SME
Engineering Thermodynamics · Problem type
Ideal Rankine cycle from table enthalpies
Use this for “A Rankine cycle runs from 15 kPa to 6 MPa; h₁ = 226, h₃ = 3300 and h₄ = 2200 kJ/kg, v₁ = 0.00101 m³/kg. Find wₚ, qᵢₙ and η.”
Diagram
Rankine cycle: the pump 1→2 and turbine 3→4 are vertical (isentropic); the boiler 2→3 heats along the high-pressure line, the condenser 4→1 runs flat under the dome.
At
4,
x₄
=
h₄ − hf
hfg
=
2,100 − 191.8
2,392
=
0.798
(hf
and
hfg
from
IAPWS
at
10
kPa).
State 3 sits above 4 (ideal turbine); the page gives h, not T, so its height is set by the area under 2→3 (qᵢₙ), not read from a table.
Net work wₙₑₜ = 1,292 kJ/kg
Thermal efficiency η = 0.4037
This is an example. Type your own values to start a new problem.
wₚ (kJ/kg)
=
v₁ (m³/kg)
×
(
P₂ (kPa)
−
P₁ (kPa)
)
wₙₑₜ (kJ/kg)
=
(
h₃ (kJ/kg)
−
h₄ (kJ/kg)
)
−
wₚ (kJ/kg)
η
=
wₙₑₜ (kJ/kg)
h₃ (kJ/kg)
−
h₁ (kJ/kg)
−
wₚ (kJ/kg)
qᵢₙ
Heat in
calculated · kJ/kg
Formulas
wₚ
=
v₁
×
(
P₂
−
P₁
)
8.0699 = 0.00101 × (8,000 − 10)
qᵢₙ
=
h₃
−
h₁
−
wₚ
3,199.9301 = 3,400 − 192 − 8.0699
wₙₑₜ
=
(
h₃
−
h₄
)
−
wₚ
+1,291.9301 = (3,400 − 2,100) − 8.0699
η
=
wₙₑₜ
qᵢₙ
0.4037
=
1,291.9301
3,199.9301
Assumptions
•
Ideal Rankine cycle: saturated liquid leaves the condenser (1); the pump and turbine are isentropic.
•
Read h₁, h₃, h₄ and v₁ from your steam tables; the dome is drawn from IAPWS-IF97.
•
Kinetic and potential energy changes are neglected; m³/kg × kPa = kJ/kg.
Step-by-step
WE KNOW
v₁ = 0.00101 m³/kg, P₁ = 10 kPa, P₂ = 8,000 kPa, h₁ = 192 kJ/kg, h₃ = 3,400 kJ/kg, h₄ = 2,100 kJ/kg
FIND
pump work (wₚ), heat in (qᵢₙ), net work (wₙₑₜ), thermal efficiency (η)
Step 1 · Find pump work (wₚ)
Use wₚ = v₁ × (P₂ − P₁)
The pump pushes nearly incompressible water: its work is v times the pressure rise.
wₚ = 0.00101 × (8,000 − 10)
wₚ = 0.00101 × 7,990
wₚ = 8.0699 kJ/kg
Step 2 · Find heat in (qᵢₙ)
Use qᵢₙ = h₃ − h₁ − wₚ
The
boiler
heats
the
water
from
h₂
=
h₁
+
wₚ
up
to
h₃
.
qᵢₙ = 3,400 − 192 − 8.0699
qᵢₙ = 3,199.93 kJ/kg
Step 3 · Find net work (wₙₑₜ)
Use wₙₑₜ = (h₃ − h₄) − wₚ
The
turbine’s
drop
h₃
−
h₄
,
less
the
pump’s
work.
wₙₑₜ = (3,400 − 2,100) − 8.0699
wₙₑₜ = +1,291.93 kJ/kg
Step 4 · Find thermal efficiency (η)
Use η = wₙₑₜ ÷ qᵢₙ
The share of the heat in that comes out as net work.
η
=
1,291.93
3,199.93
η
=
0.4037
Check
Put the numbers back in. Both sides should match.
8.0699 = 0.00101 × (8,000 − 10) ✓
3,199.93 = 3,400 − 192 − 8.0699 ✓
+1,291.93 = (3,400 − 2,100) − 8.0699 ✓
0.4037
=
1,291.93
3,199.93
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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