9.5. The drag on the disk:
For circular disk:
A = πr2
dA = 2πr dr
[kN]
The integral can be calculated numerically from the data in the table:
|
r [m] |
p(r) [kPa] |
r p(r) |
integral |
|
0.00 |
4.34 |
0 |
0 |
|
0.05 |
4.28 |
0.214 |
0.00535 |
|
0.10 |
4.06 |
0.406 |
0.02085 |
|
0.15 |
3.72 |
0.558 |
0.04495 |
|
0.20 |
3.10 |
0.62 |
0.07440 |
|
0.25 |
2.78 |
0.695 |
0.10728 |
|
0.30 |
2.37 |
0.711 |
0.14243 |
|
0.35 |
1.89 |
0.6615 |
0.17674 |
|
0.40 |
1.41 |
0.564 |
0.20738 |
|
0.45 |
0.74 |
0.333 |
0.22980 |
|
0.50 |
0.00 |
0 |
0.23813 |
[kN]
Therefore, FD = 5.42 [kN]
9.42. The free-body force diagram:
Because
at terminal velocity the forces are in equilibrium:
FD = w sin θ
while the drag:
Therefore, CD = 1.4
From Fig 9.30 for given CD and A, the rider is upright.
9.52. The free-body force diagram:
At
terminal velocity the forces are in equilibrium:
FB = w + FD
This results in:
CDv2 = 0.455 [m2/s2]
CD and v is related to each other through Re such as described in Fig 9.21(a).
Iteration with some initial CD to calculate Re and getting back CD from Fig 9.21(a) we get: CD = 0.4 for Re = 3.62 ´ 104 or v = 1.06 [m/s].
9.66. Energy loss due to aerodynamic drag is responsible for more fuel consumption in a 2ℓ trip.
Energy loss in no-wind round-trip:
Energy loss in windy round-trip:
(upwind)
(downwind)
W = Wu + Wd
Thus, it is clear that the windy round-trip needs more fuel as W > W0.
9.83. The power needed is to provide the stirrer a torque M at an angular speed of ω:
P = Mω = MvC/R
where R = 1.5 + 7/16 = 2 1/16 [in] and vC is the velocity of the disk center.
The torque has to overcome the drag caused by the paint on both disks:
M
= 2R FD = 
by neglecting the variation of velocity over the disk area.
Therefore,
To determine CD we need the Reynolds number:
Re
= 
10.5
which falls between two cases available in the book, i.e., Re ≤ 1 (Table 9.4) and Re > 103 (Fig 9.29). To be safe, we’ll use the larger value of CD obtained from those cases: CD = 20.4/Re = 1.94 (Table 9.4) and CD = 1.1 (Fig 9.29), so we use CD = 1.94 in our case → P = 7.78 ´ 10-5 [hp]
9.95. To maintain level flight there is an equilibrium between the weight of the aircraft and the lift force:
w = FL = CL˝ρv2A
The power of the flight is to overcome the drag loss:
P = FDv = CD˝ρv3A
Combining the equations above:
= 65.9 [hp]
9.98. The lift: FL = CL˝ρU2A
With the same CL and A to produce the same FL we have:
The data for air density at sea level and 10000-m altitude can be obtained from Table C.2. Then we have:
U » 1.72 U0