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|College of Engineering and Computer Science

Mechanical Engineering Department

Mechanical Engineering 370

Thermodynamics | |

| |Spring 2003 Ticket: 57010 Instructor: Larry Caretto |

March 25, 2003 Homework Solutions

5-17 A 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, [pic], we can compute [pic]. From the cycle relationship between |QH|, |QL|, and |W|, |QH| = |QL| + |W|, we can write, for heat rates and power, [pic]. Thus we find that the heat rejected to the river is [pic]=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.

5-21 An 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, [pic], 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.

[pic]

We divide this into the power (work rate) of 60 kW to compute the efficiency.

[pic] η = 15.8%

5-24 Conventional coal fired power plants cost $1,300 per kW to construct and have an efficiency of 34%. Integrated Coal Gasification Combined Cycle plants cost about $1,500 per kW to construct but their efficiency is about 45%. The average heating value of coal is about 28,000,000 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 power is determined by the efficiency of the plant and the heating value of the coal. For the conventional plant,

[pic]

For the IGCC plant, the total coal use over five years is

[pic]

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 first six months of 2002 was about $33 per ton.[1])

5-30 Consider a 3 kW hooded electric open burner in an area where the unit costs of electricity and natural gas as $0.07/kWh and $0.60/therm, respectively. The efficiency of open burners can be taken to be 73% for the electric burners and 38% for the gas burners. Determine the rate of energy consumption and the unit cost of utilized energy for both electric and gas burners.

Here we are comparing two different heat sources with two different efficiencies and two different costs. To have a common basis of comparison we have to compute the actual energy input required to provide 3 kW of useful heat. For the electric burner, where the efficiency is 74%, we have to provide (3 kW)/(78%) = 3.846 kW of electrical energy. Since this energy costs $0.07/kW, the total cost to provide 3 kW of useful energy is ($0.07/kW∙h)(3.846 kW) = $0.269/h.

For the gas burner, where the efficiency is 38%, we have to provide (3 kW)/(0.38%) = 7.895 kW of energy from the gas. The cost of natural gas, which is $0.60 per therm can be converted to a cost per kWh using the appropriate unit conversion factors: ($0.60/therm)(1 therm/105 Btu)(3412 Btu/kWh) = $0.02047/kWh. Thus the cost for the natural gas that has to supply 7.895 kW is ($0.0204/W∙h)(7.895 kW) = $0.1616/h. Thus the natural gas will be cheaper even though it has a lower efficiency.

The cost of each fuel per unit of useful energy produced in the burner can be found by dividing the fuel cost (in dollars per hour) by the 3 kW of useful energy from the burner. For electricity, the cost of the useful energy is ($0.269/h)/(3 kW) = $0.096/(useful kWh) For natural gas, the cost of the useful energy is ($0.1616/h)/(3 kW) = $0.054/(useful kWh)

5-50 A household refrigerator with a COP of 1.5 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, COP or β,[pic], we can compute [pic]= 0.667 kW..

From the cycle relationship between |QH|, |QL|, and |W|, |QH| = |QL| + |W|, we can write, for heat rates and power, [pic]. Thus we find that the heat rejected to the kitchen air is [pic]= 1.667 kW.

5-62 A 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 form the definition for the refrigerator. Here we are interested in the amount of heat added top the house which is the high temperature heat rejection from the refrigeration cycle. The COP for a heat pump is defined as follows: [pic]. 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

[pic]= 6.22 kW

5-84E A heat pump is operating on a Carnot cycle and has an 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, [pic], and the cycle relationship between |QH|, |QL|, and |W|, [pic], to eliminate [pic].

[pic]

For this problem, where[pic], and η = 55%, we can find the power output as follows.

[pic] = 23.1 hp.

For a Carnot cycle, the efficiency is given by the equation, ηCarnot = 1 – TL/TH. We can rearrange this equation and substitute the given data of η = 55% and TL = 60oF = 519,67 R to obtain the temperature of the heat source.

[pic]= 1155.6 R.

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