MOLLIER DIAGRAM

Dew Point - Tdp• The Dew Point is the temperature at which water

vapor starts to condense out of the air, the temperature at which air becomes completely saturated. Above this temperature the moisture will stay in the air.

• If the dew-point temperature is close to the air temperature, the relative humidity is high, and if the dew point is well below the air temperature, the relative humidity is low.

• The Dew Point temperature can be measured by filling a metal can with water and ice cubes. Stir by a thermometer and watch the outside of the can. When the vapor in the air starts to condensate on the outside of the can, the temperature on the thermometer is pretty close to the dew point of the actual air.

• The dew point temperature can be read by following a vertical line from the state-point to the saturation line. Dew point is represented along the 100% relative humidity line in the Mollier diagram.

Dry-Bulb Temperature - Tdb• Dry bulb temperature is usually referred to as air

temperature, is the air property that is most common used. When people refer to the temperature of the air, they are normally referring to its dry bulb temperature. Dry-bulb temperature - Tdb, can be measured by using a normal thermometer. The dry-bulb temperature is an indicator of heat content and is shown along the left axis of the Mollier diagram. The horizontal lines extending from this axis are constant-temperature lines.

Wet-Bulb Temperature - Twb• Wet bulb temperature is associated with the moisture

content of the air. Wet bulb temperature can be measured with a thermometer that has the bulb covered with a water-moistened bandage with air flowing over the thermometer. Wet bulb temperatures are always lower than dry bulb temperatures but they will be identical with 100% relative humidity in the air (the air is at the saturation line). On the Mollier diagram, the wet-bulb lines slope a little upward to the left (dotted lines).

Heating of Air

Cooling and Dehumidfying Air

Mixing of Air of different Conditions

The heat balance for the mix can be expressed as:LA hA + LC hC = (LA + LC)hBwhereL = mixing rateh = enthalpy of the airThe moisture balance for the mix can be expressed as:LA xA + LC xC = (LA + LC) xBwherex = water content in the airCalculating the mixture variables xB and hB makes it possible to calculate the mixing temperature tB.

Humidifying ,Adding Steam or Water (liquid)

Psychrometric ChartThe psychrometric chart is a variant of the Mollier diagram used in some parts of the world. The process transforming a Mollier diagram to a psychrometricchart is shown below. First it has to be reflected in a vertical mirror, then rotated 90 degrees.

Evaporation from Water Surfaces

The amount of evaporated water can be expressed as:

( )c sevapp

A h x xcm

⋅

∞

⋅≈ −

c

s

amount of evaporated water (kg/s)

A = water surface area (m2)h = heat transfer coefficient (W/m2 K)

= mean specific heat for moist air (J/kg K) = humidity ratio in the air (kg/kg)x = humi

evap

pcx

m⋅

∞

=

dity ratio in saturated air at the same temperature as the water surface (kg/kg)

Problem 6 (page 22)

An indoor pool evaporates a certain amount of water, which is removed by a dehumidifier to maintain +25ºC, φ=70% RH in the room (state 1 in figure). The dehumidifier, shown in figure, is a refrigeration cycle in which moist air flowing over the evaporator cools such that liquid water drops out, and the air continues flowing over the condenser. The air after the evaporator (state 2) has a temperature of +14ºC. For an air flow of 0,10 kg/s dry air the unit has a coefficient of performance COPR =3,0. Total pressure in the room is constant 101325 Pa.Calculatea) the amount of water that evaporates from the pool ( steady state) b) the compressor work input c) the absolute humidity and enthalpy (kJ/kg of dry air) for the air as it returns

to the room (state 3 in figure)

1 3

Water liquid

Evaporator Condenser

2

Problem dryer

Wood dryerCapacity: 500 kg/h

Heating coil

Outdoor air

T=+14 °CΦ=60% RH

T=+40 °CΦ=90% RH

+

Volume flow of moist air: 20100 m3/h

A B C

D

14 , 60 %1599 , 0.60 1599 959.4

959.40.622 0.00595 /101325 959.4

1.01 14 0.00595(2502 1.84 14) 29.17 /

40 , 90 %7375 , 0.90 7375 6637.5

6637.50.622101325 663

A A

ws w

A

A

D D

ws w

D

T C RHp Pa p Pa

x kg kg

h kJ kg

T C RHp Pa p Pa

x

ϕ

ϕ

= == = ⋅ =

= =−

= ⋅ + + ⋅ =

= == = ⋅ =

=−

o

o

3

3

0.04360 /7.5

1.01 40 0.04360(2502 1.84 40) 152.70 /

8314.510.04360 (273.15 40)18.02 0.94914

6637.51.0 0.04360 1.09952 /

0.94914

20100 1.09952 22100.3 /

D

u

moist air

dry air

kg kg

h kJ kg

Rm TMV mp

m kg mV

m V kg h

m

ρ

ρ⋅ ⋅

⋅

=

= ⋅ + + ⋅ =

+= = =

+= = =

= ⋅ = ⋅ =

22100.3 21177.0 /1.0 1.0 0.04360

( )

5000.04360 0.01999 /21177.0

152.70 /

moist air

C D

water in dryer dry air D C

water in dryerC D

dry air

C D

m kg hx

adiabatic conditions for the dryer h h

m m x x

mx x kg kgm

h h kJ kgHeating coi

⋅

⋅ ⋅

⋅

⋅

= = =+ +

⇒ =

= −

= − = − =

= =

,

0.01999 /

( )( ) (1 )

0.01999 0.04360 0.6270.00595 0.04360

: 62.7%

0.627 21177.0 13280.2 /( ) (1 )

