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

Transcription

1 MOLLIR IGRM

2 ew Point - Tdp Te ew Point is te temperature at wic water vapor starts to condense out of te air, te temperature at wic air becomes completely saturated. bove tis temperature te moisture will stay in te air. If te dew-point temperature is close to te air temperature, te relative umidity is ig, and if te dew point is well below te air temperature, te relative umidity is low. Te ew Point temperature can be measured by filling a metal can wit water and ice cubes. Stir by a termometer and watc te outside of te can. Wen te vapor in te air starts to condensate on te outside of te can, te temperature on te termometer is pretty close to te dew point of te actual air. Te dew point temperature can be read by following a vertical line from te state-point to te saturation line. ew point is represented along te 100% relative umidity line in te Mollier diagram.

3 ry-bulb Temperature - Tdb ry bulb temperature is usually referred to as air temperature, is te air property tat is most common used. Wen people refer to te temperature of te air, tey are normally referring to its dry bulb temperature. ry-bulb temperature - Tdb, can be measured by using a normal termometer. Te dry-bulb temperature is an indicator of eat content and is sown along te left axis of te Mollier diagram. Te orizontal lines extending from tis axis are constant-temperature lines. Wet-Bulb Temperature - Twb Wet bulb temperature is associated wit te moisture content of te air. Wet bulb temperature can be measured wit a termometer tat as te bulb covered wit a water-moistened bandage wit air flowing over te termometer. Wet bulb temperatures are always lower tan dry bulb temperatures but tey will be identical wit 100% relative umidity in te air (te air is at te saturation line). On te Mollier diagram, te wet-bulb lines slope a little upward to te left (dotted lines).

4 Heating of ir

5 Cooling and eumidfying ir

6 Mixing of ir of different Conditions Te eat balance for te mix can be expressed as: L + LC C = (L + LC)B were L = mixing rate = entalpy of te air Te moisture balance for te mix can be expressed as: L x + LC xc = (L + LC) xb were x = water content in te air Calculating te mixture variables xb and B makes it possible to calculate te mixing temperature tb.

7 Humidifying, dding Steam or Water (liquid)

8 Psycrometric Cart Te psycrometric cart is a variant of te Mollier diagram used in some parts of te world. Te process transforming a Mollier diagram to a psycrometric cart is sown below. First it as to be reflected in a vertical mirror, ten rotated 90 degrees.

9 vaporation from Water Surfaces Te amount of evaporated water can be expressed as: m evap c ( x s x ) c p m evap c s = amount of evaporated water (kg/s) = water surface area (m2) = eat transfer coefficient (W/m2 K) c p x = mean specific eat for moist air (J/kg K) = umidity ratio in te air (kg/kg) x = umidity ratio in saturated air at te same temperature as te water surface (kg/kg)

10 Problem 6 (page 22) n indoor pool evaporates a certain amount of water, wic is removed by a deumidifier to maintain +25ºC, φ=70% RH in te room (state 1 in figure). Te deumidifier, sown in figure, is a refrigeration cycle in wic moist air flowing over te evaporator cools suc tat liquid water drops out, and te air continues flowing over te condenser. Te air after te evaporator (state 2) as a temperature of +14ºC. For an air flow of 0,10 kg/s dry air te unit as a coefficient of performance COP R =3,0. Total pressure in te room is constant Pa. Calculate a) te amount of water tat evaporates from te pool ( steady state) b) te compressor work input c) te absolute umidity and entalpy (kj/kg of dry air) for te air as it returns to te room (state 3 in figure) 1 3 vaporator Condenser 2 Water liquid

11 Problem dryer Heating coil Outdoor air B C + T=+14 C Φ=60% RH Wood dryer Capacity: 500 kg/ T=+40 C Φ=90% RH Volume flow of moist air: m 3 /

12 o T = 14 C, ϕ = 60 % RH p = 1599 Pa, p = = Pa x ws w = = kg/ kg = ( ) = kj / kg o T = 40 C, ϕ = 90 % RH p = 7375 Pa, p = = Pa x ws w = = kg / kg = ( ) = kj / kg Ru m T ( ) V = M = = m p m ρ = = = kg / m V mmoist air = V ρ = = kg/ m dry air mmoist air = = = kg / x adiabatic conditions for te dryer = m = m ( x x ) water in dryer dry air C mwater in dryer 500 xc = x = = kg/ kg m dry air = = kj / kg C Heating coil x = x = kg/ kg B C C Mixing rate of outdoor air ( mix) x = ( mix) x + (1 mix) x B xb x mix = = = x x mount of outdoor air :62.7% m dry air, = = kg/ = ( mix) + (1 mix) B B = ( ) = kj / kg Qeating coil = mdry air ( C B) = ( ) = kw 3600

13 Problem dryer wit eat pump Cooling coil Outdoor air T=+14 C Φ=60% RH B Condenser C - Wood dryer Capacity: 500 kg/ F T=+28ºC vaporator T=+40 C Φ=90% RH Volume flow of moist air: m 3 /

14 o T = 14 C, ϕ = 60 % RH p = 1599 Pa, p = = Pa x ws w = = kg/ kg = ( ) = kj / kg o T = 40 C, ϕ = 90 % RH p = 7375 Pa, p = = Pa x ws w = = kg / kg = ( ) = kj / kg o T = 28 C, ϕ = 100 % RH ( assumed) F F p = 3780 Pa, p = 3780 Pa ( < Pa condensation occur) x ws F F w 3780 = = kg/ kg = ( ) = kj / kg Ru m T ( ) V = M = = m p m ρ = = = kg / m V mmoist air = V ρ = = kg/ m dry air mmoist air = = = kg/ x adiabatic conditions for te dryer = m = m ( x x ) water in dryer dry air mwater in dryer 500 x = x = = kg/ kg m dry air = = kj / kg Cooling coil x = x = kg / kg ( no condensation occur) C Heat transfer to eat pump ( evaporator) QL = mdry air ( F) = ( ) = kw 3600 Condenser x = x = kg / kg B C

15 Mixing rate of outdoor air ( mix) x = ( mix) x + (1 mix) x B F xb xf mix = = = x x mount of outdoor air : 22.6% m dry air, F = = kg/ = ( mix) + (1 mix) B F B = ( ) = kj / kg Q = m ( ) = need to calculate condenser dry air C B C From (log P ) diagram for R717 we read after evaporator : = 1780 kj / kg (1460 kj / kg from CTT 2) 1 after compressor : = 2100 kj / kg (1776 kj / kg from CTT 2) 2 after condenser : = 1010 kj / kg (700.6 kj / kg from CTT 2) R 717 R comp = R 717( before evaporator : = = 1010 kj / kg (700.6 kj / kg from CTT 2) Q = m ( ) m L 4 3 QL = = = kg/ s ( ) ( ) Work input to compressor : W m ) = ( ) = kw Heat transfer from condenser : Q = m ( ) = ( ) = kw H R tis eat will eat te mixed air flow Q = m ( ) H dry air C B QH C = + B = = kj / kg mdry air 3600 diabatic conditions in dryer = = kj / kg Qcooling coil = mdry air ( C ) = ( ) = kw 3600 Wood dryer using only outdoor air consumes about 456 kw of eat Wood dryer wit a mecanical eat pump consumes electricity 154 kw and deliviers 75 kw of eat ps: no efficiency ave been included.