Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 155

A single-pass cross-flow heat exchanger uses hot air (mixed) to heat water (unmixed), flowing with a mass flow rate of \(3 \mathrm{~kg} / \mathrm{s}\), from \(30^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\). The hot air enters and exits the heat exchanger at \(220^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\), respectively. If the overall heat transfer coefficient is \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the required surface area.

Short Answer

Expert verified
Answer: The required surface area of the single-pass cross-flow heat exchanger is approximately 30.86 m².

Step by step solution

01

Calculate the heat transfer rate (Q)

To find the heat transfer rate (Q), we can use the specific heat capacity of water (Cpw = 4.18 kJ/kg.K) and the mass flow rate (m) of water, as well as the temperature difference in water from the inlet to the outlet, as follows: Q = m * Cpw * (T_outlet_water - T_inlet_water) where m = 3 kg/s (mass flow rate of water) Cpw = 4.18 kJ/kg.K (specific heat capacity of water) T_outlet_water = 80°C T_inlet_water = 30°C Q = 3 kg/s * 4.18 kJ/kg.K * (80°C - 30°C) Converting kJ to J: Q = 3 kg/s * 4180 J/kg.K * (50 K) Q = 627000 W The heat transfer rate at which water is being heated is 627000 W.
02

Calculate the log-mean temperature difference (ΔTm)

Next, we need to find the log-mean temperature difference (ΔTm) which can be used to determine the surface area. The log-mean temperature difference can be determined by using the inlet and outlet temperatures of both the hot air and the water: ΔT₁ = T_inlet_hotair - T_outlet_water ΔT₂ = T_outlet_hotair - T_inlet_water To find ΔTm, use the formula: ΔTm = (ΔT₁ - ΔT₂) / ln(ΔT₁ / ΔT₂) where T_inlet_hotair = 220°C T_outlet_hotair = 100°C T_inlet_water = 30°C T_outlet_water = 80°C ΔT₁ = 220°C - 80°C = 140 K ΔT₂ = 100°C - 30°C = 70 K ΔTm = (140 K - 70 K) / ln(140 K / 70 K) ΔTm ≈ 101.46 K The log-mean temperature difference is approximately 101.46 K.
03

Calculate the required surface area (A)

Now that we have found the heat transfer rate (Q) and the log-mean temperature difference (ΔTm), we can determine the required surface area (A) of the heat exchanger using the formula: Q = U * A * ΔTm where U = 200 W/m².K (overall heat transfer coefficient) Rearranging the formula for A and substituting known values: A = Q / (U * ΔTm) A = 627000 W / (200 W/m².K * 101.46 K) A ≈ 30.86 m² The required surface area of the single-pass cross-flow heat exchanger is approximately 30.86 m².

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.

Start your free trial

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 condenser unit (shell and tube heat exchanger) of an HVAC facility where saturated refrigerant \(\mathrm{R} 134 \mathrm{a}\) at a saturation pressure of \(1318.6 \mathrm{kPa}\) and a rate of \(2.5 \mathrm{~kg} / \mathrm{s}\) flows through thin-walled copper tubes. The refrigerant enters the condenser as saturated vapor and it is desired to have a saturated liquid refrigerant at the exit. The cooling of refrigerant is carried out by cold water that enters the heat exchanger at \(10^{\circ} \mathrm{C}\) and exits at \(40^{\circ} \mathrm{C}\). Assuming initial overall heat transfer coefficient of the heat exchanger to be \(3500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the surface area of the heat exchanger and the mass flow rate of cooling water for complete condensation of the refrigerant. In practice, over a long period of time, fouling occurs inside the heat exchanger that reduces its overall heat transfer coefficient and causes the mass flow rate of cooling water to increase. Increase in the mass flow rate of cooling water will require additional pumping power making the heat exchange process uneconomical. To prevent the condenser unit from under performance, assume that fouling has occurred inside the heat exchanger and has reduced its overall heat transfer coefficient by \(20 \%\). For the same inlet temperature and flow rate of refrigerant, determine the new flow rate of cooling water to ensure complete condensation of the refrigerant at the heat exchanger exit.

A shell-and-tube heat exchanger is used for cooling \(47 \mathrm{~kg} / \mathrm{s}\) of a process stream flowing through the tubes from \(160^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). This heat exchanger has a total of 100 identical tubes, each with an inside diameter of \(2.5 \mathrm{~cm}\) and negligible wall thickness. The average properties of the process stream are: \(\rho=950 \mathrm{~kg} / \mathrm{m}^{3}, k=0.50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=3.5 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) and \(\mu=2.0 \mathrm{mPa} \cdot \mathrm{s}\). The coolant stream is water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at a flow rate of \(66 \mathrm{~kg} / \mathrm{s}\) and an inlet temperature of \(10^{\circ} \mathrm{C}\), which yields an average shell-side heat transfer coefficient of \(4.0 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\). Calculate the tube length if the heat exchanger has \((a)\) a 1 -shell pass and a 1 -tube pass and (b) a 1-shell pass and 4-tube passes.

A shell-and-tube heat exchanger with 1-shell pass and 14-tube passes is used to heat water in the tubes with geothermal steam condensing at \(120^{\circ} \mathrm{C}\left(h_{f g}=2203 \mathrm{~kJ} / \mathrm{kg}\right)\) on the shell side. The tubes are thin-walled and have a diameter of \(2.4 \mathrm{~cm}\) and length of \(3.2 \mathrm{~m}\) per pass. Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes at \(22^{\circ} \mathrm{C}\) at a rate of \(3.9 \mathrm{~kg} / \mathrm{s}\). If the temperature difference between the two fluids at the exit is \(46^{\circ} \mathrm{C}\), determine (a) the rate of heat transfer, \((b)\) the rate of condensation of steam, and \((c)\) the overall heat transfer coefficient.

Steam is to be condensed on the shell side of a 2-shell-passes and 8-tube- passes condenser, with 20 tubes in each pass. Cooling water enters the tubes a rate of \(2 \mathrm{~kg} / \mathrm{s}\). If the heat transfer area is \(14 \mathrm{~m}^{2}\) and the overall heat transfer coefficient is \(1800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the effectiveness of this condenser is (a) \(0.70\) (b) \(0.80\) (c) \(0.90\) (d) \(0.95\) (e) \(1.0\)

In a 1-shell and 2-tube heat exchanger, cold water with inlet temperature of \(20^{\circ} \mathrm{C}\) is heated by hot water supplied at the inlet at \(80^{\circ} \mathrm{C}\). The cold and hot water flow rates are \(5000 \mathrm{~kg} / \mathrm{h}\) and \(10,000 \mathrm{~kg} / \mathrm{h}\), respectively. If the shelland-tube heat exchanger has a \(U A_{s}\) value of \(11,600 \mathrm{~W} / \mathrm{K}\), determine the cold water and hot water outlet temperatures. Assume \(c_{p c}=4178 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and \(c_{p h}=4188 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Modern Physics

Read Explanation

Astrophysics

Read Explanation

Physical Quantities And Units

Read Explanation

Fluids

Read Explanation

Atoms and Radioactivity

Read Explanation
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