A hydraulic jack has two pistons. The diameters of the larger piston and smaller piston are 15 cm and 1.25 cm respectively.It is necessary to lift a 2000 pound car. How large a force must be applied to the small piston? Now assume that a small piston moves through a distance of 10 cm. By what distance will the large piston move?© BrainMass Inc. brainmass.com October 16, 2018, 3:56 pm ad1c9bdddf
The hydraulic jack works on the principle of Pascal's law:
pressure (p1) = pressure (p2)
=>F1/A1 = F2/A2
where F1 and F2 are the forces and A1 and A2 are the corresponding sides cross-sectional ...
The solution shows the formulas and calculations with explanations to arrive at the answer.
2nd Law of Thermodynamics
Please see attachment for original question format.
A combined air filled, spring loaded cylinder has a frictionless piston of area 0.012 m^2 that rests against the spring. The spring loaded end of the cylinder is open to atmosphere. The spring force in terms of piston movement is given by:
Spring Force = k X
Show that i) the change in pressure is proportional to change in volume and spring position X such that:
dP/dV = k/A^2 where A is the piston cross section area
The air temperature is increased by means of an electric heater so that the piston moves by 0.04 m with the pressure in the cylinder rising to 5 bars. The piston. The cylinder is then allowed to cool down to a temperature of 300K at which the pressure in the cylinder has dropped to 1 bar. ii) Calculate the volumetric change in the cylinder due to heating. If the air in the cylinder had an initial volume of 0.0005 m^3, iii) calculate the mass of air.
The cylinder is heated again so the volume becomes 1.3 times the original volume. Calculate iv) the final pressure, v) final temperature, vi) the work done if it is assumed that when the pressure-volume graph is extended, it would pass through origin, vii) the change in internal energy and the heat transfer.
Note: Cp = 1005J/kgK , Cv = 718J/kgK for air.View Full Posting Details