Answer:
1.96 kg/s.
Explanation:
So, we are given the following data or parameters or information which we are going to use in solving this question effectively and these data are;
=> Superheated water vapor at a pressure = 20 MPa,
=> temperature = 500°C,
=> " flow rate of 10 kg/s is to be brought to a saturated vapor state at 10 MPa in an open feedwater heater."
=> "mixing this stream with a stream of liquid water at 20°C and 10 MPa."
K1 = 3241.18, k2 = 93.28 and 2725.47.
Therefore, m1 + m2= m3.
10(3241.18) + m2 (93.28) = (10 + m3) 2725.47.
=> 1.96 kg/s.
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:
William Shockley, Walter Houser Brattain and John Bardeen.
Explanation:
It was built in 1947 and they won the novel peace prize in 1956
Answer:Antifreeze/coolant
Explanation: keeps your engine cool in warm weather and keeps it from freezing up in the winter. A 50-50 mix of full strength coolant and water generally protects to around -30 degrees Fahrenheit. Make sure you check with the supplier or your owner's manual for the correct formulation
