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NeTakaya
3 years ago
9

Air flows into a jet engine at 70 lbm/s, and fuel also enters the engine at a steady rate. The exhaust gases, having a density o

f 0.1 lbm/ft3 , exit through a circular cross section with a radius of 1 ft at 1450 ft/s relative to the engine. Find the mass of fuel which is supplied to the engine each minute.
Physics
1 answer:
Andrews [41]3 years ago
8 0

Answer:

1387908 lbm/h

Explanation:

Air flowing into jet engine = 70 lbm/s

ρ = Exhaust gas density = 0.1 lbm/ft³

r = Radius of exit with a circular cross section = 1 ft

v = Exhaust gas velocity = 1450 ft/s

Exhaust gas mass (flow rate)= Air flowing into jet engine + Fuel

Q = (70+x) lbm/s

Area of exit with a circular cross section = π×r² = π×1²= π m²

Now from energy balance

Q = ρ×A×v

⇒70+x = 0.1×π×1450

⇒70+x = 455.53

⇒ x = 455.53-70

⇒ x = 385.53 lbm/s

∴ Mass of fuel which is supplied to the engine each minute is 1387908 lbm/h

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lina2011 [118]
A) the periodic time is given by the equation;
 T= 2π√(L/g)
For the frequency will be obtained by 1/T (Hz)
T = 2 × 3.14 √ (0.66/9.81)
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    = 1.6289 Seconds
Frequency = 1/T = f = 1/1.6289
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b)  The vertical distance, the height is given by
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Applying conservation of energy
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  v² = 2gΔh = 2×9.81 × 0.01 
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v = 0.443 m/s

c) total energy = KE + GPE = KE when GPE is equal to zero (at the lowest point possible)
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E = 1/2mv²
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3 years ago
The mass of the bicycle and rider is 60 kg and the total area of the tyres in contact with
AleksAgata [21]

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a little

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a projectile is lunched with an initial speed of 60.0mm/s at an angle of 30.0° above the horizontal.The projectile lands on a hi
alexandr402 [8]

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52 mm/s (approximately)

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Angle of projection is, \theta=30.0\°

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In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.

So, the velocity in the horizontal direction always remains the same.

The horizontal component of initial velocity is given as:

u_x=u\cos\theta\\u_x=60\times \cos(30)\\u_x=30\sqrt3\approx52\ mm/s

Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.

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3 years ago
Place the gears in order from highest to lowest torque?
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60, 12, 24,48- ddfjjvdd
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3 years ago
For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate
faust18 [17]

Answer with Explanation:

We are given that

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Length of sprue=h_1=0.200 m

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Let 1 for the top and 2 for the bottom

d_=0.030 m

h_2=0

A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}

A_1=7.065\times 10^{-4} m^2

v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}

v_1=0.708 m/s

Pressure at the top and bottom of the sprue is atmospheric

h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}

Substitute the values

0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}

v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264

v_2=\sqrt{4.421264}=2.1 m/s

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A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2

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\eta=0.004 N.s/m^2

\rho=2700 kg/m^3

Substitute the values then we get

Reynolds number=\frac{2.1\times 0.03\times 2700}{0.004}

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0
3 years ago
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