Student
Tutor
Course
Dates
 
Introduction

  1. Lab description
  2. Objective of the lab

The objective of this experiment was to find out the amount of pressure needed to make a cap at the bottom of a pipe to open.

  1. Scope of the work done to meet each objective

The experiment was done by measuring the amount of water that was able to make a cap at the bottom of two vertical pipes to open. These caps were of 2” and 4” diameter. The density of the water was also determined as 64lbs/ft3 .

  1. Relevance and application to future as a practicing engineer

The knowledge in pressure and force is important since it can be used to evaluate the amount of water pressure required in supplying water to individuals leaving in a hilly area[1]. Similarly, it can be used to determine the amount of pressure needed to generate enough force that would slow or stop a car that uses a hydraulic brake system.

  1. Project description
    1. Scope of the lab

We placed two pipes vertically on the wall of the main block of the laboratory. The pipes were 4” and 2” wide. Moreover, both pipes had a length of 20 ft. These pipes were sealed with a cap in one end, the end that was facing down. We slowly put water into the pipes as we observed the amount of water needed to make the caps fail. We also measured the volume of water that we put in each pipe. The 4” pipe was able to hold 11 gallons of water before the cap failed.
Volume of 11 gallons in liters (Volume cm3 )
1 gallon= 3.78541 liters= 37.8541cm3
11 gallons= 416.6395cm3
 
The 2” pipe did not fail even after it was full to the brim. This pipe took 3.5 gallons of water. 3.5* 37.8541cm3 = 132.489cm3
Since this pipe was at its maximum potential head of the system, we were unable to slope it

  1. Methodology

            The hydrostatic pressure enacted by a liquid is determined by the height of the liquid in the container, the gravitational force, and the density of the liquid. The greater the height of the liquid, the more pressure it enacts. The denser the liquid is, the more pressure it enacts. Finally, the gravitational force influences the amount of force that the liquid enacts on a surface. As a result, hydrostatic pressure is calculated using the following formula

  1. Formula and Equation

Pressure= Density*Gravity* Height[2]
P=ρ*g*h
Where
P= pressure in fluid (N/m2 , Pa, lb/ft2 , psf)
ρ= density of the liquid (kg/m3 )
g=acceleration of gravity (9.81m/s2, 32.17405ft/s2)
h= height of fluid column or depth at which the pressure is measured (m, ft)
 
 
 
 
 
 
 
 
 
 
 
 
 
 

  1. Equipment Used for the Experiment

 
Black pipe (2” diameter), Brown piper (4” diameter)

  • Software

The excel software was used in the calculation of the pressure in each pipe.

  1. Fieldwork

The volume of water put in each pipe was correctly tabulated and was later used to calculate the pressure in each pipe. The 2” piper was filled with 3.5 gallons of water. The 4” pipe used only 11 gallons before the cap failed.

  1. Results of Analysis

Brown Piper (4” diameter)
1 inch = 2.54 cm
Radius= 5.08 cm
Area of the cap is 2
22/7*(5.08)2 = 81.1058cm2
1 gallon=3785.41cm3
11 gallons= 41,639.51cm3
Volume= Base Area*Height
Height=41639.51 cm3 /81.1068cm2
Height= 513.391cm Also, 5.13391m,
1 meter= 3.28084
Height= 5.13391*3.28084= 16.843537ft
Density of water (ρ)= 64lb/ft
P (Pressure) =ρ*g*h
Where
P= pressure in fluid (N/m2 , Pa, lb/ft2 , psf)
ρ= density of the liquid (kg/m3 )
g=acceleration of gravity (9.81m/s2, 32.17405ft/s2)
h= height of fluid column or depth at which the pressure is measured (m, ft)[3]
Pressure= 64*32.17405*16.843537
Pressure= 34,683.187 lb/ft2
PSI= (34,683.187 lb/ft2 * 0.00694444)= 240.855 psi
Since this pressure was a result of water of height 16.844, then each foot of water enacted the following force
Pressure from each additional foot of water was 34,684.186/16.843537= 2,059.187lb/ft2
(2,059.187lb/ft2 *0.00694444)= 14.30 psi
 
Table 1
1 lb/ft2 =0.00694444psi
Pressure Changes With Increase in Height of Water

