A 300 MWe (electrical power output) Power Plant having a thermal efficiency of 40% is cooled by sea water. Due to environmental regulations the seawater can only increase temperature by 5 C during the process. How much sea water (minimum) must be used in kg/s for cooling if the plant operates at it's rated capacity?

m = 22,877 kg / s

Explanation:

Let's solve this exercise in parts, first look for the amount of heat generated by the plant and then the amount of water to dissipate this heat

The plant generates a power of 300 MW at a rate of 40%, let's use a direct ratio rule to find the heat. If the power is 400 MW it corresponds to 40%, what heat (Q) corresponds to the other 60%

           Q = 300 60% / 40%

           Q = 450 MW

having the amount of heat generated we can use the calorimeter equation,

           Q = m

            m = Q / c_{e} (T_{f} - T_{o})

let's use the maximum temperature change allowed

           (T_{f} - T_{o}) = 5

the specific heat of sea water is 3934 J / kg ºC, note that it is less than that of pure water, due to the salts dissolved in sea water

power and energy are related

              W = Q / t

               Q = W t

let's calculate

             m = 450 10⁶ / (3934 5)

             m = 22,877 kg / s

answer: i'm not sure which is which for i get confused on it all the time but i can tell you that the answer lies with the convex and concave devices. i hope this you. have a good day and best of luck to you!

explanation:

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.