Problem 6A.4: Motion of a sphere in a liquid
A hollow sphere, 5.00 mm in diameter, with a mass of 0.0500 g is released in a column of liquid and
attains a terminal velocity of 0.500 cm/s. The liquid density is 0.900 g/cm 3. The local gravitational
acceleration is 980.7 cm/s2. The sphere is far enough from the containing walls so that their effect can
be neglected.
(a) Compute the drag force on the sphere in dynes.
Solution
Force of gravity on the sphere is:
𝐹𝑔𝑟𝑎𝑣𝑖𝑡𝑦 = 𝑚𝑔 = 0.0500 𝑔
980.7
𝑐𝑚
= 49.0 𝑑𝑦𝑛𝑒𝑠
𝑠2
The buoyant force of the fluid on the sphere is:
4
4
𝐹𝑏𝑢𝑜𝑦𝑎𝑛𝑡 = 𝜋𝑅 3 𝜌𝑔 = 𝜋 0.25 𝑐𝑚
3
3
3
0.900 𝑔/𝑐𝑚3
980.7
𝑐𝑚
= 57.7 𝑑𝑦𝑛𝑒𝑠
𝑠2
So, the resultant upward force is:
𝐹𝑏𝑢𝑜𝑦𝑎𝑛𝑡 − 𝐹𝑔𝑟𝑎𝑣𝑖𝑡𝑦 = 57.7 𝑑𝑦𝑛𝑒𝑠 − 49.0 𝑑𝑦𝑛𝑒𝑠 = 8.7 𝑑𝑦𝑛𝑒𝑠
(b) Compute the friction factor
Solution
The friction factor is defined by:
𝐹𝑘 = 𝜋𝑅 2
1 2
𝜌𝑣 𝑓
2 ∞
So, for this system,
𝑓=
𝜋𝐷 2
4
𝐹𝑘
8𝐹𝑘
=
=
1 2
𝜋𝐷 2 𝜌𝑣∞2 𝜋 0.5 𝑐𝑚
𝜌𝑣∞
2
2
8 8.7 𝑑𝑦𝑛𝑒𝑠
0.900 𝑔/𝑐𝑚3 0.5 𝑐𝑚/𝑠
2
= 393.8
(c) Determine the viscosity of the liquid
From figure 6.3-1, f is very close to its creeping flow asymptote, 24/Re, so Re=24/f,
𝐷𝑣∞ 𝜌 24
≈
𝜇
𝑓
Solving for viscosity,
𝜇=
𝐷𝑣∞ 𝜌𝑓
0.5 𝑐𝑚 0.5 𝑐𝑚/𝑠 0.900 𝑔/𝑐𝑚3
𝑔
=
393.8 = 3.7
= 3.7𝑥102 𝑐𝑝
24
24
𝑐𝑚 ∙ 𝑠
Problem 6B.2: Friction factor for flow along a flat plate.
(a) An expression for the drag force on a flat plate, wetted on both sides, is given in Eq. 4.4-30. This
equation was derived by using laminar boundary layer theory and is known to be in good agreement
with experimental data. Define a friction factor and Reynolds number, and obtain the f versus Re
relation.
Solution
Eq. 4.4-30
laminar flow.
𝐹𝑥 = 1.328 𝜌𝜇𝐿𝜔 2 𝑣∞3 gives the kinetic force acting on both sides of the flat plate for
An appropriate Reynolds number for a plate of length L is: 𝑅𝑒𝐿 =
𝐿𝑣∞ 𝜌
.
𝜇
Using the definition of the friction factor in Equation 6.1-1, FK=AKf,
𝑓=
𝐹𝑘
1.328 𝜌𝜇𝐿𝜔 2 𝑣∞3
𝜇
=
= 1.328
2
𝐴𝐾
𝐿𝑣∞ 𝜌
2𝑊𝐿 0.5𝜌𝑣∞
So,
𝑓=
1.328
𝑅𝑒𝐿
(b) For turbulent flow, an approximate boundary layer treatment based on the 1/7 power velocity
distribution gives
𝐹𝑘 = 0.72𝜌𝑣∞2 𝑊𝐿
𝐿𝑣∞ 𝜌 −1/5
𝜇
(6B.2-1)
When 0.072 is replaced by 0.074, this relation describes the drag force within experimental error for
𝐿𝑣 𝜌
5𝑥105 < ∞ < 2𝑥107 . Express the corresponding friction factor as a function of the Reynolds
𝜇
number.
