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:
From first law of thermodynamics(energy conservation)
Qa= Qr+W
Qa=Heat added to the engine
Qr=heat rejected from the engine
W=work output from the engine
Second law:
It is impossible to construct a heat engine that will deliver the work with out rejecting heat.
In other word ,if engine take heat then it will reject some amount heat and will deliver some amount of work.
1.
QH=6 kW,
QL=4 kW,
W=2 kW
6 KW= 4 + 2 KW
It satisfy the first law.
Here heat is also rejected from the engine that is why it satisfy second law.
2.
QH=6 kW, QL=0 kW, W=6 kW
This satisfy first law but does not satisfy second law because heat rejection is zero.
3.
QH=6 kW , QL=2 kW, W=5 kW
This does not satisfy first as well as second law.Because summation of heat rejection and work can not be greater than heat addition or we can say that energy is not conserve.
4.
QH=6 kW, QL=6 kW, W=0 kW
This satisfy first law only and does not satisfy second law.
Answer:
It's job is to block the flow of coolant to the radiator until the engine has warmed up. Once the engine reaches its operating temperature (generally about 200 degrees F, 95 degrees C), the thermostat opens. By letting the engine warm up as quickly as possible, the thermostat reduces engine wear, deposits and emissions.
