Mechanical Engineering Department
Mechanical Engineering 483
Alternative Energy Engineering II
Spring 2010 Number: 17724 Instructor: Larry Caretto
Jacaranda (Engineering) 3333Mail CodePhone: 818.677.6448
E-mail: 8348Fax: 818.677.7062
Solution to first midterm examME 483, L. S. Caretto, Spring 2010 Page 1
Solution to First Midterm Exam
1.A gas turbine is to be installed at a landfill to produce electricity from landfill gas. The landfill has the following properties.
Species / Mole percent / Molar mass / Heat of combustion (kJ/kmol)CH4 / 80% / 16.04 / 802,802
CO2 / 15% / 44.01 / 0
N2 / 5% / 28.01 / 0
The compressor of the gas turbine has an inlet pressure and temperature of 100 kPa and 290 K. Its outlet pressure is 1 MPa and its isentropic efficiency is 87%. The combustor has a pressure drop of 30 kPa. The inlet temperature of the turbine is 1450 K; its outlet pressure is 110 kPa and its isentropic efficiency is 90%.
(a)Determine the mass flow rates of air and fuel to provide 50 MW of electric power from the gas turbine engine if the efficiency of the electric generator is 94% (electrical power out divided by mechanical power in).
The heat of combustion for the mixturecan be found from the data for the individual components in the mixture.
This solution uses the equations in Hodge for an ideal gas with constant heat capacity. Hodge accounts for the temperature of the heat capacity by using different values of heat capacity different components in the system. The Hodge equations assume that each component in the gas turbine cycle is steady flow with negligible kinetic and potential energy changes. The turbine and the compressor each have one inlet and one outlet so that the general first law for these components is q = wu + hout – hin. We assume all devices are adiabatic. In the combustor, all the chemical energy of the fuel is converted to thermal energy in the exhaust stream, with no external heat loss.
We can use equation (5-14) in Hodge to find the outlet temperature of the compressor. The pressure ratio, P2/P1 = (1000 kPa) / (100 kPa) = 10.
Similarly, equation (5-20) in Hodge gives the outlet temperature of the turbine. Here we use Hodge’s value of k = 1.334 for the turbine. From the given data on the combustor pressure loss we find P3 = P2 – Pcomb = 1000 kPa – 30 kPa = 970 kPa.
T4 = 901.7 K
The specific work for the compressor and turbine are found, respectively, from equations (5-12) and (5-18) in Hodge, again using the values for mean heat capacity.
The fuel air ratio can be found from equation (5-22) in Hodge for an adiabatic combustor.
To provide 50 MW of electrical power we have to provide (50 MW)/0.94 = 53.191 MW = 53,191 kW of mechanical power. The mass flow of air is now found to meet this desired net power output is found as follows.
Multiplying this air flow rate by the fuel/air ratio of 0.02966 gives
(b)What is the oxygen concentration in the turbine exhaust?
The oxygen concentration in the turbine exhaust can be found from the equation that relates this concentration to the relative air/fuel ratio.
We know that the fuel/air ratio is 0.02966 so the air/fuel ratio is 1/0.02966 = 31.71, but we have to compute the stoichiometric air/fuel ratio to compute . This ratio requires the computation of the stoichiometric moles of O2 = x + y/4 + z – w/2. We have to compute these coefficients for our fuel components.
x = (0.8)(1) + (0.15)(1) + (0.05)(0) = 0.95
y = (0.8)(4) + (0.15)(0) + (0.05)(0) = 3.20
w = (0.8)(0) + (0.15)(2) + (0.05)(0) = 0.30
With these values of x, y, and w (z = 0 since there is no sulfur in the fuel) we find
A = x + y/4 – w/2 = 0.95 + 3.20/4 – 0.30/2 = 1.6
We can now use the relationship between the air fuel ratio and l.
%O2 = 14.71%
2.You are designing a landfill that will produce a waste gas stream; you have to choose between (i) purchasing a gas collection system that would simply burn the waste gas and (ii) purchasing a gas turbine power generation system that would generate electricity from the waste gas. The preliminary design indicates that the waste gas could operate a turbine that would generate 50 MW of electricity with a capacity factor of 80%. If the maintenance cost for the gas turbine system is $0.007/kWh, how much extra would you be willing to pay for this system if the electricity sales price were $0.06/kWh and the system lifetime were 23 years. Assume that you want a return rate of 9% on your investment.
