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professor190 [17]
4 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]4 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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A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a
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Q_{cv}=-339.347kJ

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u_1=210.49kJ/kg

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u_2=250.02kJ/kg

At the end we make a Energy balance, so

U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e

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Q_{cv}=-339.347kJ

The sign indicates that the tank transferred heat<em> to</em> the surroundings.

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Explanation:

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