SME
Heat Transfer · Problem type
Effectiveness–NTU: the heat without the outlet temperatures
Use this for “A counterflow exchanger has UA = 841 W/K. The oil’s Cₘᵢₙ is 1000 W/K and Cᵣ = 0.6. Oil enters at 120 °C, water at 20 °C. Find the heat rate.”
Diagram
Counterflow: ΔT₁ = Tₕᵢ − Tco = 70 K and ΔT₂ = Tₕₒ − Tci = 50 K.
ΔTₗₘ
=
ΔT₁ − ΔT₂
ln(ΔT₁ ÷ ΔT₂)
=
59.45
K,
at
the
dashed
bracket.
Cₘᵢₙ is the oil’s: it changes 50 K for the water’s 30 K on the same heat.
Outlets
from
q
Cₘᵢₙ
and
Cᵣ:
Tₕₒ
=
70,
Tco
=
50
°C.
ε
=
q
qₘₐₓ
=
0.5:
the
Cₘᵢₙ
stream
changes
0.5
×
(Tₕᵢ
−
Tci)
=
50
K
of
the
100
K
it
could.
q = 49.99 kW
NTU = 0.841
Overall conductance UA = 841 W/K
This is an example. Type your own values to start a new problem.
NTU
=
UA (W/K)
Cₘᵢₙ (W/K)
=
1
−
e
−NTU
×
(1
−
)
1
−
C
r
×
e
−NTU
×
(1
−
C
r
)
q (kW)
=
ε
×
C
min
×
(
Tₕᵢ (°C)
−
Tci (°C)
)
Tₕₒ
Oil outlet temperature
worked out (not typed) · °C
Tco
Water outlet temperature
worked out (not typed) · °C
Formulas
NTU
=
UA
Cₘᵢₙ
0.841
=
841
1,000
ε
=
(1
−
e
−NTU × (1 − Cᵣ)
)
÷
(1
−
Cᵣ
×
e
−NTU × (1 − Cᵣ)
)
0.4999
=
(1
−
e
−0.841 × (1 − 0.6)
)
÷
(1
−
0.6
×
e
−0.841 × (1 − 0.6)
)
q
=
ε
×
Cₘᵢₙ
×
(Tₕᵢ
−
Tci)
49,993.6787 = 0.4999 × 1,000 × (120 − 20)
Tₕₒ
=
Tₕᵢ
−
q
Cₘᵢₙ
70.0063
=
120
−
49,993.6787
1,000
Tco
=
Tci
+
Cᵣ
×
q
Cₘᵢₙ
49.9962
=
20
+
0.6
×
49,993.6787
1,000
Formulas are worked in W. The step-by-step shows the conversions.
Assumptions
•
Counterflow; the oil has the smaller capacity rate ṁcₚ, so Cᵣ is below 1.
•
Steady flow; no heat lost to the surroundings.
•
U is the same all along the exchanger; no phase change.
•
Each stream’s specific heat is constant.
Step-by-step
WE KNOW
Tₕᵢ = 120 °C, Tci = 20 °C, Cₘᵢₙ = 1,000 W/K, Cᵣ = 0.6, UA = 841 W/K
FIND
number of transfer units (NTU), effectiveness (ε), heat rate (q), oil outlet temperature (Tₕₒ), water outlet temperature (Tco)
Step 1 · Find number of transfer units (NTU)
Use NTU = UA ÷ Cₘᵢₙ
NTU measures the exchanger’s size against the smaller capacity rate.
NTU
=
841
1,000
NTU = 0.841
Step 2 · Find effectiveness (ε)
Use ε = (1 − e−NTU × (1 − Cᵣ)) ÷ (1 − Cᵣ × e−NTU × (1 − Cᵣ))
The counterflow effectiveness: the share of the largest possible heat this exchanger passes.
ε
=
(1
−
e
−0.841 × (1 − 0.6)
)
÷
(1
−
0.6
×
e
−0.841 × (1 − 0.6)
)
ε
=
1 −
e
−0.841 × 0.4
1 − 0.6 ×
e
−0.841 × 0.4
ε
=
1 −
e
−0.3364
1 − 0.6 ×
e
−0.3364
ε
=
1 − 0.7143
1 − 0.6 × 0.7143
ε
=
0.2857
1 − 0.4286
ε
=
0.2857
0.5714
ε
=
0.4999368
Step 3 · Find heat rate (q)
Use q = ε × Cₘᵢₙ × (Tₕᵢ − Tci)
The
largest
heat
is
Cₘᵢₙ
times
the
inlets’
difference;
ε
of
it
passes.
q
=
0.49993679
×
1,000
×
(120
−
20)
q
=
0.4999
×
1,000
×
100
q
=
499.9
×
100
q
=
49,993.6787
W
Step 4 · Find oil outlet temperature (Tₕₒ)
Use Tₕₒ = Tₕᵢ − q ÷ Cₘᵢₙ
The
oil
carries
the
smaller
capacity
rate,
so
it
changes
by
q
Cₘᵢₙ
.
Tₕₒ
=
120
−
49,993.68
1,000
Tₕₒ = 120 − 49.99
Tₕₒ = 70.0063 °C
Step 5 · Find water outlet temperature (Tco)
Use Tco = Tci + Cᵣ × q ÷ Cₘᵢₙ
The
water’s
capacity
rate
is
Cₘᵢₙ
Cᵣ
,
so
it
changes
by
Cᵣ
×
q
Cₘᵢₙ
.
Tco
=
20
+
0.6
×
49,993.68
1,000
Tco
=
20
+
29,996.208
1,000
Tco = 20 + 30
Tco = 49.9962 °C
Step 6 · Convert the answers
Change each answer back to the unit you chose.
q = 49,993.6787 W = 49.9937 kW (1 kW = 1000 W)
Check
Put the numbers back in. Both sides should match.
0.841
=
841
1,000
✓
0.4999368
=
(1
−
e
−0.841 × (1 − 0.6)
)
÷
(1
−
0.6
×
e
−0.841 × (1 − 0.6)
)
✓
49,993.68 = 0.4999368 × 1,000 × (120 − 20) ✓
70.0063
=
120
−
49,993.68
1,000
✓
49.9962
=
20
+
0.6
×
49,993.68
1,000
✓
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The same oil cooler in parallel flow
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