/*! 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 62 A thin-walled concentric tube he... [FREE SOLUTION] | 91Ó°ÊÓ

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A thin-walled concentric tube heat exchanger of \(0.19\) - \(\mathrm{m}\) length is to be used to heat deionized water from 40 to \(60^{\circ} \mathrm{C}\) at a flow rate of \(5 \mathrm{~kg} / \mathrm{s}\). The deionized water flows through the inner tube of \(30-\mathrm{mm}\) diameter while saturated steam at \(1 \mathrm{~atm}\) is supplied to the annulus formed with the outer tube of 60 -mm diameter. The thermophysical properties of the deionized water are \(\rho=982.3 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4181 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=0.643 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\mu=548 \times 10^{-6} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\), and \(P r=3.56\). Estimate the convection coefficients for both sides of the tube and determine the inner tube wall outlet temperature. Does condensation provide a fairly uniform inner tube wall temperature equal approximately to the saturation temperature of the steam?

Short Answer

Expert verified
In summary, we first calculated the Reynolds number and the convection coefficient for the deionized water side. Then, we calculated the convection coefficient for the steam side and determined the log mean temperature difference (LMTD). By using these values, we found the inner tube wall outlet temperature and compared it with the saturation temperature of the steam to evaluate the uniformity of the inner tube wall temperature. If the difference is relatively small, condensation provides a fairly uniform inner tube wall temperature.

Step by step solution

01

Calculate Reynolds number for deionized water

First, let's find the Reynolds number for the deionized water flow inside the inner tube. We can use the following formula: \[Re_{D} = \frac{\rho uD}{\mu}\] Where \(Re_{D}\) is the Reynolds number, \(\rho\) is the density of the fluid (\(\mathrm{kg/m^3}\)), \(u\) is the flow velocity (\(\mathrm{m/s}\)), \(D\) is the diameter of the tube (\(\mathrm{m}\)), and \(\mu\) is the dynamic viscosity of the fluid (\(\mathrm{N \cdot s / m^{2}}\)). The flow velocity can be calculated as: \[u = \frac{\dot{m}}{\rho A}\] Where \(\dot{m}\) is the mass flow rate (\(\mathrm{kg/s}\)), and \(A\) is the cross-sectional area of the tube.
02

Calculate convection coefficient for deionized water side

We can find the convection coefficient \(h_{i}\) using the Dittus-Boelter equation for the inner tube: (assuming the flow is turbulent) \[h_{i} = k \cdot \frac{0.023Re_{D}^{0.8}Pr^{0.4}}{D}\] Where \(h_{i}\) is the convection coefficient (\(\mathrm{W/m^2 \cdot K}\)), and \(Pr\) is the Prandtl number, and \(k\) is the thermal conductivity of the fluid.
03

Calculate convection coefficient for steam side

The steam condenses on the outer surface of the inner tube, which means it supplies latent heat and maintains a constant temperature (saturation temperature of steam at 1 atm). The convection coefficient is given by: \[h_{o} = \frac{q}{\Delta T}\] Where \(q\) is the heat transfer rate, and \(\Delta T\) is the temperature difference between the inner tube wall and the bulk temperature of the deionized water.
04

Determine the log mean temperature difference (LMTD)

We can now evaluate the log mean temperature difference (LMTD) as follows: \[\Delta T_{lm} = \frac{\Delta T_{1} - \Delta T_{2}}{\ln{(\Delta T_{1}/\Delta T_{2})}}\] Where \(\Delta T_{1}\) is the temperature difference between the steam condensation temperature and the water inlet temperature, and \(\Delta T_{2}\) is the temperature difference between the steam condensation temperature and the water outlet temperature.
05

Calculate the inner tube wall outlet temperature

We have all the information necessary to determine the overall heat transfer rate from the steam to the deionized water. Let's use the following equation to calculate the inner tube wall outlet temperature: \[T_{w,out} = T_{steam} - \frac{q}{h_o A_s}\] Where \(T_{w,out}\) is the inner tube wall outlet temperature (\(^{\circ}\mathrm{C}\)), \(T_{steam}\) is the saturation temperature of steam at 1 atm (\(^{\circ}\mathrm{C}\)), \(A_s\) is the surface area of the inner tube.
06

