1. PHYS207 Assignment 3 Integral Methods
Q1. CO2 (specific gas constant R = 187.8 Jkg-1
K-1
) passes point A in a 75 mm diameter pipe at a
velocity of 5 m/s. The gauge pressure at A is 2x105
Pa and the temperature is 20 o
C. At a point B
downstream (same diameter duct) the pressure is gauged as 1.4x105
Pa and the temperature is 30 o
C.
For a barometric reading of 1.03 x105
Pa, calculate the velocity at B and compare the volume
flowrates at A and B. Why are they different? What are the mass flowrates at A and B. Why are they
the same?
The density of the gas at A is P/RT where P is an absolute pressure ie includes the atmospheric
pressure:
=P /RT = (1.03 + 2)x105
/(187.8(273+20)) = 5.51 kg/m3
Similarly at B:
=P /RT = (1.03 + 1.4)x105
/(187.8(273+30)) = 4.27 kg/m3
For steady gas flow the mass flowrate is constant regardless of compression:
Q = Q
V = V kg/s
Since A is constant, V V
So V V m/s
The volumetric flowrates at A and B differ as their velocities are different for the same flow area. This
is possible whenever compression or decompression effects are significant. The mass flowrate is
constant regardless of pressure changes so long as a gas flow is steady. The decompression in this
example may be due to a range of friction losses and other heat exchanges, which do not need to be
specified so long as pressure and temperature data are available.
Q2. A 300 mm diameter pipe 200 m long carries water from A at elevation 25 m to B at elevation 37
m. The frictional shear stress between the fluid and walls is 30 Pa. Determine the pressure change in
the pipe and the lost head in metres
Start by writing Bernoulli’s equation for points A and B:
g
P
z
g
V
HeadLoss
g
P
z
g
V B
B
BA
A
A
22
22
We need to find the pressure difference:

2. g
P
g
P
HeadLosszz
g
V
g
V AB
BA
BA
22
22
Or
HeadLosszz
g
V
g
V
gPP BA
BA
AB
22
22
The pipe remains at the same diameter, so there are no velocity changes: VA = VB:
HeadLosszzgPP BAAB
The elevations are known, and a headloss is created by the shear stress on the walls. Shear stress is a
form of pressure, which can be equated to headloss by working out the shear force along 200 m of
300 mm pipe:
Pipe circumference C = D = 0.3 = 0.943 m
Pipe surface area = CL = 0.943(200) = 188.5 m2
Shear Force Ft = A = 30(188.5) = 5.66 kN
This force is equivalent to a net back pressure on the pipeline’s end face:
Ft = Pb D2
/4 = Pb (0.3)2
/4 = Pb
The net back pressure is thus:
Pb = Ft /0.0707 = 5.66x103
/0.0707 = 80.1 kPa
Which is equivalent to a headloss: Hb = Pb/ g = 80.1x103
/9800 = 8.17 m
So we have:
HeadLosszzgPP BAAB = 9800(25 – 37 – 8.17) = – 197.7 kPa
Q3. A large closed tank is filled with ammonia (gas constant R = 481.5 J kg-1
K-1
, = 1.32) at 22o
C at
a gauge pressure of 37 KPa. The ammonia is discharged via a steel pipe into a second tank at the same
elevation and atmospheric pressure. Neglecting friction losses, calculate the initial velocity of the
ammonia in the connecting pipe as it enters the second tank.
a) assuming the gas density remains constant and;
b) assuming adiabatic flow conditions (variable density)
Bernoulli’s equation provides a relationship between pressure and velocity:

3. ssespressureloP
2
V
gH
inspressuregaP
2
V
gH
2
2
2
2
1
2
1
1
Convert gauge to absolute pressure by including atmospheric pressure in all calculations:
So the pressure in the first tank initially is 37x103
+ 101.3 x103
= 138.3 kPa
The pressure in the second tank initially is 101.3 kPa (atmospheric)
The density initially is given by
RT
P
V
M
ie = 138.3x103
/(481.5*(22+273K)) = 0.98 kg/m3 and we’re told that this is constant.
The velocity in the first tank is zero (ie the gas average velocity V1 is zero in the tank, moving at
some velocity V2 in the valve)
With H1 = H2 and Z1 = Z2 (no head losses, same elevation) we have
2
2
2
1 P
2
V
P0
So
33
2 1037
980
2
1031013138
980
2
.
..
.
V = 274.8 m/s
Note that the gauge pressure, not the absolute pressure, governs the escape flow velocity into
atmosphere. But is the constant density assumption realistic?
b) Adiabatic flow has a specific relationship between pressure and density:
ttancons
P
In the first tank we have = 321
3
980
103138
.
.
.P
= 142037.7
In the second tank we have 321
3
103101
.
.P
= 142037.7 since this is constant.
Ie
32.1
1
3
2
7.142037
10x3.101
= 0.774 kg/m3
We just need to modify the density in the previous calculation of V2.

