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?
To answer these questions, we will use the following formulas:
a. Power (P) = Work (W) / Time (t)
b. Efficiency (η) = Output Power / Input Power
c. Boyle's Law: P1V1 = P2V2
Let's start with part (a):
a. Calculate the output power of the piston while it is lifting its maximum load:
We are given the lifting force (F) = 22 kN and the vertical distance (d) = 6 m. The work done by the piston can be calculated as:
Work (W) = Force * Distance
W = 22 kN * 6 m
Now, we need to convert the units to SI (International System of Units):
1 kN = 1000 N
W = (22 kN * 1000 N/kN) * 6 m
W = 132,000 N * 6 m
W = 792,000 J (Joules)
Next, we need to calculate the power. We are given the time (t) = 30 seconds:
Power (P) = Work / Time
P = 792,000 J / 30 s
P = 26,400 W (Watts)
Therefore, the output power of the piston while lifting its maximum load is 26,400 Watts.
Moving on to part (b):
b. If the efficiency of the piston is 0.9, what is the input power required?
Efficiency (η) = Output Power / Input Power
We are given the efficiency (η) = 0.9 and the output power (P) = 26,400 W. We need to rearrange the formula to solve for the input power:
Input Power = Output Power / Efficiency
Input Power = 26,400 W / 0.9
Input Power = 29,333.33 W (Watts)
Therefore, the input power required is 29,333.33 Watts.
Finally, let's move on to part (c):
c. What must be the maximum internal pressure (P2) if the total volume of oil required is 0.150 m^3?
We are given the total volume of oil (V2) = 0.150 m^3 and the outside pressure (P1) = 101,325 Pa.
To solve for the maximum internal pressure (P2), we can use Boyle's Law:
P1 * V1 = P2 * V2
We need to rearrange the formula to solve for P2:
P2 = (P1 * V1) / V2
We don't have the values for V1, but we can calculate it using the lifting force (F) and the area (A).
Area (A) = Force / Pressure
A = 22 kN / P1 (in SI units)
Now, we can express V1 as V1 = A * d:
V1 = A * d
Substituting the values, we get:
V1 = (22 kN / P1) * 6 m
Now we substitute the values back into Boyle's Law equation:
P2 = (P1 * V1) / V2
P2 = (P1 * (22 kN / P1) * 6 m) / 0.150 m^3
P2 = (22 kN * 6 m) / 0.150 m^3
P2 = 176,000 N / 0.150 m^3
P2 = 1,173,333.33 Pa (approximately)
Therefore, the maximum internal pressure (P2) must be approximately 1,173,333.33 Pa.