Hydraulics: The diagram shows two cylindrical pipes of diameters 4 cm and 1 cm,

Question

Hydraulics: The diagram shows two cylindrical pipes of diameters 4 cm and 1 cm, connected by a conical flange. Each pipe has a cylindrical piston in it. The fluid is incompressible, and has a density of 1.5 g/cm^3. Use g = 10 N/kg = 10 m/s^2.
a.) If you exert a force of 10 kN on the large piston, how much force does the fluid exert on the small piston?
b.) How much pressure does the large piston exert on the fluid?
c.) If you push the large piston 1 cm, how far will the small piston move?
d.) If you remove the pistons and flow fluid through the pipe, and the fluid moves through the large pipe at 10 cm/s, how fast will it be moving in the small pipe?
e.) If the fluid flows through the large pipe under a pressure of 2 kPa, what pressure will it have as it flows through the small pipe?

        --------------------------------

   []                                ----------------

F1 ---> []   4 cm                               [] 1 cm ---> F2

   [] fluid filled              ----------------

                 [] <----conical flange (also above the 1 cm pipe).

    --------------------------------          

                                                              supposed to be a diagonal line connecting bottom 4 cm pipe to

                                                              bottom 1 cm pipe. and a diagonal line (conical flange) connecting

                                                              top of 4 cm pipe with top of 1 cm pipe. ( [] = the cylindrical piston)

                                                              Sorry the diagram drawer wouldn't work for me..

Explanation / Answer

The diameter of the cylindrical pipe is d1 = 4 cm The area of the larger pipe A1 = r^2                                               = (2*10^-2m)^2 = 1.25*10^-3 m^2 The diameter of the cylindrical pipe is d2 = 1 cm The area of the larger pipe A2 = r^2                                               = (0.5*10^-2m)^2 = 7.855*10^-5 m^2                                               = (0.5*10^-2m)^2 = 7.855*10^-5 m^2 The density of the fluid is = 1.5 *10^3 kg/m^3 (a) From pascal law, the pressure is same through out the system so                  F1/ A1 = F2 /A2                  F2 = (A2/A1) F1                        = [(7.85*10^-5)/(1.25*10^-3 m^2 )] (10*10^3 N)                         = 628 N (b) The pressure P = F1/A1                                = 10*10^3 N / (1.25*10^-3 m^2 )                                = 8*10^6 Pa (c) When the pistion moves d1 = 1cm    then the small piston moves a distance                d2 = d1 / (A2/A1)                                        = 1*10^-2 m / [(7.85*10^-5)/(1.25*10^-3 m^2 )]                     = 0.159 m or 16 cm (d) The velocity of liquid at large pipe is v1 = 10cm/s Since the volume rate o flow is same, so by using continuity equation           A1 v1 = A2 v2 therefore the velocity at smaller pipe           v2 = A1 v1 / A2               = (1.25*10^-3 / 7.85*10^-5) (10cm/s)               = 159 cm/s (e) From Bernoullis equation        P1 + 1/2 v1^2 = P2 + 1/2 v2^2 therefore the pressure through the smaller pipe          P2 = P1 + 1/2 (v1^2 - v2^2)               = 2 kPa + (0.5)(1.5 *10^3 kg/m^3) [ (0.1m/s)^2 - (1.59m/s)^2]                          = 2 000 Pa - 1888 Pa = 112 Pa                   

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