Experiment No. 3

DETERMINATION OF AIR PROPERTIES

Course Code: MEP510L2Program: BSME

Course Title: ME LAB 3 ( THERMAL LAB. 2)Date Performed: November, 11, 2014

Section: ME51FA1Date Submitted: November 18, 2014

Student: TAGUIBAO, Yvan Rae V.Instructor: Engr. Benjamin Tiglao

1. Objective:

The activity aims to introduce the properties of air.

2. Intended Learning Outcomes (ILOs):

The students shall be able to:2.1 Explain the properties of air2.2. Determine the properties of air using the psychrometric chart2.3 Develop professional work ethics, including precision, neatness, safety and ability to follow instruction.

3. Discussion:

Atmospheric air has a volumetric composition of 20.99 % oxygen, 78.03 % nitrogen, somewhat less than 1 % argon, with small quantities of several inert gases such as water vapor, carbon dioxide, helium, hydrogen, and neon. For most engineering calculations, it is usually accurate enough to include all inert gases as nitrogen and to use the analysis: 21 % oxygen and 79 % atmospheric nitrogen by volume. Thus, in 100 moles of air, there are approximately 21 moles of O2 and 79 moles of N2, or

79 moles N2= 3.7621 moles O2

or

ft3 N23.76ft3 O2

The approximate gravimetric composition of air is 23.1 % O2, 76.9 % N2, or there are 76.9 / 23.1 = 3.32 lb N2 / lb O2.

Sling Psychrometer

Aneroid Barometer

4. Materials and Equipment:

Sling Psychrometer Barometer Cotton/Cloth Tap Water Steam Table Stop watch

5. Procedure:

1. Using the barometer, determine the atmospheric pressure by placing the barometer on the floor. Locations to conduct this experiment are as required by the manual or prescribed by your instructor. Record each reading after 5 minutes.

2. Cover the bulb of one of the thermometers in the sling psychrometer with wet cloth or cotton (if not permanently attached).

3. Rotate the sling psychrometer horizontal to the ground for five minutes.

4. Record the dry bulb and wet bulb temperature reading.

5. Plot the results on a Psychrometric Chart.

USEFUL FORMULAS:

a.) V = Ra ( Tdb + 273 ) Patm - Pv

where:Pv = Pv - (Patm - Pv ) (Tdb - Twb )1500 - 1.44 Twb

Pw = Psat @ Twb

b.) % RH = Pv Pvs

where: Pvs = Psat @ Tdb

h = Cpa (Tdb 0) + whg

hg = hg @ Tdb

c.) W = 0.622 Pv Patm Pv

where:V = specific volume; m3/kgRa = gas constant of air; KJ/kg-kTdb = dry bulb temperature; CPatm = atmospheric pressurePv = vapor pressurePw = pressure @ wet bulb temperatureTwb = wet bulb temperatureRH = Relative HumidityPvs = saturated vapor pressureh = enthalpyw = Humidity ratiohg = enthalpy @ saturated vapor

6. Data and Results:

Trial / Location1 Ice Plant (A -102)2 Boiler Room (A -101)3 Cooling Tower 14 Cooling Tower 25 5th Floor

Atmospheric Pressure101.325 kPa101.325 kPa101.325 kPa101.325 kPa101.325 kPa

Wet Bulb Temperature24.5C25C24C24.5C24.5C

Dry Bulb Temperature29C29C28C28C29C

% Relative Humidity69.24%73.37%71.83%75.15%69.24%

Humidity Ratio0.0175 kgv / kgda0.0183 kgv / kgda0.0171 kgv / kgda0.0179 kgv / kgda0.0175 kgv / kgda

Enthalpy (KJ/kg Dry Air)74 kJ / kg75.92 kJ / kg71.82 kJ / kg73.86 kJ / kg74 kJ / kg

Vapor Pressure2.7752 kPa2.9008 kPa2.7166 kPa2.8422 kPa2.7752 kPa

Average Results:

Atmospheric Pressure: 101.325 kPa

Wet Bulb Temperature: 24.4 C

Dry Bulb Temperature: 28.6 C

% Relative Humidity: 72.46%

Humidity Ratio: 0.01766 kgv / kgda

Enthalpy: 73.92 kJ / kg

Vapor Pressure: 2.802 kPa

7. Computation, Analysis and Interpretation of Data:

For Ice Plant A 102Given:

Required:RH=?

