where Rn = (R* vel)/v Reynolds number; R = hydraulic radius (= A/P_w) (given in
ID: 804610 • Letter: W
Question
where Rn = (R* vel)/v Reynolds number; R = hydraulic radius (= A/P_w) (given in m); vel = mean flow velocity (m/s); and v (i.e., nu) = kinematic viscosity (m^2/s) [note that there are different types of viscosity; kinematic viscosity = v = mu/rho (that is, kinematic viscosity equals the dynamic viscosity (mu) in Pascal seconds divided by density in kg per m^3)]. For streams with non-turbid aqueous conditions, the v value varies only slightly with temperature and can be treated as a constant (v_water = 1 * 10^-6 m^2/s = 0.000001 m^2/s; v_castor oil = 1 * 10^-3 m^2/s = 0.001 m^2/s). Hydraulic radius (R) is sometimes replaced by stream depth (h) to simplify Rn calculations, although in this lab exercise we'll use R. A small stream has a rectangular cross section with width of 5 m and depth of 2 m. The mean velocity of flow of the stream is 2 m/s. Using the Reynolds number as a guide (equation 3), is the water flow of the stream relatively laminar in nature or turbulent? Please explain, b) Keeping all other factors constant, what minimum viscosity would the flow in this stream need to have in order to be characterized by laminar flow (i.e., Rn = 500). Please show your work, c) How does this viscosity compare with that of castor oil (vegetable oil derived from the castor bean; 0.001 m^2/s) and honey (0.01 m^2/s)?Explanation / Answer
Given , width = 5m
depth = 2m
mean velocity = 2m/s
Rn = (R * vel ) / viscosity
= R = hydraulic radius = A / P
area = width *depth = 2*5 = 10m2
perimeter = 2( width + depth )
= 2 (2 + 5)
14
R = 10 /14 = 0.71 m
Rn = (0.71 * 2 ) / 0.000001
= 1420000
hence it is greater than 4000 then it is turbulant flow.
b) Rn = 500
R = 0.71
vel = 2m
viscosity = ?
500 = (0.71 * 2 ) / viscosity
viscosity = (0.71 * 2) / 500
= 0.00284m2 /s
c)this viscosity is in middle of the both castor oil and honey.
0.001 <0.0028
0.01>0.0028
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