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Chapter 6: Problem 97

Evaluate the Prandtl number from the following data: \(c_{p}=0.5 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}, k=2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot \mathrm{R}, \mu=0.3 \mathrm{lbm} / \mathrm{ft} \cdot \mathrm{s}\).

Short Answer

Expert verified
Question: Calculate the Prandtl number using the following data: specific heat at constant pressure (\(c_p\)) is 0.5 Btu/lbm路R, thermal conductivity (\(k\)) is 2 Btu/h路ft路R, and dynamic viscosity (\(\mu\)) is 0.3 lbm/ft路s. Answer: The Prandtl number for the given data is approximately 276.47.

Step by step solution

01

Convert the units

First, we'll convert the given units to SI units. For \(c_p\), we'll convert Btu/lbm路R to J/kg路K (1 Btu = 1055.06 J, 1 lbm = 0.453592 kg, and (9/5) R = 1 K): $$c_{p}=0.5\frac{\mathrm{Btu}}{\mathrm{lbm}\cdot\mathrm{R}} \times \frac{1055.06\mathrm{J}}{\mathrm{Btu}} \times \frac{0.453592\mathrm{kg}}{\mathrm{lbm}} \times \frac{9}{5}$$ For \(k\), we'll convert Btu/h路ft路R to W/m路K (1 Btu/h路ft路R = 1.73073 W/m路K and (9/5) R = 1 K): $$k=2\frac{\mathrm{Btu}}{\mathrm{h}\cdot\mathrm{ft}\cdot\mathrm{R}} \times \frac{1.73073\mathrm{W}}{\mathrm{m}\cdot\mathrm{K}}$$ For \(\mu\), we'll convert lbm/ft路s to kg/m路s (1 lbm = 0.453592 kg, and 1 ft = 0.3048 m): $$\mu=0.3\frac{\mathrm{lbm}}{\mathrm{ft}\cdot\mathrm{s}} \times \frac{0.453592\mathrm{kg}}{\mathrm{lbm}} \times \frac{1}{0.3048\mathrm{m}}$$
02

Calculate the values in SI units

Now we'll calculate the values of \(c_p\), \(k\), and \(\mu\) in SI units: $$c_p = 0.5 \times 1055.06 \times 0.453592 \times \frac{9}{5} = 2140.52\, \mathrm{J/kg\cdot K}$$ $$k = 2 \times 1.73073 = 3.46146\, \mathrm{W/m\cdot K}$$ $$\mu = 0.3 \times 0.453592 \times \frac{1}{0.3048} = 0.4476\, \mathrm{kg/m\cdot s}$$
03

Calculate the Prandtl number

Using the formula, we can now calculate the Prandtl number: $$Pr = \frac{c_p \cdot \mu}{k} = \frac{2140.52 \cdot 0.4476}{3.46146}$$ Calculate the value of the Prandtl number: $$Pr \approx 276.47$$ The Prandtl number for the given data is approximately 276.47.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity measures how easily heat flows through a material due to a temperature gradient. It is crucial in applications involving heat transfer. In simpler terms, it describes how well a material conducts heat.

Thermal conductivity (\( k \)) has units of watts per meter Kelvin (W/m路K) in the International System of Units (SI). Higher values indicate a material that is more efficient at conducting heat. A high thermal conductivity means that heat travels rapidly through the material.
  • Metals such as copper and silver have high thermal conductivity, making them excellent conductors of heat.
  • Materials like wood and rubber have low thermal conductivity, which is why they often act as insulators.
Understanding thermal conductivity helps us manage and optimize processes in engineering and environmental science. Whether designing thermal insulation in buildings or gadgets that efficiently dissipate heat, it's a key parameter.
Specific Heat Capacity
Specific heat capacity is the amount of heat required to change the temperature of one kilogram of a substance by one Kelvin. It is a crucial property used in thermodynamics, as it affects how substances heat up and cool down.

In SI units, specific heat capacity (\( c_p \)) is measured in joules per kilogram Kelvin (J/kg路K). This concept is essential in calculating energy changes in physical and chemical processes.
  • Water has a high specific heat capacity, which means it requires more energy to change its temperature. This makes it useful for cooling systems and thermal storage applications.
  • Metals generally have lower specific heat capacity compared to water. They heat up and cool down faster.
Designing systems for heating, cooling, or energy storage often involves calculating the specific heat capacity. This ensures efficiency and control in processes requiring thermal management.
Viscosity
Viscosity is the measure of a fluid's resistance to deformation or flow. In essence, it determines how "thick" or "thin" a fluid is. Understanding viscosity is crucial in engineering and fluid dynamics.

