Question

Our power plant generates electricity using a steam turbine. At the exit from our power generation process, we end up with  978.0 kg/s of liquid water at  323.0 K. We cannot release this water directly into a nearby river if the temperature is over  305.00 K, without causing significant thermal pollution that would damage fish populations. Our plan is to mix this warm water with another cooler water stream at  282.00 K. What is the mass flow rate of this cool water stream needed to meet the regulation? Assume the heat capacity for all streams is 4.184 J/gK.

123Warm

Water fromTurbineCoolWaterWater to RiverMixer

(HINT: |X| is near an order of magnitude of 10² kg/s ).

Answer 1

Mass of hot water mh = 978 kg/s

Specific heat capacity of hot water Cph = 4.184 J/gK

Initial temperature of hot water Th1 = 323 K

Final temperature of hot water Th2 = 305 K

Mass of cold water mc =?

Specific heat capacity of cold water Cpc = 4.184 J/gK

Initial temperature of cold water Tc1 = 282 K

Final temperature of cold water Tc2 = 305 K

Energy released by hot water = energy absorbed by cold water

mh x Cph x (Th1 - Th2) = mc x Cpc x (Tc2 - Tc1)

978 kg/s x 4.184 J/gK x (323 - 305)K = mc x 4.184 J/gK x (305 - 282)K

17604 = mc x 23

mc = 765.39 kg/s