4. 3
2 1037
774.0
2
xV = 309.2 m/s
So the constant density assumption is not too far wrong: about 10% error, not bad as a first guess.
Q4. A horizontal air duct reduces from 0.75 m2
to 0.20 m2
.
a) Assuming no losses and an incompressible fluid, what pressure change will occur when 6 kg/s
of air flows through the duct? Use a density of 3.2 kg/m3
for these conditions.
b) Is the incompressible assumption justified?
c) If the jet exits the smaller pipe and strikes a vertical wall, what force will it transmit?
a) The mass flowrate of 6 kg/s = Q where 3.2 kg/m3
Thus Q = 6/3.2 = 1.875 m3
/s
In the 0.75 m2
duct this equates to a velocity of V = Q/A = 1.875/0.75 = 2.5 m/s
In the 0.20 m2
duct this equates to a velocity of V = Q/A = 1.875/0.20 = 9.375 m/s
With no losses or elevation changes we can easily write Bernoulli’s equation:
g
P
g
V
g
P
g
V 2
2
21
2
1
22
Or
g
P
g
P
g
V
g
V 12
2
2
2
1
22
(2.52
-9.3752
)/19.62 = – 4.16 = (P2 – P1)/ g
So P2 – P1 = – 4.16(3.2)9.8 = – 130.5 Pa is the pressure difference
b) The velocities involved are small compared to the speed of sound in air, about 340 m/s , a lot faster
than 9.375 m/s . We can ignore compressibility if V < 0.3c where c is the speed of sound.
c) The reaction force is equal to Q V where Q is the mass flowrate = 6 kg/s and V is the velocity
change from 9.375 m/s to zero:
F= 6(9.375) = 56.25 N
Q5. Determine the reaction force of water flowing from a clean 2 cm hole in a tank at depth 3 m.
The flowrate depends on the contraction coefficient of the hole. We can take this as Cd = 0.6:

5. gH
D
CQ d 2
4
2
= 0.6 (0.02)2
/4 √(19.62(3)) = 0.0014 m3
/s or 1.4 l/s
The flow velocity assuming no losses is
gHV 2 = √(19.62(3)) = 7.67 m/s
The reaction force is given by
VF m
Where m is the mass flowrate Qm
And V is the change in velocity as the jet exits the tank:
Thus F = 1000(0.0014)7.67 = 10.74 N
Q6. A 500 mm diameter turbine develops 5400 kW of electrical power at 200 RPM for a head of 240
m. The power conversion from fluid to electrical is 82% efficient.
a) What is the flowrate at this head?
b) For the same type of turbine, what diameter is required to deliver 2850 kW under 183 m of head?
a) Let’s work out the fluid power first. We are told the conversion is 82% efficient, ie the electrical
power is 82% of the fluid power lost in the turbine. The fluid power is thus:
Power = Electrical Power/0.82 = 5400/0.82 = 6585 kW
The fluid power is given by
Power = g HQ where H is the headloss of 240 m. This leaves only the flowrate unknown:
Q = Power/ g H = 6585x103
/(9800(240)) = 2.8 m3
/s
b) This requires the use of two dimensionless numbers that are constant for the homologous series:
Specific Speed 45 /s
H
PowerN
N
is constant for all units of this type:
For the first case turbine we have 45
3
240
106585200
/sN = 543.3 RPM
We can use this for the second (design) case to work out the required speed:
Power = 2850/0.82 = 3476 kW
Power
HN
N
/
s
45
= 543.3(183)5/4
/√3476x103
) = 196.1 RPM

6. We also know that the Head relationship is constant for similar units:
ttancons
DN
H
22
For the first case,
2222
50200
240
.DN
H
= 0.024
Which equals the design case:
2222
1196
183
D.DN
H
= 0.024
Thus 02401196
183
2
..
D
= 0.445 m