HW4 Solution

, 1
3 ,
,
3
3,
2
3 ,
4
3
,
4
3
,
Note that this problem did not specifically ask for the minimum power required at sea level. Hence, other
selected values of velocity can be selected and will yield different power requirements. These will be plotted in
the Numerical Substitution section. The required power is given by,
Substituting the equation derived in Problem 6.1 for the required thrust, , into this equation for required
power yields,
2
/
1
2
,
The maximum velocity, , will occur when the aircraft it at full power. The maximum power available is given
by,
This equation can be substituted into the equation for the required power, , and then solved for . However,
the resulting equation is a fourth-order polynomial and, although closed-form solutions to this problem do exist,
the simplest way to solve this problem is by graphical means. This will be done in the Numerical Substitution
section. Finally, to find the required power and corresponding velocity at a different altitude, the following
equations are used:
, ,
/
/
where , is the required power at sea level, which we found in part (a), is the sea level air density, and is
the maximum velocity at sea level. To solve for the maximum velocity at a different altitude, the available power
must first be adjusted. It is given that the power available is proportional to the air density and so the following
equation must hold: