Hydraulic principles: pressure, flow and force

Priya Sandhu, Northcrest's senior adjuster, ran the evening block for second-year apprentices again this term, and she opened it with a pressure gauge on the classroom table. "Every hydraulic elevator you will ever adjust obeys three rules," she said. "Pressure is force divided by area. Flow is area times speed. Power is pressure times flow. If you can do those three calculations in your head, you can tell from a gauge reading whether a car is overloaded, whether a pump is worn and whether a valve is set right." This lesson works through those rules with the numbers from Holloway car 2, and adds the losses that turn some of the pump's work into heat on the way to the jack.

Pressure in a liquid

Pressure is force spread over an area. A liquid in a closed system passes pressure equally in all directions, a principle named after Blaise Pascal, who described it in the seventeenth century. If a pump pushes oil into a pipe connected to a jack, the pressure in the pipe, the valve block and the cylinder is the same everywhere, apart from small differences caused by flow and by the height of the oil. The pressure that develops is whatever is needed to hold up the load on the plunger. The pump does not decide the pressure; the load does. With no load, the pressure would be almost zero; with the car full, it rises.

Here p is the pressure, F the force on the plunger and A the area of the plunger. The area that matters is the area of the plunger's full outside diameter, because the oil presses on the plunger's whole cross-section, even if the plunger is a hollow tube. For Holloway car 2, the plunger is 6 in in diameter, so its area is 28.27 in².

The force on the plunger is the weight of everything it carries: the car and sling, the plunger itself, because a direct-acting plunger is lifted with the car, and the load in the car. With the car empty, the plunger carries 5,000 + 1,500 = 6,500 lb. With the full rated load of 4,500 lb it carries 11,000 lb.

# Jack pressure for Holloway car 2, empty and with rated load
import math
d = 6.0                      # plunger diameter, in
area = math.pi * d ** 2 / 4  # in2
car_sling = 5000            # lb
plunger = 1500
rated = 4500
for name, load in (("empty car", 0.0), ("half load", rated / 2), ("rated load", rated), ("125 % load", 1.25 * rated)):
    force = (car_sling + plunger + load) * 1      # lb
    p = force / area
    print(f"{name:<11} force {force:>9,.0f} lb   pressure {p / 1:6.0f} psi")

empty car   force     6,500 lb   pressure    230 psi
half load   force     8,750 lb   pressure    309 psi
rated load  force    11,000 lb   pressure    389 psi
125 % load  force    12,125 lb   pressure    429 psi

The results are the static pressures, the pressures with the car standing still. The empty car needs about 230 psi and the fully loaded car about 389 psi. The code calls the full-load static pressure the working pressure, and many other numbers are set from it: the setting of the pressure relief valve, the test pressures for the cylinder and piping, and the rating of the hose. Ray keeps the empty-car and full-load pressures for every hydraulic on Route 7 in his notebook, because a gauge reading that does not match the expected value is often the first clue to a problem.

Several pressure units are in use. In the United States, gauges, valve data and code values for hydraulic elevators are usually given in pounds per square inch, psi. Some imported equipment uses the bar, about 14.5 psi, or the megapascal, about 145 psi. Gauges read gauge pressure, the pressure above atmospheric pressure, sometimes written psig. A gauge reads the pressure above atmospheric pressure, which is what matters for the forces on a jack.

Flow and speed

Flow is the volume of oil moving per unit of time. To raise the plunger at a given speed, the pump must fill the space the plunger leaves behind at the same rate. The volume of that space per second is the plunger area times its speed. The relation between flow and speed is the second of Priya's rules.

For Holloway car 2, with a plunger area of 28.27 in² and a rated speed of 100 ft/min, the flow at rated speed is 28.27 × 1,200 = 33,900 in³/min, and since a US gallon is 231 in³, about 147 gpm. The power unit's pump is chosen to deliver at least that flow at the working pressure, and the valve's flow rating must match.

The relation also works the other way. If the car runs slower than rated speed on an up run, either the pump is delivering less flow than it should or some of the flow is escaping somewhere else, usually back to the tank through a valve that is not closing fully. A worn pump delivers less flow at high pressure than at low pressure, because more oil slips past its internal clearances, so a car that runs normal speed empty but slows down when loaded points to the pump. A timed run with a stopwatch, empty and loaded, is one of the simplest tests a mechanic can make.

On a direct-acting jack, the car moves at the plunger's speed. On a roped hydraulic, with 2:1 roping, the car moves at twice the plunger speed, but the plunger carries twice the load. The flow and pressure for the same car are therefore the same as for a direct-acting jack of twice the stroke, apart from the extra friction of the sheave and ropes, which is why a roped hydraulic's jack is larger in diameter than a direct jack for the same car.

Power and energy

Hydraulic power is the rate at which the oil does work. It equals the pressure times the flow, Priya's third rule, and it is the power the pump must deliver to the oil. The motor must supply more than this, because the pump and motor are not perfectly efficient.