Height Pressure lb/ft2   (A) Pressure in psi   (A*0.0069444)
1 2,059.19 14.2998182
2 4118.374 28.59963641
3 6177.561 42.89945461
4 8236.748 57.19927281
5 10295.935 71.49909101
6 12355.122 85.79890922
7 14414.309 100.0987274
8 16473.496 114.3985456
9 18532.683 128.6983638
10 20591.87 142.998182
11 22651.057 157.2980002
12 24710.244 171.5978184
13 26,769.43 185.8976366
14 28828.618 200.1974548
15 30887.805 214.497273
16 32946.992 228.7970912
17 35006.179 243.0969094
18 37065.366 257.3967277
19 39124.553 271.6965459
20 41183.74 285.9963641

 
The cap in the 4” pipe could not hold water beyond the pressure of 34,683.187 lb/ft2 (240.855 psi) that is at level 16.843537 feet. However, the 2” pipe could hold water till a height of 20 feet at a pressure of 41183.74 lb/ft2 (285.9963641). This shows that the cap in 2” tight had a higher capacity to hold pressure than that in the 4” pipe.

  1. Discussion
    1. Analysis of the Results

These results showed that the 4” pipe had a less firm cap than the 2” pipe. Basically, the 2” cap was able to hold pressure of up to 285.9963641, while the cap for the 4” pipe gave in at 240.855 psi. In addition, the results showed that the pressure acting at the base of each pipe was determined by the height of the column of water and not the area of the pipe[4].

  1. Legitimacy of the results

These results give legitimate answers of the pressure that acts at the base of a cylinder that has water. Basically, the results of these two different pipes indicate the amount of pressure that each cap can hold. Further, since there was no other external factors such as shaking of the pipes, the results are accurate and illustrate the manner the pipes would react in a real life scenario.
The results of this project are important in indicating the type of materials that engineers can use in various projects[5]. Specifically, the amount of pressure that each cap can handle enables engineers to determine the type of materials to use in various projects[6].

  1. Problems Encountered

            We were unable to accurately pour all the water in the pipe. Some of it leaked and this may have led to inaccuracies in the amount of water needed to make the cap to give in. Therefore, these results may be inaccurate in determining the amount of pressure needed to make each cap to give in.
 
 
References
Lamb, H. Hydrodynamics (6th Ed.) New York, NY: Dover Publications. (1945).
Landau, L. and Lifshitz, E. Fluid Mechanics (2nd Ed.). Oxford, UK: Butterworth-Heinemann. (1987).
McLeod, E. Introduction to Fluid Mechanics. New York, NY:  Dover Publications. (2016)
Meyer, D. Introduction to Mathematical Fluid Dynamics. New York, NY: Dover Publications. (2016).
Parr, A. Hydraulic and Pneumatics. Mumbai, India: Jaico (2005)
Shinbrot, M. Lectures on Fluid Mechanics. New York: NY: Dover Publications. (2012).
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Appendix
Equipment Used for the Experiment
 
 
 
 
Pressure Changes With Increase in Height of Water

Height Pressure lb/ft2   (A) Pressure in psi   (A*0.0069444)
1 2,059.19 14.2998182
2 4118.374 28.59963641
3 6177.561 42.89945461
4 8236.748 57.19927281
5 10295.935 71.49909101
6 12355.122 85.79890922
7 14414.309 100.0987274
8 16473.496 114.3985456
9 18532.683 128.6983638
10 20591.87 142.998182
11 22651.057 157.2980002
12 24710.244 171.5978184
13 26,769.43 185.8976366
14 28828.618 200.1974548
15 30887.805 214.497273
16 32946.992 228.7970912
17 35006.179 243.0969094
18 37065.366 257.3967277
19 39124.553 271.6965459
20 41183.74 285.9963641

