## QUESTION:

A 3-meters height hydraulic elevator consists of electric motor on the

Yi

Ki

Ci

Fi

M2

Mi

## Quarter car suspension

# Spool valve

ground, oil tank and 4 meters long, 25 mm diameter cast iron pipes on the ground, oil (hydraulic stiffness = 1.5 x 107 N/m, kinetic vis- cosity = 100 cSt, specific gravity S.G= 0.9, damping coefficient 1000 N.s/m), hydraulic pump (efficiency = 85%) on the ground near the tank, hydraulic linear actuator (150 mm bore and 100 mm rod di- ameter) on the ground and cabin (500 kg mass with 0.1 m/s required speed), determine:

a- The volume of the tank (best design)

b- Pressure and flow rate in the actuator

c- Pressure and flow rate of the pump (include friction losses) d- Power of the electric motor

e- Draw block diagram to represent the whole system with pump dis- placement as the main input and the actuators displacement as the output

f- Find the transfer function between the pumps displacement and

actuators displacement

g- Draw Bode plot (by hand) for the transfer function in the previous point and find the natural frequency and damping ratio

h- Draw the hydraulic circuit with all necessary components in sym- bolic form

## ANSWER:

**a) Determine the volume of the tank (best design)**

The volume of the tank depends on the desired speed and acceleration of the elevator. A larger tank will provide more capacity for the hydraulic fluid, which can improve the speed and acceleration of the elevator. However, a larger tank will also be more expensive and heavier.

A good rule of thumb is to use a tank that is about 2-3 times the volume of the hydraulic fluid in the system. In this case, the volume of the hydraulic fluid can be calculated as follows:

```
Volume of hydraulic fluid = Area of piston * Stroke length
```

Where:

- Area of piston = πr² = π(0.15 m)² = 0.0707 m²
- Stroke length = 3 m

Therefore, the volume of the hydraulic fluid is 0.212 m³.

Using a rule of thumb of 2-3 times the volume of the hydraulic fluid, the volume of the tank should be:

```
Volume of tank = 2-3 * Volume of hydraulic fluid
```

Therefore, the volume of the tank should be between 0.424 m³ and 0.636 m³.

**b) Determine the pressure and flow rate in the actuator**

The pressure and flow rate in the actuator can be calculated using the following equations:

```
P = F/A
Q = V/t
```

Where:

- P = pressure (Pa)
- F = force (N)
- A = area of piston (m²)
- Q = flow rate (m³/s)
- V = volume of fluid displaced by the piston (m³)
- t = time (s)

The force required to move the cabin is equal to the weight of the cabin plus the frictional force. The frictional force can be calculated using the following equation:

```
F_friction = μv
```

Where:

- F_friction = frictional force (N)
- μ = coefficient of friction (N·s/m²)
- v = velocity (m/s)

The coefficient of friction between cast iron and oil is approximately 0.1. Plugging in the values, we get:

```
F_friction = (0.1)(0.1) = 0.01 N
```

The total force required to move the cabin is:

```
F = m*g + F_friction = (500 kg)(9.81 m/s²) + (0.01 N) = 4950.5 N
```

Plugging in the values for the actuator, we get:

```
P = F/A = (4950.5 N) / (0.0707 m²) = 70000 Pa
Q = V/t = (0.15 m)(0.1 m/s) / 0.001 s = 15 m³/s
```

Therefore, the pressure in the actuator is 70,000 Pa and the flow rate is 15 m³/s.

**c) Determine the pressure and flow rate of the pump (include friction losses)**

The pressure and flow rate of the pump can be calculated using the following equations:

```
P_pump = P_actuator + ΔP_friction
Q_pump = Q_actuator
ΔP_friction = f(Q, ρ, v, d)
```

Where:

- P_pump = pump pressure (Pa)
- P_actuator = actuator pressure (Pa)
- ΔP_friction = pressure loss due to friction (Pa)
- Q_pump = pump flow rate (m³/s)
- Q_actuator = actuator flow rate (m³/s)

The pressure loss due to friction can be calculated using the Darcy-Weisbach equation:

```
ΔP_friction = fLρv²/2d
```

Where:

- f = friction factor
- L = length of pipe (m)
- ρ = density of fluid (kg/m³)
- v = velocity of fluid (m/s)
- d = diameter of pipe (m)

The friction factor can be calculated using the Moody chart. For a cast iron pipe with a diameter of 25 mm and a Reynolds number of 10,000, the friction factor is approximately 0.04.

Plugging in the values, we get:

`ΔP_friction = (0.04)(4 m)(0.9)(15 m/s)² / (2)(0.025 m) = 8640 Pa`