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Chapter Tw o P sychrom etrics of A ir C ondition Processes Chapter Two Psychrometrics of Air Condition Processes 2.1- Psychrometric Chart It is a graphical representation of various thermodynamic properties of moist air. The psychrometric chart is very useful for finding out the properties of air (w...

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Psychrometrics of Air Condition Processes VENKITARAJ KONERY PURUSHOTHAMAN

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A new t heoret ical formulat ion of dew point t emperat ures applicable for comfort air-cooling s… Mridul Sarkar Chapt er 1. Review of T hermodynamics and Heat Transfer pedro badillo PSYCHROMET RIC ANALYSIS OF AN AIR-CONDIT IONING SYST EM OPERAT ING UNDER GIVEN AMBIENT … Equat orial Journals

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P sychrom etrics of A ir C ondition Processes

Chapter Two Psychrometrics of Air Condition Processes 2.1- Psychrometric Chart

Fig.(2.1) Psychrometric chart

It is a graphical representation of various thermodynamic properties of moist air. The psychrometric chart is very useful for finding out the properties of air (which are required in the field of air condition) and eliminate lot of calculations. There is a slight variation in the charts prepared by different air-conditioning manufactures but basically they are all alike. The psychrometric chart is normally drawn for standard atmospheric pressure of 760 mm of Hg (or 1.01325 bar).

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In a psychrometric chart, dry bulb temperature is taken as abscissa and specific heat i.e. moisture content as ordinate, as shown in Fig.(2.1). Now the saturation cure is drawn by plotting the various saturation points at corresponding dry bulb temperatures. The saturation curve represent 100% relative humidity at various dry bulb temperatures. It also represents the wet bulb and dew point temperatures. Though the psychrometric chart has a number of details, yet the following lines are important from subject point of view: 1. Dry bulb temperature lines: The dry bulb temperature lines are vertical i.e. parallel to the ordinate and uniformly spaced as shown in Fig.(2.2). Generally, the temperature range of these lines on psychrometric chart is from -6ºC to 45ºC. The dry bulb temperature lines are drawn with difference of every 5ºC and up to the saturation curve as shown in the figure. The values of dry bulb temperatures are shown on the saturation curve. 2. Specific humidity or moisture content lines: The specific humidity (moisture content) lines are horizontal i.e. parallel to the abscissa and are also uniformly spaced as shown in Fig.(2.3). Generally, moisture content range of these lines on psychrometric chart is from 0 to 30 g / kg of dry air (or from 0 to 0.030 kg / kg of dry air). The moisture content lines are drawn with a difference of every 1 g (0.001 kg) and up to the saturation curve as shown in the figure. 3. Dew point temperature lines: The dew point temperature lines are horizontal i.e. parallel to the abscissa and non-uniformly spaced as shown in Fig.(2.4). At any point on the saturation curve, the dry bulb and dew point temperatures are equal. The values of dew point temperatures are generally given along the saturation curve of the chart as shown in the figure.

Fig.(2.2) Dry bulb temperature lines

Fig.(2.3) Specific humidity lines

Fig.(2.4) Dew point temperature lines

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4. Wet bulb temperature lines: The wet bulb temperature lines are inclined straight lines and nonuniformly spaced as shown in Fig.(2.5). At any point on the saturation curve, the dry bulb and wet bulb temperatures are equal. The values of wet bulb temperatures are generally given along the saturation curve of the chart as shown in the figure. 5. Enthalpy (total heat) lines: The enthalpy (or total heat) lines are inclined straight lines and uniformly spaced as shown in Fig.(2.6). These lines are parallel to the wet bulb temperature lines, and are drawn up to the saturation curve. Some of these lines coincide with the wet bulb temperature lines also. The values of total enthalpy are given on a scale above the saturation curve as shown in the figure. 6. Specific volume lines: The specific volume lines are obliquely inclined straight lines and uniformly spaced as shown in Fig.(2.7). These lines are drawn up to the saturation curve. The values of volume lines are generally given at the base of the chart. 7. Relative humidity lines: The relative humidity lines are curved and follow the saturation curve. Generally, these lines are drawn with values of relative humidity 10%, 20%, 30% etc. and up to 100%. The saturation curve presents 100% relative humidity. The values of relative humidity lines are generally given along the lines themselves as shown in Fig.(2.8).

