Brake horsepower calculation

Power. There are two main ways to measure the power deUvered by the driver to the pump. The first method is to install a torque meter between the pump and the driver. A torque meter is a rotating bat having a strain gauge to measure shear deformation of a torqued shaft. Discussion of the principle of torque meter operation is available (16). The benefit of this method is direct and accurate measurements. The power deUveted to the pump from the driver is calculated from torque, T, and speed (tpm) in units of brake horsepower, ie, BHP (eq. 4a) when Tis in lbs-ft, and kW (eq. 4b) when T is N-m. 

Drive motors should be of the high-starting-torque type and selected for 1.33 times maximum rotational speed. For two- or three-diameter Idlns, the brake horsepower for the several diameters should be calculated separately and summed. Auxiliaiy drives should be provided to maintain shell rotation in the event of power failure. These are usually gasohne or diesel engines. 

Involved in producing the curves for Figs. 29-53 and 29-55 is a calculation of the so-called balance point at which the flow and revolutions per minute required by the recovery unit match those provided by the pump. If the recovery turbine is the sole driver (as for the lean pump of Fig. 29-54), both the speed and the brake horsepower of the recoveiy turbine and its driven pump must be the same at the so-called balance point. If there is a makeup driver and the recovery unit has available to it just the flow from the pump that it is driving, as for the pump of Fig. 29-56, then the speea ana capacity must match at the balance point. 

Mdien dscous liquids are handled in centrifugal pumps, the brake horsepower is increased, the head is reduced, and the capacity is reduced as compared to the performance with water. The corrections may be negligible for viscosities in the same order of magnitude as water, but become significant above 10 centistokes (10 centipoise for SpGr = 1.0) for heavy materials. While the calculation m.ethods are accepta oly good, for exact performance charts test must be run using the pump in the service. 

Alternate brake horsepower calculation Using Figure 12-63 for air. 

The 100-psig machine for at least 1500-cfm output, unit (A) has 1660 actual capacity and requires 309 brake horsepower (bhp) at full load. The minimum capacity for units Bl and B2 must be calculated, since these can be compressing the full 60 minutes each hour. 

Brake horsepower, centrifugal pumps, 200 Driver horsepower, 201 Burst pressure, 405, 456 Cartridge filters, 274-278 Capture mechanism, 279 Edge filler, 278 Filter media, table, 278 Micron ratings, 277 Reusable elements, 281 Sintered metal, 280 Types, 276, 277, 279 Wound vs. pleated, 276, 277 Centrifugal pumps, operating characteristics, 177-180 Calculations, see hydraulic performance Capacity, 180... 

Manufacturers performance curves, such as those in Fig. 10, contain a great deal of useful information. Actual average head-capacity curves are shown for a number of impeller diameters. Also superimposed on these head curves are curves of constant efficiency. A third set of cur ves superimposed on the head curves are the NPSH requirement curves (dashed line in Fig. 10), which indicate the required NPSH at any given condition of operation. A fourth set of curves sometimes included are the BHP (brake horsepower) curves. BHP is the actual horsepower calculated in the previous HGL method illustration. It is the HHP divided by the effciency. 

Use Eq. (24) in Chap. 14 of this text to calculate the theoretical compressor power. The brake horsepower efficiency of the electric motor... 

For purposes of example, assume a flow of 8.71 mVmin (2300 gal/min) through the tower The maximum head available to the recovery turbine was calculated to be 604 m (1982 ft) this value will be slightly in error when part of the flow is bypassed since frictional losses into and out of the recovery unit will change. First, assume the lean pump to be at 3.03 mVmin (800 gahmin) running at 3900 r/min with the semilean pump at 5.68 mVmin (1500 gal/min) to get the total flow of 8.71 mVmin (2300 gal/min). At 3.03 mVmin (800 gal/min) and 3900 r/min the available head of the lean pump is read from the curve. This must be greater than the required head, and the excess is plotted as in Fig. 29-60. The brake horsepower of the lean pump is also read. 

After calculating the head, then calculate the power supplied to punp by the shaft of the pun driver, i.e., the brake horsepower, which is given by... 

Use Equation 8.35 to calculate the shaft or brake horsepower. The pxmq) efficiency, taken from Figure 8.4, is 62%. 

Exam ple,—The theoretical water horsepower required to pump 500 gal. per minut against 300 ft. head is, as calculated from the indicator diagram, 38.5 horsepower at 80 per cent efficiency the brake horsepower required will be 48 hp. 

The computer program PROG62 sizes the centrifugal pump for the given flow rate and fluid characteristics. The program calculates the hydraulic brake horsepower required by the pump and the actual brake horsepower. In addition, the program computes the available net positive suction head (NPSH) and the pump efficiency. Table 6-7 shows the input data and results of the pump hydraulic design calculation. The available NPSH is 64 ft, and the actual brake horsepower required for the pump operation is 7.0hp, with a pump efficiency of 68%. 

Using the affinity law (Equation 6-76), the corresponding values for the head at varying flow rates are shown in Table 6-8. The computer program PROG63 calculates both the brake horsepower and pump efficiencies of the 6- and 8-inch pumps. Tables 6-9 and 6-10 show the input data and the computer results of these pumps. The characteristic curves are shown in Figure 6-6. 

Efficiency and Brake Horsepower Calculations for the 6-inch Centrifugal Pump Program Inputs and Outputs... 

An electric motor is to be used to drive a compressor of 1,119 brake horsepower (BHp), The efficiency of the motor is 95%. Therefore, the electrical input to the motor must be l,119(0.7457)/0.95 = 878 kW. This is equivalent to 3,000,000 Btu/hr, which is the basis for the previous example. Calculate the kW-hr required per year for the motor if the plant-operating factor is 0.9, and calculate the cost of electricity per year. 

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