   # Questions and Answers on Fluid Mechanics

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BPH – Tut # 2

Fluid Mechanics – University of Technology, Sydney – Engineering & IT
Tutorial # 2 – Questions and Solutions

Q.1
For the situation shown, calculate the water level d inside the inverted container, given the manometer reading h = 0.7 m, and depth L = 12 m.

Q.2 (from Street, Watters, and Vennard)
The weight density γ = ρg of water in the ocean may be calculated from the empirical relation γ = γo +
K(h1/2), in which h is the depth below the ocean surface, K a constant, and γo weight density at the ocean surface. Derive an expression for the pressure at any depth h in terms of γo, h and K. Calculate the weight density and pressure at a depth of 3.22 km, assuming γo = 10 kN/m3, h in meters, and K =
7.08 N/m7/2. Calculate the force on a surface area of 2 m2 (of a submarine, for example).

Question 3
The sketch shows a sectional view through a submarine. Atmospheric pressure is 101 kPa, and sea water density is 1020 kg/m3. Determine the depth of submergence y.

1

BPH – Tut # 2

Question 4
In the dry adiabatic model of the atmosphere, in addition to the ideal gas equation, air is also assumed to obey the equation p/ρn = constant where n = 1.4. If the conditions at sea level are standard (101.3 kPa, 15°C), determine the pressure and temperature at an altitude of 4000 m above sea level.

Solutions.
Q.1
For the situation shown, calculate the water level d inside the inverted container, given the manometer reading h = 0.7 m, and depth L = 12 m.

2

BPH – Tut # 2

Solution

B

A
F

D

E

3

BPH – Tut # 2

ρwater = 1000 kg/m3

;

ρ Hg = 13600 kg/m3

PA = PB + ρHg g h

;

PB = 0 (gauge)

PA ≈ PD

(pressure change due to air column AD is negligible due to low air density)

PE = PD + ρwater g d = PF + ρwater g L ;

PF = 0 (gauge)

∴ ρwater g L = ρHg g h + ρwater g d d = (ρwater L − ρ Hg h) / ρwater = (1000×12 − 13600×0.7) / 1000 = 2.48 m

Q.2 (from Street, Watters, and Vennard)
The

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