Answer:
a.m=7596679.24kg b. Vf=527.77777m3/s c. ds=6064755J
Explanation:
The precursor to this question would have been
a. how much coal is needed by the plant
fm the relation Qh/t* 1day(86400)
Qh=2330MW*86400
Qh=MJ
201.3*10^6MJ/2.65×10^7J/kg
m=7596679.24kg
b. firstly heat from the river is given as 1330MW
Recall that Q=mcΔ∅
Q=Quantity of required
m=mass flow rate of the river
c= specific eat capacity of water
Δ∅=change in temperature
divide bot sides bt time t
Qc/t=(m/t)cΔ∅
1330=(m/t)*4200*(18.6-18)
(m/t)=1330*10^6/2520
(m/t)=527777.77kg/s
recall = density=mass rate/volumetric rate
1000kg/m3=527777.77kg/s/Vf
Vf=527.77777m3/s
3.ΔS=mcIn(t2/t1)
ΔS=527777.77*4200In(219.6/219)
=6064755J
ΔS= Entropy change