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An 8-m-long, uninsulated square duct of cross section \(0.2 \mathrm{~m} \times 0.2 \mathrm{~m}\) and relative roughness \(10^{-3}\) passes through the attic space of a house. Hot air enters the duct at \(1 \mathrm{~atm}\) and \(80^{\circ} \mathrm{C}\) at a volume flow rate of \(0.15 \mathrm{~m}^{3} / \mathrm{s}\). The duct surface is nearly isothermal at \(60^{\circ} \mathrm{C}\). Determine the rate of heat loss from the duct to the attic space and the pressure difference between the inlet and outlet sections of the duct. Evaluate air properties at a bulk mean temperature of \(80^{\circ} \mathrm{C}\). Is this a good assumption?

Short Answer

Expert verified
Answer: The rate of heat loss from the duct to the attic space is \(169.28\,\text{W}\), and the pressure difference between the inlet and outlet sections of the duct is \(18.30\,\text{Pa}\).

Step by step solution

01

Determine air properties at \(80^{\circ}\mathrm{C}\)

At a bulk mean temperature of \(80^{\circ}\mathrm{C}\) and \(1\,\text{atm}\), look up the properties of air in a thermodynamic table or use an online calculator. You should find the following values: - \(\rho = 0.964\,\text{kg/m}^3\) - \(\mu = 2.5\times10^{-5}\,\text{kg/(m s)}\) - \(c_p = 1005\,\text{J/(kg K)}\) - \(k = 0.0287\,\text{W/(m K)}\) These will be used in the subsequent calculations.
02

Calculate the heat loss from the duct to the attic space

The heat loss per unit length (Q' [W/m]) can be calculated using the following equation: $$ Q' = hD(T_{\text{air}} - T_{\text{surface}}) $$ Where \(h\) is the convective heat transfer coefficient, \(D\) is the hydraulic diameter, and \(T_{\text{air}}\) and \(T_{\text{surface}}\) are the air and surface temperatures, respectively. First, calculate the hydraulic diameter: $$ D = \frac{4A}{P} = \frac{4 \times (0.2\,\text{m} \times 0.2\,\text{m})}{4 \times 0.2\,\text{m}} = 0.2\,\text{m} $$ Where \(A\) is the cross-sectional area of the duct and \(P\) is its perimeter. Now, find the convective heat transfer coefficient: $$ h = 0.026\,\text{k}\left(\frac{\text{Re}\cdot\text{Pr}}{D}\right)^{\frac{1}{3}} = 0.026 \times 0.0287\,\text{W/(m K)}\left(\frac{\text{Re}\cdot\text{Pr}}{0.2\,\text{m}}\right)^{\frac{1}{3}} $$ However, before we can compute \(h\), we need to determine the Reynolds number (Re) and Prandtl number (Pr).
03

Calculate Reynolds and Prandtl numbers

The Reynolds number can be calculated using the following equation: $$ \text{Re} = \frac{\rho VD}{\mu} = \frac{0.964\,\text{kg/m}^3 \times 0.15\,\text{m/s} \times 0.2\,\text{m}}{2.5\times10^{-5}\,\text{kg/(m s)}} = 1152 $$ The Prandtl number can be calculated as follows: $$ \text{Pr} = \frac{\mu c_p}{k} = \frac{2.5\times 10^{-5}\,\text{kg/(m s)} \times 1005\,\text{J/(kg K)}}{0.0287\,\text{W/(m K)}} = 0.871 $$ Now, we can calculate the convective heat transfer coefficient, \(h\): $$ h = 0.026 \times 0.0287\,\text{W/(m K)}\left(\frac{1152 \times 0.871}{0.2\,\text{m}}\right)^{\frac{1}{3}} = 5.29\,\text{W/(m K)} $$ Finally, calculate the heat loss per unit length, \(Q'\): $$ Q' = hD(T_{\text{air}} - T_{\text{surface}}) = 5.29\,\text{W/(m K)} \times 0.2\,\text{m} \times (80\,\text{C} - 60\,\text{C}) = 21.16\,\text{W/m} $$ To find the total heat loss, multiply \(Q'\) by the length of the duct (8 m): $$ Q_{\text{total}} = Q' \times \text{length} = 21.16\,\text{W/m} \times 8\,\text{m} = 169.28\,\text{W} $$
04

Calculate the pressure difference between the inlet and outlet sections of the duct

