Asteam power plant produces 500 mw of electricity with an overall thermal efficiency of 35%. determine (a) the rate at which heat is supplied to the boiler and (b) the waste heat that is rejected by the plant. (c) if the heating value of coal (heat that is released when 1 kg of coal is burned) is 30 mj/kg, determine the rate of consumption of coal in tons(us)/day. assume that 100% of heat released goes to the cycle. (d) what-if scenario: what would the fuel consumption rate be if the thermal efficiency were to increase to 40%?

Explanation:

500 MW = 500 X 10⁶ J / s

If R be the rate of heat supplied and 35 % is the efficiency of conversion

R x .35 = 500 X 10⁶

R =  500 X 10⁶ / .35

= 1428.57 x 10⁶ J / s

b )

Rate of heat going waste

= .65 x 1428.57 x 10⁶

= 928 .57 x 10⁶ J / s

c )

Heat content of coal = 30 x 10⁶ J /kg

1428.57 x 10⁶ J of heat will come from

= 1428.57 x 10⁶ J  / 30 x 10⁶ J /kg

= 47.62 kg per second

= 47.62 x 24 x 60 x 60 =  4114368 kg in one day

= 4114368 / 907.185 ton per day  

= 4535.31 ton per day

d ) Fuel consumption rate  will decrease by 12.5  % if thermal efficiency is increased to 40 % .  

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz