Mechanical Engineering - IC Engines and Nuclear Power Plants - Discussion
Discussion Forum : IC Engines and Nuclear Power Plants - Section 1 (Q.No. 5)
5.
The thermal efficiency of a standard Otto cycle for a compression ratio of 5.5 will be
Discussion:
20 comments Page 1 of 2.
Vivek said:
8 years ago
Efficiency of Otto cycle is 1- (1/r^y-1).
Where r is the compression ratio : volume before compression to volume after compression and why is the ratio of specific heats which is for air is 1.4.
Where r is the compression ratio : volume before compression to volume after compression and why is the ratio of specific heats which is for air is 1.4.
Ashutosh kumar said:
1 decade ago
Efficiency of otto cycle is 1-(1/r^y-1).
Where r is the compression ratio : volume before compression to volume after compression and y is the ratio of specific heats which is for air is 1.4.
Where r is the compression ratio : volume before compression to volume after compression and y is the ratio of specific heats which is for air is 1.4.
Parthiban said:
7 years ago
Formula:
= 1-(1/c.r)^0.4,
= 1-(1/5.5)^0.4,
= 1-(0.1818)^0.4,
= 1-0.5056,
= 0.4944,
= 49.44%=50%(approximately).
= 1-(1/c.r)^0.4,
= 1-(1/5.5)^0.4,
= 1-(0.1818)^0.4,
= 1-0.5056,
= 0.4944,
= 49.44%=50%(approximately).
(8)
SHRAVANI said:
1 decade ago
Otto cycle efficiency n(neta)=1-(1/(r^(y-1)).
Compression ratio r=5.5.
T.E=1-(1/((5.5^(1.4-1))).
=0.4943.
=50%.
Compression ratio r=5.5.
T.E=1-(1/((5.5^(1.4-1))).
=0.4943.
=50%.
Zammu said:
1 decade ago
Otto cycle efficiency n(neta)=1-(1/(r^(y-1)).
Compression ratio r=5.5.
T.E = 1-(1/((5.5^(1.4-1))).
= 0.4943.
= 50%.
Compression ratio r=5.5.
T.E = 1-(1/((5.5^(1.4-1))).
= 0.4943.
= 50%.
KK Rajawat said:
1 decade ago
Efficiency of otto cycle is : 1-(1/(r^(y-1)))
So after resolve it, the efficiency of otto cycle is = 0.4943~50%.
So after resolve it, the efficiency of otto cycle is = 0.4943~50%.
Ritesh sigh said:
10 years ago
T.E = 1-(1/r(5.5))^(1.4-1).
= 1-0.5056 = 0.49.
Approximately 0.50. Hence the correct answer is 0.50*100 = 50%.
= 1-0.5056 = 0.49.
Approximately 0.50. Hence the correct answer is 0.50*100 = 50%.
(1)
Rathod Shwetketu said:
1 decade ago
Formula for finding efficiency is
= 1 - (1/C.R)^0.4
So answer will be near about 49% so its ~50%.
= 1 - (1/C.R)^0.4
So answer will be near about 49% so its ~50%.
Sujit said:
9 years ago
n = 1/(r)^(1.4 - 1), where r = 5.5.
Can we calculate without a calculator in the competitive exam?
Can we calculate without a calculator in the competitive exam?
(1)
Ravibhange said:
1 decade ago
This is only approximately you can't calculate actual but it is nearly equal to 50%.
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