Evaluate the uniformity of inner tube wall temperature

To check if the condensation process provides a fairly uniform inner tube wall temperature, compare the outlet temperature obtained in Step 5 with the saturation temperature of the steam. If the difference is relatively small (a threshold could be 1-2 degrees Celsius), then condensation is providing a fairly uniform inner tube wall temperature. This step-by-step solution shows how to calculate the convection coefficients for the inner and outer sides of the thin-walled concentric tube heat exchanger and estimate the uniformity of inner tube wall temperature due to the condensation process.

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Key Concepts

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

Convection Coefficient
In a heat exchanger, the **convection coefficient** is a crucial parameter that indicates the heat transfer capability between a fluid and a surface. It is represented by the symbol \( h \) and is expressed in watts per square meter per Kelvin (\( \, \mathrm{W/m^2 \cdot K} \)). This coefficient is essential for determining how efficiently heat is transferred within the heat exchanger.

For the deionized water flowing inside the tube of the heat exchanger, the convection coefficient \( h_i \) can be calculated using empirical correlations, such as the Dittus-Boelter equation. This equation assumes turbulent flow conditions and utilizes parameters like the Reynolds and Prandtl numbers along with the fluid's thermal conductivity:

  • **Thermal Conductivity (\( k \))**: Measures a fluid's ability to conduct heat.
  • **Reynolds Number (\( Re \))**: Determines the flow regime, impacting convection intensity.
  • **Prandtl Number (\( Pr \))**: Relates viscous diffusivity and thermal diffusivity of the fluid.

The convection coefficient on the steam side \( h_o \) is derived differently as it involves steam condensation. Here, \( h_o \) reflects how the latent heat of steam is transferred to the tube surface, usually maintaining the steam's saturation temperature.
Reynolds Number
The **Reynolds number** \( Re \) is a dimensionless value that plays an instrumental role in characterizing the flow regime within a heat exchanger. It is a key factor influencing the convection coefficient, as it determines whether the fluid flow is laminar or turbulent.

The Reynolds number is calculated using the formula:
\[Re = \frac{\rho u D}{\mu}\]
Where:
  • **\( \rho \)**: Density of the fluid (\( \mathrm{kg/m^3} \))
  • **\( u \)**: Flow velocity (\( \mathrm{m/s} \))
  • **\( D \)**: Diameter of the tube (\( \mathrm{m} \))
  • **\( \mu \)**: Dynamic viscosity (\( \mathrm{N \cdot s / m^2} \))

In the context of a heat exchanger, a higher Reynolds number typically indicates turbulent flow, which enhances mixing and improves heat transfer. Conversely, a lower Reynolds number suggests laminar flow, which may reduce heat transfer efficiency. The transition between these states commonly occurs around a Reynolds number of 2300. Understanding this number helps in determining appropriate designs and operating conditions for the heat exchanger.
Log Mean Temperature Difference
The **log mean temperature difference** (LMTD) is a vital metric when assessing the performance of a heat exchanger. It represents the potential driving force for heat transfer along the length of the heat exchanger, providing a more accurate assessment of temperature change than a simple arithmetic mean.

The formula for LMTD is:
\[\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln{(\Delta T_1/\Delta T_2)}}\]
Where \( \Delta T_1 \) and \( \Delta T_2 \) are the temperature differences at the two ends of the heat exchanger.

  • **\( \Delta T_1 \)**: Difference between the steam condensation temperature and water inlet temperature.
  • **\( \Delta T_2 \)**: Difference between the steam condensation temperature and water outlet temperature.

LMTD provides a more representative average temperature gradient across the exchanger, which is crucial for estimating its thermal performance effectively. The higher the LMTD, the greater the driving force for heat transfer, meaning more heat is transferred at a quicker rate. Calculating the LMTD helps in predicting the effectiveness of the heat exchanger and deciding the necessary dimensions and materials for its design.

