Air-Conditioning

Air-conditioning is the process of simultaneously controlling the temperature, humidity, cleanliness and motion (distribution) of air in an enclosed space to meet the requirements of the occupants or of a process.

Functions

1. Control of temperature (heating or cooling).
2. Control of moisture (humidification or dehumidification).
3. Filtration: removing dust, smoke, pollen and odours.
4. Air movement and distribution without draught.
5. Ventilation with the correct amount of fresh air.

Classification and scope

Moist air is a mixture of dry air and water vapour; psychrometry deals with its properties.

Relations: For unsaturated air, DPT $<$ WBT $<$ DBT. For saturated air all three are equal and $\phi = \mu = 100\%$.

A psychrometric chart is a graph of the properties of moist air at a constant total pressure (normally 101.325 kPa). From any two independent properties, all the others can be read.

Axes

• Horizontal axis: dry bulb temperature (DBT).
• Vertical axis (right side): specific humidity $\omega$ (or vapour pressure).

Lines on the chart

1. Constant DBT lines: vertical.
2. Constant specific humidity lines: horizontal. The dew point is read where the horizontal through the state meets the saturation curve.
3. Saturation curve (100 % RH): the upper curved boundary; DBT, WBT and DPT are equal on it.
4. Constant RH curves: curves below the saturation curve (10 %, 20 %, ...).
5. Constant WBT lines: inclined straight lines running down to the right, ending on the saturation curve.
6. Constant enthalpy lines: nearly parallel to the WBT lines; read from a scale left of the saturation curve.
7. Constant specific volume lines: steeper inclined lines (m³/kg dry air).
8. Sensible heat factor (SHF) scale and protractor at the upper left for process lines.

 w ^                         .-' saturation
   |                   .--''   (100 % RH)
   |  DPT <--------- P *  <- w = constant
   |             .-'/ |
   |          .-'  /  |  <- DBT (vertical)
   |       .-'   WBT line
   +------------------------------> DBT

Reading a state P: vertical down gives DBT; horizontal left to the saturation curve gives DPT and horizontal right gives $\omega$; the inclined line through P reaching the saturation curve gives WBT; the curve through P gives RH; the enthalpy and volume lines nearest to P give $h$ and $v$.

Let $\dot m_a$ be the mass flow of dry air, and 1 and 2 the inlet and outlet states.

(a) Sensible heating

Heat is added with no change in $\omega$. The state moves horizontally to the right; DBT, WBT and enthalpy rise, RH falls.

$$Q = \dot m_a (h_2 - h_1) = \dot m_a\,c_{pm}(t_2 - t_1)$$

with humid specific heat $c_{pm} = 1.005 + 1.88\,\omega$ kJ/kg K. Done by an electric or steam heating coil.

(b) Sensible cooling

Air is cooled by a coil whose surface temperature is above the dew point, so $\omega$ stays constant. The state moves horizontally to the left; RH rises. The same equation applies with the sign reversed.

(c) Cooling and dehumidification

The coil surface is below the dew point of the entering air, so vapour condenses. The state moves down and to the left.

$$Q = \dot m_a (h_1 - h_2) - \dot m_w h_w,\qquad \dot m_w = \dot m_a(\omega_1 - \omega_2)$$

The load splits into sensible and latent parts, $Q = Q_s + Q_l$. This is the main summer air-conditioning process.

(d) Heating and humidification

Used in winter: air is heated, then water or steam is added. Specific humidity and temperature both rise.

$$\dot m_w = \dot m_a(\omega_2 - \omega_1),\qquad Q = \dot m_a(h_2 - h_1) - \dot m_w h_w$$

 w ^
   |              o 2d  (d)
   |             /
   |  2b o<-----o 1 ---->o 2a
   |           /
   |      2c o            (c)
   +------------------------> DBT

(a) sensible heating, (b) sensible cooling, (c) cooling and dehumidification, (d) heating and humidification.

Adiabatic saturation

Unsaturated air at state 1 ($t_1, \omega_1$) flows through a long, insulated duct over a water surface. Water evaporates and the air becomes more humid. If the duct is long enough, the air leaves saturated at temperature $t^*$ and the water remains at $t^*$. Make-up water is supplied at $t^*$.

  air in  ---> [ insulated duct, water pool at t* ] ---> saturated air out
  t1, w1                                               t*, w*
                      ^ make-up water at t*

Energy balance (steady flow, no heat exchange):

$$h_1 + (\omega^* - \omega_1)\,h_{f}^{*} = h^*$$

which gives

$$\omega_1 = \frac{(2501 - 2.381\,t^*)\,\omega^* - 1.006\,(t_1 - t^*)}{2501 + 1.805\,t_1 - 4.186\,t^*}$$

The temperature $t^*$ is the thermodynamic wet bulb temperature (adiabatic saturation temperature). The air gives up sensible heat, which supplies the latent heat of evaporation.