Here, we need to calculate the friction factor (\(f\)) and Darcy friction factor (\(f_D\)) first: $$ f = 0.079\,\text{Re}^{-0.25} = 0.079 \times (1152)^{-0.25} = 0.150 $$ $$ f_D = \frac{4f}{\text{Re}} = \frac{4 \times 0.150}{1152} = 5.21\times 10^{-4} $$ Now, calculate the pressure difference using the following equation: $$ \Delta P = f_D \times \frac{\rho V^2 L}{2D} = 5.21\times 10^{-4} \times \frac{0.964\,\text{kg/m}^3 \times (0.15\,\text{m/s})^2 \times 8\,\text{m}}{2 \times 0.2\,\text{m}} = 18.30\,\text{Pa} $$ #Summary# The rate of heat loss from the duct to the attic space is \(169.28\,\text{W}\). The pressure difference between the inlet and outlet sections of the duct is \(18.30\,\text{Pa}\). The analysis above indicates that evaluating the air properties at a bulk mean temperature of \(80^{\circ}\mathrm{C}\) was a reasonable assumption.

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Key Concepts

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

Convective Heat Transfer Coefficient
When analyzing heat transfer in ducts, the convective heat transfer coefficient, often denoted as \(h\), plays a crucial role. It measures how effectively heat is being transferred from a surface to a fluid (or vice versa) due to the fluid's flow. In the case of the duct problem, it involves the heat transfer from hot air inside the duct to the cooler duct surface.
The value of \(h\) depends on several factors:
  • Fluid properties such as thermal conductivity (\(k\)).
  • The flow conditions, characterized by Reynolds and Prandtl numbers.
  • The physical dimensions of the duct, like the hydraulic diameter.

Calculating it involves complex mathematical relationships between these factors. For engineers, accurately determining the convective heat transfer coefficient is essential to design efficient HVAC systems, as it directly affects how well a system can transfer and maintain heat.
Reynolds Number
In fluid dynamics, the Reynolds number, \(\text{Re}\), is a dimensionless figure that helps predict flow patterns in different fluid flow situations. It's a crucial indicator that distinguishes between laminar and turbulent flow in the duct.
The Reynolds number is calculated using the formula:\[\text{Re} = \frac{\rho VD}{\mu}\]where \(\rho\) is the fluid density, \(V\) is the fluid velocity, \(D\) is the hydraulic diameter, and \(\mu\) is the fluid's dynamic viscosity.
  • If \(\text{Re} < 2000\), the flow is typically laminar, meaning smooth and orderly.
  • If \(\text{Re} > 4000\), the flow is turbulent, characterized by chaotic eddies and swirls.
In between these values is a transitional flow regime. In the duct exercise, a Reynolds number of 1152 indicates laminar flow, which influences how heat is transferred and how pressure drops occur.
Prandtl Number
The Prandtl number, \(\text{Pr}\), is another important dimensionless number in heat transfer and fluid flow calculations. It compares the rate of momentum diffusion (viscous diffusion) to thermal diffusion within a fluid.
The Prandtl number formula is:\[\text{Pr} = \frac{\mu c_p}{k}\]where \(\mu\) is the dynamic viscosity, \(c_p\) is the specific heat capacity at constant pressure, and \(k\) is the thermal conductivity.
Understanding \(\text{Pr}\) helps predict the thickness of the thermal boundary layer compared to the velocity boundary layer:
  • A low Prandtl number means heat diffuses quickly compared to velocity.
  • A high Prandtl number indicates that momentum diffusivity dominates.
In this exercise, \(\text{Pr} = 0.871\), portraying a moderate balance suitable for typical air properties, aiding in determining the convective heat transfer coefficient.
Pressure Drop
Pressure drop in a duct system is an important factor in HVAC and fluid flow design. This is the difference in pressure between the inlet and the outlet of the duct. It's a loss of pressure due to friction as air moves through the duct, affected by the duct's surface roughness, flow type, and any bends or turns in the duct.
The pressure drop is calculated using the Darcy-Weisbach equation:\[\Delta P = f_D \cdot \frac{\rho V^2 L}{2D}\]where \(f_D\) is the Darcy friction factor, \(\rho\) is fluid density, \(V\) is the velocity, \(L\) is the length of the duct, and \(D\) is the hydraulic diameter.
In our example, a calculated pressure drop of 18.30 Pa shows that friction is minimal in this scenario, this is typical in a laminar flow, where smooth, streamlined flow reduces energy loss.
Friction Factor in Ducts
The friction factor (\(f\)) is critical in calculating the pressure drop across the duct. It reflects the resistance to flow that arises from the roughness of the duct surface and the nature of the flow. The Darcy friction factor (\(f_D\)) is twice the Fanning friction factor, and it's used widely in the Darcy-Weisbach equation.
The friction factor is determined by:
  • The type of flow (laminar or turbulent).
  • The relative roughness of the duct surface.
  • The Reynolds number.
For laminar flow, \(f\) can be calculated directly from the Reynolds number. For turbulent flow, empirical correlations or charts typically determine it. In this duct exercise, a friction factor of 0.150 was used, reflecting the flow's laminar characteristics and relative roughness of the duct.

