In this problem, we are given the following quantities: We want to find: Equations and basic calculations:

Transcription

1 .1 It takes. million tons of oal per year to a 1000-W power plant that operates at a apaity fator of 70%. If the heating value of the oal is 1,000 /lb, alulate the plant s effiieny and the heat rate. In this problem, we are given the following quantities: We want to find: W max 1000 W max 70% 6.x10 tons / year 1,000 / lb η (dim ensionless) and HeatRate ( / kwh) Equations and basi alulations: max (1000 W)(70%) 700 W 700,000kW max 6.x10 tons 1, lb year day x10 / hr year lb ton 65 days 4 hours With these values for power output and heat input we an alulate the desired final results. HeatRate W x10 hr 700,000 kw 8,611 kwh η 700,000 kw 6.074x10 hr 41 kwh 9 9.6%

2 .4 A syntheti ontains 50% by weight CO and 50% by weight H. Calulate its heating value, J/kg and J/kg produt, and stoihiometri air- ratio using the data of Table.1. In this problem, we are given the omposition of a syntheti on a mass basis and we want to find its heating value and its stoihiometri air/ ratio. Sine the data in Table.1 are on a mass basis, we an easily find the heating value per unit mass of synthesis gas as the weighted average of the heating values of the individual omponents from Table kg H J 0.5 kg CO J 65.0 J wh H + wcoco + kg kg H kg kg CO kg To find the heating value per unit mass of produts we have to determine the mass of produts per unit mass of. This requires a speifiation of the air/ ratio, whih is not given here. For onservation of mass the produt mass equals the mass of air plus the mass of. Thus the ratio of produt mass to mass is found by the following equation. m m prod mair + m mair m m A F We use the symbol A to denote the air/ ratio on a mass basis. We do not F have an air ratio speified in this problem, nor do we have any data from whih we an alulate suh a ratio. However, we are asked to ompute the air/ ratio for stoihiometri ombustion. We an then use this air/ ratio to ompute the heating value per unit mass of produt. We an use the data in Table [.1] on stoihiometri air ratios. The overall air/ ratio for the syntheti is simply the weighted average of the air/ ratios for eah. A F w [ A ] + w [ A CO ] 0.5 kg H kg 4.8 kg air kg H H + F H F CO A kg air F kg 0.5 kg CO.467 kg air kg kg CO With an air/ ratio of 18.7, the produt/ ratio is 19.7 and we an ompute the heating value per unit of mass of produts as follows prod 1 + A F 65.0 J kg kg 19.7 kg prod.57 kg kg prod

3 5.8 A oal has a heating value of 1,000 /lb and the following moleular omposition: C 100 H 100 S 1 N 0.5. It is burned in air with 0% exess air. (a) Calulate the emission rate of and NO in lb/btu. One more assume that the emission rate of NO is twie the emission rate of NO. In this problem, we are given the heat of ombustion, 1,000 /lb, and the general formula C x H y S z O w N v, with n 100, m 100, z 1, v 0.5, and w 0. We are also told that the total NOx emissions is a fator, f, times the NOx emissions that would our if all the nitrogen in the reated to form NO. For the given data that we have 0% exess air, we know that the air- equivalene ratio, λ, is 1.. If we assume that all the z atoms of sulfur in the burn to we have the following equation for the emission rate of per unit heat input. m n z n z n z n We an find the moleular weight of the formula from equation (1) in the ombustion notes x + y + z + w + v C H S O N (1) Substituting the data for the formula and the atomi weights of the elements into this formula gives as shown below. 100( ( ) + 1(.065) + (0.5)( ) We now have all the data neessary to solve for m. m z 1lbmol S lb lbmol lbmol S.98x10 lb lb lbmol 1, lb lb.98 million 6 The alulation for NOx is similar exept for the fator, f, by whih we multiply the NOx emissions to get the total NOx emissions. We an thus use the equation that we derived for if we use the appropriate moleular weight differenes and introdue the f fator.

