An air-water vapour mixture has a dry bulb temperature of 60°C and a dew point temperature of 40°C. The total pressure is 101.3 kPa and the vapour pressure of water at 40°C and 60°C are 7.30 kPa and 19.91 kPa respectively.The humidity of air sample expressed as kg of water vapour/kg of dry air is
A. 0.048
B. 0.079
C. 0.122
D. 0.152
Answer: Option A
Solution (By Examveda Team)
Given:Dry bulb temperature = 60°C
Dew point temperature = 40°C
Total pressure = 101.3 kPa
Vapour pressure of water at 40°C = 7.30 kPa
Vapour pressure of water at 60°C = 19.91 kPa
Humidity (H) is calculated using:
H = 0.622 × (Pv / (P - Pv))
Where:
Pv = partial pressure of water vapour = vapour pressure at dew point = 7.30 kPa
P = total pressure = 101.3 kPa
Now substituting the values:
H = 0.622 × (7.30 / (101.3 - 7.30))
H = 0.622 × (7.30 / 94.0)
H ≈ 0.622 × 0.07766
H ≈ 0.0482 kg water/kg dry air
Therefore, the correct answer is Option A: 0.048
This Question Belongs to Chemical Engineering >> Mass Transfer
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We are given:
Dry bulb temperature (T_db) = 60°C
Dew point temperature (T_dp) = 40°C
Total pressure (P) = 101.3 kPa
Vapour pressure of water at 40°C = 7.30 kPa
Vapour pressure of water at 60°C = 19.91 kPa
Required: Specific humidity (ω) = kg water vapour / kg dry air
✅ Step 1: Use the formula for specific humidity:
𝜔
=
0.622
⋅
𝑃
𝑣
𝑃
−
𝑃
𝑣
ω=0.622⋅
P−P
v
P
v
Where:
𝜔
ω = specific humidity (kg water vapour / kg dry air)
𝑃
𝑣
P
v
= partial pressure of water vapour, which is the saturation pressure at the dew point
𝑃
P = total pressure (101.3 kPa)
From the data:
𝑃
𝑣
=
7.30
P
v
=7.30 kPa (since dew point is 40°C)
✅ Step 2: Plug in the values:
𝜔
=
0.622
⋅
7.30
101.3
−
7.30
=
0.622
⋅
7.30
94.0
ω=0.622⋅
101.3−7.30
7.30
=0.622⋅
94.0
7.30
𝜔
=
0.622
⋅
0.07766
≈
0.0483
ω=0.622⋅0.07766≈0.0483
✅ Final Answer:
A. 0.048 ✅
H=0.622*(Pa/(Pt-Pa))
H=0.622*(19.91/(101.3-19.91))
H=0.1521
explain please