NCER Assignment 11 solution - NCER NPTEL
Today, We will discuss Assignment-11 of AKTU which is related to NCER ( Non-Conventional Energy Resources) NPTEL. Now you can find here all solution correctly ,If any answer appears to wrong then suggest me, I will improve that answer as soon as possible.So without doing any delay, Let's Start...
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👀NCER 2019 Assignment-11 Ka Solution👀
Due date for this assignment: 2019-04-17, 23:59 IST.
(1) Fuel cells are energy storage devices
👈
Ans: False
(2) An SOFC operates at approximately 70 oC
👈
Ans: False
(3) PEFCs uses gases as reactants
👈
Ans: True
(4) Partial oxidation, used for reforming fuels, is an exothermic process
👈
Ans: True
(5) Flow fields are necessary to distribute reactants across the active area of the fuel cells
👈
False
Ans: True
(6) Under constant current mode of operation, the degradation of a fuel cell may be expressed in mA/h
👈
Ans: False
⇛Note: Two fuel cells, each operating at 0.6 V, 0.5 A/cm2, are connected in series. The area of each of the cells is 30 cm2. { Q(7) & Q(8) are connected to each other. }
(7) The voltage of the stack is __________V
⃢ 👈
Ans: 1.2
Ans: 1.2
(8) The power from the stack is ____________ W
⃢ 👈
Ans: 18
⇛Note: The theoretical voltage of a fuel cell is 1.23 V. {Q(9) & Q(10) are connected to each other.}
(9) If the operating point of the fuel cell is 0.6 V, the overpotential losses are _________ V.
⃢ 👈
Ans: 0.63
(10) If the impedance (ohmic) due to the membrane and contact resistance of the cell is 0.01Ω and the current in the cell is 15 Amps the polarization losses due to reasons other than ohmic are _______ V.
⃢ 👈
Ans: 0.48
Ans: 0.48
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🙏Note:
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For All Assignment Solution,Please Stay with us.
👍Also Read: Non-Conventional Energy Resources: AKTU Final Exam Pattern 2019
💥⇛Few More:
- Non-Conventional Energy Resources - NPTEL Assignment 9 Answers:👉 Click Here
- Non-Conventional Energy Resources - NPTEL Assignment 10 Answers:👉 Click Here
- Non-Conventional Energy Resources - NPTEL Assignment 12 Answers:👉 Click Here
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17 Comments
Thanks a lot.
ReplyDeleteYour welcome.
DeleteAttention Friends!!!,
ReplyDeleteCorrection hai, Dhyan Dijiyega:
Q(2): False
Q(3): True
Q(4): True
Q(6): False
Note: Aur hoga to update kr diya jayega, Last date ke pahle jarur match kara len.
bhaiya dekh lo tumpe Vishwas kar ke is bar submit kar rhe h..
ReplyDeleteUmeed karte sab shubh ho Koi locha na ho
Namaste ����
sorry! please do correction.solution are mentioned in comment.
Deletehow you got 2.05v for overpotential losses . can you please share the calculation behind it ?
ReplyDeletesorry! please do correction.solution are mentioned in comment.
DeleteThanks
DeleteSorry friends! again few correction:
ReplyDeleteQ(9): 0.63
Q(10): 0.48
Solution:
In Q(9)-
theoretical voltage - 1.23
operating voltage - 0.6
so, over potential losses = 1.23 - 0.6 = 0.63
Then in Q(10)-
Resistence= 0.01 ohm
Current = 15 amps
ohmic loss = 0.01 * 15 = 0.15
Then polarization losses other than ohmic loss =
0.63 - 0.15 = 0.48
Sir 7 or 8 ka answer
DeleteIn 9th and 10th, are they asking for single fuel cell or the whole stack?
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