Answer to Question #211374 in Mechanical Engineering for Mary

Question

Need help with question 3 & 4

2. Convert all data to SI (metric) units: Determine the volume in liters and

in gallons for the pool volume. The pool has dimensions of 18’X35’, where

(’) is a symbol indicating a foot. The pool has a depth of 1 meter at one

end and 2 meters at the other end. We will assume that 50% of the pool

has a depth of 1 meter and the other 50% has a depth of 2 meters. All

height dimensions are from the pool deck surface and you can

approximate 3.3’ to be equal to 1 meter.

3. Assuming that the water level in the ground surrounding the pool is just

30 cm below the pool deck surface (a reasonable assumption during the

rainy season in South Florida), determine the hydraulic pressure for the 2

halves of the pool.

4. Using the surface areas of the 2 halves, calculate the hydraulic force on

each half.

The density of concrete is 2320 kg/m

Answer 1

Part 2

The dimension of the pool in SI

18' = 5.46 m

35' = 10.61 m

Base area of the pool = $5.46*10.31=57.93 m^2$

The volume of the pool $0.5*57.93*1+0.5*57.93*2=86.90 m^3$

We know 1 m³ = 1000L= 254.1721 gallons

$86.90*254.1721 = 22087.55549 gallons$

Part 3

Hydraulic pressure =$\frac{A (Surface Area)}{F (Force) }=\frac{A (Surface Area)}{density * volume }$

$\frac{57.93}{2320*86.90}=0.00028733 N/m$

Part 4

F (Force) = P (Pressure) x A (Surface Area)

F=$0.00028733*57.93= 0.0166 N$