With the car fully loaded going up, the hydraulic power for Holloway car 2 is 389 psi × 147 gpm. In US units the hydraulic power in horsepower is psi × gpm / 1,714, so the power is 389 × 147 / 1,714 = 33.3 hp. With a pump and motor efficiency of about 80 percent, the motor needs about 42 hp, and the new power unit's motor is rated 40 hp, which is enough because hydraulic elevator motors are rated for the short, intermittent runs of elevator service rather than continuous running.

# Pump flow, hydraulic power and motor power for Holloway car 2
import math
area = math.pi * 6.0 ** 2 / 4        # in2
speed = 100                          # rated speed, ft/min
flow = area * speed * 12 / 231          # gpm
eta = 0.80                           # pump and motor together
for name, p in (("empty car", 230), ("rated load", 389)):
    hyd = p * flow / 1714
    print(f"{name:<10} flow {flow:5.0f} gpm  hydraulic power {hyd:5.1f} hp  motor input {hyd / eta:5.1f} hp")
# energy delivered to the oil in one full-load up trip
travel = 40.0
t_run = travel / speed * 60
print(f"one full-load up trip takes about {t_run:.0f} s at rated speed")

empty car  flow   147 gpm  hydraulic power  19.7 hp  motor input  24.6 hp
rated load flow   147 gpm  hydraulic power  33.3 hp  motor input  41.7 hp
one full-load up trip takes about 24 s at rated speed

The pump's flow depends very little on pressure, because a positive-displacement pump pushes a fixed volume of oil each revolution. The motor current, in contrast, rises with pressure, because the power the motor delivers is proportional to the pressure. That gives the mechanic two easy readings on an up run: the car speed, which shows the flow, and the motor current, which shows the pressure. A car that runs at rated speed with the motor current above the nameplate is overloaded or has a high back-pressure; a car that runs slow with normal current has lost flow.

Losses: friction in pipes and valves

Oil moving through pipes, fittings, valves and hoses loses pressure to friction. When the car is going up, the pump must produce the static pressure needed to hold the load plus the pressure lost on the way to the jack, so the gauge on the power unit reads higher while the car is running up than when it is standing. Going down, the losses work the other way: the pressure at the power unit is lower than in the cylinder. These differences are called the running pressures, and the difference between running and static pressure is a measure of the losses.

The pressure lost in a pipe rises with the square of the oil velocity, so the pipe size matters. A rule of thumb used by many manufacturers is to keep the oil velocity in the main line at no more than about 13 to 16 ft/s. Above that, the losses, noise and heating grow quickly and the valve's control becomes rougher. The velocity in the pipe is the flow divided by the pipe's inside area.

This is the Darcy-Weisbach equation. The pressure drop depends on a friction factor f, the length of pipe L, its inside diameter D, the oil density ρ and the velocity v. The friction factor in turn depends on whether the flow is smooth and layered, called laminar, or turbulent. A dimensionless number, the Reynolds number, decides which: below about 2,000 the flow is laminar, above about 4,000 it is turbulent. Hydraulic oil is much more viscous than water, so elevator pipe flows are often near the boundary.

# Oil velocity and friction loss in the main line to the jack (illustrative)
import math
flow = 147                 # gpm
d_in = 2.469                   # inside diameter of the pipe, in (2-1/2 in schedule 40)
length = 60.0                  # pipe length, power unit to jack, ft
nu = 46e-6                    # kinematic viscosity, m2/s (an ISO VG 46 oil at 40 C)
rho = 870.0                   # kg/m3
q = flow * 3.785e-3 / 60                # m3/s
d = d_in * 0.0254                # m
v = q / (math.pi * d ** 2 / 4)
re = v * d / nu
f = 64 / re if re < 2000 else 0.316 / re ** 0.25
dp = f * (length * 0.3048) / d * rho * v ** 2 / 2       # Pa
print(f"velocity {v / 0.3048:.2f} ft/s, Reynolds number {re:,.0f}")
print(f"friction factor {f:.3f}, pressure loss in the straight pipe {dp / 6895:.1f} psi")
print("fittings, the valve and the jack inlet add more; the total is found from the running pressure")

velocity 9.85 ft/s, Reynolds number 4,093
friction factor 0.040, pressure loss in the straight pipe 6.6 psi
fittings, the valve and the jack inlet add more; the total is found from the running pressure

The friction loss in the straight pipe at Holloway is small compared with the working pressure, which is the aim. The valve itself takes the biggest share of the losses, especially on the down run, where the valve deliberately throttles the flow to control the speed. Every loss turns pressure into heat in the oil.

Head pressure and the height of the oil

A column of liquid exerts a pressure at its base proportional to its height and density. For hydraulic oil, a column 1 ft high exerts about 0.38 psi. The difference is small compared with the working pressure, but it matters in two places. When a power unit is on a floor above the pit, as at Dunmore Square, the gauge at the power unit reads a little less than the pressure at the cylinder because of the oil column in the pipe. And when a mechanic opens a fitting at the bottom of a system, the column of oil above it keeps pushing oil out even with the pump stopped and the car landed on its stops, so isolation must include closing the shut-off valve and draining or lowering the pressure.