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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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Answer:

P_2-P_1=27209h

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For pressure gage we can determine this by saying:

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2 years ago
. Two rods, with masses MA and MB having a coefficient of restitution, e, move
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V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}....................(4)

From equation (2):

V_B = V_A + e(U_A -U_B)......(5)

Substitute equation (5) into (1)

M_AU_A + M_BU_B = M_AV_A + M_B(V_A + e(U_A -U_B))..........(6)

Solving for V_A in equation (6) above:

V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}.........(7)

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M_A = 2 kg\\M_B = 1 kg\\U_B = -3 m/s( negative x-axis)\\e = 0.65\\U_A = ?

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I = (2*0) + (1*4.32) - ((2*3.66) + (1*-3))

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Answer:

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Due to these above limitations practical cycles have low efficiency as compare to ideal cycle.

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