Highly Question All Students Should Answer<br> Why do you depend on Brainly?

. Consider the single-engine light plane described in Prob. 2. If the specific fuel consumption is 0.42 lb of fuel per horsepowe

Trava [24]

Answer:

Hence the Range and Endurance of single engine plane is given by

650.644 miles and 5.3528 hrs at standard sea level.

Explanation:

Given :

A single engine light plane with ,

Specific fuel consumption 0.42lb/hr/hp.

Fuel capacity =44 gal.

Gross weight =3400 lb.

To find :

Range and Endurance of the plane.

Solution:

Consider all standard measures of standard single engine propeller plane

as

Wing span =35.8 fts.

Wing swing area=174 sq ft

parasite drag coefficient =Cd.o.=0.025

Oswald's eff. factor= 0.8

ρ=0.002377= corresponds to standard sea level constant.

Now

Formula for Range is given by, Breguent formula.

R=(η/c) *(Cl/Cd)*ln(W1/W0)

here η is Oswald's constant,

Now calculating lift(Cl) and drag coefficient (Cd)

Cl=W/(1/2*ρ*v^2*S)

W=Gross weight

ρ=0.002377

Assume v=200 ft/sec normally,

S=174 Sq .ft.

CI=3400/(1/2*0.002377*200*200*174)

=6800/16543.9

=0.4110

Now calculating drag constant,

AR=(wing span)^2/wing swing area

=(35.8)^2/174

=7.37

Now

Drag Coefficient

Cd=Cd.o.+ (Cl^2)/(pie*e*AR)

=0.025+(0.4110)^2/(3.142*0.8*7.36)

=0.0342

Given that 44 gal fuel capacity and in Aviation weight of fuel is 5.64 lb/gal

hence weight of fuel=W1=3400- (44*5.64)

=3151.84

Now

for specific fuel consumption=0.42 lb/hp/hr

=0.42 lb*(1/550 ft)*(1/3600)sec

=2.12 *10^-7 lb/ft/sec

Now further calculating range

R=(η/c) *(Cl/Cd)*ln(W1/W0)

={0.8/(2.12*10^-7)}*(0.4110/0.0342)*ln(3151.84/3400)

=0.024908/0.072504

=0.34354*10^7

=3.4353 *10^6 fts.

1mi =5280 ft

=(3.4353/5280)*10^6

=650.644 miles

Now

For Endurance

E=(η/c)*{(Cl^3/2)/Cd}*(2*ρ*S)^1/2*[1/(W1)^1/2 -1/(W0)^1/2].

=(0.8/2.12*10^-7)*{(0.4110^3/2)/0.0342}*(2*0.002377*174)^1/2*[1/(3151.84)^1/2 -1/(3400)^1/2]

=3.7735*10^6*7.7043*0.8272*0.0006629

=0.01927*10^6

=1.927*10^4 sec

here 1hr =3600 sec

E=(1.927/3600)*10^4

=5.3528 hrs

7 0

3 years ago

A friend would like you to build an "electronic eye" for use as a fake security device. The device consists of three lights line

mars1129 [50]

Answer and explanation:

The graphical representation of the electronic eye

The state table showing

the present state

input

Next state and

the output

are shown in the attached file

8 0

3 years ago

Read 2 more answers

Define and discuss the difference between micronutrients and macronutrients. Also, discuss their importance in the body at rest

almond37 [142]

Answer:

Macronutrients are simply nutrients the body needs in a very high amount e.g Carbohydrate.

MicroNutrients are simply nutrients the body needs but in little amount e.g Minerals.

Explanation:

So for further breakdown:

What are nutrients? Nutrients are essential elements that nourish the body in different capacities. We as humans get most of out nutrients from the food and water we ingest.

Now about Macro Nutrients: From the prefix "Macro" which means large, we can infer that macro nutrients are elements need by the body for the fundamental processes of the body, deficiency in this nutrients are very easy to spot. Examples are: Carbohydrates, Protein, Fats amd Water.

Micro Nutrients: In relation to macro nutrients this are elements that the body needs but are not needed in Large quantities. They mostly work like supporting nutrients. Most chemical activities like reaction that occur in the body are a function of micro nutrients. Defiencies in micrp nutrients may take some time to spot e.g Minerals and Vitamins

In regards to exercise: Macro nutrients are the essential ones here since they are the ones that generate energy. PS: micro nutrients dont generate energy.

In regards to rest: Both the Macro and Micro Nutrients are essentail for the overall well being of the body.

5 0

3 years ago

The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in

Mazyrski [523]

Explanation:

A.

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

B.

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

5 0

3 years ago

Several different loads are going to be used with the voltage divider from Part A. If the load resistances are 300 kΩkΩ , 200 kΩ

harina [27]

Answer:

attached below

Explanation:

7 0

3 years ago

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