An evacuated tank with 1 m3capacity is initially empty with no fluid inside.Water in the amount of 2 L and at 25 °C is transferred into the tank. AtMidday, thermal equilibrium is assumed to be attained and fluid temperatureof 60 °C is uniform throughout the tank. At this condition do we find water inthe tank as a mixture of liquid and vapor or only water vapor? If only as watervapor how much add
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An evacuated tank with 1 m3capacity is initially empty with no fluid inside.
Water in the amount of 2 L and at 25 °C is transferred into the tank. At
Midday, thermal equilibrium is assumed to be attained and fluid temperature
of 60 °C is uniform throughout the tank. At this condition do we find water in
the tank as a mixture of liquid and vapor or only water vapor? If only as water
vapor how much additional water we have to add so that the tank exist only as
saturated water vapor?
Penyelesaian
ρair=1 kg/L
v
25 °C = 1,003⋅10-3m3/kg
m
air=ρair V=1 kg
L
× 2L=2kg
m1=
Vv
=
1m3
1,003⋅10-3 m3/kg =2kg
Volume spesifik fluida pada tangki (T = 60°C) adalah
v=
Vm
=
1m3
2kg =0,5 m3/kg
Untuk air padasuhu 60°C, nilai vL= 1,017 x 10-3m3/kg, vg = 7,679m3/kg
Karena nilai v berada di antara vLdan vg, maka sistem tersebut terdiri dari
campuran cairan dan uap air atau berada dalam keadaan jenuh (saturated).
Lebih lanjut, kita dapat menghitung kualitas (x) dari campuran tersebut, yaitu
v=x v
g+(1-x)vL
0,5=7,679 x+(1-x )1,017⋅10-3
x=0,063
Jika hanya terdapat uap air di dalam tangki, maka
m
uap=
Vv
=
1m3
7,679 kg
m
3
=0,130 kg
Sehingga massa air yang harus dikurangi adalah
m=2 kg-0,13 kg=1,87 kg
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