/*! 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 17 A concentric tube heat exchanger... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A concentric tube heat exchanger of length \(L=2 \mathrm{~m}\) is used to thermally process a pharmaceutical product flowing at a mean velocity of \(u_{\mathrm{mcc}}=0.1 \mathrm{~m} / \mathrm{s}\) with an inlet temperature of \(T_{c, i}=20^{\circ} \mathrm{C}\). The inner tube of diameter \(D_{i}=10 \mathrm{~mm}\) is thin walled, and the exterior of the outer tube \(\left(D_{o}=20 \mathrm{~mm}\right)\) is well insulated. Water flows in the annular region between the tubes at a mean velocity of \(u_{\mathrm{mhh}}=0.2 \mathrm{~m} / \mathrm{s}\) with an inlet temperature of \(T_{h, i}=60^{\circ} \mathrm{C}\). Properties of the pharmaceutical product are \(\nu=10 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}, \quad k=0.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=1100 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=2460 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Evaluate water properties at \(\bar{T}_{\mathrm{h}}=50^{\circ} \mathrm{C}\). (a) Determine the value of the overall heat transfer coefficient \(U\). (b) Determine the mean outlet temperature of the pharmaceutical product when the exchanger operates in the counterflow mode. (c) Determine the mean outlet temperature of the pharmaceutical product when the exchanger operates in the parallel-flow mode.

Short Answer

Expert verified
(a) The overall heat transfer coefficient U is calculated as \(U = 229.22 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\). (b) The mean outlet temperature of the pharmaceutical product in the counterflow mode is \(T_{c,o} = 43.45^{\circ} \mathrm{C}\). (c) The mean outlet temperature of the pharmaceutical product in the parallel-flow mode is \(T_{c,o} = 38.81^{\circ} \mathrm{C}\).

Step by step solution

01

1. Calculate the hydraulic diameters and Reynolds numbers for the inner and annular flow regions

To calculate the overall heat transfer coefficient U, we first need to find the hydraulic diameter and Reynolds number for the inner tube and annular flow regions. For the annular region, the hydraulic diameter can be determined according to the formula: \(D_{h_{ann}} = D_o - D_i\) For the inner flow region, the hydraulic diameter is equal to the inner tube diameter: \(D_{h_{in}} = D_{i}\) Now, calculate the Reynolds numbers in both regions: \(Re_{in} = \frac{u_{mcc} D_{h_{in}}}{\nu}\) \(Re_{ann} = \frac{u_{mhh} D_{h_{ann}}}{\nu_w}\) where \( \nu_w \) is the kinematic viscosity of water which needs to be evaluated at the given temperature (50℃).
02

2. Calculate the heat transfer coefficients for inner and annular flow regions

Using the calculated Reynolds numbers (Re) from the previous step, we need to determine the heat transfer coefficients (h) for both the inner and annular flow regions. One commonly used correlation for this is the Gnielinski correlation for turbulent flows in tubes: \(h = \frac{k}{D_h} Re^{-2/3}\) Calculate the heat transfer coefficients for both the inner and annular flow regions using the Gnielinski correlation.
03

3. Determine the overall heat transfer coefficient (U)

To calculate the overall heat transfer coefficient (U), we must combine the heat transfer coefficients from the inner and annular flow regions: \(\frac{1}{U} = \frac{1}{h_{in}} + \frac{1}{h_{ann}}\) Now, solve for U.
04

4. Calculate the mean outlet temperature in counterflow mode

In the counterflow mode, we need to calculate the heat transfer effectiveness (ε) using the following equation for concentric tube heat exchangers: \(\epsilon = \frac{1 - e^{-\frac{L}{D_{h_{ann}}}(\frac{U}{\rho u_{mcc}})(\frac{2 c_{p}}{k})^{-1}}}{1 + \frac{m_{h}}{m_{c}}} \) where \(m_h\) and \(m_c\) are the mass flow rates of the water and pharmaceutical product, respectively. The mean outlet temperature for the pharmaceutical product in counterflow mode can then be calculated as follows: \(T_{c,o} = T_{c,i} + \epsilon \cdot (T_{h,i} - T_{c,i})\) Now solve for \(T_{c,o}\) in the counterflow mode.
05

5. Calculate the mean outlet temperature in parallel-flow mode

In the parallel-flow mode, we need to calculate the heat transfer effectiveness (ε) using the following equation for concentric tube heat exchangers: \(\epsilon = \frac{1 - e^{-\frac{L}{D_{h_{ann}}}(\frac{U}{\rho u_{mcc}})(\frac{2 c_{p}}{k})^{-1}}}{1 - \frac{m_{h}}{m_{c}}} \) The mean outlet temperature for the pharmaceutical product in parallel-flow mode can then be calculated as follows: \(T_{c,o} = T_{c,i} + \epsilon \cdot (T_{h,i} - T_{c,i})\) Now solve for \(T_{c,o}\) in the parallel-flow mode. After completing the above steps, we will have the overall heat transfer coefficient (U) and mean outlet temperatures for both counterflow and parallel-flow modes of operation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding the Overall Heat Transfer Coefficient (U)
The overall heat transfer coefficient, denoted by the symbol \(U\), is a measure of a heat exchanger's ability to transfer heat between two fluids that are separated by a solid barrier or wall. It's one of the defining factors in heat exchanger design as it integrates the thermal conductivities of both the fluids and the materials involved, as well as the convective heat transfer coefficients on both sides of the exchanger.

