The document contains 6 physics questions regarding properties of fluids. Question 1 asks about pressure in a water pipe using a manometer. Question 2 involves using the ideal gas law to determine pressure and mass of air in a tire at different temperatures. Question 3 calculates residual pressure in a tank with two chambers connected by a sluice opening.

1. PHYS207 Assignment 1 Properties of Fluids
Q1. Determine the gauge pressure at point A in the water pipe due to the deflection of mercury in the
u-tube manometer. The liquid flowing in the pipe is water. The end of the manometer is open to the
atmosphere.
( H2O = 1,000 kg m-3
Hg = 13,570 kg m-3
)
We can work from one side of the system (where the pressure is known to be atmospheric), to the
other (unknown) end, adding static pressure as we progress downwards into depth and subtracting it
upwards:
“The pressure at D, plus 0.9 m of mercury, less 0.6 m of water, gives the pressure at A”
Pressure changes are given by P = g H
so PA = PD + 13570(9.8)(3.9 – 3.0) – 1000(9.8)(3.6 – 3.0)
PD is atmospheric pressure; the gauge pressure is what’s left after we subtract it:
PA(Gauge) = 13,570(9.8)0.9 – 1000(9.8)0.6 = 113.8x103
Pa = 113.8 kPa
A
B C 3.0 m
D 3.9 m
3.6 m
X-section through
pressurised pipe

2. Q2. The pressure in an automobile tyre depends on the temperature of the air in the tyre. When the air
temperature is 25 o
C the pressure gauge reads 210 kPa. If the volume of the tyre is effectively constant
at 0.025 m3
, determine the pressure in the tyre when the air temperature inside the tyre rises to 50 o
C
due to road heating. Also determine the mass (in kg) of air that must be bled off to restore pressure to
its original value at this temperature. Assume atmospheric pressure to be 100 kPa.
We can use the ideal gas law PV = nRT or its formulation in terms of density P = RT to relate
pressure to temperature and amount of gas. It is important to recall that the ideal gas law uses absolute
temperatures and pressures, thus atmospheric pressure must be added to gauge pressures and 273 K
must be added to centigrade temperatures:
P1 = 210x103
+ 100x103
= 310x103
Pa
T1 = 25 + 273 = 298 K
Specific Gas Constant R (Air) = 287 J/kg/K (Table page 15 Study Guide)
We can now calculate the density of air in the tyre at 25 o
C:
P1 = RT1
= P1 / RT1 = 310x103
/ (287 x 298) = 3.625 kg/m3
Since the volume of the tyre is effectively constant, the density does not change as the temperature
increases to 50 o
C:
T2 = 50 + 273 = 323 K
P2 = RT2 = 3.625(287)323 = 336x103
Pa (absolute, or 236 x103
Pa gauge)
Restoring the pressure to 310x103
Pa at 50 o
C reduces the density:
P3 = RT2 = 310x103
P3 / RT2 = 310x103
/(287x323) = 3.344 kg/m3
The amount of gas removed can be calculated by expressing the density as mass by multiplying out
the volume:
M =
M1 = 3.625(0.025) = 90.6 x10-3
kg
M2 = 3.344(0.025) = 83.6 x10-3
kg
The difference is 7 g of air.
Q3. A tank consists of two chambers, one open to the atmosphere and another closed and filled with a
gas. The two chambers are connected by a sluice opening as shown. What is the residual pressure in
the gas, in gauge and absolute terms, if the barometric pressure reads 742 mm Hg?

3. The barometric pressure is P = gH = 13570(9.8)0.742 = 98.68 kPa
The pressure at the outside water surface is atmospheric Po = 98.68 kPa
The absolute pressure at the bottom of the tank is P = 98680 + waterg(1.200 – 0.700) = 103.6 kPa
The absolute pressure at the gas is Pabs = 103600 – waterg(1.200) = 91.8 kPa
The gauge pressure in the gas subtracts the atmospheric pressure : Pgauge = 91.8 – 98.68 = – 6.88 kPa
Q4. The viscosity of a fluid is to be measured by a viscometer constructed of two 75 cm long
concentric cylinders. The outer diameter of the inner cylinder is 15 cm, and the gap between the two
cylinders is 1 mm. With the outer cylinder fixed the inner cylinder is rotated at 300 RPM, and the
torque is measured by a spring gauge to be 0.8 Nm. Determine the viscosity of the fluid.
The torque is produced by a force at the radius of the inner cylinder times the radius: T = F.ro
ro = 0.15/2 = 0.075 m
The force F is a shear force due to viscous shear stress multiplied by the surface area of the inner
cylinder:
F = A
A = DoL = (0.15)0.75 = 0.3534 m2
The shear stress is given by Newton’s law of viscosity
dy
dV
dy is the gap between the cylinders: dy = 0.001 m
dV is the linear velocity between the cylinders: this can be found from the angular velocity:
= 300 RPM = 300x2 rads/minute = 300x2 rads/sec
= 31.42 r/s
This gives a linear velocity of dV = r = 0.075(31.42) = 2.357 m/s
dV/dy is thus 2.357/.001 = 2357 /s
gas
water
Barometer
atmosphere
700 mm
Connecting sluice
1200 mm

4. Assembling these terms:
T = 0.8 N.m = Fro = F(0.075) = A(0.075) = dV/dy(0.3534)0.075 = (2357)(0.3534)(0.075)
Thus = 0.8/(2357(0.3534)(0.075)) = 0.0128 Pa.s
Q5. Molten rock 0.5 m thick flows down a 30o
slope with an apparent velocity of 2 m/s. a) Assuming
a linear velocity profile as a function of depth (V = ky) and a density of 2,500 kg/m3
, calculate the
viscosity of the lava.
The apparent velocity Va is the surface velocity, since it is the only part of the flow that you can see.
The linear velocity profile is V = Va(y/0.5) where y varies from 0 to 0.5, the depth of the flow.
In steady flow the weight forces acting downslope are balanced by friction shear forces acting on the
flow bed. Frictional shear is given by Newton’s law as
dy
dV
Where
Y
V
dy
dV a
A shear force is expressed over a given area of the flow bed:
T = A = wL where w is the width (into the page) of a section of flow
The weight of that section of the flow is W = mg sin = Vg sin = wLYg sin
The shear force balances the weight force: T = W
ie T = A = wL = W = wLYg sin
or = Yg sin
Thus sinYg
Y
Va
So 3089
2
50
2500
22
sin.
.
sing
V
Y
a
= 1531 Pa.s
L
mg
V = f(y)=Va(y/Y)
Va
Y = 0.5

5. Q6. A droplet of water on a non-wetting surface forms a contact angle of 90o
with the surface.
Estimate the maximum thickness h of the droplet assuming static pressure, a density of 1000 kg/m3
and a surface tension of 0.07 N/m
The maximum droplet depth is a minimum energy situation, ie a balance between potential energy and
the contributions of surface energy due to contact between the fluid, air and solid surfaces. Young’s
equation (Lecture 4) gives a contact angle of 90o
only when the surface-vapour and surface-liquid
tensions are equal, thus sv = sl.
The minimum energy condition (also derived in Lecture 4) gives a formula in which these equal terms
cancel out:
gg
h lvlvsvsl 22
(90o
contact angle only)
ie h = √(2 g) = √(2(0.072)/9800) = 3.8 x 10-3
= 3.8 mm.
A droplet of water on glass is not far from this depth.