Question

A circular tank (diameter = 1.5 m) holds water with height of water in it is h=4.5 m, what is the resulting hydrostatic force (kN) on the bottom surface? Recall that force = PA. Water density = 1000 kg/m Select one: O a. F=78.01 kN. O b. F=112.68 KN. O c. F=134.92 kN. O d. F=60.67 kN. O e.F=95.35 kN. What is meant by simple compressible system in the study of thermodynamics? Select one: a. A simple mechanism that includes components like springs and rubber O b. A matter being under investigation that must satisfy the equilibrium conditions O c. Any matter being under investigation including external effects such as magnetic. O d. A mechanism that includes logic controls and linkages to perform certain tasks. O e. Any matter being under investigation without including external effects such as gravity.

Answer 1

1) Diameter of tank= 1.5m

Height of tank= 4.5m

Area of tank at bottom = pi×(diameter)²/4= 3.1428× 1.5×1.5/4 = 1.7678 m²

pressure at bottom surface = density_water×height×g= 1000×9.81×4.5= 44145 Pa

So Resulting hydrostatic force at bottom = P×A= 44145×1.7678 = 78040.5 N= 78.04 KN

Option A is the nearest value of the above answer.

So option a is correct. F= 78.01 KN

2)

A system is called a simple compressible system in the absence of electrical, magnetic, gravitational, motion and surface tension effects.

At an equilibrium state, the values of the intensive properties are uniform throughout the system.

For above cause option a, c, d are discarded.

Option b is not true as equilibrium is a special condition for a simple compressible system. For all conditions it need not to be in equilibrium.

Option e is correct as it states the fact that it should not be applied with external forces like gravity.

So option e is correct.

All the best?.