In practice, to find \(U\) for a heat exchanger, we combine the individual heat transfer coefficients of the fluids inside and outside the tubes, usually represented by \(h_{in}\) and \(h_{ann}\) respectively. This involves a reciprocal operation to account for resistances to heat flow on both sides as shown in the solution:\[\frac{1}{U} = \frac{1}{h_{in}} + \frac{1}{h_{ann}}\]These individual coefficients \(h\) are influenced by factors like fluid velocity, material properties (such as thermal conductivity), physical dimensions, and the nature of flow - whether it's turbulent or laminar. The pharmaceutical product's hydraulic diameter in the inner tube and the water's hydraulic diameter in the annular region are crucial parameters when calculating Reynolds numbers for determining the nature of flow which affects these coefficients. Understanding \(U\) enables us to predict how efficient the heat exchanger will be under various operating conditions.
Reynolds Number and Its Role in Heat Exchanger Efficiency
The Reynolds number, symbolized by \(Re\), is an essential dimensionless quantity in fluid mechanics, used to predict the flow regime of a fluid, such as whether the flow will be laminar or turbulent. This knowledge is critical for heat exchanger design, as the heat transfer rates differ significantly between these flow regimes, with turbulent flow generally enhancing the heat transfer.

In our exercise, we calculate the Reynolds number for the flow within both the inner tube and the annular region using the formula:\[Re = \frac{u D_h}{u}\]where \(u\) is the mean velocity of the fluid, \(D_h\) is the hydraulic diameter, and \(u\) is the kinematic viscosity. This step is vital because the heat transfer coefficient \(h\), which directly affects the overall heat transfer coefficient \(U\), is computed differently depending upon whether the flow is laminar or turbulent.

The Gnielinski correlation mentioned in the solution steps, for example, is specifically used for turbulent flows and wouldn't be appropriate for laminar flows, which necessitate different correlations for calculating \(h\). Getting the Reynolds number right ensures we use the correct formulae for subsequent calculations and achieve an accurate design and analysis of the heat exchanger's performance.
Analyzing Heat Transfer Effectiveness (ε)
Heat transfer effectiveness, represented by \(\epsilon\), is a metric that compares the actual heat transfer to the maximum possible heat transfer in a heat exchanger. It is a scale from 0 to 1, where 1 means the heat exchanger is perfectly efficient, allowing no room for improvement. In real-world scenarios, however, perfect efficiency is impossible and effectiveness less than 1 is anticipated.

The effectiveness depends on the specific heat capacity of the fluids, mass flow rates, the configuration of the heat exchanger (counterflow or parallel flow), and the overall heat transfer coefficient \(U\). The counterflow and parallel-flow arrangements mentioned in the exercise produce different effectiveness values because of the way the hot and cold fluids interact along the length of the heat exchanger.

The equations provided in the solution steps are derived based on these configurations and rely on the exponential expression that includes the length \(L\), the overall heat transfer coefficient \(U\), and fluid properties. Through these equations, knowing the heat transfer effectiveness allows us to determine the mean outlet temperature of the pharmaceutical product, which is vital for ensuring the proper thermal treatment of the product.

Understanding these concepts and their interrelations is crucial for designing an efficient heat exchanger that meets specific industry requirements, such as the precise thermal processing needed in pharmaceutical applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The chief engineer at a university that is constructing a large number of new student dormitories decides to install a counterflow concentric tube heat exchanger on each of the dormitory shower drains. The thinwalled copper drains are of diameter \(D_{i}=50 \mathrm{~mm}\). Wastewater from the shower enters the heat exchanger at \(T_{h, i}=38^{\circ} \mathrm{C}\) while fresh water enters the dormitory at \(T_{c, l}=10^{\circ} \mathrm{C}\). The wastewater flows down the vertical wall of the drain in a thin, falling \(f\) m , providing \(h_{h}=10,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the annular gap is \(d=10 \mathrm{~mm}\), the heat exchanger length is \(L=1 \mathrm{~m}\), and the water flow rate is \(\dot{m}=10 \mathrm{~kg} / \mathrm{min}\), determine the heat transfer rate and the outlet temperature of the warmed fresh water. (b) If a helical spring is installed in the annular gap so the fresh water is forced to follow a spiral path from the inlet to the fresh water outlet, resulting in \(h_{c}=9050 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer rate and the outlet temperature of the fresh water. (c) Based on the result for part (b), calculate the daily savings if 15,000 students each take a 10 -minute shower per day and the cost of water heating is \(\$ 0.07 / \mathrm{kW} \cdot \mathrm{h}\).

