Air ( \(1 \mathrm{~atm})\) enters into a 5 -cm-diameter circular tube at \(20^{\circ} \mathrm{C}\) with an average velocity of \(5 \mathrm{~m} / \mathrm{s}\). The tube wall is maintained at a constant surface temperature of \(160^{\circ} \mathrm{C}\), and the outlet mean temperature is \(80^{\circ} \mathrm{C}\). Estimate the length of the tube.

Short Answer

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Question: Estimate the length of a tube, given the following conditions: the air pressure is 1 atm, the air enters the tube at 20°C with an average velocity of 5 m/s, the tube wall is maintained at a constant temperature of 160°C, and the outlet air temperature is 80°C. Answer: The length of the tube is approximately 0.851 m.

Step by step solution

01

Calculate the temperature difference

First, let's calculate the temperature difference between the inlet air and the tube wall \(Delta T_{i} = T_{wall} - T_{in} = 160 - 20 = 140^{\circ}\mathrm{C}\) Similarly, the temperature difference between the outlet air and the tube wall, \(Delta T_{o} = T_{wall} - T_{out} = 160 - 80 = 80^{\circ}\mathrm{C}\)
02

Calculate the logarithmic mean temperature difference

Now we calculate the logarithmic mean temperature difference (LMTD) using the formula \(LMTD = \frac{\Delta T_{i} - \Delta T_{o}}{ln(\frac{\Delta T_{i}}{\Delta T_{o}})}\) \(LMTD = \frac{140 - 80}{ln(\frac{140}{80})} = 106.27^{\circ}\mathrm{C}\)
03

Determine the heat transfer

Now, we can calculate the heat transfer rate from the mass flow rate, specific heat capacity of air (\(c_p\)), and the temperature difference between the inlet and outlet. First, let's calculate the mass flow rate using the air density \(\rho\) (assuming the air density at 20°C), and the flow velocity, \(\rho = 1.204\,\mathrm{kg/m^3}\) \(A = \frac{\pi (5 \times 10^{-2})^2}{4} = 1.9635 \times 10^{-3}\,\mathrm{m^2}\) \(\dot{m} = \rho \times A \times 5\,\mathrm{m/s} = 1.204 \times 1.9635 \times 10^{-3} \times 5 = 0.01183\,\mathrm{kg/s}\) Now we calculate the heat transfer, \(\dot{q} = \dot{m} \times c_p \times (T_{out} - T_{in})\) \(\dot{q} = 0.01183 \times 1005 (80 - 20) = 711.41\,\mathrm{W}\)
04

Determine the overall heat transfer coefficient (U)

To find the overall heat transfer coefficient (U), we need to consider the convective heat transfer coefficients of the air on the inside (hi) and outside (ho) the tube, as well as the thermal conductivity of the tube material (k), and the tube thickness (delta). Since we don't have tube material information or a method to determine the heat transfer coefficients, for the sake of estimation, we assume an overall heat transfer coefficient, U. \(U = 50\,\mathrm{W/m^2K}\) (Example value that can vary depending on the system)
05

Calculate the heat transfer area A and the tube length

Now, we can use the heat transfer formula to calculate the area \(A\) required for heat transfer. \(\dot{q} = UA \times LMTD\) Now, we can find the area A, \(A = \frac{\dot{q}}{U \times LMTD}\) \(A = \frac{711.41}{50 \times 106.27} = 0.1337\,\mathrm{m^2}\) Finally, we can find the length of the tube, L, using the heat transfer area A and the inner diameter of the tube, D. \(A = \pi D \times L.\) Now, we can find the length L, \(L = \frac{A}{\pi \times 5 \times 10^{-2}}\) \(L = \frac{0.1337}{\pi \times 5 \times 10^{-2}} = 0.8507\,\mathrm{m}\) The length of the tube is approximately 0.851 m.

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Key Concepts

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

Logarithmic Mean Temperature Difference
Understanding the logarithmic mean temperature difference (LMTD) is crucial when studying heat transfer in tubes as it offers a way to measure the average temperature driving force along the length of the tube. This average is important since the temperature difference between the fluids inside and outside of the tube usually changes along the tube's length.

