Homework hints, fifth problem set:11.15,18,28,39

11.15 A lawn sprinkler can be used as a simple turbine. Flow enters normal to the plane of the sprinkler in the center and splits evenly in Q/2 and V_rel leaving each nozzle. The arms rotate at angular velocity \omega and do work on a shaft. Draw the velocity diagram for this turbine. Neglecting friction, find an expression for the power delivered to the shaft. Find the rotation rate for which the power is a maximum.

Apply the Euler turbine formula:

P = \rho Q (u_2 V_t2 - u_1 V_t1 ) = \rho Q [u (W - u) - 0]

or: P = \rho Q \omega R ( V_rel - \omega R)

dP/du = \rho Q (V_rel - 2u) = 0 if \omega = V_rel / 2R

P_max = \rho Q u (2u - u) = \rho Q (\omega R)^2

11.18 A jet of velocity V strikes a vane which moves to the right at speed V_c . The vane has a turning angle \theta. Derive an expression for the power delivered to the vane by the jet. For what vane speed is the power maximum?

The jet approaches the vane a relative velocity V - V_c . Then the force is F = \rho A (V-V_c)^2 (1 - cos \theta), and Power = F V_c = \rho A V_c (V-V_c)^2 (1 - cos \theta)

Maximum power occurs when dP/dV_c = 0, or:

V_c = (1/3) V_jet ; P = (4/27) \rho A V^3 [1 - cos \theta ]

11.28 Tests by the Byron Jackson Co. of a 14.62 in centrifugal water pump at 2134 rpm yield the data below. What is the BEP? What is the specific speed? Estimate the max discharge.

Q, ft^3 /s	0	2	4	6	8	10
H, ft	340	340	340	330	300	220
bhp	135	160	205	255	330	330

The efficiences are computed from \eta = \rho g Q H / (550 bhp) and are:

Q =	0	2	4	6	8	10
\eta =	0	0.482	0.753	0.881	0.825	0.756

Thus the BEP is close to Q = 6 ft^3 / s . The specific speed is

N_s = n Q*^1/2 / H*^3/4 = 1430

For estimating Q_max the last three points fit a power law to within 0.5%:

H = 340 - 0.00168 Q^485 = 0 if Q = 12.4 ft^2 / s = Q_max