Question

A water electrolyzer draws 1000 A of current.

a. How many mol/h of Hydrogen are produced?

b. What is the corresponding mass flux of Hydrogen (g/h)?

c. The Hydrogen is pumped into a storage tank whose total volume is 100 L. If the pressure at a given point in time is 50 bar (absolute), what is the pressure 10 h later? Assume a constant temperature of 298 K.

d. How many total grams of Hydrogen does the tank contain if its pressure is raised to 350 bar (5000 psi) at 298 K?

e. What is the power (in kW) consumed by the electrolyzer, assuming ideal (reversible) conditions? Assume that the electrolyzer is operating at 1 atm and 298 K.

f. Due to losses, the voltage supplied in practice must exceed the ideal value. Assume the electrolyzer is operating at 1.8 V. What is the corresponding power in kW?

Answer 1

a) The required voltage is 1.23 V

Let the H₂ flow rate be $\dot m$ kg/s

Assuming 100% efficient electrolysis, the energy required per kg H₂ is

$e=142\times10^6\: J/kg$

For current of 1000 A

$1.23\times1000=142\times10^6\dot m$

Hence

$\dot m=8.662\times10^{-6}\: kg/s$

1 kgmol of H₂ contains 2 kg of H₂

Hence mol produced per hour is

$\dot M=\frac{\dot m}{2}(3600)=0.0156\: kgmol/h=15.6\: gmol/h$

b) The mass flow rate is calculated above

$\dot m=8.662\times10^{-6} \: kg/s=8.662\times10^{-3}\: g/s$

c) Characteristics gas constant of H₂ is

$\bar R=\frac{R}{M}=\frac{8314}{2}=4157\: J/kg-K$

Initial density is

$\rho_1=\frac{p_1}{RT_1}=\frac{50\times10^5}{(4157)(298)}=4.0362\: kg/m^3$

Initial mass inside tank is

$m_1=V_1\rho_1=100\times10^{-3}(4.0362)=0.40362\: kg$

Assuming constant flow rate of 8.661x10^-6 kg/s, the mass inside tank after 10h is

$m_2=m_1+\dot m\Delta t=0.40362+(8.662\times10^{-6})(10\times3600)=0.71545\: kg$

Density after 10h is

$\rho_2=\frac{m_2}{V}=\frac{0.71545}{0.1}=7.1545\: kg/m^3$

Therefore

$p_2=\rho_2RT_2=(7.1545)(4157)(298)\: Pa=88.63\: bar$

d) If p₃= 350 bar

$\rho_3=\frac{p_3}{RT_1}=\frac{350\times10^5}{(4157)(298)}=28.2535\: kg/m^3$

Hence mass inside tank is

$m_3=\rho_3V=28.2535(0.1)=2.82535\: kg=2825.35\: g$

(Please note that 4 subquestions need to be answered as per HOMEWORKLIB RULES)