/*! 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 99 Ethylene glycol-distilled water ... [FREE SOLUTION] | 91Ó°ÊÓ

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Ethylene glycol-distilled water mixture with a mass fraction of \(0.72\) and a flow rate of \(2.05 \times 10^{-4} \mathrm{~m}^{3} / \mathrm{s}\) flows inside a tube with an inside diameter of \(0.0158 \mathrm{~m}\) and a uniform wall heat flux boundary condition. For this flow, determine the Nusselt number at the location \(x / D=10\) for the inlet tube configuration of \((a)\) bell-mouth and \((b)\) re-entrant. Compare the results for parts \((a)\) and \((b)\). Assume the Grashof number is Gr \(=60,000\). The physical properties of ethylene glycol- distilled water mixture are \(\operatorname{Pr}=33.46, \nu=3.45 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) and \(\mu_{v} / \mu_{s}=2.0\).

Short Answer

Expert verified
Question: Determine the difference in heat transfer performance between bell-mouth and re-entrant inlet configurations for a flow of ethylene glycol-distilled water mixture inside a tube with uniform wall heat flux boundary condition. Answer: To compare the heat transfer performance between bell-mouth and re-entrant inlet configurations, we need to calculate the Nusselt numbers for both cases using the given empirical correlations and the calculated Reynolds number. After finding the Nusselt numbers, we can analyze the difference in heat transfer performance by comparing these values. Higher Nusselt numbers indicate better heat transfer performance.

Step by step solution

01

Determine the Reynolds number

First, we need to find the Reynolds number (Re) using the given values for flow rate, diameter, and kinematic viscosity: \( Re = \frac{u D}{\nu} \) where \(u\) is the flow velocity, \(D\) is the inside diameter of the tube, and \(\nu\) is the kinematic viscosity of the mixture. We are given the flow rate \(Q = 2.05 \times 10^{-4} \mathrm{~m}^{3}/\mathrm{s}\), so we can find the flow velocity by dividing the flow rate by the cross-sectional area of the tube: \( u = \frac{Q}{A} = \frac{Q}{(1/4)\pi D^{2}} \) Plugging in the given values, we can now find the Reynolds number: \( Re = \frac{ u D }{ \nu } \)
02

Determine the Nusselt number for bell-mouth configuration

For the bell-mouth inlet configuration, we can use the following empirical correlation for the Nusselt number (Nu): \( Nu = 0.021 Re^{0.8} Pr^{n} \) where \(Re\) is the Reynolds number, \(Pr\) is the Prandtl number and \(n = 0.4\) for flow with the given Grashof number (Gr = 60,000). By plugging in the values of Reynolds number and Prandtl number, we can find the Nusselt number for the bell-mouth inlet configuration.
03

Determine the Nusselt number for re-entrant configuration

For the re-entrant inlet configuration, we can use a different empirical correlation for the Nusselt number: \( Nu = 0.021 Re^{0.8} Pr^{n} (\frac{\mu_{v}}{\mu_{s}})^{0.14} \) where \(Re\) is the Reynolds number, \(Pr\) is the Prandtl number, \(n = 0.4\), and the ratios of dynamic viscosities at the bulk mean (v) and wall (s) temperature are given as \(\frac{\mu_{v}}{\mu_{s}} = 2.0\). Using the given values, we can find the Nusselt number for the re-entrant inlet configuration.
04

Compare the results

After calculating the Nusselt numbers for both inlet configurations, we can compare the results to analyze how the heat transfer performance changes due to the inlet tube configuration. Higher Nusselt numbers indicate better heat transfer performance.

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Key Concepts

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

Nusselt Number
Nusselt number is an important dimensionless parameter in heat transfer used to characterize the efficiency of convective heat transfer relative to conductive heat transfer. It is a ratio that helps us understand the performance of a fluid as it moves through a conduit, like a pipe or tube. The basic formula for Nusselt number is given by:\[ Nu = \frac{hD}{k} \]where:
  • \(h\) is the convective heat transfer coefficient.
  • \(D\) is the characteristic length, such as the inside diameter of a tube.
  • \(k\) is the thermal conductivity of the fluid.
A higher Nusselt number indicates more effective convective heat transfer. In the context of the problem provided, different empirical correlations exist for calculating the Nusselt number for different inlet configurations, such as bell-mouth and re-entrant. For each configuration, parameters like the Reynolds number and Prandtl number help adjust and define the Nusselt number to reflect the conditions accurately.
Reynolds Number
Reynolds number is a dimensionless quantity used to predict flow patterns in different fluid flow situations. It helps us understand how a fluid moves through a space and whether the flow is laminar or turbulent. The formula for Reynolds number is:\[ Re = \frac{uD}{u} \]where:
  • \(u\) is flow velocity.
  • \(D\) is the diameter of the pipe.
  • \(u\) is the kinematic viscosity of the fluid.
Reynolds number classifies the flow into laminar (smooth and orderly), transitional, or turbulent (irregular and chaotic). This classification is crucial because it influences heat transfer calculations through effects on the Nusselt number. For example, higher Reynolds numbers typically lead to higher Nusselt numbers, indicating stronger convective heat transfer.
Prandtl Number
Prandtl number is another dimensionless number used in fluid dynamics and heat transfer, named after Ludwig Prandtl. It relates the rate of momentum diffusion to the rate of thermal diffusion and is given by:\[ Pr = \frac{u}{\alpha} \]where:
  • \(u\) is the kinematic viscosity of the fluid.
  • \(\alpha\) is the thermal diffusivity, defined as \(\alpha = \frac{k}{\rho c_{p}}\).
For fluids with a high Prandtl number, the heat transfer in terms of conduction is relatively low compared to the momentum diffusion, which means the thermal boundary layer is thinner. In the example problem, the Prandtl number aids in calculating the Nusselt number with the given empirical relations, playing a key role in determining the overall efficiency of heat transfer.
Grashof Number
The Grashof number is vital in natural convection, where fluid flow occurs without external forces, driven instead by buoyancy due to temperature differences. Its equation is:\[ Gr = \frac{g\beta(T_s - T_\infty)D^3}{u^2} \]where:
  • \(g\) is the acceleration due to gravity.
  • \(\beta\) is the thermal expansion coefficient.
  • \(T_s\) and \(T_\infty\) are the surface and fluid temperatures respectively.
  • \(D\) is the characteristic length (like diameter of a pipe).
  • \(u\) is the kinematic viscosity.
High Grashof numbers indicate that natural convection currents are strong, enhancing heat transfer. In the given problem, the fixed Grashof number allows integration into the generalized Nusselt calculations for comparing the effectiveness of different inlet tube configurations, factoring in the impact of buoyancy-driven flow on heat transfer performance.

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Most popular questions from this chapter

Water at \(15^{\circ} \mathrm{C}\left(\rho=999.1 \mathrm{~kg} / \mathrm{m}^{3}\right.\) and \(\mu=1.138 \times\) \(10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\) ) is flowing in a 4-cm-diameter and \(25-\mathrm{m}\)-long horizontal pipe made of stainless steel steadily at a rate of \(7 \mathrm{~L} / \mathrm{s}\). Determine \((a)\) the pressure drop and \((b)\) the pumping power requirement to overcome this pressure drop. Assume flow is fully developed. Is this a good assumption?

How does surface roughness affect the pressure drop in a tube if the flow is turbulent? What would your response be if the flow were laminar?

What does the logarithmic mean temperature difference represent for flow in a tube whose surface temperature is constant? Why do we use the logarithmic mean temperature instead of the arithmetic mean temperature?

Water enters a \(5-\mathrm{mm}\)-diameter and \(13-\mathrm{m}\)-long tube at \(15^{\circ} \mathrm{C}\) with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\), and leaves at \(45^{\circ} \mathrm{C}\). The tube is subjected to a uniform heat flux of \(2000 \mathrm{~W} / \mathrm{m}^{2}\) on its surface. The temperature of the tube surface at the exit is (a) \(48.7^{\circ} \mathrm{C}\) (b) \(49.4^{\circ} \mathrm{C}\) (c) \(51.1^{\circ} \mathrm{C}\) (d) \(53.7^{\circ} \mathrm{C}\) (e) \(55.2^{\circ} \mathrm{C}\) (For water, nse \(k=0.615 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=5.42, v=0.801 \times\) \(\left.10^{-6} \mathrm{~m}^{2} / \mathrm{s} .\right)\)

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.

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