57:020 Mechanics of Fluids and Transport
December 3, 2014
NAME
Fluids-ID
Quiz 12. The pump shown in Figure delivers a head of 250 ft to the water. The differene in elevation of
the two ponds is 200 ft. (P =ρgQhp; ρ = 1.94 slugs/ft3; μ= 2.34×10-5 lb⋅s/ft2; g = 32.2 ft/s2; 1 hp = 550
ft⋅lbf/s; Reynolds number, Re = ρVD/µ)
Energy Equation
𝑝𝑝2 𝑉𝑉22
𝑉𝑉 2 𝑓𝑓ℓ
𝑝𝑝1 𝑉𝑉12
+
+ 𝑧𝑧1 + ℎ𝑝𝑝 =
+
+ 𝑧𝑧2 +
� + � 𝐾𝐾𝐿𝐿 �
𝜌𝜌𝜌𝜌 2𝑔𝑔
𝜌𝜌𝜌𝜌 2𝑔𝑔
2𝑔𝑔 𝑑𝑑
Friction Factor Equation (The Haaland eq.)
1
�𝑓𝑓
= −1.8 log ��
1.11
𝜀𝜀 ⁄𝑑𝑑
�
3.7
+
6.9
�
𝑅𝑅𝑅𝑅
Note: Attendance (+2 points), format (+1 point)
a) Simplify energy equation using the given conditions and determine velocity, V, as a function of
friction factor, f.
b) Use the given conditions and determine Reynolds number, Re, as a function of velocity, V.
c) Determine velocity V by following the steps listed below
1) Assume f = 0.02 as your first guess and find V using the equation from (a)
2) Find Re using the equation from (b) and the V from the previous step
3) Find a new f using the Haaland equation and Re from step 2)
4) Find a new V using the f from step 3) and the equation from (a)
5) Repeat the steps 2) through 4) until f is converged to the thousandth decimal point
d) Determine the power P that pump adds to the water.
Solution:
Where p1 = p2,V1 = V2 ≈ 0, the energy equation becomes
𝑧𝑧1 + ℎ𝑝𝑝 = +𝑧𝑧2 +
𝑉𝑉 2 𝑓𝑓ℓ
� + � 𝐾𝐾𝐿𝐿 �
2𝑔𝑔 𝑑𝑑
𝑓𝑓ℓ
2𝑔𝑔 × (ℎ𝑝𝑝 + 𝑧𝑧1 − 𝑧𝑧2 ) = 𝑉𝑉 2 � + � 𝐾𝐾𝐿𝐿 �
𝑑𝑑
2 × 32.2(250 − 200) = 𝑉𝑉 2 �𝑓𝑓
500
+ 0.8 + 4 × 1.5 + 5 + 1�
0.75
(667𝑓𝑓 + 12.8)𝑉𝑉 2 = 3220
(+2 point)
57:020 Mechanics of Fluids and Transport
December 3, 2014
3220
𝑉𝑉 = �(667𝑓𝑓+12.8) (1)
(+1 point)
Reynolds number
𝑅𝑅𝑅𝑅 =
𝜌𝜌𝜌𝜌𝜌𝜌 1.94 × 𝑉𝑉 × 0.75
=
2.34 × 10−5
𝜇𝜇
𝑅𝑅𝑅𝑅 = 6.22 × 104 𝑉𝑉 (2)
(+1 point)
Rearranging friction factor equation
6.9
𝑅𝑅𝑅𝑅
−2
𝑓𝑓 = �−1.8 log � �� (3)
Solving for velocity iteratively using equations (1), (2) and (3)
Assume 𝑓𝑓 = 0.02 → 𝑉𝑉 = 11.1
𝑓𝑓𝑓𝑓
𝑠𝑠
Assume 𝑓𝑓 = 0.012 → 𝑉𝑉 = 12.4
Thus
→ 𝑅𝑅𝑅𝑅 = 6.9 × 105 → 𝑓𝑓 = 0.012
𝑓𝑓𝑓𝑓
𝑠𝑠
→ 𝑅𝑅𝑅𝑅 = 7.7 × 105 → 𝑓𝑓 = 0.0121
𝑉𝑉 = 12.4
𝑓𝑓𝑓𝑓
𝑠𝑠
(+2 point)
Calculating pump power
𝑃𝑃 = 𝜌𝜌𝜌𝜌𝜌𝜌ℎ𝑝𝑝 = 𝜌𝜌𝜌𝜌𝜌𝜌𝜌𝜌ℎ𝑝𝑝
𝑃𝑃 = 1.94 × 32.2 × 12.4 ×
𝑓𝑓𝑓𝑓 𝑙𝑙𝑙𝑙
𝜋𝜋
(0.75)2 × 250 = 8.55 × 104
= 155 ℎ𝑝𝑝
𝑠𝑠
4
(+1 point)