Measurement of WBT

A thermometer bulb is covered with a wet wick and exposed to air moving at 2.5 to 5 m/s. Evaporation cools the bulb until heat gained by convection equals the heat lost by evaporation. For air-water vapour mixtures at normal pressures the reading is very nearly equal to the thermodynamic WBT.

A sling psychrometer has a dry bulb and a wet bulb thermometer mounted on a frame that is whirled by hand to give the required air velocity. The DBT and WBT read are used with the psychrometric chart to find RH, DPT and $\omega$. An aspirated (Assmann) psychrometer uses a small fan.

Relation

In a real coil a fraction BPF of the entering air (state 1) remains unchanged and a fraction $(1 - \text{BPF})$ is cooled to the ADP (state A). The leaving air (state 2) is their mixture, so it lies on the straight line 1-A.

  w ^   1 o
    |      \
    |    2  o         line 1-2-A
    |         \
    |        A o----- saturation curve
    +--------------------> DBT

Energy and mass balance on the mixture:

$$t_2 = \text{BPF}\,t_1 + (1 - \text{BPF})\,t_A$$ $$\text{BPF} = \frac{t_2 - t_A}{t_1 - t_A} = \frac{\omega_2 - \omega_A}{\omega_1 - \omega_A} = \frac{h_2 - h_A}{h_1 - h_A}$$

Contact factor $= \dfrac{t_1 - t_2}{t_1 - t_A}$. BPF decreases with more coil rows, closer fin spacing and lower air velocity.

Given

$t = 32$ °C, $\phi = 0.55$, $p = 101.325$ kPa, $p_s = 4.759$ kPa.

(a) Partial pressure of vapour

$$p_v = \phi\,p_s = 0.55 \times 4.759 = 2.618\ \text{kPa}$$

(b) Specific humidity

$$\omega = 0.622\,\frac{p_v}{p - p_v} = 0.622 \times \frac{2.618}{101.325 - 2.618} = 0.01649\ \text{kg/kg dry air}$$

(c) Dew point

The dew point is the saturation temperature at $p_v = 2.618$ kPa. From steam tables (interpolating between 20 °C, 2.339 kPa and 22 °C, 2.645 kPa):

$$t_{dp} \approx 21.8\ ^\circ\text{C}$$

(d) Degree of saturation

$$\omega_s = 0.622\,\frac{4.759}{101.325 - 4.759} = 0.03066,\qquad \mu = \frac{\omega}{\omega_s} = \frac{0.01649}{0.03066} = 0.538$$

(e) Enthalpy

$$\begin{aligned} h &= 1.005\,t + \omega\,(2501 + 1.88\,t)\\ &= 1.005 \times 32 + 0.01649\,(2501 + 1.88 \times 32) = 74.4\ \text{kJ/kg dry air} \end{aligned}$$

(f) Specific volume

$$v = \frac{R_a T}{p - p_v} = \frac{0.287 \times 305.15}{101.325 - 2.618} = 0.887\ \text{m}^3/\text{kg dry air}$$

(The wet bulb temperature is about 24.6 °C from the adiabatic saturation equation.)

Answer: (a) 2.62 kPa; (b) 0.01649 kg/kg; (c) about 21.8 °C; (d) 0.538 (53.8 %); (e) 74.4 kJ/kg; (f) 0.887 m³/kg.

Sensible heating means $\omega$ is constant.

