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 2 of 2.
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)
Shankar said:
10 years ago
It is not possible to calculate without formula.
Ritesh said:
10 years ago
Can we calculate without the use of a calculator?
(1)
Shubham gupta said:
9 years ago
This depends upon the cycle as per the petrol engine.
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)
Ajay said:
9 years ago
1-(1/C.R.^1.4) = 49.23 approximately.
Mahesh said:
8 years ago
1 - (1/C.R. Ress to the power 1.4) = 49.26 approximately.
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.
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)
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