拜求流體力學解答--拜求

1.
PV =nRT
PV= (W/M) RT (W :質量 M分子量)
PM =ρRT
700k×44=ρ×8.314k ×(90+273) (R=8.314k J/kgmol -K)
ρ=10.20549 kg/m^3
v(比容積) =V/W =1/ρ =9.80×10^-2 m^3/kg

比重 = ρCO2/ ρH2O
ρH2O at 90°C & 700KPa =?
查表ρH2O at 90°C & 100KPa(≒1 atm) =0.96534g/cm^3

Density and change in Pressure
ρ1 = ρ0 / (1 - (p1 - p0) / E)
where
E = bulk modulus fluid elasticity (N/m2) ； For water : 2.15×10^9 (N/m^2)
ρ1 = final density (kg/m^3)
ρ0 = initial density (kg/m^3)
p1 = final pressure (N/m^2)
p0 = initial pressure (N/m^2)
ρ1 =0.96534/ {1-(700-100)×10^3/(2.15 ×10^9) } =0.96561 g/cm^3
=> ρH2O at 90°C & 700KPa =0.96561 g/cm^3 =965.61kg/m^3
比重 = ρCO2/ ρH2O = 10.20549/965.61=0.01057
ref. 密度與溫度壓力關係
http://www.engineeringtoolbox.com/fluid-density-temperature-pressure-d_309.html

2.
PM =ρRT=(1/v)RT
200k×M=(1/0.65)×8.314k×(273+40)
M=20.02 kg/kgmole= 20.02 g/mole
工程氣體常數=? 8.314 J/gmol-K- 都已知的，為啥要求?

2014-10-24 10:58:01 補充：
更正 工程氣體常數 Rg ==> 每kg理想氣体的氣体常数
PV=nRT =(m/M) RT =mRgT
Rg= R/M = 8.314/20.02=0.4153 J/gmol-K

到下面的網址看看吧

▶▶http://qaz331.pixnet.net/blog

1.計算二氧化碳在T=90°C=363k及P=700KPa=7*10^5(N/m^2)狀況下(1) D=密度=?M=分子量=44(g/mole)=0.044(kg/mole)
D=P/RT=P*M/Ru*T=7*10^5*0.044/8.314*363=10.2055(kg/m^3)
(2) r=比重=氣體/空氣=44/29=1.517 (3) v=比容積=1/D=0.097986(m^3/kg)
2 在T=40°C=313k及P=200KPa=2*10^5(N/m^2)之狀況下,某理想氣體之比容積為v=0.65m3/kg,試計算其(1) 及工程氣體常數R=?R=P*v/T=2*10^5(N/m^2)*0.65(kg/m^3)/313(k)=415.3(J/kg.k)=0.4153(J/g.k)
(2) 分子量M=?Ru=Universal Gas Costant=8.314(J/mole.k)M=Ru/R=8.314(J/mole.k)/0.4153(J/g.k)=20(g/mole)