what is it ill know your question??!!
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:
T = 20.42 N
Explanation:
given data
standard altitude = 30,000 ft
velocity Ca = 500 mph = 0.4 m/s
inlet areas Aa = 7 ft² = 0.65 m²
exit areas Aj = 4.5 ft² = 0.42 m²
velocity at exit Cj = 1600 ft/s = 487.68 m/s
pressure exit
j = 640 lb/ft² = 0.3 bar
solution
we get here thrust of the turbojet that is express as
thrust of the turbojet T = Mg × Cj - Ma × Ca + (
j Aj -
a Ag ) .............1
here Ma = Mg
Ma =
a × Ca Aa = 0.042 kg/s
put value in equation 1 we get
T = 0.042 × (487.68 -0.14) + ( 0.3 × - 0.3 × 0.65 )
T = 20.42 N
Answer:
A continuity test
Explanation:
A continuity test is used to verified that current will flow in an electrical circuit, it performed by placing a small voltage across the chosen path. continuity test ensure that the equipment grounding conductor is electrically continuous and this test is perform on all the cord sets, receptacles that aren't part of a building or structure's permanent wiring, and cord-and-plug connected equipment required to be grounded. example of equipment used in testing current flow in continuity test are Analog multi-meter, voltage/continuity tester etc.
Continuity test and terminal connection test are the two test required by OSHA on all electrical equipment