• Be able to understand the basic terminology and definitions related to air properties;

• Be able to read psychrometric chart; and

• Be able to apply the psychrometric charts on air-conditioning or drying applications.

Relative Humidity can be calculated with a wet bulb thermometer and a dry bulb thermometer, using a psychrometric chart as below. Wet-bulb temperature is measured using a thermometer, with the glass bulb or sensor tip wrapped in a wick, which is kept wet. The evaporation of water from the wet wick has a cooling effect on the thermometer and the rate of evaporation from the wet-bulb thermometer depends on the humidity of the air. Therefore, less humidity means more cooling, which means a lower wet bulb temperature, and a bigger difference between wet and dry bulb temperatures.
The difference in the temperatures (wet bulb depression) indicated by the two thermometers gives a measure of atmospheric humidity. This chart allows humidity to be calculated from wet and dry bulb readings. (The dry bulb temperature is another name for air temperature). It also explains common atmospheric behaviour that we see daily. We will break the chart down for ease of explanation:

• The vertical lines represent dry bulb temperatures.

• The diagonal lines represent wet bulb temperatures.

• Lines of constant humidity are also shown on the chart.

• Read the relative humidity by finding where the wet bulb temperature and the dry bulb temperature readings intersect. The nearest constant humidity line is the relative humidity for the given wet and dry bulb temperatures.

• You can also read the wet bulb temperature from the chart, when you know the air temperature and relative humidity, using the same principle of finding the intersection of the relevant lines.

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On Mollier chart, flow through turbine is represented by: (SSC JE ME Paper 4- Jan 2018 Morning)

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On Mollier chart, the constant pressure lines: (JKSSB JE ME 2015)

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On Mollier chart, free expansion, or throttling process from high pressure to atmosphere is represented by:

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In a psychrometric chart, specific humidity (moisture content) lines are: (ESE Mech 2016- Paper I)

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A room contains 35 kg of dry air and 0.5 kg of water vapor. The total pressure and temperature of air in the room are 100 kPa and 25$^{\circ}\mathrm{C}$ respectively. Given that the saturation pressure for water at 25$^{\circ}\mathrm{C}$ is 3.17 kPa, the relative humidity of the air in the room is: (GATE ME 2012)

Solution:
\begin{align*} P_{total} &=100\ kPa\\ P_{saturated} &=3.17\ kPa\\ \end{align*} $RH=\dfrac {P_{water}}{P_{saturated}} \times 100$ $P_{water}=x_{water}\times P_{total}$ where, x$_{water}$ = mole fraction of water $x_{water} =\dfrac {n_{H_{2}O}}{n_{H_{2}O} + n_{air}} =\dfrac {\dfrac{0.5}{18}}{\dfrac{0.5}{18}+\dfrac{30}{29}} =0.0265$ Therefore, $P_{water} = 0.265\times 100 = 2.65\ kPa$ $RH =\dfrac {2,65}{3.17}\times 100 =83.5\%$

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The pressure, dry bulb temperature and relative humidity of air in a room are 1 bar, 30$^{\circ}\mathrm{C}$ and 70%, respectively. If the saturated steam pressure at 30$^{\circ}\mathrm{C}$ is 4.25 kPa, the specific humidity of the room air in kg water vapor/kg dry air is: (GATE ME 2013)

Solution:
Specific humidity is given by, $w = 0.622 \times \dfrac{p_v}{p_a - p_v}$ where, p$_v$ = Relative humidity $\times$ Saturated steam pressure $p_v = \phi \times p_s = 0.7 \times 0.0425 = 0.02975\ \mathrm{bar}$ p$_a$ =1\ bar\\ So that from the above equation, we have \begin{align*} w &= 0.622 \times \dfrac{0.02975}{1 - 0.02975}\\ &= 0.0191\ \mathrm{kg/kg\ of\ dry\ air} \end{align*}

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If a mass of moist air contained in a closed metallic vessel is heated, then its:

Solution:
Given, that moist air is contained in a closed vessel, so we can say that mass of moist air is constant.
$\omega$ = specific humidity = $\dfrac{m_v}{m_{dry\ air}}$ = constant
When the container is heated, temperature increases due to which saturation pressure increases, hence mass at saturation pressure increases.
So, $\phi$ = $\dfrac{m_v}{m_{saturated\ vapor}}$ = decreases

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If a steam sample is nearly in dry condition, then its dryness fraction can be most accurately determined by:

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The specific heat of superheated steam in kcal/kg is generally of the order of:

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The dry saturated steam at very low pressure, (5-10 kg/cm$^2$) when throttled to atmosphere will become:

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Water at pressure of 4 kg/cm$^2$ and 160$^\circ$C temperature when exposed to atmosphere will:

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The dry saturated steam at very high pressure (150-200 kg/cm$^2$) when throttled to atmosphere will become:

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In a throttling process:

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In an experiment to determine dryness fraction of steam, the mass of water separated was 1.2 kg in 15 minutes and the mass of steam passed out in same time was 4.8 kg. Dryness fraction is: (NPCIL Stipendiary Trainee ME 2018)