Answer Happy

In a large steam power plant, the combustion of coal provides a
heat rate of 𝑞𝑖𝑛 = 2500𝑀𝑊at a flame temperature of 𝑇ℎ = 1000𝐾. Heat
is rejected from the plant to a river flowing at 𝑇𝑐 = 300𝐾. Heat is
transferred from the combustion products to the exterior of large
tubes in the boiler by way of radiation and convection, through the
boiler tubes by conduction, and then from the interior tube surface
to the working fluid (water) by convection. On the cold side, heat
is extracted from the power plant by condensation of steam on the
exterior condenser tube surfaces, through the condenser tube walls
by conduction, and from the interior of the condenser tubes to the
river water by convection. Hot and cold side thermal resistances
account for the combined effects of conduction,
convection, and radiation and, under design
conditions, they are 𝑅𝑡,ℎ = 8 ∗ 10−8𝐾/𝑊 and 𝑅𝑡,𝑐 = 2 ∗
10−8𝐾/𝑊 , respectively.
Products of combustion in = 2500 MW = Th = 1000 K = Rih 8 x 10-8 K/W (clean) Rih = 9 x 10 th 9 K/W (fouled) -8 = Thi Power plant W W T. ci = -8 R,c = 2 x 10 K/W (clean) -8 = 2.2 x 10 K/W (fouled) Ric = Yout Cooling water To = 300 K =