An unloaded helicopter of mass 10,000 kg hovers at sea level while it is being loaded. In the unloaded hover mode, the blades rotate at 400 rpm. The horizontal blades above the helicopter cause a 18-m-diameter air mass to move downward at an average velocity proportional to the overhead blade rotational velocity (rpm). A load of 14,000 kg is loaded onto the helicopter, and the helicopter slowly rises. Determine (a) the volumetric airflow rate downdraft that the helicopter generates during unloaded hover and the required power input and (b) the rpm of the helicopter blades to hover with the 14,000-kg load and the required power input. Take the density of atmospheric air to be 1.18 kg/m3. Assume air approaches the blades from the top through a large area with negligible velocity and air is forced by the blades to move down with a uniform velocity through an imaginary cylinder whose base is the blade span area.
• SpeedingUpIndex o Somethingpersomethingelse=index o something/somethingelse=speed/time=acceleration • Tofindacceleration…(twoways) 1. DataTableWay Example: t v Δt(t 2t 1) Δv(v 2v )1 a(Δv/Δt) 0s 5mph -- -- -- 2s 25mph 2 20 10 4s 45mph 2 20 10 2. AccelerationEquation:(a)=Δv/Δt (0s,5mph) (4s,45mph) a=Δv/Δt=(45mph-5mph)/(4s-0s)=40mph/4s=10mph/s v=v 0+at v 0=v–at a=v–v /0 t=v–v /0 Example#1:Ifat0seconds,atruckismovingat10mphandisat30mph10secondslater, whatistheaccelerationofthecar a=v 1–v /0 v 0=10mph v 1=30mph a=30mph–10mph/10s-0s t0=0s t1=10s a=20mph/10s a=2mph/s Example#2:Acarstartsat5mphandacceleratesatarateof3mph/s.Howlongdoesittake togetat30mph t=v–v 0aàt=(30mph-5mph)/(3mph/s)=25/3=8.33seconds • “Falling” o velocity