Mechanical Engineering Department
Mechanical Engineering 370
Thermodynamics
Fall 2010 Course Number: 14319 Instructor: Larry Caretto
Jacaranda (Engineering) 3519Mail CodePhone: 818.677.6448
E-mail: 8348Fax: 818.677.7062
Unit seven homework solutionsME 370, L. S. Caretto, Fall 2010Page 1
Unit Seven Homework Solutions, October 26, 2010
1A 600 MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 40%. Determine the rate of heat transfer to the river water. Will the actual heat transfer rate be higher or lower than this value? Why?
From the basic definition of cycle efficiency, , we can compute . From the cycle relationship between |QH|, |QL|, and |W|, |QH| = |QL| + |W|, we can write, for heat rates and power, . Thus we find that the heat rejected to the river is =1500 MW – 600 MW = 900 MW.
In the actual power plant there will be other sources of heat loss. These include heat transfer to the surrounding air from the power plant and leaks of the working fluid. In addition, power plant efficiency is usually determined as the power output divided by the equivalent heat from the fuel. There is a significant contribution to the inefficiency from the exhaust gases.
2An automobile engine consumes fuel at a rate of 28 L/h and delivers 60 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.8 g/cm3, determine the efficiency of the engine.
We start with the basic definition of cycle efficiency, , and we have to compute the heat input rate as the product of the fuel rate, and the heating content of the fuel. (We need the density to convert from a volume flow rate of fuel to a mass flow rate of fuel. Thus, we compute the heat input as follows.
We divide this into the power (work rate) of 60 kW to compute the efficiency.
= 21.9%
3A household refrigerator with a COP of 1.2 removes heat from the refrigerated space at a rate of 60 kJ/min. Determine (a) the electric power consumed by the refrigerator and (b) the rate of heat transfer to the kitchen air.
From the basic definition of coefficient of performance,, we can compute = 0.83 kW..
From the cycle relationship between |QH|, |QL|, and |W|, |QH| = |QL| + |W|, we can write, for heat rates and power, . Thus we find that the heat rejected to the kitchen air is = 1.83kW.
4A heat pump is used to maintain a house at a constant temperature of 23oC. The house is losing heat to the outside air through the walls and windows at a rate of 60,000 kJ/h while the energy generated within the house from people, lights and appliances amounts to 4,000 kJ/h. For a COP of 2.5, determine the required power input to the heat pump.
The COP definition for a heat pump is different from the definition for the refrigerator. Here we are interested in the amount of heat added to the house which is the high temperature heat rejection from the refrigeration cycle. The COP for a heat pump is defined as follows:. We can solve this equation for the power input and plug in the given data to compute the power required. Here we note that the heat supplied by the heat pump to keep the room temperature constant is just the difference between the heat loss and the heat generation. Thus, QH = 60,000 kJ/h – 4000 kJ/h = 56,000 kJ/h. The power consumption is then found to be
= 6.22 kW
5A heat engine is operating on a Carnot cycle and has a thermal efficiency of 55%. The waste heat from this engine is rejected to a nearby lake at 60oF at a rate of 800 Btu/min. Determine (a) the power output of the engine and (b) the temperature of the heat source.
We can combine the basic definition of cycle efficiency, , and the cycle relationship between |QH|, |QL|, and |W|, , to eliminate .
For this problem,where, and = 55%, we can find the power output as follows.
= 23.1hp.
For a Carnot cycle, the efficiency is given by the equation,Carnot = 1 – TL/TH. We can rearrange this equation and substitute the given data of = 55% and TL = 60oF = 519,67 R to obtain the temperature of the heat source.
= 1156 R.
6A homeowner is trying to decide between a high-efficiency natural gas furnace with an efficiency of 97% and a ground-source heat pump with a COP of 3.5. The unit costs of electricity and natural gas are $0.092/kWh and $1.42/therm (1 therm = 105 Btu = 105,505.6 kJ). Determine which system will have a lower energy cost.
The cost of each fuel can be compared for an equivalent amount of energy delivered to the house. With the furnace, 97% of the energy in the natural gas will be delivered to the house. Thus the cost to provide 1 kJ of heating with the gas furnace is
For the heat pump, the amount of heat delivered to the house is 3.5 times the electric power input. The cost to provide 1 kJ of heating with the heat pump is
Thus, the energy cost of the heat pump is lower. However, the heat pump will have a higher initial cost and the homeowner would have to do a financial analysis to see if the savings was worth the extra cost of the heat pump.
7Water, 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
8The 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 hout – hin = 93.155 kJ/kg – 128.05 kJ/kg = –34.90 kJ/kg
9Conventional 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])
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