B C

B A D

B D

A D

dry air A

B A

l x x kg kg

Mixing rate of outdoor air mixx mix x mix x

x xmixx x

Amount of outdoor air

m kg hh mix h mix

⋅

⇒ = =

= ⋅ + − ⋅− −

= = =− −

= ⋅ == ⋅ + − ⋅

0.627 29.17 (1 0.627) 152.70 75.23 /21177.0( ) (152.70 75.23) 455.7 456

3600

D

B

dry air C Bheating coil

hh kJ kg

Q m h h kW⋅ ⋅

= ⋅ + − ⋅ =

= − = − = ≈

Problem dryerwith heat pump

Wood dryerCapacity: 500 kg/h

Cooling coil

Outdoor air

T=+14 °CΦ=60% RH

T=+40 °CΦ=90% RH

-

Evaporator

CondenserAB

F E

C D

Volume flow of moist air: 20100 m3/hT=+28ºC

14 , 60 %1599 , 0.60 1599 959.4

959.40.622 0.00595 /101325 959.4

1.01 14 0.00595(2502 1.84 14) 29.17 /

40 , 90 %7375 , 0.90 7375 6637.5

6637.50.622101325 663

A A

ws w

A

A

E E

ws w

E

T C RHp Pa p Pa

x kg kg

h kJ kg

T C RHp Pa p Pa

x

ϕ

ϕ

= == = ⋅ =

= =−

= ⋅ + + ⋅ =

= == = ⋅ =

=−

o

o

0.04360 /7.5

1.01 40 0.04360(2502 1.84 40) 152.70 /

28 , 100 % ( )3780 , 3780 ( 6637.5 )

37800.622 0.02410 /101325 3780

1.01 28 0.02410(2502 1.84 2

E

F F

ws w

F

F

kg kg

h kJ kg

T C RH assumedp Pa p Pa Pa condensation occur

x kg kg

h

ϕ

=

= ⋅ + + ⋅ =

= == = <

= =−

= ⋅ + + ⋅

o

3

3

8) 89.82 /

8314.510.04360 (273.15 40)18.02 0.94914

6637.51.0 0.04360 1.09952 /

0.94914

20100 1.09952 22100.3 /

22100.3 21177.0 /1.0 1.0 0.04360

u

moist air

moist airdry air

kJ kg

Rm TMV mp

m kg mV

m V kg h

mm kg hx

ρ

ρ⋅ ⋅

⋅⋅

=

+= = =

+= = =

= ⋅ = ⋅ =

= = =+ +

( )

5000.04360 0.01999 /21177.0

152.70 /0.01999 / ( )

D E

water in dryer dry air E D

water in dryerD E

dry air

D E

C D

adiabatic conditions for the dryer h h

m m x x

mx x kg kgm

h h kJ kgCooling coil x x kg kg no condensation occurHeat tr

⋅ ⋅

⋅

⋅

⇒ =

= −

= − = − =

= =⇒ = =

( )21177.0( ) (152.70 89.82) 369.89

36000.01999 /

dry air E FL

B C

ansfer to heat pump evaporator

Q m h h kW

Condenser x x kg kg

⋅ ⋅

= − = − =

⇒ = =

,

( )( ) (1 )

0.01999 0.02410 0.2260.00595 0.02410

: 22.6%

0.226 21177.0 4795.5 /( ) (1 )0.226 29.17 (1 0

B A F

B F

A F

dry air A

B A F

B

Mixing rate of outdoor air mixx mix x mix x

x xmixx x

Amount of outdoor air

m kg hh mix h mix hh

⋅

= ⋅ + − ⋅− −

= = =− −

= ⋅ == ⋅ + − ⋅= ⋅ + −

1

2

.226) 89.82 76.11 /

( )(log ) 717

: 1780 / (1460 / 2): 2100 / (1776 / 2)

dry air C B Ccondenser

kJ kg

Q m h h need to calculate hFrom P h diagram for R we readafter evaporator h kJ kg kJ kg from CATTafter compressor h kJ kg kJ kg from CATTafte

⋅ ⋅

⋅ =

= − =

− −==

3

4 3

717 1 4

717

1 4

717 2

: 1010 / (700.6 / 2): 1010 / (700.6 / 2)

( )

369.89 0.4804 /( ) (1780 1010)

:

(

RL

LR

comp R

r condenser h kJ kg kJ kg from CATTbefore evaporator h h kJ kg kJ kg from CATT

Q m h h

Qm kg sh h

Work input to compressor

W m h h

⋅ ⋅

⋅⋅

⋅ ⋅

=

= =

= −

= = =− −

= − 1

717 2 3

) 0.4804(2100 1780) 153.72 154

:

( ) 0.4804(2100 1010) 523.61

( )

523.61 76.11 165.121177.03600

RH

dry air C BH

HC B

dry air

kW

Heat transfer from condenser

Q m h h kWthis heat will heat the mixed air flow

Q m h h

Qh hm

⋅ ⋅

⋅ ⋅

⋅

⋅

= − = ≈

= − = − =

= −

= + = + = 2 /

152.70 /21177.0( ) (165.12 152.70) 73.07 73

3600

Wood dryer using only outdoor air consumes about 456 kW of heatWood dryer with a mec

D E

dry air C Dcooling coil

kJ kg

Adiabatic conditions in dryer h h kJ kg

Q m h h kW⋅ ⋅

⇒ = =

= − = − = ≈

hanical heat pump consumeselectricity 154 kW and deliviers 75 kW of heat ps: no efficiency have been included.

MOLLIER DIAGRAMHeating of AirCooling and Dehumidfying AirMixing of Air of different ConditionsHumidifying , Adding Steam or Water (liquid)Psychrometric ChartEvaporation from Water SurfacesProblem 6 (page 22)Problem dryerProblem dryerwith heat pump