Fig.(2.5) Wet bulb temperature lines

Fig.(2.7) Specific volume lines

Fig.(2.6) Enthalpy lines

Fig.(2.8) Relative humidity lines

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Example 2.1. For a sample of air having 22ºC DBT, relative humidity 30 percent at barometric pressure of 760 mm of Hg, calculate: 1.Vapour pressure, 2. Humidity ratio, 3. Vapour density, and 4.Enthalpy. Verify your result by psychrometric chart. Solution: Given :td =22ºC ; φ =30%=0.3 ; pb=760 mm of Hg=760×133.4= 101384N/m2=1.01384 bar 1.Vapour pressure Let pv=Vapour pressure From steam tables (table 1.4), we find that the saturation pressure of vapour corresponding to dry bulb temperature of 22ºC is ps=0.026448 bar We know that relative humidity ( φ ), p pv 0 .3 = v = ps 0.026448 ∴ pv=0.3×0.02642=0.007934 bar 2.Humidity ratio We know that humidity ratio, pv 0.622 × 0.007934 = W = 0.622 × pb − pv 1.01384 − 0.007934 = 0.0049 kg / kg of dry air 3.Vapour density We know that vapour density, W ( pb − pv ) 0.0049(1.01384 − 0.007934 )10 5 = ρv = 287(273 + 22 ) Ra Td 3 = 0.00582 kg / m of dry air 4.Enthalpy We know that enthalpy, h = (1.007t d − 0.026 ) + W (2501 + 1.84t d ) = (1.007×22-0.026) + 0.0049× (2501+1.84×22) =34.58 kJ/kg dry air Verification from psychrometric chart The initial condition of air i.e. 22ºC dry bulb temperature and 30% relative humidity is marked on the psychrometric chart at point A as shown in Fig.(2.9) From point A, draw a horizontal line meeting the humidity ratio line at C. From the psychrometric chart, we find that the humidity ratio at point C, W 5 g/kg of dry air 0.005 kg/kg of dry air

Fig.(2.9)

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We also find from psychrometric chart that the specific volume at point A is 0.843 m3/kg of dry air.

∴ Vapour density, v=W/ a=0.005/0.843=0.0059 kg/m3 of dry air Now from point A, draw a line parallel to the wet bulb temperature line meeting the enthalpy line at point E. Now the enthalpy of air as read from the chart is 34.8 kJ/kg of dry air.

2.2 Psychrometric Processes The various psychrometric processes involved in air conditioning to vary psychrometric properties of air according to the requirement are as follows: 1. Adiabatic mixing of air streams 2. Sensible heating 3. Sensible cooling 4. Humidification and dehumidification 5. Cooling and adiabatic humidification 6. Cooling and humidification by water injection 7. Heating and humidification 8. Humidification by steam injection 9. Adiabatic chemical dehumidification

2.2.1 Adiabatic Mixing of Two Air Streams When two quantities of air having different enthalpies and different specific humidities are mixed, the final condition of the air mixture depends upon the masses involved, and on the enthalpy and specific humidity of each the constituent masses which enter the mixture. Now consider two air streams 1 and 2 mixing adiabatically as shown in Fig.(2.10)(a) Let m1= Mass of air entering at 1, h1= Enthalpy of air entering at 1, W1= Specific humidity of air entering at 1, m2, h2, W2=Corresponding values of air entering at 2, and m3, h3, W3=Corresponding values of air mixture at 3.

