SME
Engineering Thermodynamics
Gas mixtures and psychrometrics
Use this for “Air at 30 °C and 101.325 kPa has 50% relative humidity (Pg = 4.25 kPa). Find the vapor pressure, the humidity ratio and the enthalpy.”
Chart
ω(50) ≈ 0.01331
This is an example. Type your own values to start a new problem.
Pᵥ (kPa)
=
φ
÷
100
×
Pg (kPa)
ω
=
0.622
×
Pᵥ (kPa)
÷
(
P (kPa)
−
Pᵥ (kPa)
)
h (kJ/kg)
=
1.005
×
T (°C)
+
ω
×
(2501
+
1.82
×
T (°C)
)
Put Pᵥ = φ ÷ 100 × Pg into the second line to get the graph’s curve, ω against φ.
Formulas
Pg
=
Pₛₐₜ
of
saturated
water
at
T
4.2467 = Pₛₐₜ of saturated water at 30
Pᵥ
=
φ
100
×
Pg
2.1233
=
50
100
×
4.2467
ω
=
0.622
×
Pᵥ
P
− Pᵥ
0.01331
=
0.622
×
2.1233
101.325 − 2.1233
h
=
1.005
×
T
+
ω
×
(2,501
+
1.82
×
T
)
64.174 = 1.005 × 30 + 0.01331 × (2,501 + 1.82 × 30)
Assumptions
•
Dry air and water vapor are ideal gases, each acting alone at its own partial pressure.
•
Pg is the saturated-water table’s Pₛₐₜ at the dry-bulb T (IAPWS-IF97); a typed table value works too.
•
ω is kg of water per kg of dry air; h is per kg of dry air, from 0 °C for both.
•
The vapor’s enthalpy is hg ≈ 2501 + 1.82T kJ/kg, T in °C.
Step-by-step
WE KNOW
T = 30 °C, φ = 50%, P = 101.325 kPa
FIND
saturation pressure (Pg), vapor pressure (Pᵥ), humidity ratio (ω), enthalpy (h)
Step 1 · Find saturation pressure (Pg)
Use Pg = Pₛₐₜ of saturated water at T
Read it off the saturated-water table at this state: the IAPWS-IF97 equations give the row a printed table lists.
Pg = Pₛₐₜ of saturated water at 30 °C
Pg = 4.2467 kPa
Step 2 · Find vapor pressure (Pᵥ)
Use Pᵥ = φ ÷ 100 × Pg
Relative
humidity
is
the
vapor
pressure
as
a
share
of
the
most
the
air
could
hold
at
T
.
Pᵥ
=
50
100
×
4.2467
Pᵥ = 0.5 × 4.2467
Pᵥ = 2.1233 kPa
Step 3 · Find humidity ratio (ω)
Use ω = 0.622 × Pᵥ ÷ (P − Pᵥ)
Moles
of
vapor
over
moles
of
dry
air,
times
18.015
28.97
=
0.622
to
turn
moles
into
kilograms.
ω
=
0.622
×
2.1233
101.325 − 2.1233
ω
=
0.622
×
2.1233
99.2017
ω
=
1.3207
99.2017
ω
=
0.013313
Step 4 · Find enthalpy (h)
Use h = 1.005 × T + ω × (2,501 + 1.82 × T)
The
dry
air’s
cpT
plus
the
vapor’s
hg
for
each
of
its
ω
kilograms.
h
=
1.005
×
30
+
0.013313
×
(2,501
+
1.82
×
30)
h
=
1.005
×
30
+
0.01331
×
(2,501
+
54.6)
h
=
1.005
×
30
+
0.01331
×
2,555.6
h
=
30.15
+
34.02
h
=
64.174
kJ/kg
Check
Put the numbers back in. Both sides should match.
4.2467 = Pₛₐₜ of saturated water at 30 °C ✓
2.1233
=
50
100
×
4.2467
✓
0.013313
=
0.622
×
2.1233
101.325 − 2.1233
✓
64.174 = 1.005 × 30 + 0.013313 × (2,501 + 1.82 × 30) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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