Answer:
<em> - 14.943 W/m^2K ( negative sign indicates cooling ) </em>
Explanation:
Given data:
Area of FPC = 4 m^2
temp of water = 60°C
flow rate = 0.06 l/s
ambient temperature = 8°C
exit temperature = 49°C
<u>Calculate the overall heat loss coefficient </u>
Note : heat lost by water = heat loss through convection
m*Cp*dT = h*A * ( T - To )
∴ dT / T - To = h*A / m*Cp ( integrate the relation )
In ( 
) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )
In ( 41 / 52 ) = 0.0159*h
hence h = - 0.2376 / 0.0159
= - 14.943 W/m^2K ( heat loss coefficient )
Answer:
Sell his crop, use his crop as food, and sell his crop
Explanation:
Answer:
The power developed by engine is 167.55 KW
Explanation:
Given that
Mean effective pressure = 6.4 bar
Speed = 2000 rpm
We know that power is the work done per second.
So
We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.
P=167.55 KW
So the power developed by engine is 167.55 KW
Answer:
R = ![\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%26cos30%26-sin30%5C%5C0%26sin30%26cos30%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcos%2060%26-sin60%260%5C%5Csin60%26cos60%2660%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
Explanation:
The mappings always involve a translation and a rotation of the matrix. Therefore, the rotation matrix will be given by:
Let 
and 
be the the angles 60⁰ and 30⁰ respectively
that is = 60⁰ and
= 30⁰
The matrix is given by the following expression:
The angles can be evaluated and left in the surd form.
