FLUID MECHANICS (ME F212)
TUTORIAL 3
BITS Pilani
Pilani
Campus Campus
Hyderabad
EXAMPLE PROBLEM 1
A piston compresses gas in a cylinder by moving at constant speed V, as in the figure below.
Let the gas density and length at t = 0 be ρ0 and L0, respectively. Let the gas velocity vary
linearly from u = V at the piston face to u = 0 at x = L. If the gas density varies only with time,
find an expression for ρ(t).
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Solution
To find 𝜌 ≡ 𝜌 𝑡
𝑑𝜌
𝜕𝑢
𝜕𝜌 𝜕 𝜌𝑢
+
𝜌
=0
+
= 0⇒
From 1D unsteady continuity equation:
𝑑𝑡
𝜕𝑥
𝜕𝑡
𝜕𝑥
u is varying linearly with x. So, 𝑢 = 𝑎 + 𝑏𝑥
At 𝑥 = 0, 𝑢 = 𝑉 and at 𝑥 = 𝐿, 𝑢 = 0.
𝑑𝜌
𝑉
𝑥
So, 𝑢 = 𝑉 1 −
⇒
+𝜌 −
=0
𝐿
𝑑𝑡
𝐿
Rearranging and integrating from 𝑡 = 0 to 𝑡 = 𝑡.
𝜌
𝑡
𝑡
𝑑𝜌
𝑑𝑡
𝑑𝑡
න
= 𝑉න
= 𝑉න
𝜌0 𝜌
0 𝐿
0 𝐿 𝑡
Initial value of L is L0 at t = 0. At any time t, L is given by 𝐿0 − 𝑉𝑡.
𝜌
𝑡
𝑑𝜌
𝑑𝑡
∴න
= 𝑉න
𝜌0 𝜌
0 𝐿0 − 𝑉𝑡
⇒ 𝜌 = 𝜌0
𝐿0
𝐿0 − 𝑉𝑡
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EXAMPLE PROBLEM 2
A fluid is flowing at a constant flow rate of 𝑄 through a divergent pipe having inlet and outlet
diameters of 𝐷1 and 𝐷2 , respectively, and a length of L. Assuming the velocity to be axial and
uniform at any section, show that the accelerations at the inlet and the outlet of the pipe are given
by
32𝑄2 𝐷2 −𝐷1
− 𝜋2 𝐿𝐷5
1
and
32𝑄2 𝐷2 −𝐷1
− 𝜋2 𝐿𝐷5
2
, respectively.
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Solution
𝐷𝑥 − 𝐷1
𝐷2 − 𝐷1
=
𝑥
𝐿
𝑄
4𝑄
Therefore, the velocity at this section can be written as: 𝑢 = 𝜋
=
2
𝐷𝑥2 𝜋 𝐷1 + 𝑥 𝐷2 − 𝐷1
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𝐿
𝜕𝑢
Acceleration at this section can be written as 𝑎 = 𝑢
𝜕𝑥
The diameter of the duct at an axial distance x from the inlet plane is:
32𝑄2 𝐷2 − 𝐷1
⇒𝑎=−
𝑥
𝜋 2 𝐿 𝐷1 + 𝐷2 − 𝐷1
𝐿
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⇒ 𝐷𝑥 = 𝐷1 +
𝐷2 − 𝐷1 𝑥
𝐿
Generic expression
32𝑄2 𝐷2 − 𝐷1
Acceleration at the inlet 𝑥 = 0 : −
𝜋 2 𝐿𝐷15
32𝑄 2 𝐷2 − 𝐷1
Acceleration at the outlet 𝑥 = 𝐿 : −
𝜋 2 𝐿𝐷25
𝐷1
𝐷𝑥
𝐷2
𝑥
𝐿
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EXAMPLE PROBLEM 3
An iceberg can be idealized as a cube of side length L, as in figure below. If seawater is denoted
by S = 1.0, then glacier ice (which forms icebergs) has S = 0.88. Determine if this “cubic”
iceberg is stable for the position shown in the figure below.
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Solution
For the iceberg to be stable 𝐺𝑀 > 0
𝐺𝑀 = 𝐵𝑀 − 𝐵𝐺
𝐵𝐺 =Distance between the C.G. and the C.B. measured from the bottom of the iceberg
min 𝐼𝑐
Now, 𝐵𝑀 =
,
∀𝑠
𝐿4 /12
𝐿2
∴ 𝐵𝑀 = 2 =
𝐿 ℎ
12ℎ
𝐿
ℎ
∴ 𝐵𝐺 =
−
2
2
min 𝐼𝑐 = minimum value of the area moment of inertia of the surface of the cube =
𝐿4
12
∀𝑠 = submerged volume of the iceberg = 𝐿2 ℎ
𝐿2
𝐿−ℎ
⇒ 𝐺𝑀 =
12ℎ
−
2
Under equilibrium, weight of the iceberg = weight of the displaced water
𝐿3 𝑆 = 𝐿2 ℎ × 1
⇒ ℎ = 𝑆𝐿
𝐿2
𝐿 − 𝑆𝐿
𝐿
𝐿 1−𝑆
∴ 𝐺𝑀 =
−
=
−
12𝑆𝐿
2
12𝑆
2
For S = 0.88,
𝐺𝑀 = 0.095𝐿 − 0.06𝐿 = 0.035𝐿 > 0 Hence the iceberg is in Stable equilibrium
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PROBLEM 4 (EVALUATIVE)
A rectangular wooden block has a width of b and is submerged with a depth of H. Its centre of
gravity is at the waterline. Find the metacentric height in terms of b/H, and hence show that for
𝑏
stable equilibrium of the block, 𝐻 ≥ 6. Assume 𝑏 < 𝐿, where L is the length of the block.
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Thank You
BITS Pilani, Pilani Campus