Answer:
Explanation:
Given:
Diameter of aluminum wire, D = 3mm
Temperature of aluminum wire, 
Temperature of air, 
Velocity of air flow 
The film temperature is determined as:
from the table, properties of air at 1 atm pressure
At 
Thermal conductivity, 
; kinematic viscosity 
; Prandtl number 
The reynolds number for the flow is determined as:
sice the obtained reynolds number is less than 
, the flow is said to be laminar.
The nusselt number is determined from the relation given by:
![Nu_{cyl}= 0.3 + \frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(\frac{0.4}{Pr})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{Re}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}](https://tex.z-dn.net/?f=Nu_%7Bcyl%7D%3D%200.3%20%2B%20%5Cfrac%7B0.62Re%5E%7B0.5%7DPr%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D%7B%5B1%2B%28%5Cfrac%7B0.4%7D%7BPr%7D%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5D%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%7D%5B1%2B%28%5Cfrac%7BRe%7D%7B282000%7D%29%5E%7B%5Cfrac%7B5%7D%7B8%7D%7D%5D%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D)
![Nu_{cyl}= 0.3 + \frac{0.62(576.92)^{0.5}(0.70275)^{\frac{1}{3}}}{[1+(\frac{0.4}{(0.70275)})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{576.92}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}\\\\=12.11](https://tex.z-dn.net/?f=Nu_%7Bcyl%7D%3D%200.3%20%2B%20%5Cfrac%7B0.62%28576.92%29%5E%7B0.5%7D%280.70275%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D%7B%5B1%2B%28%5Cfrac%7B0.4%7D%7B%280.70275%29%7D%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5D%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%7D%5B1%2B%28%5Cfrac%7B576.92%7D%7B282000%7D%29%5E%7B%5Cfrac%7B5%7D%7B8%7D%7D%5D%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D%5C%5C%5C%5C%3D12.11)
The covective heat transfer coefficient is given by:
Rewrite and solve for 
The rate of heat transfer from the wire to the air per meter length is determined from the equation is given by:
The rate of heat transfer from the wire to the air per meter length is 
Answer:
Explanation:
The pictures below shows the whole explanation for the problem
Answer:
Explanation:
Lets do this in python, our function will import an array of double numbers, then it sort the array out, drop the first and last items, aka highest and lowest score. Finally, it averages the last 3 numbers by calculate the sum and divide by the number of items
def calculate_score(scores):
scores.sort() # sort it so that the lowest is at index 0 and the highest is at last index
drop_highest_lowest = scores[1:-1] # this will drop the lowest and highest number
average_score = sum(drop_highest_lowest)/len(drop_highest_lowest)
return average_score
For efficient treatment, the dissolved oxygen concentration in the aeration tank must be kept between 1-3 mg/L. The microbial biomass will die at low DO levels, and it will take time and money to rebuild it.
The depth, which normally ranges from 3 to 4.5 meters, determines how effectively air is aerated. The breadth, which is typically maintained between 5 and 10 meters, regulates the mixing. The ideal width-to-depth ratio is 1.2 to 2.2. The distance shouldn't be shorter than 30 meters or longer than 100 meters. Bacteria used in wastewater treatment and stabilization are given oxygen by aeration. The bacteria require oxygen for biodegradation to take place.
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Answer:
What is the minimum altitude for orbit?
Technically, objects in low-Earth orbit are at an altitude of between 160 to 2,000 km (99 to 1200 mi) above the Earth's surface. Any object below this altitude will being to suffer from orbital decay and will rapidly descend into the atmosphere, either burning up or crashing on the surface.
Explanation:
orbital-mechanics orbit low-earth-orbit altitude. The altitude of a satellite is the distance between the Earth's surface and the satellite, but the Earth itself is not spherical. At the equator the Earth's radius is 21 km more than at the poles, and in fact the shape of the Earth is not even a perfect oblate spheroid.A Japan Aerospace Exploration Agency (JAXA) satellite has been certified by Guiness World Records as setting a new altitude orbit record. However, while most satellites orbit high above the Earth, the JAXA satellite Tsubame was recognized for achieving the lowest altitude, of only 167.4 kilometers.
