The half pipe coil around a reactor is full of cooling water and the reactor is heated to 145 degree C using hot oil in the internal coils. So, water in external coils is heated to 145 degree C and I want to know the pressure in the external coil at that temperature.
I have calculated vapour pressure water at 145 degree C using Antoine equation as shown below and the result is 3.06 barg pressure. This pressure is same as that of saturated steam at this temperature. The water in pipe (sealed with inlet and outlet valves shut) is liquid and not the steam. I wonder why is the pressure for water and steam at this temperature same ?
Sorry if my question sounds stupid or too simple but I am trying to learn and understand. Thanks
Log10 (P) = A - (B/(C+T))
A= 8.14019
B=1810.94
C=244.485
T= 145 degree C
P=10^A - (B/(C+T))
P= 3094.6 mmHg
P= 4.07 bara
P= 3.06 barg