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3. A hydraulic lift is shown below. The combined mass of the piston, rack, and car is 4000 lbm. The working fluid is water. There is no heat transfer to or from the water, and the internal energy of the water per unit mass is constant. The water may be considered incompressible. (a) Taking all the water in the reservoir, line, and hydraulic cylinder as the system (i.e., taking the closed-system approach), calculate the work necessary to raise the rack and car 1 ft (neglect the change in potential energy of the water in the system). (b) Repeat part (a), taking all the water plus the car and the rack as the system. (c) Repeat part (a), taking an open-system approach; choose as your system the volume of the hydraulic cylinder, excluding the piston, rack, and car. If the absolute pressure in the system is 1000 lbf/in², calculate the volume that must flow in to raise the car 1 ft.

3. A hydraulic lift is shown below. The combined mass of the piston, rack, and car is 4000 lbm. The working fluid is water. There is no heat transfer to or from the water, and the internal energy of the water per unit mass is constant. The water may be considered incompressible. (a) Taking all the water in the reservoir, line, and hydraulic cylinder as the system (i.e., taking the closed-system approach), calculate the work necessary to raise the rack and car 1 ft (neglect the change in potential energy of the water in the system). (b) Repeat part (a), taking all the water plus the car and the rack as the system. (c) Repeat part (a), taking an open-system approach; choose as your system the volume of the hydraulic cylinder, excluding the piston, rack, and car. If the absolute pressure in the system is 1000 lbf/in², calculate the volume that must flow in to raise the car 1 ft.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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please help solve for part A)

 

Q3 part a, for this system the changes in internal, potential and kinetic energies are all ≈ 0, so that  .  This seems odd, but the water is merely transferring work from the pump to the piston, rack and car.  Here the   W s the algebraic sum of the work done on the water by the pump and the work done by the water on the piston, rack and car ( i think)

 

3. A hydraulic lift is shown below. The combined mass of the piston, rack, and car is 4000 lbm. The working fluid is water. There is no heat transfer to or from the water, and the internal energy of the water per unit mass is constant. The water may be considered incompressible. (a) Taking all the water in the reservoir, line, and hydraulic cylinder as the system (i.e., taking the closed-system approach), calculate the work necessary to raise the rack and car 1 ft (neglect the change in potential energy of the water in the system). (b) Repeat part (a), taking all the water plus the car and the rack as the system. (c) Repeat part (a), taking an open-system approach; choose as your system the volume of the hydraulic cylinder, excluding the piston, rack, and car. If the absolute pressure in the system is 1000 lbf/in², calculate the volume that must flow in to raise the car 1 ft. Reservoir A) Recall Energy Balance Equation. d[m(u +92 + 1 + gz 7) system Pump =d[m (u + gz ++)lindmin - d[m(u + gz + 2 + V² Hydraulic cylinder 2)1 When dQ=0, KE and PE also become zero and equation becomes much simpler d[m(u + gz- = 0-0 +0+dWenergy flow g + +- Z loutdmout + dQ + dwe energy flow system Calc: work necessary to raise the rack & car by 1 ft (neglect PE of water. No heat transfer, H20 water is pumped from the reservoir to the vehicle, rack and piston. So pump is doing work on the H20 but H20 is also doing work on the piston, rack and automobile?.
Transcribed Image Text:3. A hydraulic lift is shown below. The combined mass of the piston, rack, and car is 4000 lbm. The working fluid is water. There is no heat transfer to or from the water, and the internal energy of the water per unit mass is constant. The water may be considered incompressible. (a) Taking all the water in the reservoir, line, and hydraulic cylinder as the system (i.e., taking the closed-system approach), calculate the work necessary to raise the rack and car 1 ft (neglect the change in potential energy of the water in the system). (b) Repeat part (a), taking all the water plus the car and the rack as the system. (c) Repeat part (a), taking an open-system approach; choose as your system the volume of the hydraulic cylinder, excluding the piston, rack, and car. If the absolute pressure in the system is 1000 lbf/in², calculate the volume that must flow in to raise the car 1 ft. Reservoir A) Recall Energy Balance Equation. d[m(u +92 + 1 + gz 7) system Pump =d[m (u + gz ++)lindmin - d[m(u + gz + 2 + V² Hydraulic cylinder 2)1 When dQ=0, KE and PE also become zero and equation becomes much simpler d[m(u + gz- = 0-0 +0+dWenergy flow g + +- Z loutdmout + dQ + dwe energy flow system Calc: work necessary to raise the rack & car by 1 ft (neglect PE of water. No heat transfer, H20 water is pumped from the reservoir to the vehicle, rack and piston. So pump is doing work on the H20 but H20 is also doing work on the piston, rack and automobile?.
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