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2 years ago

Quelles sont les types de carburant utilisés en aviation

Engineering

1 answer:

zhenek [66]2 years ago

6 0

hope it's help you ok have a good day

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A insulated vessel s has two compartments separated by a membreane. On one side is 1kg of steam at 400 degC and 200 bar. The oth

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Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

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2 years ago

You often experiment with everyday materials around you, trying to create new and unique items. You enjoy designing and creating

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answer:D. problem-solver

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2 years ago

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What is the engineering

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Engineering is the profession which deals with building the structures like bridge,house,roads etc.

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3 years ago

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Which items are NOT found on a

Alja [10]

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None of the above cause thats what i put

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2 years ago

Steam at 4 MPa, 400°C enters a steady-flow, adiabatic turbine through a 20 cm-diameter-pipe with a velocity of 20 m/s. It leaves

Furkat [3]

Answer:

28,8 m/s

Explanation:

In a steady flow system we can say that m1=m2 which means that the mass flow in the entrance in the same in the outlet. m is flow (kg/s)

we know that $m=\frac{1}{v} A*V$ where V (m/s) is velocity, A (m^2) ia area and v is specific volume (m^3/kg)

Since m1=m2 we can say

$\frac{1}{v_{1} } A_{1} *V_{1}=\frac{1}{v_{2}} A_{2} *V_{2}$

clearing the equation

$V_{2}= \frac{v_{2} }{v_{1}}*\frac{A_{1} }{A_{2}} *V_{1}$

we can specific volume (m^3/kg) from thermodynamic tables

for the entrance is 400°C and 4 MPa is superheated steam and v is : 0,7343 m^3/kg

In the outlet we have saturated vapor with quality (x) of 80%. In this case we get the specific saturated volume for the liquid (vf) and the specific volume for the saturated (vg) gas from the thermodynamic tables. we use the next equation to get (v) for the condition of interest, in this case 80% quality.

v= vf +x*(vg - vf)

where:

x: quality

vf = liquid-saturated-specific-volume

vg =steam-saturated-specific-volume.

for this problem

x = 0,8

vf = 0,00102991

vg = 3,24015

we get = 2,593 m^3/kg

The area is the one for a circle

$\pi *r^{2}$

r1 = 0,1 m^2 for area 1

r2=0,5 m^2 for area 2

A1 = 0,0314 m^2

A2 = 0,7853 m^2

we know that V1 is 20 m/s

replacing these values in the equation

$V_{2}= \frac{v_{2} }{v_{1}}*\frac{A_{1} }{A_{2}} *V_{1}$

we get V2 = 28,2 m/s.

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3 years ago