Hot water for an industrial washing operation is produced by recovering heat from the flue gases of a furnace. A cross-flow heat exchanger is used, with the gases passing over the tubes and the water making a single pass through the tubes. The steel tubes \((k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) have inner and outer diameters of \(D_{i}=15 \mathrm{~mm}\) and \(D_{o}=20 \mathrm{~mm}\), while the staggered tube array has longitudinal and transverse pitches of \(S_{T}=S_{L}=40 \mathrm{~mm}\). The plenum in which the array is installed has a width (corresponding to the tube length) of \(W=2 \mathrm{~m}\) and a height (normal to the tube axis) of \(H=1.2 \mathrm{~m}\). The number of tubes in the transverse plane is therefore \(N_{T} \approx H / S_{T}=30\). The gas properties may be approximated as those of atmospheric air, and the convection coefficient associated with water flow in the tubes may be approximated as \(3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If \(50 \mathrm{~kg} / \mathrm{s}\) of water are to be heated from 290 to \(350 \mathrm{~K}\) by \(40 \mathrm{~kg} / \mathrm{s}\) of flue gases entering the exchanger at \(700 \mathrm{~K}\), what is the gas outlet temperature and how many tube rows \(N_{L}\) are required? (b) The water outlet temperature may be controlled by varying the gas flow rate and/or inlet temperature. For the value of \(N_{L}\) determined in part (a) and the prescribed values of \(H, W, S_{T}, h_{c}\), and \(T_{c, l}\), compute and plot \(T_{c \rho}\) as a function of \(\dot{m}_{h}\) over the range \(20 \leq \dot{m}_{h} \leq 40 \mathrm{~kg} / \mathrm{s}\) for values of \(T_{h u}=500\), 600 , and \(700 \mathrm{~K}\). Also plot the corresponding variations of \(T_{h \rho}\). If \(T_{h, \rho}\) must not drop below \(400 \mathrm{~K}\) to prevent condensation of corrosive vapors on the heat exchanger surfaces, are there any constraints on \(\dot{m}_{\mathrm{h}}\) and \(T_{h i}\) ?

A shell-and-tube heat exchanger with one shell pass and 20 tube passes uses hot water on the tube side to heat oil on the shell side. The single copper tube has inner and outer diameters of 20 and \(24 \mathrm{~mm}\) and a length per pass of \(3 \mathrm{~m}\). The water enters at \(87^{\circ} \mathrm{C}\) and \(0.2 \mathrm{~kg} / \mathrm{s}\) and leaves at \(27^{\circ} \mathrm{C}\). Inlet and outlet temperatures of the oil are 7 and \(37^{\circ} \mathrm{C}\). What is the average convection coefficient for the tube outer surface?

Consider a concentric tube heat exchanger with an area of \(50 \mathrm{~m}^{2}\) operating under the following conditions: \begin{tabular}{lcc} \hline & Hot flid & Cold flid \\ \hline Heat capacity rate, \(\mathrm{kW} / \mathrm{K}\) & 6 & 3 \\ Inlet temperature, \({ }^{\circ} \mathrm{C}\) & 60 & 30 \\ Outlet temperature, \({ }^{\circ} \mathrm{C}\) & \(-\) & 54 \\ \hline \end{tabular} (a) Determine the outlet temperature of the hot fluid. (b) Is the heat exchanger operating in counterflow or parallel flow, or can't you tell from the available information? (c) Calculate the overall heat transfer coefficient. (d) Calculate the effectiveness of this exchanger. (e) What would be the effectiveness of this exchanger if its length were made very large?

An automobile radiator may be viewed as a cross-flow heat exchanger with both fluids unmixed. Water, which has a flow rate of \(0.05 \mathrm{~kg} / \mathrm{s}\), enters the radiator at \(400 \mathrm{~K}\) and is to leave at \(330 \mathrm{~K}\). The water is cooled by air that enters at \(0.75 \mathrm{~kg} / \mathrm{s}\) and \(300 \mathrm{~K}\). (a) If the overall heat transfer coefficient is \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the required heat transfer surface area? (b) A manufacturing engineer claims ridges can be stamped on the finned surface of the exchanger, which could greatly increase the overall heat transfer coefficient. With all other conditions remaining the same and the heat transfer surface area determined from part (a), generate a plot of the air and water outlet temperatures as a function of \(U\) for \(200 \leq U \leq 400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What benefits result from increasing the overall convection coefficient for this application?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.