Two identical, flat, square plates are immersed in the flow with velocity U. Compare the drag forces experienced by the SHADED areas.

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Two identical flat sqare plates are immersed in the flow with velocity U. Compare the drag forces experienced by the SHAE areas. F > F A. A B F > F B. B A C. FA = FB. It depends on whether the bondary layer is laminar or trblent.! F B F A > F B. Since the shear stress acting on the bondary decreases with increasing distance from the leading edge! w or Re x Re 5 x 5 having more area closer to the leading edge will reslt in a larger drag force on the shaded area.

Bondary layer separation is defined to occr when the shear stress at the srface is zero. Assme a polynomial representation for the laminar bondary layer of the form / U = a b! c! d! 3 where! = y/ and is the 99 laminar bondary layer thickness. Which velocity profile below is most appropriate and satisfies all the bondary conditions at the separation point. A. U = 3!!3 B. C.. U = 3!! 3 U = 3!!3 =!! U The velocity profile for the bondary layer shold satisfy the bondary conditions: no-slip at the bondary: U! = 0 = 0. All of the proposed profiles satisfy this bondary condition. continos velocity profile at the edge of the bondary layer: U! = =. All of the proposed profiles satisfy this bondary condition. d U smooth velocity profile at the edge of the bondary layer:! = d! Profiles A B and satisfy this bondary condition. d U zero wall shear stress at the separation point:! = 0 = 0. d! Only profiles B and C satisfy this condition. Hence only profile B satisfies all of the conditions. = 0.

The impeller diameter is to be scaled for a pmp operating at a constant speed. If the scaled head rise across the pmp is half of the original head rise i.e. H = H / what wold the power scale? A.! = 8! B.! =! C.! =!.! = 4! From the pmp similarity relations! =! gh gh = H = H Since it s given that H = H / we mst have / =!. Again from the pmp similarity relations W! W!! =! 3 5 * = 3 5 * W! = W! * Sbstitting in for / gives W! = W!! 5 = W!. 4 5

pmp_ Consider the pipe system containing a pmp shown in the figre below. The flid being pmped from the lake to the tank is water density = 000 kg/m 3 kinematic viscosity =.0*0-6 m /s. pstream of pmp L pmp downstream of pmp L H H diameter of both lengths of pipe = = 0 cm pstream of pmp L = 5 m downstream of pmp L = 5 m roghness of both lengths of pipe! =! =.5*0-4 m total minor loss pstream of pmp K minor =.0 total minor loss downstream of pmp K minor =.0 H = 3 m H = 0 m pmp head rise crve: H [m] = -.5*0 3 s /m 5 Q.8*0 s/m Q 6.3*0 m pmp efficiency crve: = -5.6*0 s /m 6 Q.*0 s/m 3 Q.*0 - a. etermine the operating flow rate for the system. b. What power mst be spplied to the pmp by the motor to operate at the flow rate fond in part a? Page of 3

pmp_ To determine the operating flow rate first determine the system head crve by applying the extended Bernolli eqation from point to point. pstream of pmp L pmp downstream of pmp L H H p!g V g z p =!g V g z where p = p atm p = p atm V! 0 V! 0 z = -H z = H V H L =! K i i g i = * f L L H L H S V p K minor K minor - g = f L L * K minor K minor - Q g. 4. The friction factor may be fond from the Moody diagram. Since the flow rate is nknown try assming that the flow is in the flly trblent region of the Moody diagram this assmption will need to be verified. In this region the friction factor is only a fnction of the pipe s relative roghness! =.5 *04 m =.5 *03 0.0 m From the Moody diagram f = 0.0. 3 Combining these relations gives the system head crve f H Ssystem = z! z * L L K minor K minor - Q = s 0 s Q 4 g. 4 where s 0 = z! z and 5 s = * f g! 4 L L K minor K minor -. 6 etermine the operating point by eqating the system head crve to the pmp head crve s 0 s Q = p 0 p Q p Q 7 p! s Q p Q p 0! s 0 = 0 8 Q =! p ± p! 4 p! s p! s p 0! s 0. 9 Page of 3

pmp_ Using the given data s 0 = 3 m s = 6.08*0 3 s /m 5 p 0 = 6.30*0 m p =.80*0 s/m p = -.50*0 3 s /m 5 Q = 8.3*0 - m 3 /s Check the Reynolds nmber assmption of flly trblent flow V = Q! V = 0.6 m s 0! 4 Re = V!! Re =.*0 6 Checking the Moody diagram shows that the flow is in the flly rogh zone for this Reynolds nmber and relative roghness. Ths or assmption of flly rogh flow was a good one. The power inpt to the flid by the pmp at these conditions is! =!QgH W into flid where H = 55.0 m at the operating flow rate of 8.3*0 - m 3 /s fond sing either the system or pmp head crves. Hence!! = 44.8 kw. 3 W into flid Since the pmp isn t 00 efficient the power that mst be spplied to the pmp is W! into W! flid into =! W! into = 54.6 kw. 4 pmp! pmp where the pmp efficiency at the operating point is! = 8 sing the given efficiency crve for the pmp! = -5.6*0 s /m 6 Q.*0 s/m 3 Q.*0 -. Page 3 of 3