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

An air handler is a large unmixed heat exchanger used for comfort control in large buildings. In one such application, chilled water \(\left(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters an air handler at \(5^{\circ} \mathrm{C}\) and leaves at \(12^{\circ} \mathrm{C}\) with a flow rate of \(1000 \mathrm{~kg} / \mathrm{h}\). This cold water cools air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) from \(25^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\). The rate of heat transfer between the two streams is (a) \(8.2 \mathrm{~kW}\) (b) \(23.7 \mathrm{~kW}\) (c) \(33.8 \mathrm{~kW}\) (d) \(44.8 \mathrm{~kW}\) (e) \(52.8 \mathrm{~kW}\)

Short Answer

Expert verified
Question: Calculate the rate of heat transfer between the air and water streams in the given air handler. Solution: The rate of heat transfer between the air and water streams is 8.162 kW.

Step by step solution

01

Determine the mass flow rate of the water

Convert the mass flow rate of the water from kg/hour to kg/second by dividing by 3600 (as there are 3600 seconds in an hour).\[\dot{m}_{w} = \frac{1000 \text{ kg/hr}}{3600 \text{ s/hr}} = 0.2778 \text{ kg/s}\]
02

Calculate the temperature change of the water

Subtract the inlet temperature of the water from the outlet temperature. \[\Delta T_{w} = T_{2w} - T_{1w} = 12°C - 5°C = 7°C\]
03

Calculate the heat transfer for the water

Using the equation for heat transfer, Q, find the heat transfer for the water: \[Q_w = \dot{m}_w c_{p,w} \Delta T_w = 0.2778 \text{ kg/s} \cdot 4.2 \text{ kJ/kg K} \cdot 7 \text{ K} = 8.162 \text{ kJ/s} = 8.162 \text{ kW}\]
04

Determine the mass flow rate of the air

We can use the same mass flow rate as the water since the heat transfer between the two streams must be equal under steady-state conditions. Thus, \[\dot{m}_a = \dot{m}_{w} = 0.2778 \text{ kg/s}\]
05

Calculate the temperature change of the air

Subtract the inlet temperature of the air from the outlet temperature. \[\Delta T_{a} = T_{2a} - T_{1a} = 15°C - 25°C = -10°C\]
06

Calculate the heat transfer for the air

Using the equation for heat transfer, Q, find the heat transfer for the air: \[Q_a = \dot{m}_a c_{p,a} \Delta T_a = 0.2778 \text{ kg/s} \cdot 1.0 \text{ kJ/kg K} \cdot (-10) \text{ K} = -2.778 \text{ kJ/s} = -2.778 \text{ kW}\] Since the heat transfer between the two streams must be equal under steady-state conditions, we can choose the positive value for the rate of heat transfer between the two streams. Therefore, the rate of heat transfer is 8.162 kW, which corresponds to option (a).

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!

Key Concepts

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

Heat Exchanger
A heat exchanger is a device designed to efficiently transfer heat from one medium to another. In the context of an air handler in large buildings, which is an example of a large unmixed heat exchanger, chilled water circulates through the system to cool down the air for comfort control. The basic principle involves the chilled water absorbing heat from the warmer air, thereby reducing the air's temperature.

Heat exchangers come in various types and are used in a wide range of applications, including heating systems like boilers and radiators, as well as in cooling and refrigeration. Factors influencing their efficiency include the surface area of the exchanger, the temperature difference between the media, and the properties of the fluids involved like their specific heat capacity and flow rates.
Specific Heat Capacity
Specific heat capacity, often denoted as \(c_p\), is a substance's heat capacity per unit mass. It measures how much heat energy is required to raise the temperature of one kilogram of a substance by one degree Celsius (or one Kelvin). The value of \(c_p\) depends on the material; for example, water has a high specific heat capacity, meaning it can absorb a lot of heat without significantly changing its temperature. This makes it an effective fluid for use in heat exchangers.

