Answer:
The minimum particle diameter that is removed at 85% is 1.474 * 10 ^⁻4 meters.
Solution
Given:
Length = 48 m
Width = 12 m
Depth = 3m
Flow rate = 4 m 3 /s
Water density = 10 3 kg/m 3
Dynamic viscosity = 1.30710 -3 N.sec/m
Now,
At the minimum particular diameter it is stated as follows:
The Reynolds number= 0.1
Thus,
0.1 =ρVTD/μ
VT = Dp² ( ρp- ρ) g/ 10μ²
Where
gn = The case/issue of sedimentation
VT = Terminal velocity
So,
0.1 = Dp³ ( ρp- ρ) g/ 10μ²
This becomes,
0.1 = 1000 * dp³ (1100-1000) g 0.1/ 10 *(1.307 * 10 ^⁻3)²
= 3.074 * 10 ^⁻6 = dp³ (.g01 * 10^6)
dp³=3.1343 * 10 ^⁻12
Dp minimum= 1.474 * 10 ^⁻4 meters.
Answer:
When water is surrounding T_s = 34.17 degree C
When air surrounding T_S = 1434.7 degree C
from above calculation we can conclude that air is less effective than water as heat transfer agent
Explanation:
Given data:
length = 300 mm
Outer diameter = 30 mm
Dissipated energy = 2 kw = 2000 w
Heat transfer coefficient IN WATER = 5000 W/m^2 K
Heat transfer coefficient in air = 50 W/m^2 K
we know that 
From newton law of coding we have
is surface temp.
T - temperature at surrounding
![P = hA(T_s - T_{\infity})[tex]\frac{P}{\pi hDL} = (T_s - T_{\infity})](https://tex.z-dn.net/?f=P%20%3D%20hA%28T_s%20-%20%20T_%7B%5Cinfity%7D%29%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cfrac%7BP%7D%7B%5Cpi%20hDL%7D%20%3D%20%20%28T_s%20-%20%20T_%7B%5Cinfity%7D%29)
solving for[/tex] T_s [/tex] w have
When air is surrounding we have
from above calculation we can conclude that air is less effective than water as heat transfer agent
Explanation:
Power = work / time
Power = force × distance / time
P = (35 lbf) (6 in) / (0.7 s)
P = 300 lbf in/s
I think it's false because the theory states that electrons flow from negative to positive. Since electrons are negatively charged, it follows that they are attracted by positively charged bodies and repelled by negatively charged bodies.
