Comprehensive Report

The ACME Manufacturing Co has been assigned to conduct a hydrostatics and hydraulics report. Acme Manufacturing will collect measurements of the water storage tank and the fire suppression system. The calculations gathered will be used to determine the type of valve prior to the distribution system, the pressure of fluid, the residual pressure and the type of material to be use to make repairs.

Piping System Repair

Vcylinder = (3.14) (0.75)2 x 25 ft

Vcylinder = (3.14) (0.56) x 25 ft

Vcylinder = 43.96 ft

There is 43.96 ft3 of water in this location of the pipe and water weighs 62.4 lb/ft3. Therefore, the weight of the water at the point of repairs is 2,743 Ib. Now, we must calculate the area for the pipe, which is determined by

Area = πr2

Area = (3.14)(9 in)2

Area = (3.14)(81 in)2

Area = 254.34 in2

We can now solve for the pressure, using the following equation:

P = F/a

P = 2,743.10 lb/254.34 in2

P = 10.79 lb/in2

Water Storage Tank

The velocity of the fluid at varying heights.

V = √2gh

= √(2)(32)(24)

= 39.19 ft/s at a water level of 24 feet in height.

V = √2gh

= √(2)(32)(18)

= 33.94 ft/s at a water level of 18 feet in height.

V = √2gh

= √(2)(32)(12)

= 27.71 ft/s at a water level of 12 feet in height.

V = √2gh

= √(2)(32)(6)

= 19.60 ft/s at a water level of 6 feet in height.

Fire Suppression System

v2A/2g + PA/w + zA = v2B/2g + PB/w + zB + hAB’

72ft/s / 2(32.2ft2/s) + 55psi/62lb/ft3 + 0 = 8.52ft/s /2(32.2ft2/s) + PB/62 lb/ft3 + 0 40ft

49ft/s/64.4ft2/s + 55 psi/62lb/ft3(144 in2/ft2)+ 0 = 72.25ft/s/64.4ft2 + PB/62 lb/ft3+0 + 40ft

0.761 + 0.887(144 in2/ft2) + 0 = 1.12 + PB/62lb/ft3 + 0 + 40ft

0.761 + 127.73psi + 0 = 1.12 + PB/62lb/ft3 + 0 + 40ft

128.46 psi/ft = 41.12 + PB/62 lb/ft3

128.46 psi/ft - 41.12 = 41.12 – 41.12 + PB/62 lb/ft3

87.34 = PB/62 lb/ft3

87.34(62) = (62)PB/62 lb/ft3

5,415.08 psf

5,415.08 psf / 144 in2

37.60 psi