Pump Shaft Work

The first integral on the right-hand side is the rate of work done on the fluid in the control volume by forces at the boundaiy. It includes both work done by moving solid boundaries and work done at flow entrances and exits. The work done by moving solid boundaries also includes that by such surfaces as pump impellers this work is called shaft work its rate is Ws-... 

Total pressure will only change in a fluid if shaft work or work of extraneous forces are introduced. Therefore, total pressure would increase in the impeller of a compressor or pump it would remain constant in the difmser. Similarly, total pressure would decrease in the turbine impeller but would remain constant in the nozzles. 

The turboexpander in combination with a compressor and a heat exchanger functions as a heat pump and is analyzed as follows In Fig. 29-44 consider the compressor and aftercooler as an isothermal compressor operating at To with an efficiency and assume the working fluid to be a perfect gas. Further, consider the removal of a quantity of heat by the tumoexpander at an average low temperature Ti-This requires that it dehver shaft work equal to Q. Now, make the reasonable assumption that one-tenth of the temperature drop in the expander is used for the temperature difference in the heat exchanger. If the expander efficiency is and this efficiency is mul-... 

Labyrinth seals work best when the pump is running. Centrifugal force favors the labyrinth seal s action. Earlier models were only specified for horizontal pump shafts. Later models are designed for both horizontal and vertical pump shafts and effectively perform their function whether the pump is running or off... 

The work done by the pump is found by setting up an energy balance equation. If W, is the shaft work done by unit mass of fluid on the surroundings, then —Ws is the shaft work done on the fluid by the pump. 

Because there is no pressure change and no pump or shaft work, the mechanical energy balance (Equation 2-28) reduces to... 

The work term W in equations 6.5 and 6.7 is positive if work is done on the fluid by a pump or compressor. Wis negative if the fluid does work in a turbine. W is often referred to as shaft work since it is transmitted into or out of a system by means of a shaft. 

Due to the work done on compression in the individual pumping stages, multi-stage claw pumps require water cooling for the four stages to remove the compression heat. Whereas the pumping chamber of the pump is free of sealants and lubricants, the gear and the lower pump shaft are lubricated... 

Step 6. Multiply the horsepower per pound of steam value calculated in step 5 by the turbine steam flow, in pounds per hour. This is the total shaft work that appears at the turbine s coupling. This is the amount of horsepower that is available to spin a centrifugal pump. 

The operators were not running the turbine. The turbine was spinning, because it was coupled to the pump, but there was no motive steam to the turbine. The operators reported that the turbine was not needed, as the motor was pulling only 90 percent of its maximum amperage load. The question is, dear reader, whether the pump will run faster if the motive-steam flow is opened to the turbine. And the answer is, no. While the turbine will produce shaft work, and will help... 

But suppose the amp load on the motor drops to zero, gradually increasing the steam to the turbine. Then, let s suppose we increase the motive steam to the turbine, by another notch. Will the pump, the motor, and the turbine (which are all coupled together and, hence, must run at the same speed) now run faster The answer is no. But what will happen to the increment of shaft work, generated by the turbine ... 

The required shaft work for the pump, wx, when expressed as a head, is known as the system head, Ahsysteni. This is the head required from the pump to achieve the required flow through the pipe system. 

As the shaft-work becomes more important in the system, improved system design tends in the direction of decreasing the pump work by lessening the pressure drop in the boiler. 

The system s Second Law efficiency rises as the work/heat ratio increases (Figure 7). This is partially due to improved performance of the turbine, pump, and condenser and the higher temperature steam from the boiler. This considerably decreases the available-energy destruction due to heat transfer in the boiler. Thus, the turbine can take advantage of this for the production of shaft work. 

In Fig. 13-2fl, which is particularly useful when a large temperature difference exists between the ends of the column, interreboilers add heat at lower temperatures and/or intercondensers remove heat at higher temperatures. As shown in Fig. 13-2, these intermediate heat exchangers may be coupled with a heat pump that takes energy from the intercondenser and uses shaft work to elevate this energy to a temperature high enough to transfer it to the interreboiler. 

Once the work is estimated, the pump shaft power. 

A simplified flow diagram for the process is shown in Figure 5.4.2. Table 5.4.1 list specifications for the process, obtained from Maddox and Bums [52, 53]. The shaft work for the pump will be calculated from Equation 5.48 and the shaft power from Equation 5.49. Because the process at this point is not well defined, all approximations must be made to maximize the estimated power so that the pump will not be undersized. 

Ws — shaft work, or rate of work done by the process fluid on a moving part within the system (e.g., a pump rotor)... 

The PUMP calculation would cause the pressure of the liquid stream to be raised by a specified amount the routine might also calculate the required shaft work for this operation and could go as far as to specify the type and size of pump needed for the job. The HEAT routine would calculate the heat input required to achieve the required temperature change and might also perform heat exchanger design calculations. 

A small adiabatic air compressor is used to ptunp air into a 20-ni insulated tank. The tank initially contains air at 298.15 K (25°C)and 101.33 kPa, exactly tlie conditions at wliich air enters tlie compressor. The pumping process continues until tlie pressure in tile tank reaches 1000 kPa. If tlie process is adiabatic and if compression is isentropic, wliat is tile shaft work of tlie compressor Assume air to be an ideal gas for wliich Cp = (7/2)P and Cy = (5/2)P. 

Estimate the shaft work required to pump 65 gal/min of sugar solution in water (specific gravity = 1.05) if the pump inlet pressure is 25 psig and the outlet pressure required is 155 psig. 

Equation (4.23) is known as the first law of thermodynamics for a closed system.] Keep in mind that a system may do work, or have work done on it, without some obvious mechanical device such as a pump, shaft, and so on, being present. Often the nature of the work is implied rather than explicity stated. For example, a cylinder filled with gas enclosed by a movable piston implies that the surrounding atmosphere can do work on the piston or the reverse a batch fuel cell does no mechanical work, unless it produces bubbles, but does deliver a current at a potential difference electromagnetic radiation can impinge on or leave a system and so forth. 

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