Question

This is for my fluids class - im stuck :(

please show all work by hand and I will give thumbs up

[4] 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. Note that the helicopter's motor supplies work to the helicopter in the form of kinetic energy. (a) Determine the volumetric airflow rate downdraft that the helicopter generates during the unloaded hover and the required power input. (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 p = 1.18 kg/m'. Assume that air approaches the blades from the top through a large area with negligible velocity and that air is forces by the blades to move downward with a uniform velocity through an imaginary cylinder whose base is the blade span area

Answer 1

Ansa- The weight of helicopter alone, Weg -(10000) *19.81) = 98100N The weight of load celone W = mig =(14000) (9.81) = 137340N Consider the unloaded hquee made The total force acting on helicopter is equal to weight of helicoplē Then Je=-W now "Rx - Wh =-98100 M The exit uelocity can be expressed as 98100 17 1811x139 = 18.08 mis The mass flow rate of air = 1.18 (TIX183) 18:08 - 5426.19 kq is Volumetric flow rate of cer can be calculated as y = m - 5426.19 j - 4598.47 m3/

Then Apply the consewation of energy Drinuble blow inlete lexit of combol volume We haw n a u gz.) there in P V 1.92) + mai moderne het us assume elevation diff. b/w two pofs. es negligible Wo haw - P = P2 = Pane , Z =2, 4 U, =0 What will 5426:19 x (18.08)_482886873.85W. 2 = 886.873kW Considee loaded home mode of helicopter The lotal weight of helicopter k!= klinthle - 98100+137340 = 235440 N. The exit uelocity can be gown as - VA V PWD23 7718/ Texiga V = 28 m/s m = PAV2 = 118 ( 17 x 182) 28.00 = 8405.98 kg 13 Klootor, L-in Vu 8405.98 (28) ² - 3295.14 kW 2 Then we can express for unloaded mode as Vanualnu) Similarly for loceded mode 2 an = a (2) Then 12 nuL Me - Mee (Vit) - 400(2300) m = 61.94 opm 235040 1.18/7x182)