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MEC 307
FLUID MACHINERY
Textbook: Fluid Mechanics and Hydraulic Machines by Sukumar Pati M. G. Hills
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Details of Academic Task(s)AT No. Objective Topic of the Academic Task
Nature of Academic Task (group/indiv iduals/field work
Evaluation Mode Allottmcnt / submission Week
Test ITo prepare students for VITE
Based on syllabus cov ered upto 4th week IndividualBasic concepts of the subject
4 / 5
Test 2To check the numerical solving ability of students
Based on syllabus cov ered after MTE upto 9th week Individual Studentperformance
9/ 9
Test 3 To prepare students forETE
Continuous assessment of student based on syllabus cov ered up to11th week
Individual Studentperformance
[ [ i l l
Mid Term Examination (MTE) Mar 03-11,2013
End Term Examinations(ETE) May 01-28, 2013
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JJ O V E L Y PROFESSIONALU N I V E R S I T Y
TURBINES : kapode kaba?
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JJ O V E L YPROFESSIONAL
[ U N I V E R S I T Y
Backshot Water WheelBreastshot Water Wheel
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PUMPS: Supply energy to the fluid
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JJ O V E L Y
P R O F E S S I O N A L
[ U N I V E R S I T Y
Cutter Pumps
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SUBMERSIBLE PUMP
HAND PUMP
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When applied to a single body Newton's second law can be stated as:
“The sum of forces on the body equals the rate of change of momentum of the body in the direction of the force.”
In equation form (F and V are in the same direction):
where m is the mass of the body and V is the velocity of the body and t is the time.
This also means the impulse Fdt equals the change in momentum of the body during the time dt.
Impulse Momentum Principle
dt jn
dt j
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When applied to control volume, through which the fluid is flowing, the principle can be stated as:
“The sum of forces on the fluid equals the difference between the momentum flowing in and momentum flowing out and
the change in momentum of the fluid inside the control volume, under steady flow condition the last term vanishes.
So the forces in the fluid is given by:
In other words, the net force on the fluid mass is equal to the net rate of out flow of momentum across the control surface.
The above equation can also be written as:
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in
dt jnou
t
dt j out
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= (pQV)out - (pQV)
_ poutQoutVout pinQinVin
= p2Q2V2 — piQiVi for compressible fluid
But,
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IT = . _/r for incompressible fluid
IF = I F„ k +I Fy i + I Fzk^
AV = Au i + Avj + Awkwhere u, v and w are the components of velocity in the x, y and z directions.
Using above relations we have:
I Fx k +1 Fy j +1 Fzj = pQ(AuJ + Avj + Awk)
Comparing the two sides we have:
is the impulse momentum equation in scalar form.
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IFx = pQ Au
Or, P1 A1 - P2 A2 cos 45o - Fx = pQ (u2 - u1)
Or, P1 A1 - P2 A2 cos 45o - Fx = pQ (V2 cos 45o - V1)
Or, 40000*1.2 - 30000*0.6 cos 45o - F = 1000*1.2*12 (24 cos 45o - 12)
Or, 48000- 18000cos 45o - F = 14400*4.97
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