Inlet state

$$p_{v1} = 0.70 \times 1.706 = 1.194\ \text{kPa}$$ $$\omega_1 = 0.622\,\frac{1.194}{101.325 - 1.194} = 0.00742\ \text{kg/kg}$$ $$v_1 = \frac{0.287 \times 288.15}{101.325 - 1.194} = 0.8259\ \text{m}^3/\text{kg dry air}$$

Enthalpies, $h = 1.005\,t + \omega(2501 + 1.88\,t)$:

$$\begin{aligned} h_1 &= 1.005 \times 15 + 0.00742(2501 + 28.2) = 33.83\ \text{kJ/kg}\\ h_2 &= 1.005 \times 30 + 0.00742(2501 + 56.4) = 49.12\ \text{kJ/kg} \end{aligned}$$

(a) Mass flow of dry air

$$\dot m_a = \frac{\dot V_1}{v_1} = \frac{200}{0.8259} = 242.2\ \text{kg/min} = 4.04\ \text{kg/s}$$

(b) Heat transfer

$$Q = \dot m_a (h_2 - h_1) = \frac{242.2 \times (49.12 - 33.83)}{60} = 61.7\ \text{kW}$$

(c) Final relative humidity

$$\phi_2 = \frac{p_v}{p_{s,30}} = \frac{1.194}{4.247} = 0.281$$

Answer: (a) 242 kg/min dry air; (b) about 61.7 kW; (c) 28 %.

State of each stream

(using $\omega = 0.622\,p_v/(p-p_v)$, $v = 0.287\,T(1 + 1.608\omega)/p$ and $h = 1.005t + \omega(2501 + 1.88t)$.)

Mass flow of dry air

$$\begin{aligned} \dot m_f &= \frac{1500}{0.9050 \times 3600} = 0.4604\ \text{kg/s}\\ \dot m_r &= \frac{4500}{0.8579 \times 3600} = 1.4570\ \text{kg/s}\\ \dot m_m &= 1.9174\ \text{kg/s} \end{aligned}$$

Mixture (mass and energy balances)

$$\omega_m = \frac{\dot m_f\omega_f + \dot m_r\omega_r}{\dot m_m} = 0.01153\ \text{kg/kg}$$ $$h_m = \frac{\dot m_f h_f + \dot m_r h_r}{\dot m_m} = 57.73\ \text{kJ/kg}$$

Dry bulb temperature from $h_m = 1.005\,t_m + \omega_m(2501 + 1.88\,t_m)$:

$$t_m = \frac{57.73 - 0.01153 \times 2501}{1.005 + 1.88 \times 0.01153} = 28.2\ ^\circ\text{C}$$

(The mass-weighted average, $28.1$ °C, agrees closely.) The vapour pressure is $p_{v,m} = \omega_m p/(0.622 + \omega_m) = 1.843$ kPa and the saturation pressure at $t_m$ is 3.816 kPa, so $\phi_m = 0.483$.

Answer: $t_m \approx 28.2$ °C; $\omega_m \approx 0.0115$ kg/kg; $h_m \approx 57.7$ kJ/kg; RH about 48 %.

Entering air state 1 (30 °C, 60 %)

$p_s = 4.247$ kPa, $p_v = 0.6 \times 4.247 = 2.548$ kPa.

$$\omega_1 = 0.01604\ \text{kg/kg},\quad h_1 = 71.18\ \text{kJ/kg},\quad v_1 = 0.8808\ \text{m}^3/\text{kg}$$ $$\dot m_a = \frac{120}{0.8808 \times 60} = 2.271\ \text{kg/s}$$

ADP state A (8 °C, saturated)

$p_s(8\ ^\circ\text{C}) = 1.072$ kPa, so $\omega_A = 0.00666$ kg/kg and $h_A = 24.79$ kJ/kg.

(a) Leaving state 2

$$\text{BPF} = \frac{t_2 - t_A}{t_1 - t_A} = \frac{\omega_2 - \omega_A}{\omega_1 - \omega_A} = 0.25$$ $$\begin{aligned} t_2 &= 8 + 0.25(30 - 8) = 13.5\ ^\circ\text{C}\\ \omega_2 &= 0.00666 + 0.25(0.01604 - 0.00666) = 0.00900\ \text{kg/kg}\\ h_2 &= 1.005 \times 13.5 + 0.00900(2501 + 1.88 \times 13.5) = 36.32\ \text{kJ/kg} \end{aligned}$$

Relative humidity of leaving air $\approx 0.93$ (about 93 %).

(b) Loads

Intermediate state $h'$ at ($t_2$, $\omega_1$): $h' = 1.005 \times 13.5 + 0.01604(2501 + 25.4) = 54.10$ kJ/kg.

$$\begin{aligned} Q_{total} &= \dot m_a(h_1 - h_2) = 2.271 \times (71.18 - 36.32) = 79.2\ \text{kW} = 22.5\ \text{TR}\\ Q_{sensible} &= \dot m_a(h_1 - h') = 2.271 \times (71.18 - 54.10) = 38.8\ \text{kW}\\ Q_{latent} &= \dot m_a(h' - h_2) = 2.271 \times (54.10 - 36.32) = 40.4\ \text{kW} \end{aligned}$$

(The enthalpy carried away by the condensate, about 0.9 kW, is small and neglected.)

