7 Water, initially at a temperature of 500oC and a pressure of 1 MPa is cooled at constant pressure to a mixture of liquid and vapor with a quality of 70%. What is the entropy change.
Entropy is a property. The entropy at the initial state is found from the superheat table A-6 on page 919: s(1 MPa, 500oC) = 7.7643 kJ/kg·K. At the final state we have to compute the entropy using the saturation values at the final pressure, P2, = P1 = 1 MPa. At this pressure we find sf = 2.1381 kJ/kg·K and sfg = 4.4470 kJ/kg·K. Since the final quality x2 = 0.7 we have s2 = sf + x2sfg = (2.1381 kJ/kg·K) + (0.7)(4.4470 kJ/kg·K) = 5.2510 kJ/kg·K.
The entropy change s2 – s1 = 5.2510 kJ/kg·K – 7.7643 kJ/kg·K = –2.5133 kJ/kg·K
8 The inlet to an R-134a compressor is a saturated vapor at 15 psia. The outlet of the compressor has a pressure of 300 psia. The entropy of the outlet state is the same as that of the inlet. What is the enthalpy change between the inlet and outlet states?
The properties at the inlet state are found from table A-12E on page 977: sin = sg(15 psia) = 0.22715 Btu/lbm·R and hin = hg(15 psia) = 93.155 Btu/lbm. At the final state we know Pout = 300 psia and sout = sin = 0.22715 Btu/lbm·R. At 300 psia the entropy of 0.22715 Btu/lbm·R lies between table values of 0.21745 Btu/lbm·R and 0.22802 Btu/lbm·R. Interpolating the enthalpy value at the outlet condition gives the following result.
The enthalpy change h
out – h
in = 93.155 kJ/kg – 128.05 kJ/kg =
–34.90 kJ/kg
9 Conventional coal fired power plants cost $1,300 per kW to construct and have an efficiency of 34%. Advanced plants use the clean burning Integrated Coal Gasification Combined Cycle (IGCC) in which the coal is subjected to heat and pressure to gasify it while removing sulfur and particulate matter from it. The gaseous coal is then burned in a gas turbine, and part of the waste heat form the exhaust gases is recovered to generate steam for the steam turbine. Currently the construction of IGCC plants costs about $1,500 per kW, but their efficiency is about 45%. The average heating value of coal is about 28,000,000 kJ per ton (that is, 28,000,000 kJ of heat is released when one ton of coal is burned.) If the IGCC plant is to recover its cost difference from fuel savings in five years, determine what the cost of coal should be in $ per ton.
Here we assume that the economic analysis is based on a payback period where we do not account for the time value of money. The construction cost difference is $200 per kW. The amount of electricity, in kWh, generated in five years, per kW of capacity is equal to the time that the plant is used times the fraction of its average capacity that is used. If we assume for the best case to justify the IGCC plant that both plants are operated at full capacity for five years, the total hours of operation, assuming only one leap year in five years, will be (24 hours/day)(4*365+366 days) or a total of 43,824 hours. Thus each kW of capacity will produce a total of 43,824 kWh over the five-year period.
The amount of coal to produce this energy (from each kW of generating capacity) is determined by the efficiency of the plant and the heating value of the coal. For the conventional plant, the total coal use over five years (for each kW of generating capacity) is
For the IGCC plant, the total coal use over five years (for each kW of generating capacity) is
Thus the IGCC plant will save 16.57 – 12.52 = 4.05 tons over the five year period for each kW of plant capacity. Since the IGCC plant costs an extra $200 to build, we will be able to pay off this cost difference in five years if the price of coal is at least $200/(4.05 tons) =$49.37/ton.
We see that the amount of coal saved is directly proportional to the amount of time the plant runs. If the plant only produced 75% of the maximum possible kilowatt hours over a give year period, the coal savings would be only 3.04 tons and coal would have to cost about $65.83 to make the IGCC plant pay off in five years. (The average price of coal in the US for the 2008 was $32.05 per ton.1)