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A medium-sized jet has a 3.8-mm-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to th

SCORPION-xisa [38]

Answer:

$F_{thrust}$ ≅ 111 KN

Explanation:

Given that;

A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8

mass = 85,000 kg

drag co-efficient (C) = 0.37

(velocity (v)= 230 m/s

density (ρ) = 1.0 kg/m³

To calculate the thrust; we need to determine the relation of the drag force; which is given as:

$F_{drag}$ = $\frac{1}{2}$ × CρAv²

where;

ρ = density of air wind.

C = drag co-efficient

A = Area of the jet

v = velocity of the jet

From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0

SO, $F_{drag}-F_{thrust} = 0$

We can as well say:

$F_{drag}= F_{thrust}$

We can now replace $F_{thrust} with F_{drag}$ in the above equation.

Therefore, $F_{thrust}$ = $\frac{1}{2}$ × CρAv²

The A which stands as the area of the jet is given by the formula:

$A=\frac{\pi d^2}{4}$

We can now have a new equation after substituting our A into the previous equation as:

$F_{thrust}$ = $\frac{1}{2}$ × Cρ $(\frac{\pi d^2}{4})v^2$

Substituting our data from above; we have:

$F_{thrust}$ = $\frac{1}{2}$ × $(0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2$

$F_{thrust}$ = $\frac{1}{8} (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2$

$F_{thrust}$ = 110,990N

$F_{thrust}$ in N (newton) to KN (kilo-newton) will be:

$F_{thrust}$ = $(110,990N)*\frac{1KN}{1,000N}$

$F_{thrust}$ = 110.990 KN

$F_{thrust}$ ≅ 111 KN

In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.

5 0

3 years ago

Technician A says automotive gasoline engines run on the 2 stroke principal. Technician B says Diesel engine hace less compressi

d1i1m1o1n [39]

Answer:

both are incorect

Explanation:

2 stroke principal is a mix of gasoline and engine oil a normal gasoline engine does not run on 2 stroke fuel

technician B is also wrong Diesel engines generally generate much higher compersion than gasoline engines

3 0

2 years ago

Air is saturated with water vapor at 35.0 oC and a total pressure of 1.50 atmospheres. If the molar flow rate of the dry air in

Marizza181 [45]

Answer:

11.541 mol/min

Explanation:

temperature = 35°C

Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa

note : partial pressure of water in mixture = saturation pressure of water at T = 35°c )

from steam table it is = 5.6291 Kpa

calculate the mole fraction of H $_{2}o$ ( YH $_{2}o$ )

= 5.6291 / 151.95

= 0.03704

calculate the mole fraction of air ( Yair )

= 1 - mole fraction of water

= 1 - 0.03704 = 0.9629

Now to determine the molar flow rate of water vapor in the stream

lets assume N = Total molar flow rate

NH $_{2}o$ = molar flow rate of water

Nair = molar flow rate of air = 300 moles /min

note : Yair * n = Nair

therefore n = 300 / 0.9629 = 311.541 moles /min

Molar flowrate of water

= n - Nair

= 311.541 - 300 = 11.541 mol/min

4 0

3 years ago

How does a truth tables impact the development of a completed circuit? Make sure to use as many keywords as possible to back up

WINSTONCH [101]

Answer:

it is one this so the key word are pfp tpf htf pfp

Explanation:

3 0

2 years ago

When the Challenger spacecraft exploded, which design step had not been thoroughly explored?

Llana [10]

Answer:

The escape systems not working

8 0

2 years ago