SME
Engineering Thermodynamics · Problem type
Cooling and dehumidifying coil
Use this for “1.5 kg/s of dry air enters a coil at h = 68 kJ/kg, ω = 0.015 and leaves at 30 kJ/kg, ω = 0.008; condensate leaves at 50 kJ/kg. Find the water removed and the heat.”
Diagram
This is an example. Type your own values to start a new problem.
ṁw (kg/s)
=
ṁₐ (kg/s)
×
(
ω₁
−
ω₂
)
Q̇ₒᵤₜ (kW)
=
ṁₐ (kg/s)
×
(
h₁ (kJ/kg)
−
h₂ (kJ/kg)
)
−
ṁw (kg/s)
×
hw (kJ/kg)
ṁᵥ₁
Water in
worked out (not typed) · kg/s
ṁᵥ₂
Water out
worked out (not typed) · kg/s
Q̇ᵢₙ
Heat in
worked out (not typed) · kW
Formulas
ṁw
=
ṁₐ
×
(
ω₁
−
ω₂
)
0.0105 = 1.5 × (0.015 − 0.008)
Q̇ₒᵤₜ
=
ṁₐ
×
(
h₁
−
h₂
)
−
ṁw
×
hw
+56.475 = 1.5 × (68 − 30) − 0.0105 × 50
ṁᵥ₁
=
ṁₐ
×
ω₁
0.0225 = 1.5 × 0.015
ṁᵥ₂
=
ṁₐ
×
ω₂
0.012 = 1.5 × 0.008
Q̇ᵢₙ = −Q̇ₒᵤₜ
−56.475 = −56.475
Assumptions
•
Steady flow and no work; kinetic and potential energy changes are ignored.
•
h₁ and h₂ are per kg of dry air, and ω is kg of water per kg of dry air.
•
The condensate leaves at the outlet temperature, so hw is about hf there.
•
Q̇ₒᵤₜ is the heat the coil takes out; the picture’s Q̇ᵢₙ is the same heat, negative.
Step-by-step
WE KNOW
ṁₐ = 1.5 kg/s, h₁ = 68 kJ/kg, h₂ = 30 kJ/kg, ω₁ = 0.015, ω₂ = 0.008, hw = 50 kJ/kg
FIND
water removed (ṁw), heat removed (Q̇ₒᵤₜ), water in (ṁᵥ₁), water out (ṁᵥ₂), heat in (Q̇ᵢₙ)
Step 1 · Find water removed (ṁw)
Use ṁw = ṁₐ × (ω₁ − ω₂)
A water balance: the water the air brings in, less what it carries out, drains away.
ṁw = 1.5 × (0.015 − 0.008)
ṁw = 1.5 × 0.007
ṁw = 0.0105 kg/s
Step 2 · Find heat removed (Q̇ₒᵤₜ)
Use Q̇ₒᵤₜ = ṁₐ × (h₁ − h₂) − ṁw × hw
An
energy
balance:
the
air
gives
up
ṁₐ(
h₁
−
h₂
);
the
condensate
carries
some
away,
the
coil
takes
the
rest.
Q̇ₒᵤₜ = 1.5 × (68 − 30) − 0.0105 × 50
Q̇ₒᵤₜ = 1.5 × 38 − 0.0105 × 50
Q̇ₒᵤₜ = 57 − 0.525
Q̇ₒᵤₜ = +56.475 kW
Step 3 · Find water in (ṁᵥ₁)
Use ṁᵥ₁ = ṁₐ × ω₁
The
water
the
air
brings
in:
ω₁
for
each
kilogram
of
dry
air.
ṁᵥ₁ = 1.5 × 0.015
ṁᵥ₁ = 0.0225 kg/s
Step 4 · Find water out (ṁᵥ₂)
Use ṁᵥ₂ = ṁₐ × ω₂
The
water
the
air
carries
out:
ω₂
for
each
kilogram
of
dry
air.
ṁᵥ₂ = 1.5 × 0.008
ṁᵥ₂ = 0.012 kg/s
Step 5 · Find heat in (Q̇ᵢₙ)
Use Q̇ᵢₙ = −Q̇ₒᵤₜ
Heat into the air is the heat removed with its sign turned.
Q̇ᵢₙ = −56.475 kW
Check
Put the numbers back in. Both sides should match.
0.0105 = 1.5 × (0.015 − 0.008) ✓
+56.475 = 1.5 × (68 − 30) − 0.0105 × 50 ✓
0.0225 = 1.5 × 0.015 ✓
0.012 = 1.5 × 0.008 ✓
−56.475 = −56.475 ✓
Topic check
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