t= 179.96℃，h''=2777.67 kJ/kg，s ' ' = 6.5895 kj/(kg·k)，v''=0. 1944 m^3/kg

2. Enthalpy remains unchanged before and after river closure, that is, h2=h 1=h''=2777.67 kJ/kg.

3. According to h2 and P2(=0. 1MPa), we can look up the table: (Note that it is possible to look up the map, but it is rough).

t= 150.84℃，S2 = 7.6 168 kj/(kg·k)，v= 1.9404 m^3/kg

4. Calculation of thermodynamic energy and available energy loss caused by throttling.

1) u = h-PV = 2777.67-0./kloc-0 /×10 3×1.9404 = 2758.266 kj/kg.

2)△I = To×△s = 300×(7.6 168-6.5895)= 308. 19 kJ/kg

That is, the temperature, enthalpy and thermodynamic energy after throttling and the effective energy loss caused by throttling are:

150.84℃, 2777.67 kj/kg, 2758.266 kj/kg, 308. 19 kj/kg.

end

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Sorry, I made it with saturated steam. According to saturated water, the method is the same, but the parameter data are different. I won't worry about it.

This problem is mainly because the calculation is too troublesome. But the main requirement is to master the fluid enthalpy before and after river closure.