Mechanical Engineering Lab Report
The impact force of a water jet is measured on a plane
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Abstract
This experiment is about the jet water impact, against the plane or a curved surface. Time and displacement of the water jet was recorded in order to come with conclusion. The experiment shows that the impact force against the plane is 0.8308942N and produces a volumetric flow rate of 0.8682844M ^{3}/s. Finally the experiment outcome shows that the volumetric flow rate is directly proportional to the impact force.
Introduction
The main aim of this experiment is to determine the relationship between the impact force of a water jet on a plane or a curved surface and volume flow rate and the impact force. It involves directing a jet of water against plane and measuring time and displacement of the loaded plane. Large number of data was collected to improve the accuracy of these data.
Discussion of Results:
The results from the experiment were recorded as shown in the attached excel sheet. The calculation was done as follows.
(Batchelor ,1967).
We made the following assumption.
- Water jet doesn’t spread hence the area of the jet nozzle is equal to the impact area on the plane.
- The fluid being used is not compressible.
- The final velocity is zero on impacting the plane, therefore V _{2}=0
Initial velocity =V _{1 }and is calculated as in the excel sheet
Area of the nozzle =?
Diameter =150mm
Area=75*75*22/7= 17678.6mm ^{2 }or 0.017678.6m ^{2}
Pressure = force/area
Therefore force = pressure * area
From fluid energy equation
Also mass displaced =4kg
Force =4* 9.81=39.24N
Therefore pressure acting on the plane =Force/area=39.24N/17678.6mm ^{ 2 = } 5.7N/ mm ^{2}=2219.6N/M ^{2}
Approx 2220N/M2
Given density =998.2kg/m3
From energy equation above
2220N/M ^{2} =P1 ^{2 }+1/2* 998.2(V _{1} ^{2}-V _{2} ^{2})
V _{2} ^{2}=0
V _{1} ^{2}=from calculations = 8.990364504mm/s
=9 mm/s
=0.009M/s
2220N/M ^{2} =P1 ^{2 }+1/2* 998.2(0.009 ^{2}-0 ^{2})
P _{1} ^{2 } =2220-0.000003275
P _{1}=47.10
=47N/M ^{2}
But we know that the area of nozzle =0.0176786m ^{2}
Force = pressure * area
=47*0.0176786m ^{2}
0.8308942N
=0.8308942N
Relationship between volume flow rate and the impact force
Volume flow rate =VA=Q
Where V=flow velocity
A=cross sectional area
From� � � 2220N/M ^{2} =P1 ^{2 }+1/2* 998.2(V _{1} ^{2}-V _{2} ^{2})
Force/ area = pressure
2220N/M ^{2} =( Force/ area ) ^{2 }+1/2* 998.2(V _{1} ^{2})
VA=Q
V=Q/A
2220N/M ^{2} = (Force/ area ) ^{ 2 }+1/2* 998.2((Q/A) ^{ 2})
{2220N/M ^{2} = (Force/ area ) ^{ 2 }+ 499.1((Q/A) ^{ 2})} ^{1/2}
47= Force/ area +22.3 Q/A
Area=A
47A= Force +22.3 Q
Q=-0.045F+47A and comparing with Y=ax+c
Then plotting graph of Q against F gives a straight line which shows that
Volume flow rate is directly proportional to the impact force
Solving Q=-0.045F+47A gives
Q=-0.045( 0.8308942N ) +47( 0.0176786m ^{2} )
0.037390239+0.8308942= 0 .8682844M ^{3}/s
Conclusions:
The experiment was done successful and the objectives was achieved whereby the impact was found to be 0.8308942N � while the volume flow rate was found to be 0.8682844 M ^{3}/s .Finally the graph of volumetric flow rate a against� impact force show that they are directly proportional to one another.
Reference
Batchelor, G. K. (1967). A n Introduction to Fluid Dynamics. Cambridge University Press
- Appendix 2. Use the results of your investigation to:
- Recommend a volume flow rate which will achieve a force of 8308942 N� � (insert the last two digits of your student ID number).
- Determine the force generated by a volume flowrate of 8682844 (insert the 3 ^{rd} number in your student ID).
- Comment on the validity of the above two estimates.