An air-water vapor mixture enters a steady-flow heater...

An air-water vapor mixture enters a steady-flow heater humidifier unit at state $1: 10^{\circ} \mathrm{C}, 10 \%$ relative humidity, at the rate of $1 \mathrm{m}^{3} / \mathrm{s}$. A second air-vapor stream enters the unit at state $2: 20^{\circ} \mathrm{C}$ $20 \%$ relative humidity, at the rate of $2 \mathrm{m}^{3} / \mathrm{s} .$ Liquid water enters at state $3: 10^{\circ} \mathrm{C},$ at the rate of $400 \mathrm{kg}$ per hour. A single air-vapor flow exits the unit at state $4: 40^{\circ} \mathrm{C}$ shown in Fig. P12.124. Calculate the relative humidity of the exit flow and the rate of heat transfer to the unit.

Key Concepts

Heat Transfer in Humidification Processes

This concept involves understanding how energy is added or removed from a system to change the state of the air–vapor mixture, including both sensible and latent heating. In humidification, heat transfer is crucial as it not only raises the temperature (sensible heat) but also often involves phase changes or increases in the moisture content (latent heat). The overall heat balance helps in determining the rate of heat transfer required to achieve the desired exit conditions.

Steady?Flow Process

This concept involves analyzing systems in which fluid flows continuously through a control volume while system properties do not change with time. In such processes, the energy, mass, and momentum entering and leaving the system are balanced, simplifying the application of the conservation laws and allowing for the prediction of system behavior under dynamic conditions.

Mass and Moisture Balance

This principle focuses on the conservation of mass, notably for both the total mass and the water vapor component within air–water vapor mixtures. It requires accounting for the contributions from multiple inlet streams and any added liquid water, ensuring that the sum of the mass flows and the moisture content entering a system equals that leaving it, a crucial step for accurately predicting mixing outcomes like the exit relative humidity.

Psychrometrics

Psychrometrics is the study of the thermodynamic properties of moist air, including relationships among dry-bulb temperature, wet-bulb temperature, relative humidity, humidity ratio, and enthalpy. It provides the necessary tools and charts to analyze and predict the behavior and state properties of air–vapor mixtures, especially when they undergo processes such as heating, cooling, humidification, or dehumidification.

Mixing of Air Streams

When air streams with different thermodynamic states are combined, their properties mix based on mass-weighted averages. This concept uses both the energy and mass balances to determine the final mixture state in terms of temperature and humidity. Understanding the mixing process is essential for designing and analyzing systems where multiple air flows with varying properties converge.

Transcript

0:00 Hi, everybody.

00:01 So we need to calculate the thorough humidity of exit flow and the rate of heat transfer in the unit.

00:08 Okay.

00:10 So we have pv1 equals 5 times pg1 equals 0 .1 times 1 .2276 equals 0 .1 .2276.

00:31 K pascal's.

00:33 Okay.

00:35 And so now we have absolute humidity is 0 .622 times pv1 p minus pb1 equals 0 .622 times 1 .0 .1 276 100 minus 0 .12226 equals 0 .076 equals 0 .00745.

01:14 And we can do the mass flow rate of the air, air 1, a1 equals p -a -1, v -a -1 over r -t -a -1.

01:28 Okay, and so we have 100 minus 0 .12, 276 times 1 divided by 0 .287 times 10 plus 273 .15 equals 1 .29045 kilograms per second.

01:57 Okay.

01:59 And so we need to find the mass volume of vapor, and that's humidity times the mass air flow rate, which is 0 .007645 times 1 .229045, which is 0 .0045, which is 0 .00765, which is 0 .0045, which is 0 .0045.

02:29 9396 kilograms per second.

02:33 And so now we have the enthropoe of the water at 10 degrees celsius is 2519 .74 kilojoules per kilogram.

02:48 And that's also your hg1.

02:52 And let me let's pull down here.

02:54 And now at 5 .2, we need to find the pressure of p2, b2.

03:04 So that's going to be five times pg2, and that's 52, by the way.

03:10 And make that 5 to look nice.

03:12 And so it was 0 .2 times 2 .339.

03:17 And we get 0 .4678 kilopascals.

03:23 And now we have absolute humidity is 0 .622 times 0 .4678 divided by 100 minus 0 .4678 equals 0 .02923.

03:45 And we need to find the mass of acceleration here of the air, which is going to be 100 minus 0 .1 to 276 times 2 divided by 0 .287 times 20 plus 273 .15, which equals 2 .3742 kilograms per second.

04:21 And now mv2 equals humidity of 2, m .a2, which equals 0 .02923, times 2 .2 equals 0 .306.

04:43 939 kilograms per second.

04:50 Okay.

04:51 And so for the continuing equation of error in the system, we have m -a -1 plus m -a -2 equals m -a -4.

05:05 And we have 1 .229 .45 plus 2 .3742 equals m -a -4.

05:15 A4...

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