Figure 18-46 shows a hydraulic lift operated by pumping fluid into the hydraulic system. The large cylinder is 40 cm in diameter, while the small tube leaving the pump is 1.7 cm in diameter. A total load of 2500 kg is raised 2.2 m at a constant rate. Neglect pressure variations with height, and neglect also the weight of the fluid raised.
(car lift on lift)
--------------
____---____ <- lift
..| |
..| |
...\ \__________________
......__________O_______....\
..................[....|....|.....]
................. ________________
------------------^^^^^^^^^^^^^^^^
fluid reservoir, atmospheric pressure (bottom right, above arrows)
---any "......" = fillers, pretend they're just empty space (they're to make the pic work)
---"O" = the pump, to the left of the pump (the "O" in the pic) is pressurized fluid
(a) What volume of fluid passes through the pump?
(b) What is the pressure at the pump outlet?
(c) How much work does the pump do?
(d) If the lifting takes 40 s, what is the pump power?
Pic should be fixed below:
(car lift on lift)
--------------
____---____ <- lift
...|...|
...|...|
...\...\__________________
....__________O_______.....\
..................[....|....|.....]
................. ________________
------------------^^^^^^^^^^^^^^^^
fluid reservoir, atmospheric pressure (bottom right, above arrows)
(car lift on lift)
--------------
____---____ <- lift
.........| |
.........| |
.........\ \__________________
..........__________O_______....\
....................................[....|....|.....]
................. ________________
------------------^^^^^^^^^^^^^^^^
fluid reservoir, atmospheric pressure (bottom right, above arrows)
To solve this problem, we can use the principles of fluid mechanics and work-energy theorem.
(a) To find the volume of fluid passing through the pump, we need to use the principle of continuity, which states that the volume flow rate is constant at all points in a closed fluid system. Mathematically, we can express this as:
A1 * v1 = A2 * v2
Where A1 and A2 are the cross-sectional areas of the large cylinder and the small tube, respectively, and v1 and v2 are the velocities of the fluid in the large cylinder and the small tube, respectively. Rearranging the equation, we have:
v2 = (A1/A2) * v1
Given that the diameter of the large cylinder (D1) is 40 cm and the diameter of the small tube (D2) is 1.7 cm, we can calculate the areas using the formula A = π * (D/2)^2.
A1 = π * (40/2)^2 = 1256.64 cm^2
A2 = π * (1.7/2)^2 = 2.2635 cm^2
Since the fluid passes through the pump at a constant rate, the velocity in the large cylinder (v1) is also constant. Therefore, we can say that the volume flow rate (Q) is equal to the product of the cross-sectional area and the velocity, i.e., Q = A2 * v2.
Substituting the values, we can solve for Q = A2 * ((A1/A2) * v1), and the result will be the volume flow rate passing through the pump.
(b) To find the pressure at the pump outlet, we can apply Pascal's Law, which states that the pressure in a fluid is transmitted uniformly in all directions. This means that the pressure at the pump outlet will be the same as the pressure in the large cylinder.
The pressure in a fluid can be calculated using the formula P = F/A, where P is the pressure, F is the force, and A is the cross-sectional area. In this case, the force exerted by the weight of the load is balanced by the pressure in the large cylinder, so we have:
F = mg, where m is the mass of the load and g is the acceleration due to gravity.
Using the given values of the load (2500 kg), the gravitational acceleration (9.8 m/s^2), and the cross-sectional area of the large cylinder (A1), we can calculate the pressure at the pump outlet.
(c) To calculate the work done by the pump, we can use the equation W = P * ΔV, where W is the work done, P is the pressure, and ΔV is the change in volume. Since the fluid passed through the pump at a constant rate, ΔV will be equal to the volume of fluid passing through the pump, which we calculated in part (a).
(d) To find the pump power, we can use the formula P = W/t, where P is the power, W is the work done, and t is the time taken to lift the load. Given that the lifting takes 40 s, we can calculate the pump power using the work done calculated in part (c) and the time taken.
By following these steps and applying the principles of fluid mechanics and work-energy theorem, you should be able to find the answers to each of the questions.