Compute the present worth of the electricity sales minus the maintenance cost. This is the price that you would be willing to pay for the gas turbine. For 50 MW =50,000 kW of power and a capacity factor of 80% the annual generation would be (50,000 kW)(80%)(8766 h/yr) = 3.506x108 kWh/yr. The sales price minus the maintenance cost is $0.06/kWh - $0.07/kWh = $0.053/kWh. This will produce an annual net income of (3.506x108 kWh/yr)($0.053/kWh) = $1.858x107/yr. Using the P/A formula for i = 9% and N = 23 years gives the present worth of this annual cost as
You would be willing to pay $178 million for the gas turbine system.
3.The set of discrete wind-speed and frequency data the table below shows the percent of the wind speed data for given velocity range. For example, the proportion of the wind speed data between speeds of 0 and 1 m/s is 2.8747%. These data do not fit well to a typical Weibull distribution and should be analyzed by considering the percent of wind speeds and power in each band.
Percent of Wind-Speed Data Between Lower and Upper Velocity Bounds (V in m/s)Lower / Upper / Percent / Lower / Upper / Percent / Lower / Upper / Percent
0 / 1 / 2.8747% / 10 / 11 / 4.3213% / 20 / 21 / 0.8028%
1 / 2 / 9.8109% / 11 / 12 / 4.1559% / 21 / 22 / 0.5310%
2 / 3 / 10.307% / 12 / 13 / 4.1527% / 22 / 23 / 0.3928%
3 / 4 / 9.4960% / 13 / 14 / 3.9050% / 23 / 24 / 0.2427%
4 / 5 / 8.0058% / 14 / 15 / 4.0583% / 24 / 25 / 0.1476%
5 / 6 / 6.0967% / 15 / 16 / 3.4830% / 25 / 26 / 0.1102%
6 / 7 / 5.1868% / 16 / 17 / 3.0287% / 26 / 27 / 0.0716%
7 / 8 / 4.6691% / 17 / 18 / 2.1695% / 27 / 28 / 0.0310%
8 / 9 / 4.6374% / 18 / 19 / 1.6005% / 28 / 29 / 0.0114%
9 / 10 / 4.3865% / 19 / 20 / 1.2489% / 29 / 0.0640%
Consider the application of the wind data to compute the average operating power for a 2.5 MW wind turbine with rotor diameter of 90 m and a power coefficient of 0.48. Assume an air density of 1.225 kg/m3. The following data apply: a cut-in speed of 5 m/s, a cut-our speed of 25 m/s.
(a)Within a given band, between a lower limit, a, and an upper limit, b, there is a uniform distribution of speeds. This corresponds to a probability distribution function (within one band) which is f = 1/(b – a). Show that for such a probability distribution that the mean velocity cubed within a given band is given by the following equation.
The integral is consistent with the computation of the function of a probability distribution. The actual integration gives
(b)What is the contribution of the wind speeds between the rated wind speed and the cut-out wind speed to the average operating power?
The rated speed is given by the following calculation.
The percent of the time the wind speed is between this speed and the cut out speed of 25 m/s can be found by summing the percentages for all bands, starting with the band at a lower limit of 11 m/s and ending with the band that has an upper limit of 25 m/s. This sum is 4.1559% + 4.1527% + 3.9050% + 4.0583% + 3.4830% + 3.0287% + 2.1695% + 1.6005% + 1.2489% + 0.8028% + 0.5310% + 0.3928% + 0.2427% + 0.1476% = 29.9194%. The average power during this velocity range is the total fraction of time times the maximum power of 2.5 MW which is (2.5 MW)(0.299194) = 0.748 MW.
(c)The contribution of the wind speeds between cut-in speed and the rated wind speed to the average operating power will be a summation of the average power over each band in this range times the wind frequency for the band. Write a general from for this summation equation and evaluate the first and last terms in this sum.
If fk denotes the fraction of wind data in band k, the total fraction between the cut-in speed and the rated speed will be given by the following equation.
In this equation the “Start” band will be the one with that has a lower limit of the cut-in velocity of 5 m/s and the “End” term will be for the band whose upper limit is the rated wind speed of 11 m/s. Using the equation from part (a) for the value of the mean velocity cubed we find the following value for the first term that starts with a velocity of 5 m/s.
Similarly we find the following equation for the last term that ends with a velocity of 11 m/s.
(d)If the value of the sum in part c is 0.3007 MW, what is the capacity factor for the wind turbine?
The capacity factor is the average power divided by the maximum power. The average power is the sum of the power above the cut-our speed which was found to be 0.748 MW in part (b), and the sum from part (c), which is given in the problem statement as 0.3007 MW. Thus the average power is 0.748 MW + 0.3007 MW = 1.0487 MW. The maximum power is 2.5 MW so the capacity factor is (1.047 MW) / (2.5 MW) = 41.8%