18. (II) At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate ehat input of the second. Th eoperating temeprautres of the first are 680 degrees C and 430 degrees C, and of the second 415 degrees C and 280 degrees C. If the heat of combuston of coal is 2.8 E8 J/kg, at what rate msut coal be burned if the plant is to put out 900 MW of power. Assume the efficiency of the engines is 65 percent of the ideal (Carnot) efficiency.
Ok I calculated the efficiency of the first engine to be about 17.05 % efficient and the second one to be about 14.64 % efficient.
I do not know were to go from here. Please show me the formulas you use. Thanks You!
I know that efficiency is
e = |W|/|QH|
were the H is the subscript
I know that 900 MW is a value of power which is not work
I know that |QH| is 2.8 E8 J/kg but I do not like the units for this, J/kg, as I thought Q was measured in joules only...
How do I find the work done?
I need help with this, thanks!
the work that engine one does is (using your numbers)
.65*.17Heatinput
Then the waste heat is
Heatinput- work doneinengine1
But this is used as input to engine2, so the work done in engine 2 is
Heatinput*massrate(1-.65*.17)(.65*.14)
But the total work done is
Heat input*massrate(.65*.17)+Heatinput*massrate(1-.65*.17)(.65*.14)
but we want that to be 900Mwatts,
900M=
Heat input*massrate(.65*.17)+Heatinput*massrate(1-.65*.17)(.65*.14)
Solve for mass rate in kg/sec
Do not multiply the percentages you calculated by .65, because you already multiplied it to get 17% and 14.6% :)
To solve this problem, we can start by calculating the Carnot efficiency for each engine and then compute the actual efficiency considering the Carnot efficiency. The Carnot efficiency is given by the formula:
Efficiency_carnot = 1 - (Tc / Th)
Where Tc is the cold temperature and Th is the hot temperature.
For the first engine:
Tc1 = 430 °C = 703 K
Th1 = 680 °C = 953 K
Efficiency_carnot1 = 1 - (703 K / 953 K) ≈ 0.2646 ≈ 26.46 %
For the second engine:
Tc2 = 280 °C = 553 K
Th2 = 415 °C = 688 K
Efficiency_carnot2 = 1 - (553 K / 688 K) ≈ 0.1962 ≈ 19.62 %
Now, we need to find the actual efficiency of each engine considering the Carnot efficiency and the given efficiency factor (65% = 0.65). The actual efficiency is given by the formula:
Efficiency_actual = Efficiency_carnot * Efficiency_factor
For the first engine:
Efficiency_actual1 ≈ 0.2646 * 0.65 ≈ 0.1719 ≈ 17.19 %
For the second engine:
Efficiency_actual2 ≈ 0.1962 * 0.65 ≈ 0.1274 ≈ 12.74 %
Next, we can calculate the total efficiency of the power plant. Since the engines work in pairs and the heat output of the first engine is the approximate heat input of the second, we can assume they are connected in series. The overall efficiency of the power plant is given by the product of the efficiencies of the two engines:
Total Efficiency = Efficiency_actual1 * Efficiency_actual2 ≈ 0.1719 * 0.1274 ≈ 0.0219 ≈ 2.19 %
Now, we can calculate the energy required to produce 900 MW of power using the formula:
Energy = Power / Efficiency
First, let's convert the power to watts:
900 MW = 900 * 10^6 W
Then, using the total efficiency, we can calculate the energy required:
Energy = (900 * 10^6 W) / (0.0219) ≈ 4.1105 * 10^10 J
The heat of combustion of coal is given as 2.8 * 10^8 J/kg. To find the rate at which coal must be burned, we can divide the energy by the heat of combustion:
Rate of coal burned = Energy / Heat of combustion
Rate of coal burned ≈ (4.1105 * 10^10 J) / (2.8 * 10^8 J/kg)
Rate of coal burned ≈ 147.165 kg/s
Therefore, to produce 900 MW of power, approximately 147.165 kg/s of coal must be burned at the given efficiency.