Effects of process-generated hydrogen on RPV walls

5. Hydrogen production in PWR

As the Henry constant is known, one only needs to know the concentration of H 2 in the water expressed in mol/m 3 to find the equilibrium hydrogen gas pressure above the primary water. The hydrogen concentration in the primary water is known to be between 25 and 50 STP cm 3 /kg. For the conversion to mol/m 3 , the molar volume [m 3 /mol] of hydrogen gas at STP conditions is needed. The specific volume, v , of H 2 at STP is 11.983 m 3 /kg [57]. Together with the molar mass of 2.0159 g/mol, the molar volume is found: V m = v · 10 − 3 M = 11 . 983 · 0 . 0020159 = 0 . 024156 m 3 /mol (5.3) Using this value for the molar volume of H 2 at STP conditions, the amount of moles in 1 STP cm 3 are known and further on the moles/kg in the primary water corresponding to 25 – 50 STP cm 3 /kg can be calculated. A H 2 concentration of 25 STP cm 3 /kg corresponds to 1.035 10 -3 mol/kg. Equiva- lently, 50 STP cm 3 /kg corresponds to 2.070 10 -3 mol/kg. As the density of water at a temperature of 300 ◦ C and 150 bar is 705.6 kg/m 3 (see Figure 5.2), the molar concentration of H 2 in the primary water is c H 2 = 1 . 035 · 10 − 3 × 705 . 6 = 0 . 730 mol/m 3 (5.5) The hydrogen concentration in the primary water is found to be 0.730 mol/m 3 for 25 STP cm 3 /kg and 1.460 mol/m 3 for 50 STP cm 3 /kg. Finally, both calculations of the Henry constant and the hydrogen gas concentra- tion in the primary water can be combined to find the equilibrium hydrogen pressure in the primary system in equilibrium with the dissolved hydrogen: Figure 5.4 shows the hydrogen pressure as a function of temperature for a concentration in the range of 25 to 50 STP cm 3 /kg. The hydrogen pressure in equilibrium with the primary water with a concentration of 25 and 50 STP cm 3 /kg at room temperature is 1.39 10 5 Pa and 2.78 10 5 Pa, respectively. This corresponds to 1.37 and 2.74 atm. The equilibrium pressure decreases with increasing temperature, resulting in a pressure of 2.29 10 4 Pa and 4.58 10 4 Pa for 25 and 50 STP cm 3 /kg, respectively at 300 ◦ C. This is 0.226 and 0.452 atm, respectively. 42 n = V 10 − 6 V m = 25 10 − 6 V m = 1 . 035 · 10 − 3 mol/kg (5.4) f H 2 = c H 2 H H 2 (5.6)