Safe simulation of hydraulic systems

A hydraulic system is modelled from its drawing: first find the piston areas $A_1$ and $A_2$ from the diameters, then apply Pascal's law — pressure in a confined liquid at rest is transmitted unchanged, so $p_1 = p_2 = p$. This gives the basic relation of a hydraulic machine:

$$\frac{F_1}{A_1} = \frac{F_2}{A_2}.$$

The force gain equals the area ratio, $F_2/F_1 = A_2/A_1$, but it is paid for by a loss in displacement: liquid volume is conserved, so $A_1 s_1 = A_2 s_2$, and the work is unchanged, $F_1 s_1 = F_2 s_2$. Safe simulation means that all these steps are done on paper, in a spreadsheet or in a simulator, without building a real pressurised rig. The final step is a strength check: the calculated pressure $p$ is compared with the allowable $p_{\max}$ and the safety factor $k = p_{\max}/p$ is computed; it must be at least the required value (usually $k \ge 1.5$).

In a hydraulic press drawing the small piston has area $A_1 = 5\ \text{cm}^2$ and the large piston $A_2 = 250\ \text{cm}^2$. A force $F_1 = 200\ \text{N}$ is applied to the small piston. Find the pressure in the liquid $p$ and the force $F_2$ on the large piston.

A hydraulic jack has $A_1 = 2\ \text{cm}^2$ and $A_2 = 40\ \text{cm}^2$. In one stroke the small piston moves down by $s_1 = 20\ \text{cm}$. 1) How far does the load rise? 2) What is the force gain? 3) Verify the work equality.

Before building a rig, a student analyses a press on paper only: $A_1 = 4\ \text{cm}^2$, $A_2 = 200\ \text{cm}^2$, the relief valve opens at $p_{\max} = 6.0\ \text{MPa}$, the operator pushes with at most $250\ \text{N}$, and the required safety factor is $k \ge 1.5$. (a) What pressure does the operator create? (b) What is the largest force the press can develop with this drive? (c) Is lifting a load $F_2 = 8\ \text{kN}$ safe?

Safe self-study plan. Do not assemble or load a pressurised hydraulic rig — instead model the system in a spreadsheet or in an online hydraulic press/lift simulator.

1. Build a table with separate cells for $A_1$, $A_2$, $F_1$, $s_1$, $p_{\max}$ and the required safety factor $k_{\text{req}}$.
2. Add formulas: $p = F_1/A_1$, $F_2 = p A_2$, $s_2 = A_1 s_1/A_2$, $k = p_{\max}/p$, and a "safe/unsafe" verdict via a logical function.
3. Run the model on three data sets: (1) $A_1 = 5\ \text{cm}^2$, $A_2 = 250\ \text{cm}^2$, $F_1 = 200\ \text{N}$; (2) $A_1 = 2\ \text{cm}^2$, $A_2 = 40\ \text{cm}^2$, $s_1 = 20\ \text{cm}$; (3) the same press with $A_2$ doubled.
4. For each set, record whether the relief valve would open, whether $k \ge k_{\text{req}}$, and how $F_2$ and $s_2$ changed.

Finish with a 3–4 sentence conclusion: which piston area affects safety more strongly, and why.

Show that pressure is the same throughout a liquid at rest and that the force gain $A_2/A_1$ is necessarily accompanied by a displacement loss $s_2/s_1 = A_1/A_2$.

A schematic cross-section of two connected cylinders: a small piston of area $A_1$ and a large piston of area $A_2$; an arrow $F_1$ downwards at the small piston and an arrow $F_2$ upwards at the large one; identical pressure labels $p$ at both pistons; stroke markers $s_1$ and $s_2$; dashed liquid-level lines before and after the stroke, showing that volume is conserved. Beside it, a calculation flow chart: "drawing → $A_1, A_2$ → $p = F_1/A_1$ → $F_2 = p A_2$ → check $k = p_{\max}/p$".