(c) Sensible heat factor

$$\text{SHF} = \frac{Q_s}{Q_{total}} = \frac{38.8}{79.2} = 0.49$$

(d) Condensate

$$\dot m_w = \dot m_a(\omega_1 - \omega_2) = 2.271 \times (0.01604 - 0.00900) = 0.01599\ \text{kg/s} = 57.6\ \text{kg/h}$$

Answer: (a) 13.5 °C, $\omega$ = 0.00900 kg/kg, RH about 93 %; (b) total 79.2 kW (22.5 TR), sensible 38.8 kW, latent 40.4 kW; (c) SHF = 0.49; (d) 57.6 kg/h.

Room state

$$p_v = 0.5 \times 3.169 = 1.585\ \text{kPa},\qquad \omega_r = 0.622 \times \frac{1.585}{101.325 - 1.585} = 0.00988\ \text{kg/kg}$$ $$h_r = 1.005 \times 25 + 0.00988(2501 + 47) = 50.3\ \text{kJ/kg}$$

(a) Room sensible heat factor

$$\text{RSHF} = \frac{Q_s}{Q_s + Q_l} = \frac{18}{18 + 4.5} = 0.80$$

(b) Supply air mass flow (from the sensible load)

$$Q_s = \dot m_a\,c_{pm}(t_r - t_s),\qquad c_{pm} = 1.005 + 1.88\,\omega_s \approx 1.0215$$ $$\dot m_a = \frac{18}{1.0215 \times (25 - 14)} = 1.602\ \text{kg/s} = 5767\ \text{kg/h}$$

($\omega_s$ is found in step (c) and the two steps are iterated; the values converge after two or three trials.)

(c) Supply air humidity (from the latent load)

$$Q_l = \dot m_a(\omega_r - \omega_s)\,h_{fg}$$ $$\omega_r - \omega_s = \frac{4.5}{1.602 \times 2501} = 1.123 \times 10^{-3}$$ $$\omega_s = 0.00988 - 0.00112 = 0.00876\ \text{kg/kg}$$

Relative humidity at 14 °C: saturation specific humidity at 14 °C is 0.00997 kg/kg, and

$$\phi_s = \frac{p_v}{p_s} = 0.88\ \ (\approx 88\ \%)$$

The supply enthalpy is $h_s = 36.2$ kJ/kg.

(d) Volume flow of supply air

$$v_s = \frac{0.287 \times 287.15\,(1 + 1.608 \times 0.00876)}{101.325} = 0.8248\ \text{m}^3/\text{kg}$$ $$\dot V_s = 1.602 \times 0.8248 = 79.278\ \text{m}^3/\text{s}\ (\approx 79\ \text{m}^3/\text{min})$$

Answer: (a) RSHF = 0.80; (b) 1.60 kg/s; (c) $\omega_s$ = 0.00876 kg/kg, RH about 88 %; (d) about 1.32 m³/s (79 m³/min).

Thermal comfort is the state of mind in which a person feels satisfied with the thermal environment, i.e. neither too warm nor too cold. The body rejects its metabolic heat (about 100 W at rest) by convection, radiation, evaporation and respiration; comfort exists when this heat loss equals heat production without sweating or shivering.

Factors affecting comfort

1. Dry bulb temperature of the air.
2. Relative humidity: high RH reduces evaporation of sweat; very low RH dries the skin and throat.
3. Air velocity: moving air increases the heat loss by convection and evaporation (draughts if above about 0.25 m/s).
4. Mean radiant temperature of walls, windows and ceilings.
5. Activity level (metabolic rate) and clothing.
6. Air purity: freedom from dust, odour and CO₂; adequate fresh air.
7. Age, sex and acclimatisation of the occupants.

Effective temperature

Effective temperature (ET) is the index of the combined effect of DBT, humidity and air motion. It equals the temperature of still, saturated air that gives the same feeling of warmth or cold as the actual condition.

Typical design conditions

Air velocity of 0.1 to 0.25 m/s in the occupied zone and about 0.3 to 0.5 m³/min of fresh air per person or more as per code are used for design.