SHORT QUESTION ANSWERS-PSYCHROMETRY CLASS NOTES FOR MECHANICAL ENGINEERING

PSYCHROMETRY SHORT

QUESTION ANSWERS CLASS

NOTES FOR MECHANICAL

ENGINEERING

Question answers are helpful in

understanding the psychrometry. This

also increases clarity. It is useful in design

of air conditioning equipment. This also

help in overcoming practical difficulties.

Q1. Define dry bulb temperature, wet bulb temperature and dew point temperature.

Dry bulb temperature, tdb

It is temperature of a space. It is measured by an ordinary thermometer. A vertical line represents it on the Psychrometric chart.

Wet Bulb Temperature, twb

It is temperature read from the stem of a thermometer whose bulb is surrounded by a constantly wet cloth. Air is blown over the wet cloth. Some water goes in air with water. This evaporated water causes cooling.  The reading of thermometer decreases. Heat will flow from atmosphere to bulb. This drop continues for some time till entire heat required for evaporation is supplied from the atmosphere. Now the reading of the thermometer becomes constant. It is the wet bulb temperature.

Actually it is the  adiabatic saturation temperature of air.  Air becomes saturated in the vicinity of the thermometer bulb by taking entire heat for evaporation from the atmosphere. The condition becomes steady state condition. The thermometer shows a constant reading. It is wet bulb temperature. Find it by moving along the constant enthalpy line up to the saturation point on the Psychrometric chart.

Dew point temperature

Sensibly cool the atmospheric air by removing heat. Its temperature will start decreasing. Ultimately a temperature will be reached when the first drop of water condenses. This temperature is called Dew Point Temperature. It is also a state of saturated air but its heat content has been decreased from the starting point. It is found by moving horizontally up to the saturation limit from the given condition on the Psychrometric chart.

Dew Point Depression

It is the difference between dry bulb temperature and dew point temperature.

Wet Bulb Depression

It is the difference of wet bulb temperature and the dew point temperature.

Apparatus Dew Point Temperature

It is the temperature of the cooling coil (Evaporator) in an air conditioning system.

Q2. Differentiate between specific humidity and relative humidity.

Specific humidity is the amount of water vapors per kg of dry air  i.e. say 10g/kg of dry air

=0.01 kg/kg of dry air. Its symbol is w or wh.

Example: Moist air is a mixture of dry air and water vapor. Say there is 10 g of water vapors per kg of moist air i.e. 10 g water vapor in 990 g of dry air. Then the specific

humidity will be 10 x 1000/990 = 11.11 g/kg of dry air

Relative humidity: Ratio of mass of water vapors /m3 of dry air to mass of water vapors /m3 of dry air when saturated at the same temperature. Its symbol is RH. It is expressed as %.

Example: 20 g of water vapor in 1 m3 of dry air say at 20C.  55 g of water vapor in 1 m3 of dry air at 200C when saturated. Then

RH= (20/55) x 100 = 36.36 %

Define Degree of saturation

 Ratio of specific humidity to saturated specific humidity at the same temperature.

It has no units. It is always less than 1.Its symbol is ψ.

Q3. Effect of water addition on Dew Point and Relative Humidity.

 If some water sprinkles in a room, how does it affect the followings:

(i) the dew point

(ii) the relative humidity.

When water sprinkles in a room, air gets water. Temperature of the room lowers as in a desert cooler .

(i) Adding of water to air increase the dew point temperature.

(ii) Lowering of temperature increases the relative humidity since RH=pv/pvs

pv is partial pressure of water vapor which increases by the addition of water.

pvs is saturation pressure of water vapor which decreases with lowering of room temperature. Thus RH increases.

Q4. When air temperature increases, its relative humidity  decreases. Why?

RH is=pv/pvs.

With the increase of room temperature, pv remains constant. But pvs  increases. Therefore RH decreases.

Q5. DBT and DPT of air are 160C and 7.60C respectively. If the saturated pressure is 13.5 mm of Hg at 160C and 7.8 mm of Hg at 7.60C, find the relative humidity of air.

RH is=pv/pvs.

