/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 143 Ambient air at \(20^{\circ} \mat... [FREE SOLUTION] | 91Ó°ÊÓ

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Ambient air at \(20^{\circ} \mathrm{C}\) flows over a 30-cm-diameter hot spherical object with a velocity of \(2.5 \mathrm{~m} / \mathrm{s}\). If the average surface temperature of the object is \(200^{\circ} \mathrm{C}\), the average convection heat transfer coefficient during this process is \(\begin{array}{ll}\text { (a) } 5.0 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & \text { (b) } 6.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\end{array}\) (c) \(7.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (d) \(9.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (e) \(11.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (For air, use \(k=0.2514 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \mathrm{Pr}=0.7309, v=1.516 \times\) \(\left.10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \mu_{s}=1.825 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, \mu_{s}=2.577 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right)\)

Short Answer

Expert verified
a) 3.5 W/m·K b) 6.2 W/m·K c) 7.5 W/m·K d) 9.3 W/m·K Answer: c) 7.5 W/m·K

Step by step solution

01

Convert air temperatures to Kelvin and calculate Reynold's number

First, we need to convert the given air temperature and object surface temperature to Kelvin. \(T_\infty = 20^{\circ}\mathrm{C} + 273.15 = 293.15\ \mathrm{K}\) \(T_s = 200^{\circ}\mathrm{C} + 273.15 = 473.15\ \mathrm{K}\) Next, we calculate the Reynolds number based on the diameter of the spherical object. \(Re = \frac{\rho V D}{\mu}\) Since we have air velocity (V), sphere diameter (D), and dynamic viscosity of air (\(\mu\)), we can calculate the Reynolds number. \(Re = \frac{2.5 \mathrm{~m} / \mathrm{s} * 0.3\ \mathrm{m}}{1.825 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}} = 41,095\) #Step 2: Calculate the Prandtl number#
02

Calculate the Prandtl number

We are given the Prandtl number in the exercise, so we have it directly: \(Pr = 0.7309\) #Step 3: Use the Dittus-Boelter equation to find the Nusselt number#
03

Calculate the Nusselt number using the Dittus-Boelter equation

For forced convection, we can use the Dittus-Boelter equation: $Nu = \begin{cases} 0.023 Re^{0.8} Pr^{0.3}, & \text{for heating}\\ 0.023 Re^{0.8} Pr^{0.4}, & \text{for cooling} \end{cases}$ In our case, we have a heating scenario, so we will use the heating formula: \(Nu = 0.023 \times 41095^{0.8} \times 0.7309^{0.3} \approx 350.4\) #Step 4: Calculate the convection heat transfer coefficient#
04

Calculate the convection heat transfer coefficient

Finally, we calculate the convection heat transfer coefficient (h) using the Nusselt number and the thermal conductivity of air (k): \(h = \frac{Nu\cdot k}{D}\) \(h = \frac{350.4 \times 0.2514\ \mathrm{W} / \mathrm{m} \cdot \mathrm{K}}{0.3\ \mathrm{m}} \approx 7.5\ \mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) Therefore, the average convection heat transfer coefficient during this process is approximately \(7.5\ \mathrm{W} / \mathrm{m} \cdot \mathrm{K}\), which corresponds to option (c).

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Key Concepts

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

Reynolds Number
The Reynolds number is a dimensionless quantity used in fluid mechanics to predict flow patterns in different fluid flow situations. It compares the inertial forces to the viscous forces within a fluid, and is defined by the formula \( Re = \frac{\rho V D}{\mu} \) where \( \rho \) is the density of the fluid, \( V \) is the velocity, \( D \) is the characteristic linear dimension (e.g., diameter of a cylinder), and \( \mu \) is the dynamic viscosity of the fluid. High Reynolds numbers indicate turbulent flow, while low numbers suggest laminar flow. In the context of our exercise, the Reynolds number lets us characterize the air flow over the hot spherical object and determine whether the proper equations for forced convection can be applied. For the given problem with air flowing over a spherical object, the value calculated was \( Re = 41,095 \) which suggests a turbulent flow regime.

Understanding Reynolds number is crucial for predicting the nature of the fluid flow and for choosing the right equations to calculate heat transfer coefficients in convection.
Prandtl Number
The Prandtl number (Pr) is another dimensionless quantity, named after the German physicist Ludwig Prandtl. It signifies the relative thickness of the momentum and thermal boundary layers. Mathematically, it's given by \( Pr = \frac{u}{\alpha} \) where \( u \) is the kinematic viscosity and \( \alpha \) is the thermal diffusivity. For the case at hand, \( Pr \) is given as 0.7309, indicating a certain relation between the momentum and thermal diffusivity for air at the specified conditions. This parameter is essential when calculating heat transfer in fluids when there is motion (convection) because it factors into correlations that yield the Nusselt number, hence affecting the heat transfer rate.

The Prandtl number, like other dimensionless numbers, allows engineers and scientists to predict the outcome of heat transfer phenomena in varying scenarios without the need to conduct new experiments for each unique case.
Nusselt Number
The Nusselt number (Nu) is a dimensionless parameter representing the enhancement of heat transfer through a fluid as a result of convection relative to conduction across a boundary. The Nusselt number is defined as \( Nu = \frac{hL}{k} \) where \( h \) is the convective heat transfer coefficient of the fluid, \( L \) is a characteristic length, and \( k \) is the thermal conductivity of the fluid. It's a crucial tool for analyzing convection and is often determined using empirical correlations, like the Dittus-Boelter equation shown in our problem. In our specific exercise, the Nusselt number was found to be approximately 350.4, signifying the degree of convective heat transfer relative to conductive heat transfer taking place from the hot spherical object to the ambient air.