LMTD is calculated using the formula:
\[ LMTD = \frac{\Delta T_{i} - \Delta T_{o}}{\ln(\frac{\Delta T_{i}}{\Delta T_{o}})} \]
where \( \Delta T_{i} \) is the temperature difference at the inlet and \( \Delta T_{o} \) is the temperature difference at the outlet. The natural logarithm (ln) accounts for the change in temperature difference in a non-linear relationship, ensuring a weighted mean based on the relative differences in temperature.

For clearer understanding, think of LMTD as the 'corrected' average difference that ensures the calculated heat transfer rate is equivalent to that of a counterflow heat exchanger with constant temperatures.
Convective Heat Transfer Coefficient
Moving on to the convective heat transfer coefficient, this parameter is essential because it quantifies the ease with which heat transfer occurs between a surface and a fluid flowing past it. Designations like \( h_i \) for internal or \( h_o \) for external flow coefficients highlight the context of the flow, whether it's inside or outside the tube.

The coefficient is influenced by several factors, including the fluid's properties (like its thermal conductivity and viscosity), the flow conditions (like velocity and turbulence), and the surface characteristics (like roughness and temperature). Higher values indicate more efficient heat transfer capabilities, which is why understanding and maximizing this coefficient is a key aim in heat exchanger design.

A simple way to improve the predictive power of heat transfer calculations is by investing more effort in accurately determining the convective heat transfer coefficient, whether through empirical correlations or advanced computational methods.
Overall Heat Transfer Coefficient
Lastly, the overall heat transfer coefficient, \( U \), ties the previous concepts together by representing the total resistance to heat transfer between the hot and cold fluids in the tube setup. It encapsulates both the conductive and convective aspects of heat transfer. In the formula given:
\[ \dot{q} = UA \times LMTD \]
The coefficient \( U \) represents how well heat is transferred through the tube wall and across the boundary layers of fluid on each side. The value of \( U \) can vary based on material properties, flow characteristics, and dimensions, reflecting the composite effect of multiple thermal resistances in series.

Improving the overall heat transfer coefficient often entails optimizing material selection, surface treatments to enhance convection, and selecting appropriate configurations that boost heat transfer. It's a key design parameter in engineering applications involving heat exchangers and can significantly affect their size, cost, and performance.

Understanding how to balance these factors, along with the pursuit of precise estimations for \( U \), can yield more efficient designs for real-world applications.

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Most popular questions from this chapter

A 10 -m-long and 10 -mm-inner-diameter pipe made of commercial steel is used to heat a liquid in an industrial process. The liquid enters the pipe with \(T_{i}=25^{\circ} \mathrm{C}, V=0.8 \mathrm{~m} / \mathrm{s}\). A uniform heat flux is maintained by an electric resistance heater wrapped around the outer surface of the pipe, so that the fluid exits at \(75^{\circ} \mathrm{C}\). Assuming fully developed flow and taking the average fluid properties to be \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=\) \(4000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \mu=2 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, k=0.48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\operatorname{Pr}=10\), determine: (a) The required surface heat flux \(\dot{q}_{s}\), produced by the heater (b) The surface temperature at the exit, \(T_{s}\) (c) The pressure loss through the pipe and the minimum power required to overcome the resistance to flow.

Determine the convection heat transfer coefficient for the flow of \((a)\) air and \((b)\) water at a velocity of \(2 \mathrm{~m} / \mathrm{s}\) in an \(8-\mathrm{cm}\) diameter and 7-m-long tube when the tube is subjected to uniform heat flux from all surfaces. Use fluid properties at \(25^{\circ} \mathrm{C}\).

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 \(50^{\circ} \mathrm{C}\).

Hot water at \(90^{\circ} \mathrm{C}\) enters a \(15-\mathrm{m}\) section of a cast iron pipe \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are 4 and \(4.6 \mathrm{~cm}\), respectively, at an average velocity of \(1.2 \mathrm{~m} / \mathrm{s}\). The outer surface of the pipe, whose emissivity is \(0.7\), is exposed to the cold air at \(10^{\circ} \mathrm{C}\) in a basement, with a convection heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the walls of the basement to be at \(10^{\circ} \mathrm{C}\) also, determine \((a)\) the rate of heat loss from the water and \((b)\) the temperature at which the water leaves the basement.

Water is flowing in fully developed conditions through a 3 -cm-diameter smooth tube with a mass flow rate of \(0.02 \mathrm{~kg} / \mathrm{s}\) at \(15^{\circ} \mathrm{C}\). Determine \((a)\) the maximum velocity of the flow in the tube and \((b)\) the pressure gradient for the flow.

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