Where pv is saturation pressure at Dew point temperature which is 7.8 mm of Hg

pvs is saturation pressure at Dry bulb temperature which is 13.5 mm of Hg.

Therefore RH=7.8/13.5= 59 %

 Q6. Given  the DBT is 35C and WBT is 230C. Calculate

( a) relative humidity
( b) humidity ratio
(c) dew point temperature
(d) density
(e) enthalpy of atmospheric air

Plot the DBT and WBT on the Psychrometric chart and read the values:
RH= 36%,
humidity ratio=15.25 g/kg of d.a.
DPT =17.80C
Density = (1/0.89) kg/m3 of d.a.
h=68 kJ/kg
Q7. WBT of saturated air is 280C. What will be its DBT, DPT and RH?

 For saturated air, DBT=WBT=DPT=280C

And RH=100%

Q8. Define SHF, By pass factor(BPF), Contact factor and apparatus Dew point.

SHF=(SHL/(SHL+LHL))

BPF for a cooling coil =temperature drop not achieved/max temperature drop achievable

= (t2—t3)/(t1—t3)

Where  t1 is temperature of hot air entering the cooling coil

t2 is the temperature of air after the cooling coil

tis the temperature of the cooling coil.

Contact  Factor= 1—BPF

Apparatus Dew Point Temperature= temperature of the cooling coil

= temperature of the evaporator

This point lies on 100% RH line of the Psychrometric chart.

It is obtained (i) By the intersection of ERSHF line with 100 % RH curve

(ii) By the intersection of GSHF line with 100 % RH curve.

Q9. What are psychrometric processes? Explain briefly all the psychrometric processes in which moisture is removed or added using psychrometric chart.

Psychrometric processes change the existing condition of air to the desired condition. It may be comfort air conditioning or industrial air conditioning.

Various psychrometric processes

Fig. Psychrometric Processes

(i) Sensible heating, OA: Air is heated to raise its temperature alone keeping moisture constant

(ii) Sensible cooling, OB: Air is heated to decrease its temperature alone keeping moisture constant

(iii) Humidification, OC: adding moisture keeping temperature constant

(iv)         Dehumidification, OD: Removing moisture keeping temperature constant

(v)          Heating and humidification, OE: Increase of temperature and increase of moisture

(vi)         Cooling and dehumidification, OF: decrease of temperature and decrease of moisture

(vii)        Cooling and humidification, OG: decrease of temperature but increase of moisture

(viii)       Heating and dehumidification, OH: Increase of temperature but decrease of moisture.

(ix)         Adiabatic saturation: Adding moisture at constant enthalpy till the air becomes fully saturated.

Q10. What do you mean by adiabatic saturation temperature?

From the given condition of air, move along constant enthalpy line to RH of 100 %.

Adiabatic saturation is also the wet bulb temperature. Adding moisture at constant enthalpy till the air becomes saturated i.e. no more moisture addition possible after saturation. The temperature thus achieved is adiabatic saturation temperature. This is wet bulb temperature (WBT). There is process of adiabatic saturation  means saturating the air at constant enthalpy.

  Q11. Explain the importance of comfort chart.


Importance

It suggests that a range of temperatures and relative humidity  gives the same degree of comfort. Hence system becomes more flexible. 

Q13. Define a psychrometric chart and explain its utility.

 Psychrometric Chart

Fig. Psychrometric Chart

It is a chart to study the psychrometric properties of moist air.

Firstly read Dry bulb temperature on X-axis.

Secondly read Wet bulb temperature on X-axis

Thirdly read Dew point temperature on X-axis.

Specific humidity is along y-axis

Sensible heat factor is along y-axis

Firstly there are constant Relative Humidity lines, MR and QP. These are  10 %, 30 % , 50 and 100%. 100 % RH line is on extreme left.

Secondly Constant enthalpy lines AB, CD and EF,

Thirdly Constant specific volume lines, GH, IJ and KL.

In order to use the chart, any two properties are given.

Locate a point from the given conditions and find the properties.

Mark a red point at DBT of 950F and WBT of 700F. Then find properties from the Chart.

Determine specific humidity on y- axis by moving horizontally.

