SME
Heat Transfer · Problem type
Evaporation by the heat–mass transfer analogy
Use this for “Air flows over a wet 0.5 m² surface with h = 25 W/(m²·K) (ρ = 1.18 kg/m³, cₚ = 1007 J/(kg·K), Le = 0.85). The vapor density is 0.0230 kg/m³ at the surface and 0.0070 in the air. Find hₘ and the evaporation rate.”
Chart
Heat coefficient h = 25 W/(m²·K)
Air density ρ = 1.18 kg/m³
Specific heat cₚ = 1,007 J/(kg·K)
Lewis number Le = 0.85
Mass coefficient hₘ = 2.3447 cm/s
Wet area A = 0.5 m²
Evaporation rate ṁ = 0.1876 g/s
This is an example. Type your own values to start a new problem.
hₘ (cm/s)
=
h (W/(m²·K))
ρ (kg/m³)
×
cₚ (J/(kg·K))
÷
Le
2
3
Δρᵥ (kg/m³)
=
ρv,s (kg/m³)
−
ρᵥ,∞ (kg/m³)
ṁ (g/s)
=
h
m
×
A (m²)
×
Δρᵥ
Formulas
hₘ
=
h
ρ
× cₚ × L
e
2 ÷ 3
0.02345
=
25
1.18 × 1,007 ×
0.85
2 ÷ 3
Δρᵥ = ρv,s − ρᵥ,∞
0.016 = 0.023 − 0.007
ṁ
=
hₘ
×
A
×
Δρᵥ
1.8757 ×
10
-4
=
0.02345
×
0.5
×
0.016
Formulas are worked in m/s, kg/s. The step-by-step shows the conversions.
Assumptions
•
The same flow carries heat and vapor (0.6 < Pr, Sc < 60), so Chilton–Colburn links h and hₘ.
•
Low mass-transfer rates: the evaporating vapor barely changes the flow.
•
Le = α ÷ DAB, the air’s thermal over its vapor diffusivity (about 0.85 for water vapor in air); the surface vapor is saturated.
Step-by-step
WE KNOW
h = 25 W/(m²·K), ρ = 1.18 kg/m³, cₚ = 1,007 J/(kg·K), Le = 0.85, A = 0.5 m², ρv,s = 0.023 kg/m³, ρᵥ,∞ = 0.007 kg/m³
FIND
mass coefficient (hₘ), vapor density drop (Δρᵥ), evaporation rate (ṁ)
Step 1 · Find mass coefficient (hₘ)
Use hₘ = h ÷ (ρ × cₚ × Le2 ÷ 3)
Chilton–Colburn:
the
same
flow
moves
heat
and
vapor,
so
hₘ
is
h
over
ρcₚ,
scaled
by
L
e
2
3
.
hₘ
=
25
1.18 × 1,007 ×
0.85
2 ÷ 3
hₘ
=
25
1.18 × 1,007 × 0.89732
hₘ
=
25
1,188.26 × 0.89732
hₘ
=
25
1,066.2
hₘ = 0.02345 m/s
Step 2 · Find vapor density drop (Δρᵥ)
Use Δρᵥ = ρv,s − ρᵥ,∞
The driving difference: the vapor at the surface less the vapor in the air.
Δρᵥ = 0.023 − 0.007
Δρᵥ
=
0.016
kg/
m
3
Step 3 · Find evaporation rate (ṁ)
Use ṁ = hₘ × A × Δρᵥ
Vapor leaves like heat by convection: the coefficient times the area times the driving difference.
ṁ
=
0.02345
×
0.5
×
0.016
ṁ
=
0.011725
×
0.016
ṁ
=
0.0001876
kg/s
Step 4 · Convert the answers
Change each answer back to the unit you chose.
hₘ = 0.02345 m/s = 2.3447 cm/s (1 m/s = 100 cm/s)
ṁ = 0.0001876 kg/s = 0.1876 g/s (1 kg/s = 1000 g/s)
Check
Put the numbers back in. Both sides should match.
0.02345
=
25
1.18 × 1,007 ×
0.85
2 ÷ 3
✓
0.016 = 0.023 − 0.007 ✓
1.8757 ×
10
-4
=
0.0234467
×
0.5
×
0.016
✓
Topic check
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