An elevator operates on a very slow hydraulic jack. Oil is pumped into a cylinde

Question

An elevator operates on a very slow hydraulic jack. Oil is pumped into a cylinder that contains the piston. The piston lifts the elevator as the internal cylinder pressure (P2) rises in comparison to the outside pressure (P1=101,325 Pa or about 14.7 psi). If the piston can lift 22 kN (about 5000 lb) a vertical distance of 6 m (about 20 ft) in 30 seconds then:

a. Calculate the output power of the piston while it is lifting its maximum load

b. If the efficiency of the piston is 0.9, what is the input power required?

c. What must be the maximum internal pressure (P2) if the total volume of oil required is 0.150 m^3?

Explanation / Answer

Sorry that the equations are not done in the equation editor, but it kept crashing on me and I wanted to get you an answer as soon as possible.

This question can be solved by looking at the definition of power. The units of power are Force*Distance/Time or in this case, N*m/s (W). Since the piston can lift a force of 22 kN a distance of 6 m in 30 seconds, the work done by the system is

P = (22 kN)*(6 m)/(30 s) = 4.4 kW (1)

where P is the power output.

a. 4.4 kW (ans)

Part b takes into account the efficiency of the piston. The efficiency is equal to

= Pout/Pin (2)

where is the efficiency, Pout is the power the hydraulic jack releases into the system, and Pin is the total power used to run the jack. Therefore, using the constants = 0.9, and Pout = 4.4 kW, Pin is calculated to be 4.89 kW.

b. 4.9 kW (ans)

For part c, using gauge pressure is a simpler approach, as atmospheric conditions can be assumed to have a pressure of 0 Pa. Knowing that volume is equal to the cross-sectional area of a the cylinder multiplied times the height, the cross-sectional area can be calculated using the equation

V = A*H (3)

where V is the volume needed, A is the cross-sectional area, and H is the needed head of the hydraulic jack. Using the constants V = 0.150 m^3, and H = 6 m, the cross sectional area is calculated to be 0.025 m^2. Knowing that a pressure is equal to a Force/Area, taking the maximum force and dividing it by this cross-sectional area will give the maximum internal pressure.

Pi = (22 kN)/(0.025 m) = 880 kPa gauge

or Pi = 981 kPa absolute.

c. 880 kPa gauge, or 981 kPa abs.

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