The unit of dynamic viscosity (\( \mu \)) is the Pascal-second (kg/m路s) in SI units. Viscosity affects how substances are mixed and transported. It plays an essential role in applications like lubrication, pipeline flow, and food processing.
  • Liquids like honey and syrup have high viscosity, meaning they flow slowly.
  • Fluids like water and air have low viscosity, allowing them to flow easily.
  • Increased temperature often reduces a liquid's viscosity, causing it to flow more easily.
Controlling viscosity is vital for optimizing processes in various industries, from maintaining the proper flow of engine oils in vehicles to ensuring the quality of food products.
Unit Conversion
Unit conversion involves changing a quantity from one unit of measurement to another. This process is mandatory for comparing and using data consistently across various systems.

Understanding unit conversion is especially important in fields like science and engineering, where measurements in different units need to be compared.
  • Converting from one temperature scale to another involves understanding the relationship between units, like Celsius, Fahrenheit, and Kelvin.
  • Length, mass, and volume conversions are common when working with international standards or comparing ancient historical data.
  • When calculating energy, work, or power, conversions between systems like imperial and SI are necessary for accurate results.
Mastering these conversions ensures clarity and accuracy, forming the basis of quantitative science and engineering.

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Most popular questions from this chapter

Air at \(5^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), is used for cooling metal plates coming out of a heat treatment oven at an initial temperature of \(300^{\circ} \mathrm{C}\). The plates \((k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=2800 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) ) have a thickness of \(10 \mathrm{~mm}\). Using EES (or other) software, determine the effect of cooling time on the temperature gradient in the metal plates at the surface. By varying the cooling time from 0 to \(3000 \mathrm{~s}\), plot the temperature gradient in the plates at the surface as a function of cooling time. Hint: Use the lumped system analysis to calculate the plate surface temperature. Make sure to verify the application of this method to this problem.

Liquid water at \(15^{\circ} \mathrm{C}\) is flowing over a \(0.3\)-m-wide plate at \(65^{\circ} \mathrm{C}\) a velocity of \(3.0 \mathrm{~m} / \mathrm{s}\). Using EES, Excel, or other comparable software, plot (a) the hydrodynamic boundary layer and \((b)\) the thermal boundary layer as a function of \(x\) on the same graph for the range of \(x=0.0 \mathrm{~m}\) to \(x=x_{\text {cr. }}\) Use a critical Reynolds number of 500,000 .

Friction coefficient of air flowing over a flat plate is given as \(C_{f}=0.664(V x / \nu)^{-0.5}\), where \(x\) is the location along the plate. Using EES (or other) software, determine the effect of the air velocity \((V)\) on the wall shear stress \(\left(\tau_{w}\right)\) at the plate locations of \(x=0.5 \mathrm{~m}\) and \(1 \mathrm{~m}\). By varying the air velocity from \(0.5\) to \(6 \mathrm{~m} / \mathrm{s}\) with increments of \(0.5 \mathrm{~m} / \mathrm{s}\), plot the wall shear stress as a function of air velocity at \(x=0.5 \mathrm{~m}\) and \(1 \mathrm{~m}\). Evaluate the air properties at \(20^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

A rectangular bar with a characteristic length of \(0.5 \mathrm{~m}\) is placed in a free stream flow where the convection heat transfer coefficients were found to be \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) when the free stream velocities were \(25 \mathrm{~m} / \mathrm{s}\) and \(5 \mathrm{~m} / \mathrm{s}\), respectively. If the Nusselt number can be expressed as \(\mathrm{Nu}=C \operatorname{Re}^{m} \operatorname{Pr}^{n}\), where \(C, m\), and \(n\) are constants, determine the convection heat transfer coefficients for similar bars with (a) \(L=1 \mathrm{~m}\) and \(V=5 \mathrm{~m} / \mathrm{s}\), and \((b) L=2 \mathrm{~m}\) and \(V=50 \mathrm{~m} / \mathrm{s}\).

For the same initial conditions, one can expect the laminar thermal and momentum boundary layers on a flat plate to have the same thickness when the Prandtl number of the flowing fluid is (a) Close to zero (b) Small (c) Approximately one (d) Large (e) Very large
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