Name :_________________ Quiz: No. 10 Time: 15 minutes Student ID# : ____________ Course: ME 5160, Fall 2021
--- The exam is closed book and closed notes.
In below figure the pipe entrance is sharp-edged. If the flow rate is 0.004 m
3/s, what power, in W, is extracted by the turbine?
Sharp edge entrance: K=0.5 Open globe valve: K 6.9
Water properties: 𝜌𝑤𝑎𝑡𝑒𝑟= 998 𝑘𝑔/𝑚3, 𝜇𝑤𝑎𝑡𝑒𝑟 = 0.001 Pa⋅s
ℎ𝑓 = 𝑓𝐿 𝐷
𝑉2
2𝑔, ℎ𝑚=𝑉2 2𝑔𝐾 Energy equation:
(𝑝 𝜌𝑔+𝑉2
2𝑔+ 𝑧)
1
+ ℎ𝑝= (𝑝 𝜌𝑔+𝑉2
2𝑔+ 𝑧)
2
+ ℎ𝑡+ ℎ𝑓+ ∑ ℎ𝑚
Name :_________________ Quiz: No. 10 Time: 15 minutes Student ID# : ____________ Course: ME 5160, Fall 2021
---
Solution:
Energy equation with given condition
(𝑝 𝜌𝑔+𝑉2
2𝑔+ 𝑧)
1
+ ℎ𝑝= (𝑝 𝜌𝑔+𝑉2
2𝑔+ 𝑧)
2
+ ℎ𝑡+ ℎ𝑓+ ∑ ℎ𝑚
∴
ℎ𝑡 =(
𝑧1− 𝑧2)
−𝑉222𝑔− ℎ𝑓−
∑
ℎ𝑚 ℎ𝑡= (𝑧1− 𝑧2) −𝑉222𝑔− 𝑓𝐿 𝑑
𝑉22 2𝑔−𝑉22
2𝑔(𝐾𝐸𝑛𝑡𝑟𝑎𝑛𝑐𝑒+ 𝐾𝑔𝑙𝑜𝑏𝑒 𝑣𝑎𝑙𝑣𝑒)
ℎ𝑡= (𝑧1− 𝑧2) −𝑉22
2𝑔[1 + 𝑓𝐿
𝐷+ 𝐾𝐸𝑛𝑡𝑟𝑎𝑛𝑐𝑒+ 𝐾𝑔𝑙𝑜𝑏𝑒 𝑣𝑎𝑙𝑣𝑒]
∴
ℎ𝑡 = 40 −𝑉222𝑔
[
1 + 𝑓𝐿𝐷+ 0.5 + 6.9
]
Calculate the velocity with flow rate 𝑉
2= 𝑄
𝐴 = 0.0004 𝜋
4 0.05
2= 2.037𝑚/𝑠
Calculate Reynolds number
𝑅𝑒
𝐷= 998 × 2.037 × 125
0.001 = 101,646 → 𝑓 ≈ 0.0316
∴
ℎ𝑡 = 40 − 2.03722 × 9.81
[
1 + 0.0316 × 1250.05+ 0.5 + 6.9
]
= 21.516𝑃
𝑡= 𝜌𝑔𝑄ℎ
𝑡= 998 × 9.81 × 0.004 × 21.516 = 842.6𝑊
(+2)
(+1) (+2.5)
(+1.5) (+1.5)
(+1.5)