SH = ?

Solution:

For Vapor Pressure (Pv):

By Interpolation:Temperature Pressure 24 C 0.002985 mPa 24.5 C Pwb 25 C 0.003169 mPa

For Relative Humidity (RH); Ps @ dry bulb temp. 29 C = 4.008 kPa

For Humidity Ratio (W):

For Enthalpy (h):

For Boiler Room A 101Given:

Required:RH=?

SH = ?

Solution:

For Vapor Pressure (Pv):

For Relative Humidity (RH); Ps @ dry bulb temp. 29 C = 4.008 kPa

For Humidity Ratio (W):

For Enthalpy (h):

For Cooling Tower 1Given:

Required:RH=?

SH = ?

Solution:

For Vapor Pressure (Pv):

For Relative Humidity (RH); Ps @ dry bulb temp. 28 C = 3.782 kPa

For Humidity Ratio (W):

For Enthalpy (h):

For Cooling Tower 2Given:

Required:RH=?

SH = ?

Solution:

For Vapor Pressure (Pv):

By Interpolation:Temperature Pressure 24 C 0.002985 mPa 24.5 C Pwb 25 C 0.003169 mPa

For Relative Humidity (RH); Ps @ dry bulb temp. 28 C = 3.782 kPa

For Humidity Ratio (W):

For Enthalpy (h):

For 5th FloorGiven:

Required:RH=?

SH = ?

Solution:

For Vapor Pressure (Pv):

By Interpolation:Temperature Pressure 24 C 0.002985 mPa 24.5 C Pwb 25 C 0.003169 mPa

For Relative Humidity (RH); Ps @ dry bulb temp. 29 C = 4.008 kPa

For Humidity Ratio (W):

For Enthalpy (h):

For Mean of Atmospheric PressureGiven:Trial 1 / Ice plant (A - 102): 101.325 kPaTrial 2 / Boiler Room (A - 101): 101.325 kPaTrial 3 / Cooling Tower 1: 101.325 kPaTrial 4 / Cooling Tower 2: 101.325 kPaTrial 5 / 5th floor: 101.325 kPaRequired:

Solution:

For Mean of Wet bulb TemperatureGiven:Trial 1 / Ice plant (A - 102): 24.5CTrial 2 / Boiler Room (A - 101): 25CTrial 3 / Cooling Tower 1: 24CTrial 4 / Cooling Tower 2: 24.5CTrial 5 / 5th floor: 24.5CRequired:

Solution:

For Mean of Dry bulb TemperatureGiven:Trial 1 / Ice plant (A - 102): 29CTrial 2 / Boiler Room (A - 101): 29CTrial 3 / Cooling Tower 1: 28CTrial 4 / Cooling Tower 2: 28CTrial 5 / 5th floor: 29CRequired:

Solution:

For Mean of Relative HumidityGiven:Trial 1 / Ice plant (A - 102): 69.24%Trial 2 / Boiler Room (A - 101): 73.37%Trial 3 / Cooling Tower 1: 71.83%Trial 4 / Cooling Tower 2: 75.15%Trial 5 / 5th floor: 69.24%Required:

Solution:

For Mean of Humidity RatioGiven:Trial 1 / Ice plant (A - 102): 0.0175 kgv / kgdaTrial 2 / Boiler Room (A - 101): 0.0183 kgv / kgdaTrial 3 / Cooling Tower 1: 0.0171 kgv / kgdaTrial 4 / Cooling Tower 2: 0.0179 kgv / kgdaTrial 5 / 5th floor: 0.0175 kgv / kgdaRequired:

Solution:

