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Psychrometry is the science that deals with the thermodynamic properties of moist air and the physical and thermodynamic processes involving moist air. It is fundamental to understanding and designing heating, ventilation, and air conditioning (HVAC) systems, which control indoor air quality and comfort. Moist air is a mixture of dry air and water vapor, and its properties determine how air behaves when heated, cooled, humidified, or dehumidified.
In India, where climatic conditions vary widely from hot and humid to dry and cold, psychrometry helps engineers design efficient air conditioning systems that maintain comfortable indoor environments while minimizing energy consumption. Understanding psychrometric processes enables precise control over temperature and humidity, which affects human comfort, health, and productivity.
This chapter introduces the key properties of moist air, explains common air conditioning processes, and demonstrates how to use the psychrometric chart-a powerful graphical tool-to analyze and solve practical problems.
Before exploring air conditioning processes, it is essential to understand the fundamental properties of moist air. These properties describe the state of air and water vapor mixture and are used to analyze changes during conditioning.
The dry bulb temperature is the temperature of air measured by a regular thermometer freely exposed to the air but shielded from radiation and moisture. It represents the sensible heat content of air and is expressed in degrees Celsius (°C).
The wet bulb temperature is measured by a thermometer covered with a wet wick and exposed to airflow. Evaporation of water from the wick cools the thermometer, so the wet bulb temperature is always lower than or equal to the dry bulb temperature. It reflects the combined effect of temperature and moisture content and is useful for determining humidity and enthalpy.
Relative humidity is the ratio of the partial pressure of water vapor present in the air to the saturation pressure of water vapor at the same temperature, expressed as a percentage:
Relative humidity indicates how close the air is to being saturated with moisture. 100% RH means the air is fully saturated and cannot hold more moisture without condensation.
The dew point temperature is the temperature at which air becomes saturated (100% RH) when cooled at constant pressure. Below this temperature, moisture begins to condense out of the air as dew or water droplets. It is a direct measure of moisture content.
Also called the moisture content or specific humidity, the humidity ratio is the mass of water vapor per unit mass of dry air, typically expressed in kg of water vapor per kg of dry air. It is calculated from the partial pressure of water vapor and atmospheric pressure:
These properties are interconnected. For example, knowing the dry bulb and wet bulb temperatures allows calculation of humidity ratio and relative humidity. The psychrometric chart visually represents these relationships, making it easier to analyze air conditioning processes.
Psychrometric processes describe how the properties of moist air change during heating, cooling, humidification, dehumidification, and mixing. These processes are represented as paths or lines on the psychrometric chart, helping visualize the changes in temperature and moisture content.
Sensible heating or cooling changes the dry bulb temperature of air without altering its moisture content. For example, when air passes over a heater, its temperature rises but humidity ratio remains constant. On the psychrometric chart, this process moves horizontally (constant humidity ratio) to the right (heating) or left (cooling).
Humidification adds moisture to the air, increasing the humidity ratio and often the wet bulb temperature, while the dry bulb temperature may remain constant or change slightly. Dehumidification removes moisture, decreasing the humidity ratio. These processes involve latent heat transfer and are represented by vertical or near-vertical lines on the psychrometric chart.
Mixing occurs when two air streams with different temperatures and humidity ratios combine. The resulting air properties are weighted averages based on the mass flow rates of the streams. On the psychrometric chart, the mixed air state lies on the straight line connecting the two initial states.
graph TD A[Start: Initial Air State] --> B[Sensible Heating/Cooling] B --> C[Change in Dry Bulb Temp only] A --> D[Humidification/Dehumidification] D --> E[Change in Humidity Ratio] A --> F[Mixing of Air Streams] F --> G[Weighted Average Properties]
The psychrometric chart is a graphical representation of the thermodynamic properties of moist air at a constant pressure (usually atmospheric pressure). It is an indispensable tool for engineers to analyze air conditioning processes quickly and accurately.
To plot a point on the chart, you need the dry bulb temperature and one other property such as relative humidity, wet bulb temperature, or humidity ratio. Once plotted, other properties like dew point temperature and enthalpy can be read directly.
Different air conditioning processes follow characteristic paths on the chart:
Using the chart, you can find unknown properties by locating the intersection of known parameters. For example, knowing dry bulb and wet bulb temperatures allows determination of relative humidity and humidity ratio.
Psychrometry is applied extensively in designing and analyzing HVAC systems to maintain indoor air quality and comfort.
Cooling load is the amount of heat energy that must be removed from a space to maintain desired conditions. It includes sensible heat (temperature reduction) and latent heat (moisture removal). Psychrometric principles help calculate these loads accurately.
