A rigid tank contains 5 kg of saturated vapor steam at 100°C. The steam is cooled to the ambient temperature of 25°C. (a) Sketch
the process with respect to the saturation lines on a T-vdiagram. Indicatepressure valuesat each state. (5pt) (b) Determinethe entropy change of the steam, in kJ/K. (10 pt) (c) Determinethe totalentropy change associated with this process, in kJ/K. (15pt)
1 answer:
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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:
Explanation:
The schedule using shortest remaining time, non-preemptive priority and round Robin with quantum number 30 is shown in the attached file, please kindly go through it to access the answer.
