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Use of Property Table Example
T Psat sp. Vol(v) sp int energy(u) sp.Enthalpy(h) sp.
Entropy(s)
(C) kPa m!kg k"!kg k"!kg k"!kg #
subs$ript f for li%ui& i.e vf 'sp vol for saturate&
li%ui&
subs$ript g for vapor i.e ug 'sp int. energy for saturate&
vapor
ote vfg* vg' vf
a) + rigi& tank $ontains ,- kg of saturate& li%ui&
ater at /0oC.
1etermine the pressure in the tank an& the volume in the
tank.
P * Psat2/0C* 3.03 kPa (Table +',)
V * Vf2/0C* -.--4-56 m!kg (Table +',)
Total volume * (,- kg)( -.--4-56 m!kg) * -.-,4-, m
7n general8 Compresse& li%ui& is $hara$teri9e&
by
:igher pressures (P ; Psatat a given T)
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1etermine the missing properties an& phase &es$riptions
in the folloing
table for ater
T8o
C P8 kPa u8 k"!kg x Phase1es$ription
i) 3-- -.-
ii) 60 ,--
iii) 4--- ---
iv) 5-- -.6
v 450 4--
i) x = 0. So we have saturated liquid at the specified
pressureof 800 kPa.
T = 170.41
u = uf!800kPa = 7"0."" k#$k%
ii) &t 400 kPa' Tsat=14(.(o
T * Tsat Therefore' we have co+pressed liquid
u uf!, = "7".0- k#$k%
ualit/ has o +eai% i co+pressed liq re%io.
iii) &t 1Pa' uf = 71.(- k#$k% ad u% = ",8".8 k#$k%
u 2 u% Therefore' we have superheated vapor
ualit/ has o +eai% for superheated vapor
To fid te+p' eed to iterpolate3
Tale a5
Te+p u 6k#$k%)
400 "-,7.(
T (000 6%ive i pro)
,00 (1"4.4
T ,-- * --- 5>0/.>
0-- ,-- 450 5>0/.>
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T *
iv) x = 0. +eas 0 of +ass is vapor' 40 liq. Therfefore
saturated liq5vap at "00 kPa. Therefore te+p is saturatio
te+pT = Tsat!"00kPa = 1"0."(
o
&lso' uf = ,04.,0 k#$k% ad uf% = "0"4. k#$k%
u= uf xuf% = ,04.,0 60.)6"0"4.)
= 171-." k#$k%
v) &t 1",o' uf = ,"4.8( k#$k% ad u% = ",(4.( k#$k%
saturated liquid5vap +ixture ecause uf* u * u%Saturatio P =
Psat!1",= "("."( kPa
u= uf xuf%
f%
f
u
uux
= = 00.-0.5-->
3.05,4--=
7n summary
T8 oC P8 kPa u8 k"!kg x Phase
1es$ription
i) 4/-.,4 3-- /4>.>/ -.- ?aturate&
li%ui&
ii) 60 ,-- 5/5.-> (ot
appli$able)
$ompresse&
li%ui&
iii) 4--- --- +(ot
appli$able)
superheate&vapor
iv) 45-.5 5-- 4/4>.56 -.6 ?aturate&
li%ui&'
vapor
mixture
V 450 55.5 4-- -.00 ?aturate&
li%ui&'
vapormixture
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