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3 years ago

Why would an aerospace engineer limit the maximum angle of deflection of the control surfaces?

Engineering

1 answer:

WITCHER [35]3 years ago

4 0

Answer:

You can create high drag which allows a steeper angle without increasing your air speed on landing. you can reduce the length of landing role. Flaps are also used to increase the drag they are retracted when they are not needed. it is adviseable to down he flaps during the time of take off.

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3 years ago

A mass weighing 22 lb stretches a spring 4.5 in. The mass is also attached to a damper with Y coefficient . Determine the value

Dominik [7]

Answer:

Cc= 12.7 lb.sec/ft

Explanation:

Given that

m = 22 lb

g= 32 ft/s²

$m = \dfrac{22}{32}=0.6875\ s^2/ft$

x= 4.5 in

1 in = 0.083 ft

x= 0.375 ft

Spring constant ,K

$K=\dfrac{m}{x}=\dfrac{22}{0.375}$

K= 58.66 lb/ft

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3 years ago

Determine the following parameters for the water having quality x=0.7 at 200 kPa:

ra1l [238]

Solution :

Given :

Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa

The phase diagram is provided below.

a). The phase is a standard mixture.

b). At pressure, p = 200 kPa, T = $T_{saturated}$

Temperature = 120.21°C

c). Specific volume

$v_{f}= 0.001061, \ \ v_g=0.88578 \ m^3/kg$

$v_x=v_f+x(v_g-v_f)$

$=0.001061+0.7(0.88578-0.001061)$

$=0.62036 \ m^3/kg$

d). Specific energy ( $u_x$ )

$u_f=504.5 \ kJ/kg, \ \ u_{fg}=2024.6 \ kJ/kg$

$u_x=504.5 + 0.7(2024.6)$

$=1921.72 \ kJ/kg$

e). Specific enthalpy $(h_x)$

At $h_f = 504.71, \ \ h_{fg} = 2201.6$

$h_x=504.71+(0.7\times 2201.6)$

$= 2045.83 \ kJ/kg$

f). Enthalpy at m = 0.5 kg

$H=mh_x$

$= 0.5 \times 2045.83$

= 1022.91 kJ

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