Homework hints, tenth week:11.8,17,23,42,45

11.8 A centrifugal pump delivers 550 gal/min of water at 20 deg C when the brake horsepower is 22 and the efficiency is 71%. a. Estimate the head rise in ft and the pressure rise in psi. b. Also estimate the head rise and horsepower if instead the delivery is 550 gal/min of gasoline at 20 deg C.

Soln.: Density \rho = 998 kg/m^3 = 1.9 slug/ft^3. Power =22(550) = 12100 ft lbf/s = \rho g Q H/\eta , H = 112 ft.

Pressure rise \delta p = \rho g H = 49 psi.

Power = \rho g Q H/\eta = 1.69(32.2)(117/449)(50)/0.84 = 1.54 hp.

**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

11.17 A centrifugal pump has d_1 = 13 in, b_1 = 4 in, b_2 = 3 in, \beta_1 = 40 degrees, and \beta_2 = 40 degrees and rotates at 1160 r/min. If the fluid is gasoline at 20 deg C and the flow enters the blades radially, estimate the theoretical a. flow rate in gal/min, b. horsepower, and c. head in ft.

Soln.: For gasoline take \rho = 1.32 slug/ft^3. Compute \omega = 121.5 rad/s. u_1 = \omega r_1 = 121(3.5/12) = 35.4 ft/s.

V_n1 = u_1 tan \beta_1 = 35.4 tan 25 deg = 16.5 ft/s

Q= 2 \pi r_1 b_1 V_n1 = 10 ft^3/s

V_n2 = Q/2 \pi r_2 b_2 = 11.9 ft/s

u_2 = \omega r_2 = 65.8 ft/s

V_t2 = u_2 - V_n2 cot 40 deg = 51.7 ft/s.

P_ideal = \rho Q u_2 V_t2 = 1.32(10)(65.8)(51.7) = 44900/550 = 82 hp.

H = P/\rho g Q = 106 ft.

**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 ]

11.23 If the 38 in pump from fig. 11.7 is used to deliver 20 deg C kerosine at 850 rpm and 22000 gal/min, what a. head; and b. brake horsepower will result?

Soln.: For kerosine \rho = 1.56 and water 1.94 slug/ft^3. Use the scaling laws, 11.28: Q_1/Q_2 = (n_1/n_2)(D_1/D_2)^3 = Q_1/22000 = 710/850 so Q_1 = 18400 gpm.

Read h_1 = 235 ft and P_1 = 1175 bhp

D_1 = D_2 ; H_2 = 235 (850/710)^2 = 340 ft

P_2 = P_1(\rho_2/\rho_1)(n_1/n_2)^3 = 1175(1/56/1/94)(850/710)^3 = 1600 bhp.

**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

11.42 An 8 in model pump delivering water at 180 deg F at 800 gal/min and 2400 rpm begins to cavitate when the inlet pressure and velocity are 12 psia and 20 ft/s, respectively. Find the required NPSH of a prototype which is 4 times larger and runs at 1000 rpm.

Soln.: Water, \rho g = 60.6 lbf/ft^3 and p_v = 1600 psfa at 180 deg F. From 11.19,

NPSH_model = (p_i - p_v)/\rho g + V_i^2/2 g = 12x144 - 1600 / 60.6 + 20^2 / 2(32.2) = 8.32 ft

Similarity:

11.45 Determine the specific speeds of the seven Taco, Inc. pump impellers in fig. 11.24. Are they appropriate for centrifugal designs? Are the approximately equal within experimental uncertainty? If not, why not?

Soln.: Read the BEP values for each impeller and make a little table for 1160 rpm:

D, in	10.0	10.5	11.0	11.5	12.0	12.5	12.95
Q*, gpm	390	420	440	460	480	510	530
H*, ft	41	44	49	56	60	66	72
Sp. Sp. N_S	1414	1392	1314	1215	1179	1131	1080

These are well within the centrifugal pump range (N_S < 4000) but they are not equal because they are not geometrically similar (7 different impellers within a single housing).