determine the optimum compressor pressure ratio specific thrust fuel comsumption 2.1 220k 1700k 42000 1.004
1 answer:
Answer:
hello your question is incomplete attached below is the complete question
A) optimum compressor ratio = 9.144
B) specific thrust = 2.155 N.s /kg
C) Thrust specific fuel consumption = 1670.4 kg/N.h
Explanation:
Given data :
Mo = 2.1 , To = 220k , Tt4 = 1700 k, hpr = 42000 kj/kg, Cp = 1.004 kj/ kg.k
γ = 1.4
attached below is the detailed solution
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Answer:
n=2.32
w= -213.9 KW
Explanation:
Mass of air=1 kg
For polytropic process ,n is the polytropic constant.
n=2.32
Work in polytropic process given as
w=
w=
Now by putting the values
w=
w= -213.9 KW
Negative sign indicates that work is given to the system or work is done on the system.
For T_V diagram
We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.
Answer:
the maximum length of the specimen before deformation is 0.4366 m
Explanation:
Given the data in the question;
Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²
cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m
tensile load F = 1810 N
maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m
Now to calculate the maximum length for the deformation, we use the following relation;
= [ Δl × E × π × D² ] / 4F
so we substitute our values into the formula
= [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )
= 3161.025289 / 7240
= 0.4366 m
Therefore, the maximum length of the specimen before deformation is 0.4366 m