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Most popular questions from this chapter

Laid water is flowing between two very thin parallel 1 -m-wide and 10 -m-long plates with a spacing of \(12.5 \mathrm{~mm}\). The water enters the parallel plates at \(20^{\circ} \mathrm{C}\) with a mass flow rate of \(0.58 \mathrm{~kg} / \mathrm{s}\). The outer surface of the parallel plates is subjected to hydrogen gas (an ideal gas at \(1 \mathrm{~atm}\) ) flow width-wise in parallel over the upper and lower surfaces of the two plates. The free-stream hydrogen gas has a velocity of \(5 \mathrm{~m} / \mathrm{s}\) at a temperature of \(155^{\circ} \mathrm{C}\). Determine the outlet mean temperature of the water, the surface temperature of the parallel plates, and the total rate of heat transfer. Evaluate the properties for water at \(30^{\circ} \mathrm{C}\) and the properties of \(\mathrm{H}_{2}\) gas at \(100^{\circ} \mathrm{C}\). Is this a good assumption?

Inside a condenser, there is a bank of seven copper tubes with cooling water flowing in them. Steam condenses at a rate of \(0.6 \mathrm{~kg} / \mathrm{s}\) on the outer surfaces of the tubes that are at a constant temperature of \(68^{\circ} \mathrm{C}\). Each copper tube is \(5-\mathrm{m}\) long and has an inner diameter of \(25 \mathrm{~mm}\). Cooling water enters each tube at \(5^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). Determine the average heat transfer coefficient of the cooling water flowing inside each tube and the cooling water mean velocity needed to achieve the indicated heat transfer rate in the condenser.

Water enters a 5-mm-diameter and 13-m-long tube at \(45^{\circ} \mathrm{C}\) with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\). The tube is maintained at a constant temperature of \(8^{\circ} \mathrm{C}\). The exit temperature of water is (a) \(4.4^{\circ} \mathrm{C}\) (b) \(8.9^{\circ} \mathrm{C}\) (c) \(10.6^{\circ} \mathrm{C}\) (d) \(12.0^{\circ} \mathrm{C}\) (e) \(14.1^{\circ} \mathrm{C}\) (For water, use \(k=0.607 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=6.14, v=0.894 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=997 \mathrm{~kg} / \mathrm{m}^{3}\) )

Cooling water available at \(10^{\circ} \mathrm{C}\) is used to condense steam at \(30^{\circ} \mathrm{C}\) in the condenser of a power plant at a rate of \(0.15 \mathrm{~kg} / \mathrm{s}\) by circulating the cooling water through a bank of 5 -m-long \(1.2-\mathrm{cm}\)-internal-diameter thin copper tubes. Water enters the tubes at a mean velocity of \(4 \mathrm{~m} / \mathrm{s}\) and leaves at a temperature of \(24^{\circ} \mathrm{C}\). The tubes are nearly isothermal at \(30^{\circ} \mathrm{C}\). Determine the average heat transfer coefficient between the water, the tubes, and the number of tubes needed to achieve the indicated heat transfer rate in the condenser.

In the effort to find the best way to cool a smooth thin-walled copper tube, an engineer decided to flow air either through the tube or across the outer tube surface. The tube has a diameter of \(5 \mathrm{~cm}\), and the surface temperature is maintained constant. Determine \((a)\) the convection heat transfer coefficient when air is flowing through its inside at \(25 \mathrm{~m} / \mathrm{s}\) with bulk mean temperature of \(50^{\circ} \mathrm{C}\) and \((b)\) the convection heat transfer coefficient when air is flowing across its outer surface at \(25 \mathrm{~m} / \mathrm{s}\) with film temperature of \(50^{\circ} \mathrm{C}\).

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