Homework SourseConsider a coal fired plant converting fuel energy into mech
Consider a coal fired plant converting fuel energy into mechanical energy and then to electrical energy. In a typical day, the plant uses 800 ton of coal. The coal energy value is 25000 kj/kg. a). Only 30% of burning coal energy is converted to the electrical energy; Calculate it.
b). 80% of the wasted energy(heat) is used by cooling water taken from a nearby river. The cooling water temperature rises 12 degrees celcius during the cooling process.Compute the flow rate of the water taken from the lake.
c). The temperature of water in the river upstream and downstream of the plant are 15 and 20 degrees celcius respectively. Determine the flow rate of the river.
Solution
Mass of coal M = 800 ton
= 800 x 1000 kg
= 8 x10 5 kg
Coall energy E = 25000 kJ / kg
= 25000 x10 3 J / kg
= 25 x10 6 J / kg
Total coal energy E \' = ME
= 8 x10 5 x 25 x10 6 J
= 2 x10 13 J
(a). Electrical energy E \" = 30 % of E \'
= 0.3 E \'
= 0.3 x2 x10 13 J
= 0.6 x10 13 J
(b).wated energy Ew = 70 % of E \'
= 0.7 ( 2 x10 13 J )
= 1.4 x10 13 J
80 % of wasted energy e = 80 % of Ew
= 0.8 Ew
= 0.8 x1.4 x10 13 J
= 1.12 x10 13 J
Waste energy given to water per day = 1.12 x10 13 J
Waste energy given to water per second e \' = (1.12 x10 13 ) / (24 x60x60 ) Since 1 day = 24 x60 x60 s
= 129.62 x10 6 J/s
Temprature raise dt = 12 o C
We know e \' = mCdt
From this rate of mass m = e \' /(C dt )
Where C = Specific heat of water = 4186 J / kg o C
Substitute values you get ,
m = (129.62 x10 6 ) /(4186 x 12)
= 2580.6 kg/s
(c). dt = 5 o C
m = (129.62 x10 6 ) /(4186 x 5)
= 6193 kg/s