The Nusselt number is vital as it directly relates to the intensity of the heat transfer occurring by convection, which allows us to compute the average convection heat transfer coefficient with the known thermal conductivity.
Dittus-Boelter Equation
The Dittus-Boelter equation is an empirical relationship used to estimate the Nusselt number for turbulent flow in tubes or over surfaces. The equation is typically presented in the form \( Nu = 0.023 Re^{0.8} Pr^{n} \) where \( n \) is 0.3 for heating and 0.4 for cooling scenarios. The equation is derived from experimental observations and provides a convenient method to calculate the convective heat transfer coefficient when dealing with forced convection. In our particular problem, since we're dealing with heating, the equation simplifies to \( Nu = 0.023 Re^{0.8} Pr^{0.3} \) and when the values for the Reynolds and Prandtl numbers are substituted, we obtain the Nusselt number used to calculate the heat transfer coefficient. The Dittus-Boelter equation is helpful because it provides a relatively simple way to estimate a complex process that would otherwise require advanced numerical methods to describe.
Forced Convection
Forced convection is a mechanism of heat transfer in which fluid motion is generated by an external source, like pumps or fans, leading to the increase in heat transfer between the fluid and a surface. Unlike natural convection where the fluid motion is due to buoyancy effects, forced convection provides a more controllable and often more efficient mode of heat transfer. This principle is applied when designing systems like radiators, heat exchangers, and air-cooling systems. The appropriate calculation of heat transfer coefficients in forced convection scenarios relies on understanding the flow dynamics, encapsulated by the Reynolds, Prandtl, and Nusselt numbers. Our example problem features forced convection as the air moving over the hot spherical object is driven by its velocity, not temperature differences.

Recognizing the forced convection scenario is key, as it informs our choice of heat transfer correlation – in our case, we selected the Dittus-Boelter equation due to the turbulent flow indicated by the calculated Reynolds number.

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Most popular questions from this chapter

Air at \(15^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) flows over a \(0.3\)-m-wide plate at \(65^{\circ} \mathrm{C}\) at a velocity of \(3.0 \mathrm{~m} / \mathrm{s}\). Compute the following quantities at \(x=x_{\mathrm{cr}}\) : (a) Hydrodynamic boundary layer thickness, \(\mathrm{m}\) (b) Local friction coefficient (c) Average friction coefficient (d) Total drag force due to friction, \(\mathrm{N}\) (e) Local convection heat transfer coefficient, \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (f) Average convection heat transfer coefficient, W/m² \(\cdot \mathrm{K}\) (g) Rate of convective heat transfer, W

Steam at \(250^{\circ} \mathrm{C}\) flows in a stainless steel pipe \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(4 \mathrm{~cm}\) and \(4.6 \mathrm{~cm}\), respectively. The pipe is covered with \(3.5-\mathrm{cm}-\) thick glass wool insulation \((k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose outer surface has an emissivity of \(0.3\). Heat is lost to the surrounding air and surfaces at \(3^{\circ} \mathrm{C}\) by convection and radiation. Taking the heat transfer coefficient inside the pipe to be \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the steam per unit length of the pipe when air is flowing across the pipe at \(4 \mathrm{~m} / \mathrm{s}\). Evaluate the air properties at a film temperature of \(10^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Oil at \(60^{\circ} \mathrm{C}\) flows at a velocity of \(20 \mathrm{~cm} / \mathrm{s}\) over a \(5.0\)-m-long and \(1.0-\mathrm{m}\)-wide flat plate maintained at a constant temperature of \(20^{\circ} \mathrm{C}\). Determine the rate of heat transfer from the oil to the plate if the average oil properties are: \(\rho=880 \mathrm{~kg} / \mathrm{m}^{3}, \mu=0.005 \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\).

Air at \(20^{\circ} \mathrm{C}\) flows over a 4-m-long and 3-m-wide surface of a plate whose temperature is \(80^{\circ} \mathrm{C}\) with a velocity of \(5 \mathrm{~m} / \mathrm{s}\). The rate of heat transfer from the surface is (a) \(7383 \mathrm{~W}\) (b) \(8985 \mathrm{~W}\) (c) \(11,231 \mathrm{~W}\) (d) 14,672 W (e) \(20,402 \mathrm{~W}\) (For air, use \(k=0.02735 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7228, \nu=1.798 \times\) \(\left.10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\)

Air at \(25^{\circ} \mathrm{C}\) flows over a 5 -cm-diameter, \(1.7\)-m-long pipe with a velocity of \(4 \mathrm{~m} / \mathrm{s}\). A refrigerant at \(-15^{\circ} \mathrm{C}\) flows inside the pipe and the surface temperature of the pipe is essentially the same as the refrigerant temperature inside. Air properties at the average temperature are \(k=0.0240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.735\), \(\nu=1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). The rate of heat transfer to the pipe is (a) \(343 \mathrm{~W}\) (b) \(419 \mathrm{~W}\) (c) \(485 \mathrm{~W}\) (d) \(547 \mathrm{~W}\) (e) \(610 \mathrm{~W}\)

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