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
