![2-Oil with 𝝁 = 2. x 10−4 Ib−sec/ft
𝟐
flowing in a 4-inch (10.16 cm)
diameter pipe has a parabolic velocity profile given by:
v = 0.45[1 - (r/R)2] fps , v = 0.1372[1 - (r/R)2] mps, in which r is the radial distance
from the pipe centerline and R is the radius of the pipe.
What is the shear stress in the oil at r = 0“ , 1" (2.54 cm) and 2" (5.08 cm)?
Sol
𝝉 = 𝝁
𝒅𝒗
𝒅𝒕
= 𝟐 ∗ 𝟏𝟎−𝟒Ib−sec/ft
𝟐
∗ 𝟒. 𝟒𝟒𝟖𝟐𝟐 𝑵 ∗
𝟏
𝟎. 𝟎𝟗𝟐𝟗𝟎𝟑𝟎𝟒 𝒎𝟐
𝒅 𝟎. 𝟏𝟑𝟕𝟐 1 −
r
R
𝟐
𝒅𝒕
𝝉 = 𝟎. 𝟎𝟎𝟗𝟓𝟗𝟓 𝑵. 𝒔/𝒎𝟐
(𝟎. 𝟐𝟕𝟒𝟒 ∗ 𝒓/𝑹𝟐
)](https://image.slidesharecdn.com/examples1-240625163237-ef42ae17/85/Examples-1-pptxAdd-more-information-to-your-upload-3-320.jpg)
Examples (1).pptxAdd more information to your upload
- 1. EXAMPLE 1
- 2. 1- A container with a volume of 8.5 ft3 (0.241 m3) is filled with a liquid. The container and liquid weigh 650 lbs (2890 m) and the container weighs 55 Ibs (244.6 N). Determine the density and specific weight of the liquid 2- specific weight= weight / volume =2645.4 /0.241 =10976.7 N/m3 1- Density = (mass)/ volume = (weight/ g) / volume =(2645.4/9.81) / 0.241 =1118.9 kg/m3 weight of the liquid= (container and liquid) weigh -container weighs= 2890-224.6 = 2645.4 N
- 3. 2-Oil with 𝝁 = 2. x 10−4 Ib−sec/ft 𝟐 flowing in a 4-inch (10.16 cm) diameter pipe has a parabolic velocity profile given by: v = 0.45[1 - (r/R)2] fps , v = 0.1372[1 - (r/R)2] mps, in which r is the radial distance from the pipe centerline and R is the radius of the pipe. What is the shear stress in the oil at r = 0“ , 1" (2.54 cm) and 2" (5.08 cm)? Sol 𝝉 = 𝝁 𝒅𝒗 𝒅𝒕 = 𝟐 ∗ 𝟏𝟎−𝟒Ib−sec/ft 𝟐 ∗ 𝟒. 𝟒𝟒𝟖𝟐𝟐 𝑵 ∗ 𝟏 𝟎. 𝟎𝟗𝟐𝟗𝟎𝟑𝟎𝟒 𝒎𝟐 𝒅 𝟎. 𝟏𝟑𝟕𝟐 1 − r R 𝟐 𝒅𝒕 𝝉 = 𝟎. 𝟎𝟎𝟗𝟓𝟗𝟓 𝑵. 𝒔/𝒎𝟐 (𝟎. 𝟐𝟕𝟒𝟒 ∗ 𝒓/𝑹𝟐 )
- 4. • R=10.16 cm / 2 = 5.08 cm • 𝝉 = 𝟎. 𝟎𝟎𝟗𝟓𝟗𝟓 𝑵. 𝒔/𝒎𝟐(𝟎. 𝟐𝟕𝟒𝟒 ∗ 𝒓/𝑹𝟐) 1. 𝒂𝒕 𝒓 = 𝟎 𝝉 =. 𝟎𝟎𝟗𝟓𝟗𝟓 . 𝟐𝟕𝟒𝟒 𝟎 .𝟎𝟎𝟐𝟓𝟖𝟏 = 𝑶 𝑵𝒔𝒎 2. 𝒂𝒕 𝒓 = 𝟐. 𝟓𝟒 𝝉 =. 𝟎𝟎𝟗𝟓𝟗𝟓 (. 𝟐𝟕𝟒𝟒) (. 𝟎𝟐𝟓𝟒/. 𝟎𝟎𝟐𝟓𝟖𝟏) . 𝟎𝟐𝟓𝟖𝟓 𝑵𝒔𝒎 3. 𝒂𝒕 𝒓 = 𝟓. 𝟎𝟖 𝝉 =. 𝟎𝟎𝟗𝟓𝟗𝟓 . 𝟐𝟕𝟒𝟒 .𝟎𝟓𝟎𝟖 .𝟎𝟎𝟐𝟓𝟖𝟏 = 𝟎. 𝟎𝟓𝟏𝟕 𝑵𝒔𝒎
- 5. 3- With what force would the flowing oil in problem 2 tend to pull 1000 ft (304.8 m) length of the 4“ (10.16 cm) pipe? Relate this force to the pressure drop in the pipe. Solution 𝐹 = 𝐴 ∗ 𝜏 = (2𝜋𝑅𝐿)𝜏 = 2 𝜋 0.0508 304.8 0.0517 = 5.03 𝑁 A control volume of the 1000 ft of fluid shows that the difference in pressure forces at the ends must equal the shear force, i.e. 𝐹𝜏 = 𝑃1 − 𝑃2 𝐴 = ∆𝑃 ∗ 𝐴 & ∆𝑃 = 𝐹𝑟/𝐴 𝐴 = 𝜋 4 ∗ 𝐷2 ∆𝑃 = 𝐹𝜏 𝐴 = 5.03 8.1073 ∗ 10−3 = 620 𝑁 𝑚2
- 6. 1-2 A 6-inch (0.1524 m) pipe is connected to an 8-inch (0.2032 m) pipe. If the average velocity V= 35 fps (1O.67 mps) in the 6-inch pipe what is the average velocity in the 8-inch pipe? Determine the mass flow rate, the weight flow rate, and the volumetric flow rate. • V A inlet =V A outlet A= 𝜋∗𝐷2 4 • V1=𝑉2 𝐷2 2 𝐷1 2 = 10.67 0.1524 0.2032 2 = 6 𝑚 𝑠 • 𝐺 = 𝜌 ∗ 𝐴 ∗ 𝑉 = 1000 ∗ 10.67 ∗ 𝜋∗0.15242 4 = 194.6 𝑘𝑔/𝑠 mass flow rate 1 • 𝑊 = 9.81 ∗ 𝐺 = 1909 𝑁/𝑠 weight flow rate 2 • 𝑄 = 𝐺 𝜌 = 194.6 1000 = 0.1946 𝑚3 𝑠 volumetric flow rate 3
- 10. 1/12=1ft
- 11. 60sec *10=600sec
- 16. V3=Q/A