Fig(2.10)Adiabatic mixing of two air streams Assuming no loss of enthalpy and specific humidity during the air mixture process, we have for the mass balance of dry air, m1 + m2 = m3 ………………… (2.1)

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For the energy balance, m1 h1 + m2 h2 = m3 h3 ………………… (2.2) and for the mass balance of water vapour, m1 W1 + m2 W2 = m3 W3 ………………… (2.3) Substituting the value of m3 from equation (2.1) in equation (2.2), m1 h1 + m2 h2 = (m1 + m2 ) h3 = m1 h3 + m2 h3 or m1 h1 − m1 h3 = m2 h3 − m2 h2 m1 (h1 − h3 ) = m2 (h3 − h2 ) ∴

m2 h1 − h3 = ………………… (2.4) m1 h3 − h2

Or

h3 =

m1h1 + m2 h2 ………………… (2.5) m1 + m2

Similarly, substituting the value of m3 from equation (2.1) in equation (2.3), we have m2 W1 −W3 = ………………… (2.6) m1 W3 −W2 m W + m2W2 Or W3 = 1 1 ……………………… (2.7) m1 + m2 Now from equations (2.4) and (2.6) m2 h1 − h3 W1 − W3 ∴ = = ………………… (2.8) m1 h3 − h2 W3 − W2 The adiabatic mixing process is represented on the psychrometric chart as shown in Fig.(2.10) (b). The final condition of the mixture (point 3) lies on the straight line 1-2. The point 3 divides the line 1-2 in the inverse ratio of the mixing masses. Example 2.2. 300 m3/min of fresh air at 30 ºC (DBT) dry bulb temperature and 50% RH is to be mixed with 800 m3/min of recirculated air at 22 ºC (DBT) dry bulb temperature and 10 ºC dew point temperature. Determine the enthalpy, specific volume, humidity ratio, and dew point temperature of the mixture. Solution. Given: v1=300 m3/min; td1=30 ºC; φ1 =50%; v2=800 m3/min; td2=22 ºC; tdp2=10 ºC Enthalpy of the mixture Let h3= Enthalpy of the mixture The condition of recirculated air at 22 ºC DBT and 10 ºC dew point temperatures is marked on the psychrometric chart at point 2 as shown in Fig.(2.11). Now mark the condition of fresh air at 30 ºC dry bulb temperature and 50% relative humidity at point 1 as shown in the figure. Join 1 and 2.

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P sychrom etrics of A ir C ondition Processes h1 50%

h3

1 h2

W1

3

tdp3

W3

10 ºC

W2

2

0.876 m3/kg

vs3 0.846 m3/kg 30

22

Fig(2.11) From the psychrometric chart, we find that enthalpy of air at point 1, h1=64.6 kJ/kg of dry air Enthalpy of air at point 2, h2=41.8 kJ/kg of dry air Specific humidity of air at point 1, W1= 0.0134 kg/kg of dry air Specific humidity of air at point 2, W2= 0.0076 kg/kg of dry air Specific volume at point 1, vs1= 0.877 m3/kg of dry air and Specific volume at point 2, vs2= 0.846 m3/kg of dry air We know that mass of fresh air at point 1, v 300 m1 = 1 = = 342.07 kg / min v s1 0.877 and mass of recirculated air at point 2, v 800 m2 = 2 = = 945.6 kg / min v s 2 0.846 We know that m1 h3 − h2 342.07 h3 − 41.8 = = or 945.6 64.6 − h3 m2 h1 − h3 h3=47.86 kJ/kg of dry air Ans. ∴ Specific volume, humidity ratio, and dew point temperature of the mixture Plot point 3 on line joining the points 1 and 2 corresponding to enthalpy h3=47.86 kJ/kg of dry air, as shown in Fig.(2.11). From point 3 on psychrometric chart, we find that specific volume of the mixture at point 3, vs3= 0.855 m3/kg of dry air Ans. Humidity ratio of the mixture at point 3, W3= 0.0092 kg/kg of dry air Ans. And dew point temperature of the mixture at point 3, tdp3 12.7 ºC Ans.

