Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Under what conditions can a counterflow heat exchanger have an effectiveness of 1 ? What would your answer be for a parallel-flow heat exchanger?

Short Answer

Expert verified
Answer: A counterflow heat exchanger can have an effectiveness of 1 when the temperature difference between the hot fluid inlet and the cold fluid outlet is infinitely small, meaning the hot fluid cools down completely, transferring all its heat energy to the cold fluid, while the cold fluid heats up completely, reaching the initial temperature of the hot fluid. In contrast, a parallel-flow heat exchanger can never have an effectiveness of 1 because the temperature difference between the hot and cold fluids remains at every point throughout the heat exchanger, preventing the maximum possible heat transfer from being achieved.

Step by step solution

01

Counterflow heat exchanger

To find the condition for a counterflow heat exchanger to have an effectiveness of 1, we need to analyze the temperature profiles of the hot and cold fluids in the heat exchanger. In a counterflow heat exchanger, the hot and cold fluids move in opposite directions. The effectiveness (ε) is given by: ε = (actual heat transfer) / (maximum possible heat transfer) For an effectiveness of 1, the actual heat transfer must be equal to the maximum possible heat transfer.
02

Analyzing the temperature profiles

For the counterflow heat exchanger, the effectiveness can reach 1 under the following condition: the temperature difference between the hot fluid inlet and the cold fluid outlet must be infinitely small. In mathematical terms: \[\lim_{\Delta T \to 0}(T_{H, inlet} - T_{C, outlet}) = 0\] This means that the hot fluid must cool down completely, transferring all of its heat energy to the cold fluid, while the cold fluid must heat up completely, reaching the initial temperature of the hot fluid.
03

Parallel-flow heat exchanger

Next, we will analyze the conditions for a parallel-flow heat exchanger to have an effectiveness of 1. In a parallel-flow heat exchanger, the hot and cold fluids flow in the same direction. Using the same equation for effectiveness as in step 1, we will analyze the temperature profiles to find the condition for an effectiveness of 1 in a parallel-flow heat exchanger.
04

Analyzing the temperature profiles for a parallel-flow heat exchanger

In a parallel-flow heat exchanger, the effectiveness can never reach 1. This is because the temperature difference between the hot and cold fluids reduces as the fluids flow through the heat exchanger. At the outlet, the hot fluid will always have a higher temperature than the cold fluid, so there will always be a temperature difference between the hot and cold fluids at every point throughout the heat exchanger. Consequently, the maximum possible heat transfer can never be achieved, and the effectiveness cannot be equal to 1 in a parallel-flow heat exchanger. In conclusion, a counterflow heat exchanger can have an effectiveness of 1 under the condition that the temperature difference between the hot fluid inlet and the cold fluid outlet is infinitely small, while a parallel-flow heat exchanger can never have an effectiveness of 1.

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 Vaia!

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

Consider a water-to-water counterflow heat exchanger with these specifications. Hot water enters at \(90^{\circ} \mathrm{C}\) while cold water enters at \(20^{\circ} \mathrm{C}\). The exit temperature of the hot water is \(15^{\circ} \mathrm{C}\) greater than that of the cold water, and the mass flow rate of the hot water is 50 percent greater than that of the cold water. The product of heat transfer surface area and the overall heat transfer coefficient is \(2200 \mathrm{~W} / \mathrm{K}\). Taking the specific heat of both cold and hot water to be $c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\(, determine \)(a)\( the outlet temperature of the cold water, \)(b)$ the effectiveness of the heat exchanger, \((c)\) the mass flow rate of the cold water, and \((d)\) the heat transfer rate.

Ethanol is vaporized at $78^{\circ} \mathrm{C}\left(h_{f \mathrm{~g}}=846 \mathrm{~kJ} / \mathrm{kg}\right)$ in a double-pipe parallel-flow heat exchanger at a rate of \(0.04 \mathrm{~kg} / \mathrm{s}\) by hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(115^{\circ} \mathrm{C}\). If the heat transfer surface area and the overall heat transfer coefficients are \(6.2 \mathrm{~m}^{2}\) and $320 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively, determine the outlet temperature and the mass flow rate of oil using \((a)\) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

Geothermal water $\left(c_{p}=4250 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( at \)75^{\circ} \mathrm{C}$ is to be used to heat fresh water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(17^{\circ} \mathrm{C}\) at a rate of \(1.2 \mathrm{~kg} / \mathrm{s}\) in a double-pipe counterflow heat exchanger. The heat transfer surface area is $25 \mathrm{~m}^{2}\(, the overall heat transfer coefficient is \)480 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, and the mass flow rate of geothermal water is larger than that of fresh water. If the effectiveness of the heat exchanger must be \(0.823\), determine the mass flow rate of geothermal water and the outlet temperatures of both fluids.

Consider a double-pipe heat exchanger with a tube diameter of $10 \mathrm{~cm}$ and negligible tube thickness. The total thermal resistance of the heat exchanger was calculated to be \(0.025 \mathrm{k} / \mathrm{W}\) when it was first constructed. After some prolonged use, fouling occurs at both the inner and outer surfaces with the fouling factors $0.00045 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\( and \)0.00015 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}$, respectively. The percentage decrease in the rate of heat transfer in this heat exchanger due to fouling is (a) \(2.3 \%\) (b) \(6.8 \%\) (c) \(7.1 \%\) (d) \(7.6 \%\) (e) \(8.5 \%\)

The National Sanitation Foundation (NSF) standard for commercial warewashing equipment (ANSL/NSF 3) requires that the final rinse water temperature be between 82 and \(90^{\circ} \mathrm{C}\). A shell-and-tube heat exchanger is to heat \(0.5 \mathrm{~kg} / \mathrm{s}\) of water $\left(c_{p}=4200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( from 48 to \)86^{\circ} \mathrm{C}$ by geothermal brine flowing through a single shell pass. The heated water is then fed into commercial warewashing equipment. The geothermal brine enters and exits the heat exchanger at 98 and \(90^{\circ} \mathrm{C}\), respectively. The water flows through four thin-walled tubes, each with a diameter of $25 \mathrm{~mm}$, with all four tubes making the same number of passes through the shell. The tube length per pass for each tube is \(5 \mathrm{~m}\). The corresponding convection heat transfer coefficients on the outer and inner tube surfaces are 1050 and $2700 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. The estimated fouling factor caused by the accumulation of deposit from the geothermal brine is $0.0002 \mathrm{~m}^{2} . \mathrm{K} / \mathrm{W}$. Determine the number of passes required for the tubes inside the shell to heat the water to \(86^{\circ} \mathrm{C}\), within the temperature range required by the ANIS/NSF 3 standard.

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.

Sign-up for free