Download Problems for self

This document contains few fully solved problems that cover the whole material. Then some
problems for self-control follow
PROBLEM
Water with density ρ=1000 kg/m3 flows with flow rate Q through a pipeline. The diameter of the
pipeline changes from d1 = 100 mm to d2 = 50mm. The manometer, connected between sections 1
and 2, is filled with mercury (ρHg = 13600 kg/m3) and has a reading h = 40mm. The pipeline ends
with a nozzle with diameter d3 = 20 mm. Determine the flow rate Q through the pipe and the
reaction force R at the nozzle. Neglect losses.
d1
(1)
Q
d2
d3
• (2)
(1)
Nozzle
h
SOLUTION:
First we shall find the flow rate Q. For this purpose we need to find the velocity in either section of
the pipe. We shall find the velocity in section 2 using the Bernoulli equation between p.1 – p.2:
1
1) p 1  p 2  V22
2
Take into account that the left tube is a Pitot-tube, i.e. it measures total pressure. As such, p1 is the
total pressure and therefore equals the sum of the static and dynamic pressure in section 1. That is
why the dynamic pressure in section 1 is missing from the equation. If the left tube was a static
tube, then the dynamic pressure in section 1 must be included in the left hand side of equation 1.
Both pressures in equation 1. are unknown, but we can find their difference using the U-tube
manometer. For this purpose we rearrange equation 1) as below:
1
2) p1  p 2  V22
2
The pressure difference reported by the manometer can be found from fluid statics:
3) p1+ρgh-ρhggh=p2, or p1-p2 = gh(ρhg-ρ).
Attention: take into account that both the water and mercury density take part. Common mistake is
to express the pressure difference only through the mercury density: p1-p2 = ghρhg, which is wrong.
Now it is time to substitute the pressure difference p1-p2 from equation 3. into equation 2.:
1
4) gh ( hg )  V22
2
After rearrangement, the velocity V2 is:
2gh ( hg )
5) V2 

The flow rate Q is a product of the velocity and crossectional area:
d 2 2gh ( hg )
6) Q  2
4

The second part of the problem concerns the reaction force in the nozzle, Rx. The reaction force can
be found from the Reynolds transport theorem:
 (V2  V3 )  Fpx2  Fpx3
7) Rx  m
Remember, the rule for the velocity subtraction in the Reynolds transport theorem is the inflow
velocity minus the outflow velocity. By inflow and outflow we mean into or out of the control
volume considered, in this case, the nozzle. Velocity V2 is the inflow velocity and velocity V3 is
the outflow velocity in the present case. The pressure forces are:
d 2
Fpx2  p 2 . 2
4
8)
d 2
Fpx3  p 3 . 3
4
The sign of the forces is based on the rule that the pressure of the surroundings acts always in a
direction into the control volume. Based on this rule, pressure p2 acts from left to right (into the
volume) and pressure p3 acts from right to left (again into the volume contained within the nozzle).
It is conventional to assume that the x-axis points from left to right, and thus Fp2 points in the
positive direction and has positive sign and Fp3 points in the negative direction, and has negative
sign.
A basic rule when fluid flows as a free jet out of a pipe is that the static pressure in the
corresponding crossection (section 3 in this case) is equal to the ambient pressure (in this case the
atmospheric). It is conventional to use the atmospheric pressure as a reference, i.e. the atmospheric
pressure is set to zero in this kind of problems. So, p3=0, Fp3=0.
The pressure at section 2 is however unknown. We have to use Bernoulli equation between point 2
and point 3 (the jet discharge point):
1
1
9) p 2  V22  p 3  V32
2
2
In this equation pressure p3=0, as explained before. So the pressure p2 is
1
1
10) p 2  V32  V22
2
2
Velocities are known once the flow rate is found, and the flow rate was found from the first part of
the problem.
The pressure p2 is then substituted in the first equation 8), and then equations 8) are substituted into
7):
d 2
11) Rx  Q(4Q d 22  4Q d 32 )  p2. 2  0 , where Q is substituted from equation 6) and
4
p2 from equation 10.
Solve the same problem as above, but with both tubes being static tubes:
d1
d2
Q
• (1)
d3
• (2)
Nozzle
h
PROBLEM
Rz
ξn
(3)
30°
Rx (3)
h
(2)
g
(2)
H
P1
Z
L
(1)
(1)
X
Water is discharged from a pressurized tank through a straight pipe and nozzle. The diameter of the
pipe is d2 = 50mm, the diameter of the nozzle exit is d3 = 25mm, the height of nozzle is h = 1m, the
loss coefficient of the nozzle is ξn = 2, and the vertical distance H = 16m. The water from the tank
enters the pipe with a loss coefficients of ξin = 0.5. The friction factor and the length of the straight
pipe are λ = 0.03 and L = 20 m, respectively. If the horizontal component of the reaction force in
the nozzle is Rx = 50 N, determine:
(1) The volume flow rate of water through the nozzle, Q.
(2) The pressure p1 in the container.
(3) The vertical component of the reaction force of the nozzle, Rz, if the volume of water in the
nozzle is W = 0.01 m3.
SOLUTION
1. Find volume flow rate Q
Reynolds transport theorem for nozzle, written for X coordinate:
 V3 cos 30  0  Rx
m
4Q 2
 2 cos 30  Rx
d 3
2. Express velocity V2 and V3:
d i2
V2=Q/A2, V3=Q/A3, where Ai 
, i=2,3
4
3. Find pressure p1:
The Bernoulli equation between section 0 and section 3 looks like this:
1
L 1
1
p1  V32  gH  
V22   n V32
2
d2 2
2
The right-hand side contains two losses: one head loss and one minor loss.
4. Find Rz
To find the vertical reaction force Rz, first we have to find the pressure in point 2.
To find pressure p2, we use Bernoulli equation between point 1 and point 2:
1
L 1
p1  0  0  V22  gH  h   
V22  p 2
2
d2 2
Alternatively, we may use the Bernoulli equation between point 2 and point 3:
1
1
1
p 2  V22  V32  gh   n V32
2
2
2
P2 from the first equation is:
1
L 1
p 2  p1  V22  gH  h   
V22
2
d2 2
To Find Rz, we use the Reynolds transport theorem for the nozzle. Written for the Z component:
  Q , Fpz2  p 2 .
 V2  V3 sin 30  Fpz2  W.g , where m
Rz  m
d 22
, and W is the water
4
volume in the nozzle, with negative sign because it acts downwards (and we assume the positive z
direction to be upwards).
PROBLEM
(0)
(0)
H
Gate valve
d1
ξG
ξD
(1)
(2)
X
l
(1)
(2)
h
Water flows from a container with level H = 12m to a container with level h = 5m. The pipe that
connects the two containers has length l = 100m, diameter d1 = 0.06m, and friction factor λ = 0.03.
The water flows into the second tank through a diffuser, which expands to a diameter d2 = 0.12m.
The water flows from the diffuser as a free jet. The gate valve has loss coefficient ξ G = 3, and the
diffuser has loss coefficient ξD=2. Other minor losses are negligible. Determine:
(1) The volume flow rate of water, Q.
(2) The reaction force in the diffuser, Rx.
Hint: The pressure loss in the diffuser (between sections 1 and 2) is determined by the expression
1
p m.loss   D V22
2
SOLUTION
1. Find Q
Bernoulli between p.0 and p.2:
 L

1
1
gH  V22 1   D   gh  V12     G 
2
2
 d1

2
Vi=4Q/pi.Di
2. Find p1:
Bernoulli between p.0 and p.1:
L 1
1
1
gH  
V12  p1  V12   G V12
d1 2
2
2
or between p.1 and p.2
1
1
p1  V12  gh  V22 1   D 
2
2
3. Find Rx (Reynolds transport theorem)
 (V 2  V1)  P1  P 2  Rx
m
P1=p1.A1,P2=p2.A1, p2=ρgh, p1 from above
PROBLEM
P0
Z
(0)
(0)
X
Confusor
H
(1)
(2)
(2)
(1)
Gate valve
d2
(3) Free jet
L
(3)
Water is discharged from a tank pressurized to a gage pressure p0 = 15 kPa. The level of water
in the tank is H = 6m. The water flows through a pipe consisting of a confusor with loss
coefficient ξc = 0.4, a gate valve with loss coefficient ξV = 2, and straight pipe (between sections
2 and 3) with length L = 30m and friction factor λ = 0.04. The diameters are d1 = 120mm, d2 =
d3 = 60mm. Determine:
(1) The flow rate Q of water
(2) The reaction force Rx in the confusor
INSTRUCTIONS:
1. Find Q and velocities:
Bernoulli equation between section (0) and (3)
2. Find p1:
Bernoulli equation between section 1 and 0 or 1 and 3
3. Find p2:
Bernoulli equation between section 2 and 0 or 3
4. Find Rx:
Re tr. Theorem and final expression for Rx
PROBLEM
Water flows from a large tank, as shown in the figure. If viscous effects are neglected, at what
height, h, the pressure in section 2 will be equal to -80kPa?
PROBLEM
Two Pitot-tubes and two pressure taps are placed in the pipe contraction shown on the figure. The
flowing fluid is water (ρ=1000 kg/m3). The density of the gage fluid in the Pitot-tubes (located
below the pipe) is ρ1=1100 kg/m3 and in the static tubes (located above the pipe) ρ2=700 kg/m3.
Determine the two manometer readings, h and H.
ρ2
ρΡ11
PROBLEM
Fluid with density ρ flows through the pipe. What is the average velocity in the large crossection?
Q
D
d
h
ρm
PROBLEM
When driven initially into motion, the water in the container will continue to flow through the hose.
a) Calculate the volume flow rate of water if the hose has a constant diameter of d=0.1m.
b) Estimate the maximum allowable elevation of the top of the hose, p.A, if the vacuum there
must not exceed 25 kPa.
.A
z = 2m
d
z = 0m
Free jet