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2.2.2 Sensible Heat Factor Actually, the heat added during a psychrometric process may be split up into sensible heat and latent heat. The ratio of the sensible heat to the total heat is known as sensible heat factor (briefly written as SHF) or sensible heat ratio (briefly written as SHR). Mathematically, Qs Sensible heat SH SHF = = = ………………(2.9) Total heat SH + LH Qs + Ql where SH=Sensible heat, and LH=Latent heat 1 ! " &)

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SHR ht

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2 Sensible heat ratio line

Fig(2.12) Sensible heat ratio line

2.2.3 Sensible Heating The heating of air, without any change in its specific humidity, is known as sensible heating. Let air at temperature td1 passes over a heating coil of temperature td3, as shown in Fig.(2.13)a. It may be noted that the temperature of air leaving the heating coil (td2) will be less than td3. The process of sensible heating, on the psychrometric chart, is shown by a horizontal line 1-2 extending from left to right as shown in Fig.(2.13) b. The point 3 represents the surface temperature of the heating coil. The heat absorbed by the air during sensible heating may be obtained from the psychrometric chart by the enthalpy difference (h2-h1) as shown in Fig.(2.13) b. It may be noted that the specific humidity during the sensible heating remains constant (i.e.W1=W2). The dry bulb temperature increases from td1 to td2 and relative humidity reduces from φ1 to φ2 as shown in Fig.(2.13) b. The amount of heat added during sensible heating may also be obtained from the relation: Heat added, q = h2 − h1 = c pa (t d 2 − t d 1 ) + W c ps (t d 2 − t d 1 )

= (c pa + W c ps )(t d 2 − t d 1 ) = c pm (t d 2 − t d 1 )

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P sychrom etrics of A ir C ondition Processes

The term (c pa + W c ps ) is called humid specific heat (cpm) and its value is taken as 1.022 kJ/kg K. ∴ Heat added

q = 1.022 (t d 2 − t d 1 ) kJ / kg ………….(2.10)

Fig(2.13) Sensible Heating Notes: 1- For sensible heating, steam or hot water is passed through the heating coil. The heating coil may be electric resistance coil. 2- The sensible heating of moist air can be done to any desired temperature.

2.2.4 Sensible Cooling The cooling of air, without any change in its specific humidity, is known as sensible cooling. Let air at temperature td1 passes over a cooling coil of temperature td3, as shown in Fig.(2.14)a. It may be noted that the temperature of air leaving the heating coil (td2) will be more than td3. The process of sensible cooling, on the psychrometric chart, is shown by a horizontal line 1-2 extending from right to left as shown in Fig.(2.14) b. The point 3 represents the surface temperature of the cooling coil.

Fig(2.14) Sensible Cooling The heat rejected by the air during sensible cooling may be obtained from the psychrometric chart by the enthalpy difference (h1-h2) as shown in Fig.(2.14) b. It may be noted that the specific humidity during the sensible cooling remains constant (i.e. W1=W2). The dry bulb temperature reduces from td1 to td2 and relative humidity increases from φ1 to φ2 as shown in Fig.(2.14) b. The amount of heat rejected during sensible cooling may also be obtained from the relation: Heat rejected,

q = h1 − h2 = c pa (t d 1 − t d 2 ) + W c ps (t d 1 − t d 2 )

= (c pa + W c ps )(t d 1 − t d 2 ) = c pm (t d 1 − t d 2 )

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P sychrom etrics of A ir C ondition Processes

The term (c pa + W c ps ) is called humid specific heat (cpm) and its value is taken as 1.022 kJ/kg K.

q = 1.022 (t d 1 − t d 2 ) kJ / kg

∴ Heat rejected, Generally, Heat rejected,

q = m × (h1 − h2 ) ………………(2.11) Notes: 1- For sensible cooling, the cooling coil may have refrigerant, cooling water or cool gas flowing through it. 2- The sensible cooling can be done only upto the dew point temperature (tdp) as shown in Fig.(2.14) b. The cooling below this temperature will result in the condensation of moisture.