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Most popular questions from this chapter

Consider the situation of Problem \(10.67\) at relatively high vapor velocities, with a fluid mass flow rate of \(\dot{m}=2.5 \mathrm{~kg} / \mathrm{s}\). (a) Determine the heat transfer coefficient and condensation rate per unit length of tube for a mass fraction of vapor of \(X=0.2\). (b) Plot the heat transfer coefficient and the condensation rate for \(0.1 \leq X \leq 0.3\).

Saturated vapor from a chemical process condenses at a slow rate on the inner surface of a vertical, thinwalled cylindrical container of length \(L\) and diameter D. The container wall is maintained at a uniform temperature \(T_{s}\) by flowing cold water across its outer surface. Derive an expression for the time, \(t_{f}\), required to fill the container with condensate, assuming that the condensate film is laminar. Express your result in terms of \(D, L,\left(T_{\text {sat }}-T_{s}\right), g\), and appropriate fluid properties. Derive an expression for the time, \(t_{f}\), required to fill the container with condensate, assuming that the condensate film is laminar. Express your result in terms of \(D, L,\left(T_{\text {sat }}-T_{s}\right), g\), and appropriate fluid properties.

A dielectric fluid at atmospheric pressure is heated with a \(0.5\)-mm-diameter, horizontal platinum wire. Determine the temperature of the wire when the wire is heated at \(50 \%\) of the critical heat flux. The properties of the fluid are \(c_{p l}=1300 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, h_{f,}=142 \mathrm{~kJ} / \mathrm{kg}, k_{l}=0.075 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(v_{l}=0.32 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}, \rho_{l}=1400 \mathrm{~kg} / \mathrm{m}^{3}, \rho_{v}=7.2 \mathrm{~kg} / \mathrm{m}^{3}\), \(\sigma=12.4 \times 10^{-3} \mathrm{~N} / \mathrm{m}, T_{\text {sot }}=34^{\circ} \mathrm{C}\). Assume the nucleate boiling constants are \(C_{s f}=0.005\) and \(n=1.7\). For small horizontal cylinders, the critical heat flux is found by multiplying the value associated with large horizontal cylinders by a correction factor \(F\), where \(F=0.89+2.27\) \(\exp \left(-3.44 \mathrm{Co}^{-1 / 2}\right)\). The Confinement number is based on the radius of the cylinder, and the range of applicability for the correction factor is \(1.3 \leq \mathrm{Co} \leq 6.7[11]\).

A vertical plate \(500 \mathrm{~mm}\) high and \(200 \mathrm{~mm}\) wide is to be used to condense saturated steam at 1 atm. (a) At what surface temperature must the plate be maintained to achieve a condensation rate of \(\dot{m}=25 \mathrm{~kg} / \mathrm{h}\) ? (b) Compute and plot the surface temperature as a function of condensation rate for \(15 \leq \dot{m} \leq 50 \mathrm{~kg} / \mathrm{h}\). (c) On the same graph and for the same range of \(\dot{m}\), plot the surface temperature as a function of condensation rate if the plate is \(200 \mathrm{~mm}\) high and \(500 \mathrm{~mm}\) wide.

Consider a gas-fired boiler in which five coiled, thinwalled, copper tubes of \(25-\mathrm{mm}\) diameter and \(8-\mathrm{m}\) length are submerged in pressurized water at \(4.37\) bars. The walls of the tubes are scored and may be assumed to be isothermal. Combustion gases enter each of the tubes at a temperature of \(T_{m, i}=700^{\circ} \mathrm{C}\) and a flow rate of \(\dot{m}=0.08 \mathrm{~kg} / \mathrm{s}\), respectively. (a) Determine the tube wall temperature \(T_{s}\) and the gas outlet temperature \(T_{m, o}\) for the prescribed conditions. As a first approximation, the properties of the combustion gases may be taken as those of air at \(700 \mathrm{~K}\). (b) Over time the effects of scoring diminish, leading to behavior similar to that of a polished copper surface. Determine the wall temperature and gas outlet temperature for the aged condition.

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