power work-done

Question

5.3. the following data refers to a single-acting hydraulic press: Diameter of the ram piston= 380mm Diameter of the plunger piston= 60mm The stroke length of the plunger= 126mm Mechanical advantage of the plunger= 40 Calculate the following: 5.3.1. the effort that must be applied to the handle to lift a load of 3Mg 5.3.2. the distance the load will be raised after 120 pumping strokes of the plunger 5.3.3. the work done by the press if the efficiency is 92% 5.4. calculate the power required to pump 32m^2 of water per minute to a reservoir 14.7m above the water level if the efficiency of the pump is 85%. (the density of water is 1000 kg/m^3.)

Answer 1

(5.3.1)

Diameter of the Ram Piston = 380 mm.

Diameter of the Plunger Piston = 60 mm.

Stroke length of the Plunger = 126 mm.

Mechanical advantage on the lever = 40

Mass of the load = 3 Mg = 3000 kg

We know that Mechanical advantage = Weight lift by the Plunger / Force applied on the plunger.

Force applied on the plunger = Weight lift by the Plunger / Mechanical advantage.

F = 3000 / 40

F = 75 kN.

Force applied in order to lift the load F = 75 kN.

Answer 2

(5.3.2)

Diameter of the Ram Piston = 380 mm.

Radius of the Ram Piston = (380/2) = 190 mm = 0.19 m.

Diameter of the Plunger Piston = 60 mm.

Radius of the Ram Piston = (60/2) = 30 mm = 0.03 m.

Stroke length of the Plunger = 126 mm = 0.126 m.

Mechanical advantage on the lever = 40

Number of Pumping Stroke = 120.

Area of the Ram = πr² = π(0.19)² = 0.1134 m²

Area of the Plunger = πr² = π(0.03)² = 2.827 * 10^-3 m²

Let the distance travelled by the weight per stroke = x mts.

Volume displaced by ram = Area of the ram * Distance

Volume displaced by ram = 0.1134 x m³.

Volume displaced by Plunger = Area of the Plunger * stroke length

Volume displaced by Plunger = 2.827 * 10^-3 * 0.126

Volume displaced by Plunger = 3.09204 * 10^-4 m³

As we know that volume of the plunger and ram are equal when there is no slip.

3.09204 * 10^-4  = 0.1134 x

x = 0.002727 per stroke.

Distance travelled for 120 mm pumping stroke = 0.002727* 0.12

Distance travelled = 3.272 * 10^-4 m.

Distance travelled = 0.3273 mm.

Distance travelled to lift the wieght = 0.3273 mm.

Answer 3

(5.3.3)

The diameter of the Ram = 380 mm.

Diameter of the Plunger = 30 mm

Number of strokes = 120 mm

Stroke length of the Plunger = 126 mm =0.126 m

Efficiency = 92%

Radius of the Ram = 380 / 2 = 190 mm = 0.19 m

Radius of the Plunger = 60 / 2 = 30 mm = 0.03 mm

Area of the Ram = πr² = π(0.19)²

Area of the Ram = 0.1134 m²

Area of the Plunger = πr² = π(0.03)² = 2.827 * 10^-3 m².

Let the distance travelled by the weight per stroke = x mts.

Volume displaced by ram = Area of the ram * Distance

Volume displaced by ram = 0.1134 x m³.

Volume displaced by Plunger = Area of the Plunger * stroke length

Volume displaced by Plunger = 2.827 * 10^-3 * 0.126

Volume displaced by Plunger = 3.09204 * 10^-4 m³

As we know that volume of the plunger and ram are equal when there is no slip.

3.09204 * 10^-4  = 0.1134 x

x = 0.002727 per stroke.

Distance travelled for 120 mm pumping stroke = 0.002727* 0.12

Distance travelled = 3.272 * 10^-4 m.

Distance travelled = 0.3273 mm.

Distance travelled to lift the wieght = 0.3273 mm.

Work done = weight lift * Distance travelled

Work done = 3000 * 0.3273 = 0.9819 Nm.

Efficiency = (Output work/ input work) * 100

Input work = Output work / (Efficiency / 100)

Pin = 0.9819 / (92/100)

Work done  = 1.067 Nm.

Work done by the press to lift the load = 1.067 Nm.

Answer 4

(5.4)

Efficiency of the pump = 85%

Flow rate = 32 m² per minute.

Flow rate = 47820.2467 gallons per min.

Height of the reservoir = 14.7 m.

Power required to pump

Where Q is the flow rate

µ is the efficiency = 85% = 0.85

P is the Pressure P = ρgh

P = 1000 * 9.8 * 14.7

P = 144060 Pa.

P = 20.89 psi

Power required to pump

P = 675.76 BHP

P = 503.9 kW

Therefore the power required to pump is 503.9 kW.