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PY 2: GATE ME 2020 Official Paper: Shift 2

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

__Concept: __

For saturated air relative humidity (ϕ) is 100% i.e. P_{v} = P_{vs}

i.e. partial pressure of water vapour in the moist air is equal to the saturation pressure of the water vapour

**Specific Humidity: **It is the ratio of the mass of water vapour to the mass of air at a given temperature and volume.

Specific humidity \(\omega = \frac{{{m_v}}}{{{m_a}}} \times \frac{{{P_v}}}{{{P_t} - {P_v}}}\)

where P_{v} is the partial pressure of water vapour and P_{t} is the total pressure of moist air.

__Calculation: __

If we find the specific humidity before the cooling coil and after the cooling coil then the difference between two will give the amount of water condensed in the cooling coil.

Before entering into the cooling coil,

ϕ = 80% = 0.8, m_{v} = 18 g/mol, m_{a} = 28.94 g/mol, P_{t }= 105 kPa

At 30°C P_{vs} = 4.24 kPa

Relative humidity (ϕ) \(= \frac{{{P_v}}}{{{P_{vs}}}}\)

∴ P_{v} = ϕ × P_{vs} = 0.8 × 4.24

⇒ P_{v} = 3.392 kPa

Specific humidity (ω) \(= \frac{{{m_v}}}{{{m_a}}} \times \frac{{{P_v}}}{{{P_t} - {P_v}}}\)

\({\left( \omega \right)_{moistair}} = \frac{{18}}{{28.94}} \times \frac{{{P_v}}}{{105 - {P_v}}}\)

\(\therefore {\omega _{moistair}} = \frac{{18}}{{28.94}} \times \frac{{3.392}}{{105 - 3.392}} = \;0.0207{\rm{\;kg}}/{\rm{kg\;of\;dry\;air}}\)

Now after the exit of the cooling coil

At 15°C P_{vs} = 1.7 kPa

For saturated air ϕ = 1

⇒ P_{v} = P_{vs} = 1.7 kPa

\({\omega _{saturated\;air}} = \frac{{18}}{{28.94}} \times \frac{{1.7}}{{100 - 1.7}} = \;0.0107{\rm{\;kg}}/{\rm{kg\;of\;dry\;air}}\)

∴ Mass of water condensing = ω_{moist air} – ω_{saturated air}

⇒ Mass of water condensing = 0.0207 – 0.0107 = 0.010 kg/kg of dry air

Mass of water condensing out from duct = 10 gm/kg of dry air