Proper ventilation requires controlling temperature and humidity to prevent discomfort and health issues. Psychrometry aids in selecting appropriate ventilation rates and conditioning methods.
Optimizing air conditioning processes based on psychrometric analysis can reduce energy consumption by minimizing unnecessary heating, cooling, or humidification.
Accurate measurement of air properties is essential for psychrometric analysis.
Devices that measure dry bulb and wet bulb temperatures, enabling calculation of humidity and other properties.
Instruments that directly measure relative humidity.
Modern HVAC systems use electronic sensors and data loggers for continuous monitoring and control.
Step 1: Determine initial vapor pressure \(P_v\) using relative humidity and saturation pressure at 35°C.
Saturation pressure at 35°C, \(P_{sat}\approx 5.62\,kPa\) (from steam tables or psychrometric chart).
\(P_v = \phi \times P_{sat} = 0.40 \times 5.62 = 2.248\,kPa\)
Step 2: Calculate initial humidity ratio \(W_1\):
\[ W_1 = 0.622 \times \frac{2.248}{101.325 - 2.248} = 0.622 \times \frac{2.248}{99.077} = 0.0141\,kg/kg \]
Step 3: Since cooling is sensible only, humidity ratio remains constant:
\(W_2 = W_1 = 0.0141\,kg/kg\)
Step 4: Find saturation pressure at final dry bulb temperature 25°C:
\(P_{sat,25} \approx 3.17\,kPa\)
Step 5: Calculate final vapor pressure \(P_{v2}\) from humidity ratio:
\[ P_{v2} = \frac{W_2 \times P}{0.622 + W_2} = \frac{0.0141 \times 101.325}{0.622 + 0.0141} = \frac{1.429}{0.6361} = 2.247\,kPa \]
Step 6: Calculate final relative humidity:
\[ \phi_2 = \frac{P_{v2}}{P_{sat,25}} \times 100 = \frac{2.247}{3.17} \times 100 = 70.9\% \]
Answer: Final dry bulb temperature is 25°C, relative humidity is approximately 71%, and humidity ratio remains 0.0141 kg/kg dry air.
Step 1: Find saturation pressure at wet bulb temperature (24°C):
\(P_{sat,24} \approx 2.98\,kPa\)
Step 2: Assume psychrometric constant \(A = 0.00066 \times P = 0.00066 \times 101.325 = 0.0669\,kPa/°C\)
Step 3: Calculate vapor pressure \(P_v\) using the formula:
\[ P_v = P_{sat,wet} - A \times (T_{dry} - T_{wet}) = 2.98 - 0.0669 \times (30 - 24) = 2.98 - 0.401 = 2.579\,kPa \]
Step 4: Calculate humidity ratio \(W\):
\[ W = 0.622 \times \frac{P_v}{P - P_v} = 0.622 \times \frac{2.579}{101.325 - 2.579} = 0.622 \times \frac{2.579}{98.746} = 0.0163\,kg/kg \]
Answer: Humidity ratio is approximately 0.0163 kg of water vapor per kg of dry air.
Step 1: Calculate total mass flow rate:
\(m_{total} = 2 + 3 = 5\,kg/s\)
Step 2: Calculate mixed humidity ratio \(W_{mix}\):
\[ W_{mix} = \frac{2 \times 0.012 + 3 \times 0.008}{5} = \frac{0.024 + 0.024}{5} = \frac{0.048}{5} = 0.0096\,kg/kg \]
Step 3: Calculate enthalpy of each stream (approximate using \(h = 1.005 \times T + W \times (2501 + 1.88 \times T)\), where \(T\) in °C, \(h\) in kJ/kg dry air):
For Stream 1:
\[ h_1 = 1.005 \times 30 + 0.012 \times (2501 + 1.88 \times 30) = 30.15 + 0.012 \times (2501 + 56.4) = 30.15 + 0.012 \times 2557.4 = 30.15 + 30.69 = 60.84\,kJ/kg \]
For Stream 2:
\[ h_2 = 1.005 \times 20 + 0.008 \times (2501 + 1.88 \times 20) = 20.10 + 0.008 \times (2501 + 37.6) = 20.10 + 0.008 \times 2538.6 = 20.10 + 20.31 = 40.41\,kJ/kg \]
Step 4: Calculate mixed enthalpy \(h_{mix}\):
\[ h_{mix} = \frac{2 \times 60.84 + 3 \times 40.41}{5} = \frac{121.68 + 121.23}{5} = \frac{242.91}{5} = 48.58\,kJ/kg \]
Step 5: Estimate mixed dry bulb temperature \(T_{mix}\) by iteration or approximation:
Using inverse of enthalpy formula ignoring minor variation:
\[ T_{mix} \approx \frac{h_{mix} - W_{mix} \times 2501}{1.005 + 1.88 \times W_{mix}} = \frac{48.58 - 0.0096 \times 2501}{1.005 + 1.88 \times 0.0096} = \frac{48.58 - 24.01}{1.005 + 0.018} = \frac{24.57}{1.023} = 24.02^\circ C \]
Answer: Mixed air has dry bulb temperature approximately 24.0°C and humidity ratio 0.0096 kg/kg dry air.
