Answer:
The major effects of ice accretion on the aircraft is that it disturbs the flow of air and effects the aircraft's performance.
Explanation:
The ice accretion effects the longitudinal stability of an aircraft as:
1. The accumulation of ice on the tail of an aircraft results in the reduction the longitudinal stability and the elevator's efficacy.
2. When the flap is deflected at 
with no power there is an increase in the longitudinal velocity.
3. When the angle of attack is higher close to the stall where separation occurs in the early stages of flow, the effect of ice accretion are of importance.
4. When the situation involves no flap at reduced power setting results in the decrease in aircraft's longitudinal stability an increase in change in coefficient of pitching moment with attack angle.
I think the answer would be A if its wrong I’m sorry
Answer:
Explanation:
volume of 20.9 N
= 20.9 / 11.5 m³
= 1.8174 m³
In one hour 1.8174 m³ flows
in one second volume flowing = 1.8174 / 60 x 60
= 5 x 10⁻⁴ m³
Rate of volume flow = 5 x 10⁻⁴ m³ / s .
Answer:
h = 375 KW/m^2K
Explanation:
Given:
Thermo-couple distances: L_1 = 10 mm , L_2 = 20 mm
steel thermal conductivity k = 15 W / mK
Thermo-couple temperature measurements: T_1 = 50 C , T_2 = 40 C
Air Temp T_∞ = 100 C
Assuming there are no other energy sources, energy balance equation is:
E_in = E_out
q"_cond = q"_conv
Since, its a case 1-D steady state conduction, the total heat transfer rate can be found from Fourier's Law for surfaces 1 and 2
q"_cond = k * (T_1 - T_2) / (L_2 - L_1) = 15 * (50 - 40) / (0.02 - 0.01)
=15KW/m^2
Assuming SS is solid, temperature at the surface exposed to air will be 60 C since its gradient is linear in the case of conduction, and there are two temperatures given in the problem. Convection coefficient can be found from Newton's Law of cooling:
q"_conv = h * ( T_∞ - T_s ) ----> h = q"_conv / ( T_∞ - T_s )
h = 15000 W / (100 - 60 ) C = 375 KW/m^2K
Answer:
DIAMETER = 9.797 m
POWER = 
Explanation:
Given data:
circular windmill diamter D1 = 8m
v1 = 12 m/s
wind speed = 8 m/s
we know that specific volume is given as
where v is specific volume of air
considering air pressure is 100 kPa and temperature 20 degree celcius
v = 0.8409 m^3/ kg
from continuity equation
mass flow rate is given as
the power produced ![\dot W = \dot m \frac{ V_1^2 - V_2^2}{2} = 717.3009 [\frac{12^2 - 8^2}{2} \times \frac{1 kJ/kg}{1000 m^2/s^2}]](https://tex.z-dn.net/?f=%5Cdot%20W%20%3D%20%5Cdot%20m%20%5Cfrac%7B%20V_1%5E2%20-%20V_2%5E2%7D%7B2%7D%20%3D%20717.3009%20%5B%5Cfrac%7B12%5E2%20-%208%5E2%7D%7B2%7D%20%5Ctimes%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2Fs%5E2%7D%5D)