Understanding the concept of specific heat capacity is essential when solving problems related to temperature changes and heat transfer, as it helps determine how much energy is involved in heating or cooling a certain mass of the substance.
Mass Flow Rate
The mass flow rate, denoted by \(\dot{m}\), is the amount of mass passing through a cross-sectional area per unit time. It is commonly measured in kilograms per second (kg/s). In our exercise, the mass flow rate of the water in the air handler is given in kilograms per hour and must be converted to kilograms per second to be useful in the calculation of heat transfer.

The mass flow rate is a crucial factor in calculating the heat transfer rate in a heat exchanger because it indicates how much mass is available to absorb or transfer heat in a given period.
Temperature Change
Temperature change, represented as \(\Delta T\), is the difference between the final and initial temperatures of a substance after heat has been added or removed. In heat transfer calculations, it's one of the vital parameters as it quantifies the degree to which the temperature of a substance has been altered.

For a substance with known specific heat capacity and mass flow rate, the temperature change can be used to compute the overall heat transfer using the formula \(Q = \dot{m} c_p \Delta T\). Positive values of \(\Delta T\) indicate a gain in thermal energy, while negative values indicate a loss. Consistent with the principle of conservation of energy, the amount of heat lost by one medium in a heat exchanger should equal the heat gained by the other medium under steady-state conditions.

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

There are two heat exchangers that can meet the heat transfer requirements of a facility. Both have the same pumping power requirements, the same useful life, and the same price tag. But one is heavier and larger in size. Under what conditions would you choose the smaller one?

Hot oil \(\left(c_{p}=2.1 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(110^{\circ} \mathrm{C}\) and \(8 \mathrm{~kg} / \mathrm{s}\) is to be cooled in a heat exchanger by cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) entering at \(10^{\circ} \mathrm{C}\) and at a rate of \(2 \mathrm{~kg} / \mathrm{s}\). The lowest temperature that oil can be cooled in this heat exchanger is (a) \(10.0^{\circ} \mathrm{C}\) (b) \(33.5^{\circ} \mathrm{C}\) (c) \(46.1^{\circ} \mathrm{C}\) (d) \(60.2^{\circ} \mathrm{C}\) (e) \(71.4^{\circ} \mathrm{C}\)

Cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes of a heat exchanger with 2 -shell passes and 23 -tube passes at \(14^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\), while hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the shell at \(200^{\circ} \mathrm{C}\) at the same mass flow rate. The overall heat transfer coefficient based on the outer surface of the tube is \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the heat transfer surface area on that side is \(20 \mathrm{~m}^{2}\). Determine the rate of heat transfer using \((a)\) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

In a parallel-flow, water-to-water heat exchanger, the hot water enters at \(75^{\circ} \mathrm{C}\) at a rate of \(1.2 \mathrm{~kg} / \mathrm{s}\) and cold water enters at \(20^{\circ} \mathrm{C}\) at a rate of \(09 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient and the surface area for this heat exchanger are \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(6.4 \mathrm{~m}^{2}\), respectively. The specific heat for both the hot and cold fluid may be taken to be \(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). For the same overall heat transfer coefficient and the surface area, the increase in the effectiveness of this heat exchanger if counter-flow arrangement is used is (a) \(0.09\) (b) \(0.11\) (c) \(0.14\) (d) \(0.17\) (e) \(0.19\)

A 2-shell passes and 4-tube passes heat exchanger is used for heating a hydrocarbon stream \(\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) steadily from \(20^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\). A water stream enters the shellside at \(80^{\circ} \mathrm{C}\) and leaves at \(40^{\circ} \mathrm{C}\). There are 160 thin-walled tubes, each with a diameter of \(2.0 \mathrm{~cm}\) and length of \(1.5 \mathrm{~m}\). The tube-side and shell-side heat transfer coefficients are \(1.6\) and \(2.5 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) Calculate the rate of heat transfer and the mass rates of water and hydrocarbon streams. (b) With usage, the outlet hydrocarbon-stream temperature was found to decrease by \(5^{\circ} \mathrm{C}\) due to the deposition of solids on the tube surface. Estimate the magnitude of fouling factor.

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