Solution
For turbulent flow, the force acting on the plate is given by equation 6B.2-1, with the 0.074 replacing
0.072,
𝐹𝑘 =
Then the friction factor is:
0.74𝜌𝑣∞2 𝑊𝐿
𝐿𝑣∞ 𝜌
𝜇
−1/5
𝑓=
𝐹𝑘
=
𝐴𝐾
0.74𝜌𝑣∞2 𝑊𝐿
2𝑊𝐿
𝐿𝑣∞ 𝜌
,
𝜇
Simplifying, and using ReL=
𝑓=
0.074
1/5
𝑅𝑒𝐿
𝐿𝑣∞ 𝜌
𝜇
1 2
𝜌𝑣
2 ∞
−1/5
7A.3: Compressed gas flow in a cylindrical pipe.
Gaseous nitrogen is in isothermal turbulent flow at 25 oC through a straight length of horizontal pipe
with 3-in. inside diameter at a rate of 0.28 lbm/s. The absolute pressures at the inlet and outlet are 2 atm
and 1 atm respectively. Evaluate 𝐸𝑣 , assuming ideal gas behavior and racially uniform velocity
distribution.
Solution
For steady isothermal, horizontal pipe flow of an ideal gas at T=25 oC=536.7 R with flat velocity profiles
and 𝑊𝑚 = 0, equation 7.4-7 gives,
2
𝐸𝑣 = −
1
1
𝑣22 𝑣12
𝑅𝑇 𝑝1 1 𝑤
𝑑𝑝 −
−
=
𝑙𝑛 +
𝜌
2
2
𝑀 𝑝2 2 𝑆
2
1
1
2− 2
𝜌1 𝜌2
Let's take each of the pieces separately as this may be more straightforward.
4.9686𝑥104 𝑙𝑏𝑚 𝑓𝑡 2
536.7 𝑅
𝑅𝑇 𝑝1
𝑓𝑡 2
𝑠2 𝑙𝑏 𝑚𝑜𝑙 𝑅
𝑙𝑛 =
ln 2 = 6.60𝑥105 2
𝑙𝑏𝑚
𝑀 𝑝2
𝑠
28.01
𝑙𝑏 𝑚𝑜𝑙
and with 𝜌 = 𝑝𝑀/𝑅𝑇 in corresponding units,
𝜌1 = 2 𝑥 2116.2 𝑙𝑏𝑓 /𝑓𝑡 2 28.01 𝑙𝑏𝑚 /𝑙𝑏 𝑚𝑜𝑙 1544.3 𝑓𝑡 𝑙𝑏𝑓 /𝑙𝑏 ∙ 𝑚𝑜𝑙 ∙ 𝑅 = 0.1430 𝑙𝑏𝑚 /𝑓𝑡 3
We are given that 𝜌1 = 2𝜌2 in the problem statement.
Next,
2
1 𝑤
2 𝑆
2
1
1
1 0.28 𝑙𝑏𝑚 /𝑠
2− 2 =2
𝜋
𝜌1 𝜌2
64 𝑓𝑡 2
𝑓𝑡 2
= −2.39𝑥103 2
𝑠
1
0.1430 𝑙𝑏𝑚 /𝑓𝑡 3
2
−
1
0.0.715 𝑙𝑏𝑚 /𝑓𝑡 3
Combining these results, we obtain
𝐸𝑣 = 6.60𝑥105
𝑓𝑡 2
𝑓𝑡 2
𝑓𝑡 2
3
5
−
2.39𝑥10
=
6.58𝑥10
𝑠2
𝑠2
𝑠2
Using the information in Table F.3-3 to convert the units,
𝐸𝑣 = 6.58𝑥105
𝑓𝑡 2
𝐵𝑇𝑈
𝐽
= 26.3
= 6.11𝑥104
2
𝑠
𝑙𝑏𝑚
𝑘𝑔
2