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Ne4ueva [31]
3 years ago
7

Air enters an adiabatic compressor through 0:5m2 opening and exhausts through a 0:2m2 opening. The inlet conditions of air are 2

00 kPa and 52C with an unknown velocity and exits at 3 MPa and 427C with a velocity of 100 m/s. Determine the mass ow rate of air through the compressor, the inlet velocity and the power required to run the compressor.
Engineering
1 answer:
gladu [14]3 years ago
8 0

Answer:

i) 43.55 kg/s

ii) 40 m/s

iii)  -199.32 KW

Explanation:

To resolve the above question we have to make some assumptions :

  • mass flow  through the system is constant
  • The only interactions that are between the system and the surrounding are work and heat
  • The fluid is uniform

i) first we have to determine the mass flow rate of the air

M = (\frac{P1}{RT1} )A1v1

    = ( \frac{200}{0.287*325} ) 0.5 * V1 ---------- (1) hence  M = 43.55 kg/s

ii) using this relationship : A1V1 = A2V2 hence V1 = (0.2/0.5) * 100 = 40m/s (    inlet velocity )

input this value into equation 1

iii) Next we will determine the power required to run compressor

attached below

power required = -199.32 KW ( this value indicates that there is power supplied )

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Draw a circuit diagram connected in a reverse bias showing a p-n material and a battery
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2 years ago
A wood pipe having an inner diameter of 3 ft. is bound together using steel hoops having a cross sectional area of 0.2 in^2. The
Minchanka [31]

Answers:

31.7 inches

Explanation:

Given:

Diameter = 3ft

Let D = Diameter

So, D = 3ft. (Convert to inches)

D = 3 * 12in = 36 inches

Coss-sectional area of the steel = 0.2in²

Gauge Pressure (P) = 4psi

Stress in Steel (σ)= 11.4ksi

Force in steel = ½ (Pressure * Projected Area)

Area (A) = 2 * Force/Pressure

Also, Area (A) = Spacing (S) * Wood Pipe Diameter

Area = Area

2*Force/Pressure = Spacing * Diameter

Substitute values I to the above expression

2 * Force / 4psi = S * 36 inches

Also

Force in steel (F) = Stress in steel (σ) × Cross-sectional area of the steel

So, F = 11.4ksi * 0.2in²

F = 11.4 * 10³psi * 0.2in²

F = 2.28 * 10³ psi.in²

So, 2 * Force / 4psi = S * 36 becomes

2 * 2.28 * 10³/4 = S * 36

S = 2 * 2.28 * 10³ / (4 * 36)

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5 0
4 years ago
End A of the uniform 5-kg bar is pinned freely to the collar, which has an acceleration = 4 m/s2 along the fixed horizontal shaf
sergiy2304 [10]

Answer:

Fa = 57.32 N

Explanation:

given data

mass = 5 kg

acceleration = 4 m/s²

angular velocity ω = 2 rad/s

solution

first we take here moment about point A that is

∑Ma = Iα +  ∑Mad    ...............1

put here value and we get

so here I = ( \frac{1}{12} ) × m × L²    ................2

I = ( \frac{1}{12} ) × 5 × 0.8²

I = 0.267 kg-m²  

and

a  is =  r × α    

a  = 0.4 α

so now put here value in equation is 1

0 = 0.267 α + m r α (0.4) - m A (0.4)

0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4

so angular acceleration α = 7.5 rad/s²

so here force acting on x axis will be

∑ F(x) = m a(x)    ..............3

a(x) = m a - m rα

put here value

a(x) = 5 × 4 -  5 × 0.4 × 7.5

a(x) = 5 N

and

force acting on y axis will be

∑ F(y) = m a(y)    .............. 4

a(y) - mg = mrω²

a(y) - 5 × 9.81  = 5 × 0.4 × 2²

a(y) = 57.1 N

so

total force at A will be

Fa = \sqrt{a(x)^2+a(y)^2}    ...............5

Fa = \sqrt{(57.1)^2+(5)^2}  

Fa = 57.32 N

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4 years ago
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Lilit [14]

Answer:

See explaination

Explanation:

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