Step 1: Calculate initial vapor pressure \(P_{v1}\):
Saturation pressure at 30°C, \(P_{sat,30} \approx 4.24\,kPa\)
\(P_{v1} = 0.60 \times 4.24 = 2.544\,kPa\)
Step 2: Calculate initial humidity ratio \(W_1\):
\[ W_1 = 0.622 \times \frac{2.544}{101.325 - 2.544} = 0.622 \times \frac{2.544}{98.781} = 0.0160\,kg/kg \]
Step 3: Calculate saturation pressure at 15°C:
\(P_{sat,15} \approx 1.705\,kPa\)
Step 4: At 15°C, air can hold maximum moisture corresponding to saturation humidity ratio \(W_{sat}\):
\[ W_{sat} = 0.622 \times \frac{1.705}{101.325 - 1.705} = 0.622 \times \frac{1.705}{99.62} = 0.0106\,kg/kg \]
Step 5: Since air is cooled below dew point, moisture condenses out. Final humidity ratio \(W_2 = W_{sat} = 0.0106\,kg/kg\)
Step 6: Moisture removed per kg dry air:
\[ \Delta W = W_1 - W_2 = 0.0160 - 0.0106 = 0.0054\,kg/kg \]
Step 7: Final relative humidity is 100% at 15°C (saturated air).
Answer: Moisture removed is 0.0054 kg per kg dry air, and final air is saturated at 15°C with 100% relative humidity.
Step 1: Calculate mass flow rate of air:
Air changes per hour (ACH) = 5 means entire volume replaced 5 times per hour.
Volume flow rate \(V = 5 \times 100 = 500\,m^3/hr = \frac{500}{3600} = 0.139\,m^3/s\)
Mass flow rate \(m = \rho \times V = 1.2 \times 0.139 = 0.167\,kg/s\)
Step 2: Calculate initial and final humidity ratios.
At 35°C, \(P_{sat,35} \approx 5.62\,kPa\)
\(P_{v1} = 0.60 \times 5.62 = 3.372\,kPa\)
\[ W_1 = 0.622 \times \frac{3.372}{101.325 - 3.372} = 0.622 \times \frac{3.372}{97.953} = 0.0214\,kg/kg \]
At 24°C, \(P_{sat,24} \approx 2.98\,kPa\)
\(P_{v2} = 0.50 \times 2.98 = 1.49\,kPa\)
\[ W_2 = 0.622 \times \frac{1.49}{101.325 - 1.49} = 0.622 \times \frac{1.49}{99.835} = 0.0093\,kg/kg \]
Step 3: Calculate enthalpy of outside and inside air:
Outside air enthalpy \(h_1 = 1.005 \times 35 + 0.0214 \times (2501 + 1.88 \times 35) = 35.175 + 0.0214 \times (2501 + 65.8) = 35.175 + 0.0214 \times 2566.8 = 35.175 + 54.93 = 90.11\,kJ/kg\)
Inside air enthalpy \(h_2 = 1.005 \times 24 + 0.0093 \times (2501 + 1.88 \times 24) = 24.12 + 0.0093 \times (2501 + 45.12) = 24.12 + 0.0093 \times 2546.12 = 24.12 + 23.68 = 47.80\,kJ/kg\)
Step 4: Calculate cooling load \(Q\):
\[ Q = m \times (h_1 - h_2) = 0.167 \times (90.11 - 47.80) = 0.167 \times 42.31 = 7.06\,kW \]
Answer: The cooling load required is approximately 7.06 kW to maintain the desired indoor conditions.
When to use: When solving problems involving changes in air properties to quickly identify process types and property changes.
When to use: To simplify calculations by holding humidity ratio constant during sensible processes.
When to use: When moisture content changes during air conditioning processes, especially for latent heat calculations.
When to use: During entrance exams when exact values are not required, and quick estimation is sufficient.
When to use: To avoid calculation errors, especially when mixing temperature scales or pressure units.
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