Question

A hydraulic pump can produce a pressure of 500kPa. A system with 4 cylinders must support a load of 125,000 pounds. What diameter must the cylinders be used go maintain this load (assume 1/4" increments)

I started on the problem could someone tell me if it is right or wrong. If its wrong could you please explain why?

125,00016(cond) 500KPA P= F M² 125000 bS x 4.45N = 556250N Lo Force 500 haxlo00D0 - so9000 Pa Pressure 556.250N =11125 m² 500,000 P. + VD = V87m IxD? = 1.1125m² X T/4 %3D lme 39.37in .93m/4 =.23m 9 '4 diameler Pipe

Answer 1

Sol.

Review about the given solution:-

You initiated the solution in the correct way, but after finding the total area required ( ie. 1.1125m), the area must be divided by 4, instead of the diameter, if the diameter of all the cylinders is the same. And its also provided in the problem

Note: It is not exactly clear in the problem that they are talking about which increment. Thus the solution will be provided for two case

1. All cylinders with the same diameter.
2. Cylinders with increasing diameter.

The data provided in the problem are:-

The pressure produced by the pump = 500 kPa, Load = 125000 lbs = 556250 N, There are 4 supporting cylinders.

Now,

The pressure is defined as the force applied per unit area.

$\implies$ P = F/A

$\implies$ A= F/ P

$\therefore$  A = 556250 / 500000

A = 1.1125 m²

case 1). If all the cylinders have the same diameter ( D)

A = 4 * ( $\pi$ / 4) * D²

$\implies$ 1.1125 = 4 * ( $\pi$ / 4) * D²

$\implies$ D² = 0.354

$\therefore$ D = .595 m = 23.45 in

case 2). If there is an increment of 1/4" ( 0.00635 m) in each cylinder

The diameters of the cylinders will be given by:

D₁ = D, D₂ = D + 0.00635,  D₃ = D + 0.0127, D₄ = D + 0.01905

$\therefore$ Total area = Area 1 + Area 2 + Area 3 + Area 4

A = ( $\pi$ / 4) * { D² + ( D + 0.00635)² + ( D + 0.0127)² + ( D + 0.01905)²}

4D² + 0.0762 D + (5.64511*10^-4) = 1.1125 / ( $\pi$ / 4)

4D² + 0.0762 D - 1.41583 = 0

On solving this quadratic equation, we get:-

D = 0.585 and D = -0.0604

Since negative value of diameter is not valid.

$\therefore$ D = 0.585 m = 23.03 in

$\therefore$  D₁ = 23.03 , D₂ = 23.03 + 0.00635,  D₃ = 23.03 + 0.0127, D₄ = 23.03 + 0.01905

D₁ = 23.03 in, D₂ = 23.03635 in,  D₃ = 23.0427 in , D₄ =23.04905 in