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professor190 [17]
3 years ago
11

A jetliner flying at an altitude of 10,000 m has a Mach number of 0.5. If the jetliner has to drop down to 1000 m but still main

tain the Mach number (0.5), how much faster must the jetliner fly?
Engineering
1 answer:
andreev551 [17]3 years ago
6 0

Answer

Assuming

At 10000 m height temperature T = -55 C = 218 K

At 1000 m height temperature T = 0 C  = 273 K

\dfrac{V_1}{C_1} =\dfrac{V_2}{C_2} = 0.5

R = 287 J/kg K

C_1 = \sqrt{\gamma RT_1} = \sqrt{1.4\times 287\times 218} = 295 m/s

C_2 = \sqrt{\gamma RT_2} = \sqrt{1.4\times 287\times 273} = 331 m/s

V_2 = \dfrac{V_1}{C_1}\timesC_2

V₂ = V₁ ×1.1222

V₁ = 0.5 × C₁ = 0.5 × 295 = 147.5 m/s

V₂ = 1.1222 ×  147.5 = 165.49 m/s

so, the jetliner need to increase speed by ( V₂ -V₁  )

= 165.49 - 147.5

= 17.5 m/s

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An air compressor of mass 120 kg is mounted on an elastic foundation. It has been observed that, when a harmonic force of amplit
kupik [55]

Answer:

equivalent stiffness is 136906.78 N/m

damping constant is 718.96 N.s/m

Explanation:

given data

mass = 120 kg

amplitude = 120 N

frequency = 320 r/min

displacement = 5 mm

to find out

equivalent stiffness and damping

solution

we will apply here frequency formula that is

frequency ω = ω(n) √(1-∈ ²)      ......................1

here  ω(n) is natural frequency i.e = √(k/m)

so from equation 1

320×2π/60 = √(k/120) × √(1-2∈²)

k × ( 1 - 2∈²) = 33.51² ×120

k × ( 1 - 2∈²) = 134752.99    .....................2

and here amplitude ( max ) of displacement is express as

displacement = force / k  ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})

put here value

0.005 = 120/k   ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})  

k ×∈ × √(1-2∈²) = 1200       ......................3

so by equation 3 and 2

\frac{k\varepsilon \sqrt{1-\varepsilon^2})}{k(1-2\varepsilon^2)} = \frac{12000}{134752.99}

\varepsilon^{2} - \varepsilon^{4}  = 7.929 * 10^{-3} - 0.01585 * \varepsilon^{2}

solve it and we get

∈ = 1.00396

and

∈ = 0.08869

here small value we will consider so

by equation 2 we get

k × ( 1 - 2(0.08869)²) = 134752.99

k  = 136906.78 N/m

so equivalent stiffness is 136906.78 N/m

and

damping is express as

damping = 2∈ √(mk)

put here all value

damping = 2(0.08869) √(120×136906.78)

so damping constant is 718.96 N.s/m

7 0
3 years ago
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