For Mean of EnthaplyGiven:Trial 1 / Ice plant (A - 102): 74 kJ / kg Trial 2 / Boiler Room (A - 101): 75.92 kJ / kgTrial 3 / Cooling Tower 1: 71.82 kJ / kgTrial 4 / Cooling Tower 2: 73.86 kJ / kg Trial 5 / 5th floor: 74 kJ / kgRequired:

Solution:

For Mean of Vapor PressureGiven:Trial 1 / Ice plant (A - 102): 2.7752 kPa Trial 2 / Boiler Room (A - 101): 2.9008 kPaTrial 3 / Cooling Tower 1: 2.7166 kPaTrial 4 / Cooling Tower 2: 2.8422 kPaTrial 5 / 5th floor: 2.7752 kPa Required:

Solution:

For Trial 1 / Ice Plant (A - 102)

For Trial 2 / Boiler Room (A - 101)

For Trial 3 / Cooling Tower 1

For Trial 4 / Cooling Tower 2

For Trial 5 / 5th floor

Analysis and Interpretation of Data:

Based on the data gathered from the experiment, using the above formula of the properties of air, the values of temperatures and pressures depends on the assigned location. Wet bulb temperatures are low because its already fall and winter is just around the corner. Atmospheric pressure did not change in trial 3 and trial 5 which are different elevation. Normally atmospheric pressure drops as you go up in higher places. Pressure drops at 265 millibars at every 10 km elevation, so the atmospheric pressure in trial 3 and trial 5 remains the same.

8. Conclusion and Recommendation:

By performing this experiment it helps me understand more about the properties of air. Air is composed of dry air and water vapor. Dry air is non-condensable while water vapor is the amount of space in air that can hold water. It is non-condensable and it varies depends of the temperature and pressure of the surrounding. In determining the properties of air I used the empirical formulas and steam table to get the values of saturation pressures and enthalpies. I also try to use the Psychrometric chart, the datas I get from the chart and the computed datas has a little difference. But I think the results in using the empirical formulas are much more accurate than the Psychrometric chart in determining the properties of air.

In performing the experiment always follow the instructions carefully and make sure the equipment and materials to be used are in good condition. In our case we used two sling psychrometers, and then compare both of their readings. I think the other one is faulty because the reading in the wet bulb and the dry bulb is too high compared to the other one, plus we performed the experiment during night time and its already fall so the temperature is expected to be lower.

9. Assessment Rubric:

T I P - V P A A 0 5 4 D Revision Status/Date:0/2009 September 09TECHNOLOGICAL INSTITUTE OF THE PHILIPPINESRUBRIC FOR LABORATORY PERFORMANCECRITERIABEGINNER1ACCEPTABLE2PROFICIENT3SCORE

Laboratory Skills

Manipulative SkillsMembers do not demonstrate needed skills.Members occasionally demonstrate needed skills.Members always demonstrate needed skills.

Experimental Set-upMembers are unable to set-up the materials.Members are able to set-up the materials with supervision.Members are able to set-up the material with minimum supervision.

Process SkillsMembers do not demonstrate targeted process skills.Members occasionally demonstrate targeted process skills.Members always demonstrate targeted process skills.

Safety PrecautionsMembers do not follow safety precautions.Members follow safety precautions most of the time.Members follow safety precautions at all times.

Work Habits

Time Management/Conduct of ExperimentMembers do not finish on time with incomplete data. Members finish on time with incomplete data.Members finish ahead of time with complete data and time to revise data.

Cooperative and Teamwork Members do not know their tasks and have no defined responsibilities. Group conflicts have to be settled by the teacher.Members have defined responsibilities most of the time. Group conflicts are cooperatively managed most of the time.Members are on tasks and have responsibilities at all times. Group conflicts are cooperatively managed at all times.

Neatness and OrderlinessMessy workplace during and after the experiment.Clean and orderly workplace with occasional mess during and after the experiment.Clean and orderly workplace at all times during and after the experiment.

Ability to do independent workMembers require supervision by the teacher.Members require occasional supervision by the teacher.Members do not need to be supervised by the teacher.

Other Comments/Observations:TOTAL SCORE

RATING= x 100%