MOLLIER DIAGRAM

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)hB
where
L = mixing rate
h = enthalpy of the air
The moisture balance for the mix can be
expressed as:
LA xA + LC xC = (LA + LC) xB
where
x = water content in the air
Calculating the mixture variables xB and hB makes
it possible to calculate the mixing temperature tB.
Humidifying ,
Adding Steam or Water (liquid)
Psychrometric Chart
The psychrometric chart is a variant of the Mollier diagram used in
some parts of the world.
The process transforming a Mollier diagram to a psychrometric
chart 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:
⋅
m
⋅
m
evap
evap
A ⋅ hc
( xs − x∞ )
≈
cp
= amount of evaporated water (kg/s)
A = water surface area (m2)
h c = heat transfer coefficient (W/m2 K)
c p = mean specific heat for moist air (J/kg K)
x∞ = humidity ratio in the air (kg/kg)
x s = humidity 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.
Calculate
a) 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
Condenser
Evaporator
2
Water liquid
3
Problem dryer
Heating coil
Outdoor air
A
T=+14 °C
Φ=60% RH
B
C
+
Wood dryer
Capacity: 500 kg/h
D
T=+40 °C
Φ=90% RH
Volume flow of moist air: 20100 m
3/h
TA = 14o C , ϕ A = 60 % RH
pws = 1599 Pa,
pw = 0.60 ⋅1599 = 959.4 Pa
959.4
= 0.00595 kg / kg
101325 − 959.4
hA = 1.01 ⋅14 + 0.00595(2502 + 1.84 ⋅14) = 29.17 kJ / kg
x A = 0.622
TD = 40o C , ϕ D = 90 % RH
pws = 7375 Pa,
pw = 0.90 ⋅ 7375 = 6637.5 Pa
6637.5
= 0.04360 kg / kg
101325 − 6637.5
hD = 1.01 ⋅ 40 + 0.04360(2502 + 1.84 ⋅ 40) = 152.70 kJ / kg
xD = 0.622
8314.51
Ru
(273.15 + 40)
T 0.04360
18.02
V= M =
= 0.94914 m3
p
6637.5
m 1.0 + 0.04360
ρ= =
= 1.09952 kg / m3
V
0.94914
m
⋅
⋅
m moist air = V ⋅ ρ = 20100 ⋅1.09952 = 22100.3 kg / h
⋅
m dry air
⋅
m moist air
22100.3
=
=
= 21177.0 kg / h
1.0 + x 1.0 + 0.04360
adiabatic conditions for the dryer ⇒ hC = hD
⋅
⋅
m water in dryer = m dry air ( xD − xC )
⋅
xC = xD −
m water in dryer
⋅
= 0.04360 −
m dry air
hC = hD = 152.70 kJ / kg
500
= 0.01999 kg / kg
21177.0
Heating coil ⇒ xB = xC = 0.01999 kg / kg
Mixing rate of outdoor air (mix)
xB = (mix) ⋅ x A + (1 − mix) ⋅ xD
mix =
xB − xD 0.01999 − 0.04360
=
= 0.627
x A − xD 0.00595 − 0.04360
Amount of outdoor air : 62.7%
⋅
m dry air , A = 0.627 ⋅ 21177.0 = 13280.2 kg / h
hB = (mix) ⋅ hA + (1 − mix) ⋅ hD
hB = 0.627 ⋅ 29.17 + (1 − 0.627) ⋅152.70 = 75.23 kJ / kg
⋅
⋅
Q heating coil = m dry air (hC − hB ) =
21177.0
(152.70 − 75.23) = 455.7 ≈ 456 kW
3600
Problem dryer
with heat pump
Cooling coil
Outdoor air
A
T=+14 °C
Φ=60% RH
B
Condenser
C
D
-
Wood dryer
Capacity: 500 kg/h
F
T=+28ºC
E
Evaporator
T=+40 °C
Φ=90% RH
Volume flow of moist air: 20100 m
3/h
TA = 14o C , ϕ A = 60 % RH
pws = 1599 Pa,
pw = 0.60 ⋅1599 = 959.4 Pa
959.4
= 0.00595 kg / kg
101325 − 959.4
hA = 1.01 ⋅14 + 0.00595(2502 + 1.84 ⋅14) = 29.17 kJ / kg
xA = 0.622
TE = 40o C , ϕ E = 90 % RH
pws = 7375 Pa,
pw = 0.90 ⋅ 7375 = 6637.5 Pa
6637.5
= 0.04360 kg / kg
101325 − 6637.5
hE = 1.01⋅ 40 + 0.04360(2502 + 1.84 ⋅ 40) = 152.70 kJ / kg
xE = 0.622
TF = 28o C , ϕ F = 100 % RH (assumed )
pws = 3780 Pa,
pw = 3780 Pa (< 6637.5 Pa condensation occur )
3780
= 0.02410 kg / kg
101325 − 3780
hF = 1.01⋅ 28 + 0.02410(2502 + 1.84 ⋅ 28) = 89.82 kJ / kg
xF = 0.622
8314.51
Ru
(273.15 + 40)
T 0.04360
18.02
V= M =
= 0.94914 m3
p
6637.5
m 1.0 + 0.04360
ρ= =
= 1.09952 kg / m3
V
0.94914
m
⋅
⋅
m moist air = V ⋅ ρ = 20100 ⋅1.09952 = 22100.3 kg / h
⋅
⋅
m moist air
22100.3
=
=
= 21177.0 kg / h
1.0 + x 1.0 + 0.04360
m dry air
adiabatic conditions for the dryer ⇒ hD = hE
⋅
⋅
m water in dryer = m dry air ( xE − xD )
⋅
xD = xE −
m water in dryer
⋅
= 0.04360 −
m dry air
hD = hE = 152.70 kJ / kg
500
= 0.01999 kg / kg
21177.0
Cooling coil ⇒ xC = xD = 0.01999 kg / kg (no condensation occur )
Heat transfer to heat pump (evaporator )
⋅
⋅
21177.0
(152.70 − 89.82) = 369.89 kW
3600
Condenser ⇒ xB = xC = 0.01999 kg / kg
Q L = m dry air (hE − hF ) =
Mixing rate of outdoor air (mix)
xB = (mix) ⋅ x A + (1 − mix) ⋅ xF
xB − xF 0.01999 − 0.02410
=
= 0.226
x A − xF 0.00595 − 0.02410
mix =
Amount of outdoor air : 22.6%
⋅
m dry air , A = 0.226 ⋅ 21177.0 = 4795.5 kg / h
hB = (mix) ⋅ hA + (1 − mix) ⋅ hF
hB = 0.226 ⋅ 29.17 + (1 − 0.226) ⋅ 89.82 = 76.11 kJ / kg
⋅
⋅
Q condenser = m dry air (hC − hB ) = need to calculate hC
From (log P − h) − diagram for R717 we read
after evaporator : h1 = 1780 kJ / kg
(1460 kJ / kg from CATT 2)
after compressor : h2 = 2100 kJ / kg
(1776 kJ / kg from CATT 2)
after condenser : h3 = 1010 kJ / kg
(700.6 kJ / kg from CATT 2)
before evaporator : h4 = h3 = 1010 kJ / kg (700.6 kJ / kg from CATT 2)
⋅
⋅
⋅
QL
369.89
=
=
= 0.4804 kg / s
(h1 − h4 ) (1780 − 1010)
Q L = m R 717 (h1 − h4 )
m R 717
⋅
Work input to compressor :
⋅
⋅
W comp = m R 717 (h2 − h1 ) = 0.4804(2100 − 1780) = 153.72 ≈ 154 kW
Heat transfer from condenser :
⋅
⋅
Q H = m R 717 (h2 − h3 ) = 0.4804(2100 − 1010) = 523.61 kW
this heat will heat the mixed air flow
⋅
⋅
Q H = m dry air (hC − hB )
⋅
QH
523.61
+ 76.11 = 165.12 kJ / kg
21177.0
m dry air
3600
Adiabatic conditions in dryer ⇒ hD = hE = 152.70 kJ / kg
hC =
⋅
⋅
+ hB =
⋅
Q cooling coil = m dry air (hC − hD ) =
21177.0
(165.12 − 152.70) = 73.07 ≈ 73 kW
3600
Wood dryer using only outdoor air consumes about 456 kW of heat
Wood dryer with a mechanical heat pump consumes
electricity 154 kW and deliviers 75 kW of heat
ps: no efficiency have been included.