SME
Engineering Thermodynamics
Properties of pure substances
Use this for “Water at 200 kPa has v = 0.4 m³/kg. Find its quality and enthalpy from the saturated-water table.”
Diagram
Water’s vapor dome on the T–v plane, from the IAPWS-IF97 saturation equations; it closes at the critical point, 373.9 °C and 22,060 kPa.
At P = 200 kPa water boils at Tₛₐₜ = 120.2 °C: the tie line runs from vf to vg.
x
=
v − vf
vg − vf
=
0.4 m³/kg − 0.001061
0.8857 − 0.001061
=
0.4509:
the
state
is
under
the
dome,
45%
of
the
way
from
liquid
to
vapor.
Enthalpy h = 1,497 kJ/kg
This is an example. Type your own values to start a new problem.
x
=
(
v (m³/kg)
−
vf (m³/kg)
)
÷
(
vg (m³/kg)
−
vf (m³/kg)
)
h (kJ/kg)
=
hf (kJ/kg)
+
x
×
hfg (kJ/kg)
P
Pressure
example · kPa
Formulas
vf
=
vf
of
saturated
water
at
P
0.001061 = vf of saturated water at 200
vg
=
vg
of
saturated
water
at
P
0.8857 = vg of saturated water at 200
hf
=
hf
of
saturated
water
at
P
504.68 = hf of saturated water at 200
hfg
=
hfg
of
saturated
water
at
P
2,201.6 = hfg of saturated water at 200
x
=
v
− vf
vg − vf
0.4509
=
0.4 − 0.001061
0.8857 − 0.001061
h
=
hf
+
x
×
hfg
1,497.5 = 504.68 + 0.4509 × 2,201.6
Assumptions
•
The state is under the dome, 0 ≤ x ≤ 1: liquid and vapor together at Tₛₐₜ.
•
vf, vg, hf and hfg are the saturated-water table’s row at P, worked out by the IAPWS-IF97 equations; a row typed from a printed table works too.
•
Quality x is the vapor’s share of the mass; v and h are averages weighted by it.
•
A compressed liquid is read as saturated liquid at its T (v ≈ vf, h ≈ hf).
Step-by-step
WE KNOW
P = 200 kPa, v = 0.4 m³/kg
FIND
saturated liquid volume (vf), saturated vapor volume (vg), saturated liquid enthalpy (hf), heat of vaporization (hfg), quality (x), enthalpy (h)
Step 1 · Find saturated liquid volume (vf)
Use vf = vf of saturated water at P
Read it off the saturated-water table at this state: the IAPWS-IF97 equations give the row a printed table lists.
vf = vf of saturated water at 200 kPa
vf = 0.0010605 m³/kg
Step 2 · Find saturated vapor volume (vg)
Use vg = vg of saturated water at P
Read it off the saturated-water table at this state: the IAPWS-IF97 equations give the row a printed table lists.
vg = vg of saturated water at 200 kPa
vg = 0.88574 m³/kg
Step 3 · Find saturated liquid enthalpy (hf)
Use hf = hf of saturated water at P
Read it off the saturated-water table at this state: the IAPWS-IF97 equations give the row a printed table lists.
hf = hf of saturated water at 200 kPa
hf = 504.684 kJ/kg
Step 4 · Find heat of vaporization (hfg)
Use hfg = hfg of saturated water at P
Read it off the saturated-water table at this state: the IAPWS-IF97 equations give the row a printed table lists.
hfg = hfg of saturated water at 200 kPa
hfg = 2,201.56 kJ/kg
Step 5 · Find quality (x)
Use x = (v − vf) ÷ (vg − vf)
Quality
is
how
far
v
lies
along
the
tie
line
from
vf
to
vg:
its
distance
from
vf
over
the
line’s
length.
x
=
0.4 − 0.0010605
0.88574 − 0.0010605
x
=
0.3989
0.8847
x
=
0.45094
Step 6 · Find enthalpy (h)
Use h = hf + x × hfg
The
liquid’s
enthalpy
hf,
plus
the
share
x
of
the
mass
that
took
in
hfg
to
boil.
h
=
504.684
+
0.45094
×
2,201.56
h
=
504.684
+
992.8
h
=
1,497.5
kJ/kg
Check
Put the numbers back in. Both sides should match.
0.001061 = vf of saturated water at 200 kPa ✓
0.8857 = vg of saturated water at 200 kPa ✓
504.6838 = hf of saturated water at 200 kPa ✓
2,201.5575 = hfg of saturated water at 200 kPa ✓
0.45094
=
0.4 − 0.0010605
0.88574 − 0.0010605
✓
1,497.5 = 504.684 + 0.45094 × 2,201.56 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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