prandtl Number
The Prandtl number (Pr) is a dimensionless number named after the German physicist Ludwig Prandtl. This figure represents the ratio of momentum diffusivity (kinematic viscosity) to thermal diffusivity. It essentially compares how quickly momentum and thermal energy are transmitted through the fluid.

In our exercise, the Prandtl number is given as 6 for water at a temperature of 50 degrees Celsius. This number plays a significant role in determining the convective heat transfer for the given situation, as seen in Jakob's correlation. The Prandtl number makes it clear how the fluid properties at a specific temperature affect heat transfer, and it must be combined with the Reynolds number to compute the Nusselt number effectively.
Heat Transfer Coefficient
The heat transfer coefficient (h) is a measurable property that quantifies the convective heat transfer between a solid boundary and the adjacent fluid. It reflects the ease with which heat is transferred from the solid surface to the fluid medium or vice versa. In essence, a higher heat transfer coefficient indicates that the fluid is more effective at absorbing heat from the surface or giving it off.

Once the Nusselt number has been calculated, the heat transfer coefficient can be found using the relationship between these two, the thermal conductivity (k), and the characteristic length of the surface (L). In the square tube example involving water, the coefficient is derived to find out how effectively the tube can transfer heat to the moving water at the set conditions.
Heat Transfer Rate
The heat transfer rate (Q) describes the amount of thermal energy transferred per unit time, typically measured in watts (W) or kilowatts (kW). It is affected by various factors, including the heat transfer coefficient, the surface temperature of the heat exchanging body, and the temperature of the moving fluid. Mathematically, it is expressed as the product of the heat transfer coefficient, the surface area for heat exchange, and the temperature difference between the two exchanging mediums.

The final step of our problem requires calculating this rate using the heat transfer coefficient derived from the Nusselt number, the surface area of the tube, and the temperature difference between the tube surface and the water. By doing so, we can determine the actual heat being transferred in kilowatts, which is a practical and measurable output that engineers and scientists use to analyze and design thermal systems.

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Most popular questions from this chapter

A heated long cylindrical rod is placed in a cross flow of air at \(20^{\circ} \mathrm{C}(1 \mathrm{~atm})\) with velocity of \(10 \mathrm{~m} / \mathrm{s}\). The rod has a diameter of \(5 \mathrm{~mm}\) and its surface has an emissivity of \(0.95\). If the surrounding temperature is \(20^{\circ} \mathrm{C}\) and the heat flux dissipated from the rod is \(16000 \mathrm{~W} / \mathrm{m}^{2}\), determine the surface temperature of the rod. Evaluate the air properties at \(70^{\circ} \mathrm{C}\).

Liquid mercury at \(250^{\circ} \mathrm{C}\) is flowing with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\) in parallel over a \(0.1-\mathrm{m}\)-long flat plate where there is an unheated starting length of \(5 \mathrm{~cm}\). The heated section of the flat plate is maintained at a constant temperature of \(50^{\circ} \mathrm{C}\). Determine \((a)\) the local convection heat transfer coefficient at the trailing edge, \((b)\) the average convection heat transfer coefficient for the heated section, and \((c)\) the rate of heat transfer per unit width for the heated section.

What does the friction coefficient represent in flow over a flat plate? How is it related to the drag force acting on the plate?

Consider a person who is trying to keep cool on a hot summer day by turning a fan on and exposing his entire body to air flow. The air temperature is \(85^{\circ} \mathrm{F}\) and the fan is blowing air at a velocity of \(6 \mathrm{ft} / \mathrm{s}\). If the person is doing light work and generating sensible heat at a rate of \(300 \mathrm{Btu} / \mathrm{h}\), determine the average temperature of the outer surface (skin or clothing) of the person. The average human body can be treated as a 1-ft-diameter cylinder with an exposed surface area of \(18 \mathrm{ft}^{2}\). Disregard any heat transfer by radiation. What would your answer be if the air velocity were doubled? Evaluate the air properties at \(100^{\circ} \mathrm{F}\).

What is drag? What causes it? Why do we usually try to minimize it?
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