以前做过个类似的计算,原理是一样的
To use 5.5 Barg steam to heat up 30 C DI water to 0.3 Barg BFW ,how much steam and DIW are needed for 2 T BFW
LS Steam
Pressure P1 0.3 barg
Temperature T1,s 107.1 degC
Spec. enthalpy h1,vs 2687.1 kJ/kg
LS steam flow M1 251.65 kg/h
Enthalpy, HS H1,vs 676222.08 kJ/h
DIW
Pressure P2 10 barg
Temperature T2 30 degC
Spec. enthalpy h2,T 126.75 kJ/kg
DIW flow M2 1748.35 kg/h
Enthalpy, BFW H2,T 221595.0 kJ/h
BFW
Pressure P3 0.3 barg
Temperature T3,s 107.1 degC
Spec. enthalpy h3,vs 448.91 kJ/kg
BFW flow M3 2000 kg/h
Required BFW flow M3,r 2000 kg/h Target
Enthalpy,BFW H3,vs 897817.07 kJ/h
Enthalpy error dH 0.00 kJ/h
两个未知数,LS FLOW 和DI FLOW,,LS+DI=BFW ,
TOTAL 焓变等于零,
所以可以通过EXCEL的线性归划来求解.
也可以在ASPEN里面做设计规定求解 |