Repeat Prob. 14-79 for a total pressure of 88 kPa for air. |...

Name: Problem 81, Chapter 14: Thermodynamics: An Engineering Approach
Uploaded: 2022-02-24T18:54:35-08:00
Duration: 11 min 47 s
Channel: Mohammad Mehran
Description: Repeat Prob. 14-79 for a total pressure of 88 kPa for air.

Key Concepts

Ideal Gas Law

The ideal gas law, expressed as PV = nRT, relates the pressure, volume, temperature, and number of moles of a gas. It is used in solving problems involving gases under various conditions and is fundamental for calculating the behavior of gases when the total pressure or temperature is altered.

Gas Mixture Composition

Understanding the typical composition of air, which is mostly nitrogen, oxygen, and small amounts of other gases, is important when applying gas laws to air. The composition determines the relative mole fractions of the component gases and thereby affects the partial pressures when the total pressure changes.

Dalton's Law of Partial Pressures

This principle states that in a gas mixture, the total pressure is the sum of the individual partial pressures of the component gases. It is crucial for determining the pressure contribution of each gas when given an overall total pressure, and it allows the calculation to adjust with changes in total system pressure.

Transcript

00:01 All right guys, in order to solve this problem, we first need to find the dew point temperature of the air that is tdb.

00:09 So for that we will first calculate the water vapor pressure from the given relative humidity that is 70 % 51 is equal to 70 % and the pressure pg at the given temperature that is t1 is equal to 32 degrees centigrates so we know that pv1, pv1 is equal to 51 into pg1 and that is equal to 0 .7 multiplied by 4 .76, 4 .76 is equal to 3 .332 kilopascal and tdp is equal to 25 .8 degrees centigrade.

01:16 Air is t2 that is equal to 20 degrees centigrade which means there will be some condensation of water in the cooling process.

01:31 To solve this problem we will first need to calculate some of the needed parameters from the given ambient pressure that is p is equal to 88 kioskl and the calculated water vapor water pressure, pv1, we can calculate the air pressure that is pa, pa1 is equal to p minus pv1 is equal to 95 minus 3 .332 is equal to 91 .668 kilopascal.

02:12 Now we will also need the absolute humidity, w1, enthalph, h1 and the specific volume v1 for the calculation.

02:20 We will use the given temperature t1 and the pressure.

02:24 P1 and pv1 and pa1, we will also need the enthalpy of saturated vapor that is h1hg1 is equal to 2 550 kilo joules per kg.

02:41 Now guys we have w1 is equal to 0 .622 into pv1 divided by p1 minus pv1 and that is equal to 0 .622 multiplied by 3 .332 divided by 95 minus 3 .32 and it's equal to 0 .0226.

03:15 So h1 is equal to cp into t1 plus w1 into hg1 and that is equal to 1 .005 multiplied by 32 plus 0 .0 026 .0226 multiplied by 2 550 and h1 is equal to 89 .79 kios per kg.

03:57 So guys we have v1 is equal to r a into t1 divided by pa1 that is equal to 0 .287 multiplied by 305 divided by 91 .668 is equal to 0 .95 cubic meters per kg.

04:34 So from the given velocity of air v1 is equal to 120 meters per minute and the radius of the tube r is equal to 0 .2m.

04:48 And the specific volume v1 we can calculate the mass flow rate of the air that is m is equal to v over v1 so we have v1 so we have v1 so we have v1 into r square into pi divided by small v1 and that is equal to 120 multiplied by 0 .2 into pi divided by 0 .955 and it's equal to 15 .794 kilojoules per minute.

05:37 So guys, the next we need to repeat the calculation for the exit point.

05:43 For the calculations, we will need the relative humidity that is 5 .2 is equal to 100%.

05:51 And the temperature g2.

05:53 So first we need the vapor pressure, vp, pv2, and from it the absolute humidity.

06:02 So pv2 is equal to 52 into pg2, that is 1 into 2 .3392 is equal to 2 .3392 kilopascal.

06:17 And w2 is equal to 0 .622 into pv2 divided by p1 minus pv2 and that is equal to 0 .622 minus 2 .3292 divided by 95 minus 2 .3392 and the answer is 0 .0157.

06:43 So next we can use the absolute humidity w2 to calculate.

06:48 The enthalpy h2 and we can also calculate the specific volume so h2 is equal to cp into t2 plus w2 into h2 that is equal to 1 .005 multiplied by 20 plus 0 .0157 multiplied by 25 137 .4 this is 25 1307 this is 25 1303 37 .4 is equal to 59 .94 kilojoules per kg...

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