Lecture 14B - Fuel Cells - Part 2

Lecture 14B - Fuel Cells - Part 2 - 4/11/11 ECE 4364/5374G:...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
4/11/11 1 ECE 4364/5374G-14B (c) Saifur Rahman 1 Professor Saifur Rahman Electrical & Computer Engineering Dept. Virginia Tech ECE 4364/5374G: Alternate Energy Systems Lecture 14B: Fuel Cells – Part 2 ECE 4364/5374G-14B (c) Saifur Rahman 2 Electrochemistry in the Power Sec2on • Two separate reactions take place in the fuel cell which contains the electrodes and the electrolyte; an oxidation half-reaction occurring at the anode and a reduction half- reaction occurring at the cathode. • The final production of the overall reaction are electricity, water, and excess heat.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4/11/11 2 ECE 4364/5374G-14B (c) Saifur Rahman 3 In the oxidation half-reaction at the anode, gaseous hydrogen produces hydrogen ions, which travel through the ionically conducting membrane to the cathode. At the same time electrons travel through an external circuit to the cathode. In the reduction half-reaction at the cathode, oxygen supplied from air combines with the hydrogen ions and electron to form water and excess heat. Fuel Cell Opera2on Details Source: Hydrogenics Corporation, 2000 ECE 4364/5374G-14B (c) Saifur Rahman 4 Fuel Cell Opera2on Details, Cont. Anode: 2H2 4H+ + 4e- Cathode: O2 + 4H+ + 4e- 2H2O -------------------------------------------------- Overall: 2H2 + O2 2H2O
Background image of page 2
4/11/11 3 ECE 4364/5374G-14B (c) Saifur Rahman 5 Calcula2ng the fuel cell voltage and maximum cell efficiency Charge transfer q due to a kg-mole of gas (H2) where H2 + 2OH- = 2H2O + 2e- is given by: q = 2eNo = (2)(1.6x10-19)(6.023x1026) = 1.93x108 Coulombs Where N0 = Avogadro's number (No. of molecules contained in a kg-mole) This charge is equivalent to a current of 1 ampere for 1.93 x 108 seconds or about 6 years. ECE 4364/5374G-14B (c) Saifur Rahman 6 Calcula2ng the fuel cell voltage and maximum cell efficiency, cont. The useful energy delivered, W e is given by: W e = qV (function of potential achieved.) The output work (W e ) and input heat (Q) are related to the enthalpy of reactants, h 1 , and reaction products, h 2 . h 1 + Q = h 2 + W e where h 1 = enthalpy of reactants h 2 = enthalpy of reaction produced W e = h 1 - h 2 + Q Q < 0 ( since in a fuel cell heat is generated) Therefore, electrical energy is less than the enthalpy change, h 1 - h 2 .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4/11/11 4 ECE 4364/5374G-14B (c) Saifur Rahman 7 Calcula2ng the fuel cell voltage and maximum cell efficiency, cont. To find the cell voltage, we use the Gibbs function for free energy property, i.e., work is equivalent to change in Gibbs function. For a reversible, isothermal and constant pressure process at 25 0 C We = g 1 – g 2 = 56.7x10 6 cal/(kg.mole of H 2 ) = 2.37 x 10 8 Joules/(kg.mole of H 2 ) Actual voltage is much less (0.5-1.0V) due to electrode resistance, polarization of the electrolyte, and depletion of electrolyte.
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/16/2012 for the course ECE 5374G taught by Professor Srahman during the Spring '12 term at Virginia Tech.

Page1 / 12

Lecture 14B - Fuel Cells - Part 2 - 4/11/11 ECE 4364/5374G:...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online