Download Lab 3 - Jet velocity and trajectory

Lab 3 - Jet velocity and trajectory
Terminology
ρ - density of water
H - Head of water in Tank
Pi - Pressure at point i
1 - point on the streamline at free surface
V -Velocity
x-y - co-ordinates along horizontal and vertical direction respectively
A2 - Area of the orifice
g - acceleration due to gravity
Patm - atmospheric pressure
Vi Velocity at point i
2 - point on the streamline at the start of the jet
t- time
Q- flowrate
Jet Velocity
Using Bernoulli’s Equation
Applying Bernoulli’s equation to the streamline between point 1 and 2 shown in Fig. 1 ,
P2 V22
P1 V12
+
+ gy1 =
+
+ gy2 .
(1)
ρ
2
ρ
2
The ambient pressure the tank is exposed to is Patm , hence P1 = P2 = Patm . Velocity at
the water surface is zero, V1 = 0. The origin of the x-y axes is chosen to be at point 2, hence
y2 = 0 and y1 = H. Substituting the above into Eq. 1
Patm V22
Patm
+ gH =
+
.
ρ
ρ
2
Simplifying the above equation we get the velocity of the jet a point 2
V2 =
p
2gH.
(2)
(3)
Using Flow rate
By conservation of mass we have
A2 V2 = Q,
(4)
where Q is the measured flow rate and A2 is the area of cross section of the orifice. The jet
velocity is given by
V2 = Q/A2 .
1
(5)
Figure 1:
Jet trajectory
Velocity and acceleration are related as
Z
V =
adt,
V = at + C1 .
(6)
Where C1 is a constant of integration. Applying the above equation for the X component
of the jet velocity, at t = 0, V = V2 hence C1 = V2 and since there is no acceleration in the x
direction a = 0, thus we have
V =
x=
dx
= V2
Zdt
V2 dt = V2 t + C2 .
(7)
The x-y axes are chosen such that the origin is at point 2 where the jet starts, hence at
t = 0 when the jet starts x = 0 therefore C2 = 0. The x component of the trajectory of the jet
as function of time is given by
p
x = V2 t = t 2gH
(8)
Now applying Eq. 6 to the y component of the trajectory. At t = 0, when the jet starts
at point 2 it does not have any velocity along the y direction, V = 0 therefore C1 = 0 and the
acceleration due to gravity acts downwards in the negative y direction hence a = −g.
V =
y=
dy
= −gt,
Zdt
−gtdt = −
gt2
+ C3 .
2
(9)
At point 2 where the jet starts y = 0 (t=0) hence C3 = 0, thus we have
gt2
y=− .
2
2
(10)
Eliminating the time variable t from Eq. 8 and Eq. 10 will give
√ an equation for the trajectory
of the jet in terms of spatial co-ordinates. Substituting t = x/ 2gH from Eq. 8 into the above
equation gives
y=−
x2
4H
(11)
The Apparatus
Procedure
1. Insert the orifice into the orifice fitting device.
2. Start the water supply into the tank by starting the pump.
3. Adjust water head in the tank by means of the adjustable overflow.
4. Beginning on the right side adjust the trajectory probes such that the tip if the probe is
in contact with the jet.
5. Measure the height of the probes using a ruler.
6. Change the water head in the tank and repeat steps 3-5
3
Head H =
mm, Diameter of Orifice D =
Probe
x (mm)
ymeasured (mm)
1
2
3
4
5
6
7
8
Volume (litre)
time (sec)
Vmeasured (mm/sec)
VT heoretical (mm/sec)
mm
ytheoretical (mm)
Head =
mm, Diameter of Orifice D =
mm
Probe
x (mm)
ymeasured (mm) ytheoretical (mm)
1
2
3
4
5
6
7
8
Volume (litre)
time (sec)
Vmeasured (mm/sec)
VT heoretical (mm/sec)
Head =
mm, Diameter of Orifice D =
mm
Probe
x (mm)
ymeasured (mm) ytheoretical (mm)
1
2
3
4
5
6
7
8
Volume (litre)
time (sec)
Vmeasured (mm/sec)
VT heoretical (mm/sec)
4