Advanced Math A bipolar alkaline water electrolyzer stack module comprises 160 electrolytic cells that have an effective cell area of 2 m^2. At nomina

kuCAu 2020-11-16 Answered
Advanced Math
A bipolar alkaline water electrolyzer stack module comprises 160 electrolytic cells that have an effective cell area of \(\displaystyle{2}{m}^{{2}}\). At nominal operation, the current density for a single cell of the electrolyzer stack is 0.40 \(\displaystyle\frac{{A}}{{c}}{m}^{{2}}\). The nominal operating temperature of the water electrolyzer stack is \(\displaystyle{70}^{\circ}\) C and pressure 1 bar. The voltage over a single electrolytic cell is 1.96 V at nominal load and 1.78 V at 50% of nominal load. The Faraday efficiency of the water electrolyzer stack is 95% at nominal current density, but at 50% of nominal load, the Faraday efficiency decreases to 80%.
(Give your answer to at least three significant digits.)
Calculate the nominal stack voltage:
Answer in V
Calculate the nominal stack current:
Answer in A
Calculate the nominal power on the water electrolyzer stack:
Answer in kW

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Expert Answer

Sadie Eaton
Answered 2020-11-17 Author has 28373 answers

Step 1
given
nominal cell potential =1.96 V
number of cell =160
\(\displaystyle{c}{u}{r}{r}{e}{n}{t}\ {d}{e}{n}{s}{i}{t}{y}={0.40}\frac{{A}}{{c}}{m}^{{2}}={0.40}{x}×{10}^{{4}}\frac{{A}}{{m}^{{2}}}\)
\(\displaystyle{c}{e}{l}{l}\ {a}{c}{t}{i}{v}{e}\ {a}{r}{e}{a}={2}{m}^{{2}}\)
Step 2
Explanation
formula for finding
\(\displaystyle{n}{o}\ min{a}{l}\ {s}{t}{a}{c}{k}\ {v}{o}{lt}{a}ge={n}{o}\ min{a}{l}\ {c}{e}{l}{l}\ {p}{o}{t}{e}{n}{t}{i}{a}{l}\times nu{m}{b}{e}{r}\ {o}{f}\ {c}{e}{l}{l}{s}\)
substitute the values
\(\displaystyle{n}{o}\ min{a}{l}\ {s}{t}{a}{c}{k}\ {v}{o}{lt}{a}ge={1.96}{V}\times{160}\ {n}{o}\ min{a}{l}\ {s}{t}{a}{c}{k}\ {v}{o}{lt}{a}ge={313.6}{V}\)
formula for finding
\(\displaystyle{n}{o}\ min{a}{l}\ {s}{t}{a}{c}{k}\ {c}{u}{r}{r}{e}{n}{t}={c}{u}{r}{r}{e}{n}{t}\ {d}{e}{n}{s}{i}{t}{y}\times{c}{e}{l}{l}\ {a}{c}{t}{i}{v}{e}\ {a}{r}{e}{a}\)
substitute the values
Step 3
\(\displaystyle{n}{o}\ min{a}{l}\ {s}{t}{a}{c}{k}\ {c}{u}{r}{r}{e}{n}{t}={\left({0.40}×{10}^{{4}}\frac{{A}}{{m}^{{2}}}\right)}×{2}{m}^{{2}}\) nominal stack current=8000 A
formula for finding
\(\displaystyle{n}{o}\ min{a}{l}\ {p}{o}{w}{e}{r}\ {o}{n}\ {t}{h}{e}\ {w}{a}{t}{e}{r}\ {s}{t}{a}{c}{k}={n}{o}\ min{a}{l}\ {v}{o}{lt}{a}ge\times{n}{o}\ min{a}{l}\ {c}{u}{r}{r}{e}{n}{t}\)
substitute the values
\(\displaystyle{n}{o}\ min{a}{l}\ {p}{o}{w}{e}{r}\ {o}{n}\ {t}{h}{e}\ {w}{a}{t}{e}\ {r}{s}{t}{a}{c}{k}={313.6}{V}\times{8000}{A}={2508.8}{K}{W}\)

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