Given:
parasite Drag Coefficient, is reduced by 19% or 0.19 times
Solution:
Using Breguet's Endurance Eqn:
or
(1)
From eqn (1), other terms being constant, then
<u>or</u>
Therefore, the above relation shows that when drag coefficient decreases performance increases and vice-versa due to inverse relation between the.
So, 19% decrease will result in:
= 5.26 times increase in performance
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Answer:
Explanation:
(a) Swirl tangential speed = 3 x MPS ( mean piston speed)
= 3 x 420.76 inch/sec = 280.51 [ft/sec]
The other detailed steps is as shown in the attached file
Answer:
a) 358.8 KJ/kg
b) 0.0977 KJ/K- kg
c) 83.28%
Explanation:
N2 at 300 k. ( use the properties of N2 at 300 k (T1) )
Cp = 1.04 KJ/kg-k , Cv = 0.743 KJ/Kgk , R = 0.1297 KJ/kgk , y = 1.4 ,
Given data:
T2 = 645 k
P1 = 1 bar , P2 = 10.5 bar
<u>a)Determine the work input in KJ/Kg of N2 flowing </u>
Winput = h2 - h1 = Cp( T2 - T1 ) = 1.04 ( 645 - 300 ) = 358.8 KJ/kg
<u>b) Determine the rate of entropy in KJ/K- kg of N2 flowing </u>
Rate of entropy ( Δs ) = Cp*InT2/T1 - R*In P2/P1
= 1.04 * In (645/300) - 0.1297 * In ( 10.5 / 1 )
= 0.0977 KJ/K- kg
<u>c) Determine isentropic compressor efficiency </u>
Isentropic compressor efficiency = 83.28%
calculated using the relation below
( h'2 - h1 ) / ( h2 - h1 ) = ( T'2 - T1 ) / ( T2 - T1 )
where T'2 = 587.314
Carbonation is more of a healer to the engine