David wants to determine what his organization should do in the future, and set project targets. Which management function can D
1 answer:
You might be interested in
Answer:
- the mass flow rate of air is 7.53 kg/s
- the exit area is 0.108 m²
Explanation:
Given the data in the question;
lets take a look at the steady state energy equation;
m" = W" / [ (h₁ - h₂ ) -
]
Now at;
T₁ = 900K, h₁ = 932.93 k³/kg
T₂ = 500 K, h₂ = 503.02 k³/kg
so we substitute, in our given values
m" = [ 3200 kW × ] / [ (932.93 - 503.02 )k³/kg -
|
||
| ]
m" = 7.53 kg/s
Therefore, the mass flow rate of air is 7.53 kg/s
now, Exit area A₂ = v₂m" / V₂
we know that; pv = RT
so
A₂ = RT₂m" / P₂V₂
so we substitute
A₂ = {[ ×
×
] / [(1 bar)(100 m/s )]} |
||
A₂ = 0.108 m²
Therefore, the exit area is 0.108 m²
Answer:
The final pressure is 3.16 torr
Solution:
As per the question:
The reduced pressure after drop in it, P' = 3 torr =
At the end of pumping, temperature of air,
After the rise in the air temperature,
Now, we know the ideal gas eqn:
PV = mRT
So
(1)
where
P = Pressure
V = Volume
R = Rydberg's constant
T = Temperature
Using eqn (1):
Now, at constant volume the final pressure, P' is given by:
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.
Answer:
105.70 mm
Explanation:
Poisson’s ratio, v is the ratio of lateral strain to axial strain.
E=2G(1+v) where E is Young’s modulus, v is poisson’s ratio and G is shear modulus
Since G is given as 25.4GPa, E is 65.5GPa, we substitute into our equation to obtain poisson’s ratio
Original length
Where is final diameter,
is original diameter,
is final length and
is original length.
Therefore, the original length is 105.70 mm