If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will: a) increase during a climb.
<h3>What is a
ram air input?</h3>
A ram air input can be defined as an air intake system which is designed and developed to use the dynamic air pressure that is created due to vehicular motion, or ram pressure, in order to increase the static air pressure within the intake manifold of an internal combustion engine of an automobile.
This ultimately implies that, a ram air input allows a greater mass-flow of air through the engine of an automobile, thereby, increasing the engine's power.
In conclusion, indicated airspeed will increase during a climb when both the ram air input and drain hole of the pitot system become blocked.
Read more on pilots here: brainly.com/question/10381526
#SPJ1
Complete Question:
If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will
a) increase during a climb
b) decrease during a climb
c) remain constant regardless of altitude change
Answer:
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Explanation:
We are given;
T∞ = 70°C.
Inner radii pipe; r1 = 6cm = 0.06 m
Outer radii of pipe;r2 = 6.5cm=0.065 m
Electrical heat power; Q'_s = 300 W
Since power is 300 W per metre length, then; L = 1 m
Now, to the heat flux at the surface of the wire is given by the formula;
q'_s = Q'_s/A
Where A is area = 2πrL
We'll use r2 = 0.065 m
A = 2π(0.065) × 1 = 0.13π
Thus;
q'_s = 300/0.13π
q'_s = 734.56 W/m²
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Answer:
The minimum diameter for each cable should be 0.65 inches.
Explanation:
Since, the load is supported by two ropes and the allowable stress in each rope is 1500 psi. Therefore,
(1/2)(Weight/Cross Sectional Area) = Allowable Stress
Here,
Weight = 1000 lb
Cross-sectional area = πr²
where, r = minimum radius for each cable
(1/2)(1000 lb/πr²) = 1500 psi
500 lb/1500π psi = r²
r = √1.061 in²
r = 0.325 in
Now, for diameter:
Diameter = 2(radius) = 2r
Diameter = 2(0.325 in)
<u>Diameter = 0.65 in</u>