Deliverables
Calculate the pressure in psi created on the cap for each experiment conducted above
Pressure in 4” pipe
1 inch = 2.54 cm
Radius= 5.08 cm
Area of the cap is π(r)2
22/7*(5.08)2 = 81.1058cm2
1 gallon=3785.41cm3
11 gallons= 41,639.51cm3
Volume= Base Area*Height
Height=41639.51 cm3 /81.1068cm2
Height= 513.391cm Also, 5.13391m,
1 meter= 3.28084
Height= 5.13391*3.28084= 16.843537ft
Density of water (ρ)= 64lb/ft
P (Pressure) =ρ*g*h
Where
P= pressure in fluid (N/m2 , Pa, lb/ft2 , psf)
ρ= density of the liquid (kg/m3 )
g=acceleration of gravity (9.81m/s2, 32.17405ft/s2)
h= height of fluid column or depth at which the pressure is measured (m, ft)
Pressure= 64*32.17405*16.843537
Pressure= 34,683.187 lb/ft2
Pressure of 2” pipe
Area of the cap is π(r)2
22/7*(2.54)2 = 20.7646 cm2
3.5 Gallons= 13248.935cm3
Density of water (ρ)= 64lb/ft
The pipe was full, therefore the height is 20 ft
P (Pressure) =ρ*g*h
Where
P= pressure in fluid (N/m2 , Pa, lb/ft2 , psf)
ρ= density of the liquid (kg/m3 )
g=acceleration of gravity (9.81m/s2, 32.17405ft/s2)
h= height of fluid column or depth at which the pressure is measured (m, ft)
Pressure= 64*32.17405*20
Pressure= 41182.784 lb/ft2
 
 Calculate the volume of water needed for each experiment conducted above
Volume of water
4” pipe= 11 gallons of water
1 gallon= 3.78541 liters= 37.8541cm3
11 gallons= 416.6395cm3
2” pipe=3.5 gallons of water
3.5* 37.8541cm3 = 132.489cm3
 
 Explain the difference in elevation required for the vertical 2” pipe and the sloped pipe
The pressure that is enacted by a pipe is determined by its vertical height. Therefore, the pressure enacted by the same volume of water for a 2” pipe at the cap is more when it is vertical than when it is elevated.
 Explain the difference in pressure required for the vertical 2” pipe and 4” vertical pipe
The differences required to make the caps leak is due to the various force per given area that each cap. The vertical pipes for the 4” and 2” enact equal amount of pressure at the same height, However, these caps are different and have different counteractive forces.
 
 Derive the PSI increase created for every 1 foot of depth (or elevation gain) of water utilizing the nit weight of water
Pressure= 64*32.17405*16.843537
Pressure= 34,683.187 lb/ft2
Since this pressure was a result of water of height 16.844ft, then each foot of water enacted the following force
Pressure from each additional foot of water was 34,684.186/16.843537= 2,059.187lb/ft2
Pressure in psi
1lb/ft2 =0.0069444psi
(2,059.187lb/ft2 * 0.0069444)= 14.2998psi
 Explain why the difference in cross sectional area does not affect the psi generated in  a static vertical water column
Pressure measures the force acting per square area, therefore, the differences in cross sectional area does not affect the psi generated by static water.
 
 Knowing what you learned in this lab explain why some cities use elevated water tanks to provide potable water to citizens
The use of elevated water systems enables cities to provide water in regions that are at the hill tops. Basically, the pressures needed to push water to these high areas is always high, and it can only be enabled by having elevated water tanks.
 Use the information determined in this lab to answer the following items:
o How much elevation difference would be needed if we wanted to achieve a pressure of 85 psi at the bottom of a water column
Pressure from additional column= 14.2998 psi
Pressure= ρ*g*h
Therefore: Pressure/( ρ*g)= h
85/(14.2998)=h
H= 5.9441 inches Approx. 15.1 centimeters, 0.151 meters
 
o If Flagstaff wants to use an elevated water tank to provide water to all its citizens at 85 psi what is the range of pressure if the elevation of the highest home is 7150 ft and the lowest home is 6825 ft?
The differences in heights of the houses is (7150-6825= 325 ft)
Pressure= ρ*g*h
Pressure= 64*32.17405*325= 669,220.24 lb/ft2
Psi (669,220.24 lb/ft2 * 0.0069444)= 4647.33)
Psi=4647.33
 
 Each individual must redo all calculations. Provide copies of all hand calculations in the appendices
[1] Landau, L. and Lifshitz, E. Fluid Mechanics (2nd Ed.). Oxford, UK: Butterworth-Heinemann. (1987).
 
[2] Shinbrot, M. Lectures on Fluid Mechanics. New York: NY: Dover Publications. (2012).
 
[3] Parr, A. Hydraulic and Pneumatics. Mumbai, India: Jaico (2005)
 
[4] Lamb, H. Hydrodynamics (6th Ed.) New York, NY: Dover Publications. (1945).
 
[5] Meyer, D. Introduction to Mathematical Fluid Dynamics. New York, NY: Dover Publications. (2016).
[6] McLeod, E. Introduction to Fluid Mechanics. New York, NY:  Dover Publications. (2016)