Question

Thermal pollution in wastewater. A 1000-MW coal-fired power plant with an efficiency of 35% uses water from a neighboring stream (with flow of 75 m³/s and 20°C) to carry away waste heat. About 12% of waste heat goes up in the smokestack; the remaining 88% is taken away by cooling water.

(a) Calculate the amount of waste heat (inkWorkJ/s) to be removed by cooling water

(b) If the cooling water is only allowed to rise in temperature by 11°C, what is the flow rate (in m3/s) of water required for cooling purposes?

(c) Whatwouldbetherivertemperature(in°C)just after it receives the heated cooling water?

(d) Why do we need to consider waste heat in considering stream water quality? Note: in addition to

worrying about temperature sensitivity of aquatic life, we also need to consider oxygen availability.

Answer 1

Given effieciency = 35% therefore Total rate at which energy enters power plant is:- input power = output power/eta = 1000/0.35 = 2857.1428 MW Total losses = 2857.1428 � 1000 = 1857.1428 MW Given 12% of waste heat goes rightarrow smoke stack = 0.12 times 1857.1428 = 222.857 MW Remaining 88 is taken by coding water:- = 0.88 times 1857.1428 = 1634.2856 MW therefore Amount of waste heat to be removed by coding water = 1634.2856 MW = 1.63428 times 10^6 KJ/sec Given temperature increase delta t = 11 degree c therefore coding water needed to remove 1634.2856 MW is:- Rate of change in stored energy = mc delta T 1634.2856 = m degree times 4184 J/kg degree c times 11 times 1MW (10^-6) therefore m degree = 35.51 times 10^3 kg/sec since 1000kg = 1m^3 of water therefore the flow rate is 35.5 m^3/sec Given river has a flow