2.2.5 By-pass Factor of Heating and Cooling Coil We already discussed that the temperature of the air coming out of the apparatus (td2) will be less than *td3 in case the coil is a heating coil and more than td3 in case the coil is a cooling coil. Let 1 kg of air at temperature td1 is passed over the coil having its temperature (i.e. coil temperature surface) td3 as shown in Fig. (2.15). A little consideration will show that when air passes over a coil, some of it (say x kg) just bypasses unaffected while the remaining (1-x) kg comes in direct contact with the coil. This by-pass process of air is measured in terms of a by-pass factor. The amount of air that by-passes or by-pass factor depends upon the following factors: 1. The number of fins provided in a unit length i.e. the pitch of the cooling coil fins; 2. The number of rows in a coil in the direction of flow ; and 3. The velocity of flow air. It may be noted that by-pass factor of a cooling coil decreases with decrease in fin spacing and increase in number of rows.

Fig(2.15) By-pass factor Balancing the enthalpies, we get x c pm t d 1 + (1 − x )c pm t d 3 = 1× c pm t d 2

x (t d 3 − t d 1 ) = t d 3 − t d 2

...(where c pm = Specific humid heat )

td 3 − td 2 …………………..(2.12) t d 3 − t d1 where x is called by-pass factor of the coil and is generally written as BPF. Therefore, by-pass factor for heating coil, ∴

x=

BPF =

td 3 − td 2 td 3 − td1

…………………..(2.13)

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P sychrom etrics of A ir C ondition Processes

* Under ideal condition, the dry bulb temperature of the air leaving the apparatus ( t d 2 ) should be equal to that of the coil ( t d 3 ). But it is not so, because of the inefficiency of the coil. This phenomenon is known as by-pass factor.

Similarly, by-pass factor for cooling coil, t −t BPF = d 2 d 3 …………………..(2.14) td1 − td 3 The value of x (BPF) also is obtained by balancing the enthalpies as follows: h2 = x h1 + (1 − x )h3 for heating coil h − h2 x= 3 ……………………….. (2.15) h3 − h1 for cooling coil h −h x= 2 3 ………………….. (2.16) h1 − h3 Note: The performance of a heating or cooling coil is measured in terms of a by-pass factor. A coil with low by-pass factor has better performance.

2.2.6 Efficiency of Heating and Cooling Coils The term (1-BPF) is known as efficiency of coil or contact factor. ∴

Efficiency of the heating coil, t −t t −t η H = 1 − BPF = 1 − d 3 d 2 = d 2 d 1 td 3 − td1 td 3 − td1 Similarly, efficiency of the cooling coil,

ηC = 1 −

t d 2 − t d 3 t d1 − t d 2 = …………………..(2.17) t d1 − t d 3 td1 − t d 3

Example 2.3. In a heating application, moist air enters a steam heating coil at 10º C, 50% RH and leaves at 30º C. Determine the sensible heat transfer, if mass flow rate of air is 100kg of dry air per second. Also determine the steam mass flow rate if steam enters saturated at 100º C and condensate leaves at 80º C. Solution. Given: td1=10º C ; φ1 =50% ; td2=30º C; ma=100kg/s ; ts= 100º C; tC= 80º C Sensible heat transfer First, mark the initial condition of air, i.e. 10 ºC dry bulb temperature and 50% relative humidity on the psychrometric chart at point 1, as shown in Fig.(2.16). Draw a constant specific humidity line from point 1 to intersect the vertical line drawn through 30º C dry bulb temperature at point 2. the line 1-2 represents sensible heating of air. From the psychrometric chart, we find that enthalpy at point 1, h1=19.5 kJ/kg of dry air Fig(2.16) and enthalpy at point 2, h2=40 kJ/kg of dry air We know that sensible heat transfer, Q=ma(h2-h1)=100(40-19.5)=2050 kJ/s Ans.

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Steam mass flow rate From steam tables (table (1.4)), corresponding to temperature of 100º C, we find that enthalpy of saturated steam, hg=2675.44 kJ/kg and enthalpy of condensate, corresponding to 80 ºC, hf=335 kJ/kg ∴ Steam mass flow rate Q 2050 = = = 0.876 kg / s hg − h f 2675.44 − 335 = 0.876 × 3600 = 3153 kg / h

Ans.

Example 2.3. The air enters a duct at 10 ºC and 80% RH at the rate of 150 m3/min and is heated t...