Locate and read  Dew point temperature by moving horizontally towards left up to the saturated curve.

Find enthalpy by moving along constant enthalpy from the red point.

Read RH by drawing a curve parallel to nearing curve of RH through the marked point.

Find the specific volume of air.

 Q14. Utility of the Chart

(i) It decides the processes to the make atmospheric air to acquire the comfortable conditions.

(ii)Used in the design of cooling coil of the air conditioning plants.

 Q15.  What are psychrometric processes? Explain briefly all the moisture removal or addition processes. 

Psychrometric processes change the condition of air. It changes air to comfort conditions.

Various psychrometric processes are

(i) Sensible heating

Heat the air to raise its temperature alone, keeping moisture constant.

(ii) Sensible cooling

Cool to the air to decrease its temperature alone, keeping moisture constant.

(iii) Humidification: adding moisture keeping temperature constant

(iv) Dehumidification: Removing moisture keeping temperature constant

(v) Heating and humidification: Increase of temperature and increase of moisture

(vi) Cooling and dehumidification: decrease of temperature and decrease of moisture

(vii) Cooling and humidification: decrease of temperature but increase of moisture

(viii) Heating and dehumidification: Increase of temperature but decrease of moisture.

(ix)  Adiabatic saturation: Adding moisture at constant enthalpy till the air becomes fully saturated

Q16. Explain the use of Psychrometric Chart for the design of air conditioning apparatus.

Design means finding

(a)Apparatus dew point

(b) Temperature of air leaving cooling coil

These are achieved after locating different points on the psychometric chart:

(i) Inside Design Conditions

(ii) Outside Design Conditions

(iii) Condition of Mixture of re-circulated and fresh air

(iv) Location of alignment circle point ( 260C, 50 % RH point)

(v) Location of SHF on the SHF scale in the upper right corner of the chart. Join it with alignment circle point.

(v) locating supply air condition from the cooling coil

(a) Draw lines parallel to the RSHF line from inside condition. Then draw parallel line from mixture condition for GSHF line.

(vi) locating ADP (Apparatus Dew Point Temperature) i.e. Temperature of Cooling Coil where GSHF line meets the saturation curve.

Q17. At which condition the temperature of a room and the dew point temperature are equal and why?

 For fully saturated air, its RH=100 %. The room temperature and DPT are equal.  Water comes out when fully saturated air  cools. It takes place at the dew point.

Q18. Explain the physical significance of By Pass Factor for a heating coil.

Heating coils

air is not heated to  coil temperature.  Some air do not come in physical contact with the coil.  The air not coming in contact is said to be by passed.

If t1 is temperature of air entering the heating coil.= say 5oC

t3 is the temperature of air leaving the heating coil = say 40oC

tis the temperature of the heating coil= say 70oC

Temperature rise  achieved is (t3 – t1  ) = 40 –5 = 35  

temperature rise which was possible t2 – t1.= 70 –5 = 65

Contact factor for the heating coil

= temperature rise obtained/ temperature rise possible

= 35/65 = 0.54

Contact factor is always less than 1.

By pass factor (BPF) for heating coil = 1–CF

BPF is also always less than 1

CF + BPF = 1

Q13.  Explain the physical significance of By Pass Factor for a cooling coil.

Cooling coil

Air is not cooled to  coil temperature.  Some air do not come in physical contact with the coil. This air not coming in contact is said to be by passed.

If t1 is temperature of air entering the cooling coil.= say 55oC

t3 is the temperature of air leaving the cooling coil = say 100C

tis the temperature of the cooling coil= say 5oC

Temperature decrease  achieved is (t1 – t2  ) = 55 –10 = 45  

temperature decrease which was possible t1 – t2.= 55 –5 = 50

Contact factor of the cooling coil

= Actual temperature decrease/ temperature decrease possible

= 45/50 = 0.90

Contact factor is always less than 1.

Hence By pass factor (BPF) for cooling coil  = 1– CF= 1–0.9 = 0.10

BPF is always less than 1.

CF + BPF = 1

It is always advisable to have high contact factor and least by pass factor.

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