### Reading a Psychrometric Chart

• 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.

### Solved Example:

#### 77-2-01

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

### Solved Example:

#### 77-2-02

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

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#### 77-2-03

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

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#### 77-2-04

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

### Solved Example:

#### 77-2-05

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, $P_{water} = 0.265\times 100 = 2.65\ kPa$ $RH =\dfrac {2,65}{3.17}\times 100 =83.5\%$

### Solved Example:

#### 77-2-06

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*}

### Solved Example:

#### 77-2-07

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

### Solved Example:

#### 77-2-08

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

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#### 77-2-09

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

### Solved Example:

#### 77-2-10

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

### Solved Example:

#### 77-2-11

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

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#### 77-2-12

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

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#### 77-2-13

In a throttling process:

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#### 77-2-14

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)