Question

https://www.chegg.com/homework-help/questions-and-answers/cuiul-illalliol-13-de-ulima-u--211-pump-driven-1440-rev-min-displacement-125-ml-rev-volume-q40364067?trackid=b6961e1b09e1&strackid=e8792248da1c

Answer 1

Solution;

Given;

Speed of pump;

1440rev/min

Displacement;

12.5ml/rev

Volumetric efficiency;

87%

Cylinder 1;

63mm bore ×35mm rod×250mm stroke

Cylinder 2;

80mm bore ×55mm rod ×150mm stroke

Flow of fluid;

$1440×12.5=18000ml/min$

For cylinder 1;

$V_1=\frac{π}{4}D^2L$

$V_1=\frac{π}{4}×6.3^2×25=779.31cm^3$

Also;

$V_2=\frac{π}{4}(D_b^2-D_r^2)L$

$V_2=\frac{π}{4}×(6.3^2-3.5^2)×25=538.78cm^3$

Since ;

$1cm^3=1ml$

$V_1=779.31ml$

$V_2=538.78ml$

Time take for cylinder 1 to extend and retract is;

$t_{c1}=\frac{779.31+538.78}{18000}=0.0732min$

$t_{c1}=4.3936s$

For cylinder 2;

$V_1=\frac{π}{4}×8^2×15=753.98cm^3=753.98ml$

$V_2=\frac{π}{4}×(8^2-5.5^2)×15=397.61cm^3=397.61ml$

Time take for cylinder 2 to extend and retract is;

$t_{c2}=\frac{753.98+397.61}{18000}=0.06398min$

$t_{c2}=3.8386s$

Chose $t_{c1}$ which is greater since it allows both cylinder to extend and retract fully.

$t_{minimum}=4.3936s$