A mountain biker squeezes the brake lever with force $F_1=120\,\mathrm{N}$. The master-cylinder piston area is $A_1=1.5\,\mathrm{cm^2}=1.5\times10^{-4}\,\mathrm{m^2}$. The total caliper-piston area is $A_2=6.0\,\mathrm{cm^2}=6.0\times10^{-4}\,\mathrm{m^2}$. In a safe bench calculation, the pressure is $p=F_1/A_1=120/(1.5\times10^{-4})=8.0\times10^5\,\mathrm{Pa}=0.80\,\mathrm{MPa}$, so the caliper force is $F_2=pA_2=0.80\times10^6\times6.0\times10^{-4}=480\,\mathrm{N}$. The brake line has a burst pressure $p_{\mathrm{burst}}=8\,\mathrm{MPa}$, giving a safety factor $p_{\mathrm{burst}}/p=10$. The oil column from lever to caliper is about $h=0.3\,\mathrm{m}$, adding only $\rho gh\approx850\times9.81\times0.3=2.5\times10^3\,\mathrm{Pa}=0.0025\,\mathrm{MPa}$, which is negligible beside $0.80\,\mathrm{MPa}$.

Pascal's law is used twice: the same pressure $p=F_1/A_1$ acts in the closed fluid, and the output force is $F_2=pA_2$; hydrostatics contributes the height term $p_{\mathrm{hydro}}=\rho gh$ between lever and caliper. The safe calculation compares the working pressure with the line burst pressure through $p_{\mathrm{burst}}/p$.

A driver presses the brake pedal with force $F_1=200\,\mathrm{N}$ on a master-cylinder piston of area $A_1=5.0\,\mathrm{cm^2}=5.0\times10^{-4}\,\mathrm{m^2}$. The simulated pressure is $p=F_1/A_1=200/(5.0\times10^{-4})=4.0\times10^5\,\mathrm{Pa}=0.40\,\mathrm{MPa}$. Each front caliper piston has area $A_2=25\,\mathrm{cm^2}=2.5\times10^{-3}\,\mathrm{m^2}$, so the clamping force is $F_2=pA_2=4.0\times10^5\times2.5\times10^{-3}=1000\,\mathrm{N}$. The steel brake line is rated at $p_{\mathrm{rated}}=10\,\mathrm{MPa}$ with safety factor $2$, so the allowed working pressure is $5\,\mathrm{MPa}$; $0.40\,\mathrm{MPa}$ passes the check. A height difference $h=0.5\,\mathrm{m}$ between master cylinder and caliper gives $\rho gh\approx900\times9.81\times0.5=4.4\times10^3\,\mathrm{Pa}=0.0044\,\mathrm{MPa}$.

Pascal's law supplies the transmitted pressure $p$ and the force multiplication $F_2=pA_2$; hydrostatics gives the small height correction $\rho gh$ that the simulation adds for the caliper position. The safety check uses the working pressure against the rated pressure divided by the safety factor.

An excavator boom cylinder has bore diameter $D=80\,\mathrm{mm}$, so $A=\pi D^2/4=\pi(0.080\,\mathrm{m})^2/4=5.03\times10^{-3}\,\mathrm{m^2}$. The pump supplies $p=20\,\mathrm{MPa}=20\times10^6\,\mathrm{Pa}$. The simulated cylinder force is $F=pA=20\times10^6\times5.03\times10^{-3}=1.01\times10^5\,\mathrm{N}\approx101\,\mathrm{kN}$, enough to hold a mass $m=F/g\approx1.01\times10^5/9.81\approx10.3\,\mathrm{t}$. The hose has burst pressure $p_{\mathrm{burst}}=80\,\mathrm{MPa}$, so the safety factor is $80/20=4$. An oil column of height $h=2.0\,\mathrm{m}$ adds $\rho gh\approx900\times9.81\times2.0=1.8\times10^4\,\mathrm{Pa}=0.018\,\mathrm{MPa}$, small compared with $20\,\mathrm{MPa}$.

Pascal's law lets the simulation use the pump pressure $p$ unchanged in the cylinder and compute $F=pA$; hydrostatics adds $p_{\mathrm{hydro}}=\rho gh$ for the vertical hose run. The safety part checks the calculated working pressure against the burst pressure with the factor $p_{\mathrm{burst}}/p$.