Question

(9). 10. 6 (in your book) Prob 10.6 A water electrolyzer operating at RTP requires 1.83 V to produce 1 metric ton of hydrogen per day when oper- a. What current does it draw? b. How man c. How many MJ of heat does it either reject or absorb per day ating continuously. y m3 of water does it use per day? (state which)?

Answer 1

Reaction in electrolyzer:

$Anode: 2H_{2}O(l)\rightarrow O_{2}(g)+4H^{+}(aq)+4e^{-}$

$Cathode: 2H^{+}(aq) + 2e^{-}\rightarrow H_{2}(g)$

$Overall: 2H_{2}O(l)\rightarrow 2H_{2}(g) + O_{2}(g)$

To produce 1 kilomole of H₂ , 2 kilomoles of elctrons involved. 1 mole electrons = 6.02 x 10²³ electrons, therefore 1 kilomoles electrons = 6.02 x 10²⁶ electrrons and charge of one electron = 1.6 x 10^-19C

Charge of 2 kilomoles electrons = 2 x 6.02 x 10²⁶ x 1.6 x 10^-19 = 1.93 x 10⁸ C

1 kilomoles H₂ involved 1.93 x 10⁸ C current.

Production rate of hydrogen = 1 metric ton/ day = 1000/(24 x 3600) kg/sec = 1.16 x 10^-2 kg/sec = (1.16 x 10^-2)/2 kmol/sec = 0.58  x 10^-2 kmol/sec

Therefore,  0.58  x 10^-2 kmol hydrogen involved = 1.93 x 10⁸ x 0.58  x 10^-2 = 1.12 x 10⁶ C

Current = 1.12 x 10⁶ C/sec = 1.12 x 10⁶ A

To produce 1000 kg of hydrogen per day, a current of 1.12 MA (megamperes) is needed

A better design would call for a large number of cells in series so as to reduce the required current. Say, 100 cells operating under 183 V and drawing 11.2 kA.

2) 1 kilomole of hydrogen required 1 kilomole of water

Therefore, 0.58  x 10^-2 kmol hydrogen required = 0.58  x 10^-2 kmol water = 0.58  x 10^-2 x 18  kg of water = 10.44 x 10^-2 kg of water

Use of water = 10.44 x 10^-2 kg/sec = 10.44 x 10^-2 x 24 x 3600 kg/day = 9 metric ton/day

3)

The electrolyzer (per cell) voltage exceeds the 1.48 V thermal break even voltage.

Hence, heat will be released.

The overvoltage is 1.83?1.48 = 0.35 V.

The heat loss is 0.35 V ×1.12×10⁶ A or 3.9×10⁵ W.

The heat energy generated throughout a 24-hour period is 3.9 × 10⁵ × 86, 400 = 33.9 × 10⁹ J/day

Daily heat rejection is 33,900 MJ or 33.9 GJ.