4 NO f v NO 0.5lbmol N lb NO lbmol lbmol N lb 1,000 lbmol lb.86x10 lb NO lb NO.86 million NO The flue gas of the plant in Problem 5.8 is to be treated with flue gas desulfurization (FGD) using a limestone (CaCO) wet srubber to remove the. Assume that moles of CaCO are neessary for every mole of. How muh limestone is onsumed per ton of oal? In this problem, we are given molar ratio, r moles CaCO per mole that is required for the FGD unit. From problem 5.8 we know that the plant in question has an emission rate of.98 pounds of per million BTU using a oal with a heat of ombustion, 1,000 /lb. The desired mass flow rate ratio for this problem may be omputed as follows. CaCO CaCO CaCO n n CaCO CaCO r Substituting the given data for r and 1,000 /lb as well as the results from the previous problem for the emission fator, and the appropriate moleular weights gives. CaCO lbmol lb CaCO r.98x10 lbmol CaCO lbmol 6 lb lb CaCO lb lb CaCO 1, lb lbmol CaCO CaCO We thus require.149 tons 0 ton 6.1 (a) Calulate the mass defiit ( m) in atomi mass units (amu) of the following fission reation. (Use literature values for the exat masses of the isotopes and neutrons.) 5 U + n 19 Xe + 95 Sr + n (b) Calulate the energy (ev) released per one fission. () Calulate the energy released per kilogram of 5 U and ompare it to the energy released in the ombustion of 1 kg of arbon.

5 The atomi mass data and their soures are shown in the table below. Nuleus ass (amu) Referene 5 U Fay and Golomb, page 11. n Class notes (taken from NIST web site) 19 Xe Sr (a) The mass defiit for the reation given is Δm amu (b) The annihilation of 1 amu of mass produes ev. Thus, the mass defiit omputed here would produe 18.6 ev. () One ev x10-19 J and one amu x10-7 kg. So the energy release per unit mass of the 5 U atom for the fission reation shown above is found as follows. The energy released in the ombustion of 1 kg of arbon is given on page 11 of Fay and Golomb as J/kg. The fission energy release is over two million times this amount. Note that the energy for the fission reation omputed here is slightly different from that omputed for the different fission reation disussed on page 11 of the text. 6.5 The isotope 19I has a half-life of 15.7 years. In a nulear power plant aident, 1 kg of the isotope is dispersed into the surroundings of the plant. How muh of the iodine isotope will remain in the surroundings after 1, 10, and 100 years. The leture notes for September 5 have the following equation that relates the urrent onentration, N, the initial onentration, N 0, the time, t, and the half-life, t 1/. N N e 0 ln()t / t 1 We an solve this equation for the ratio, N/N 0, whih is independent of the measure of the isotope present. Thus, we an also apply this equation to determine the mass of 19 I that will be present for the various times shown. If we replae N and N 0 by m and m 0 1 kg, we obtain the following results for the mass remaining after 1, 10, and 100 years.

6 m m m 1 year 10 years 100 years (1 kg)e (1 kg)e (1 kg)e l yr ln() 15.7 yr l0 yr ln() 15.7 yr l yr ln() yr kg 0.64 kg kg Of ourse, all these alulations assume that there is no further diffusion of the 19 I during the time periods used here. 8.7 A passenger vehile diesel engine has a minimum brake speifi onsumption of 0. kg/kwh. Calulate its maximum thermal effiieny. The relationship between the brake speifi onsumption (bsf) and the thermal effiieny (η) is given by the following equation. η.6 ( bsf )(kg/kwh) LHV (J/kg) For the LHV of ombustion of diesel, we an use the value of J/kg in the seond line of page 199 of the text. This gives the following result for the effiieny. η.6 9.7% A 1.5-ton vehile has a frontal area of m, a rolling resistane oeffiient of 0.01, and a drag oeffiient of 0.. Calulate the mehanial power delivered to the wheels at a steady speed of 50 and 100 km/h, if the atmospheri density is 1. kg/m. The wheel power requirement for a vehile operating at a steady speed is given by the following equation. P C R mgv + ( C D ρv A) In this equation, m is the vehile mass, g is the aeleration of gravity, C R is the rolling frition oeffiient, V is the vehile speed, ρ is the air density, C D is the drag oeffiient and A is the frontal area of the vehile. Substituting the given data, using g 9.81 m/s, and making the appropriate unit onversions gives the desired power as follows.

7 50 10 P (0.01) ( ) (0.)( m ) P (0.01) ( ) (0.)( m ) kw 11.8kW 9.1 Compute the emission rate of (kg/s) from a large, oal-fired power plant that uses million tons of oal per year having a sulfur ontent of % by weight. In this problem, we are given the following quantities: x10 tons/year x10 kg/year 6 9 S We want to find the emission rate (kg/s). From the data given, we an ompute the average rate of, by assuming that all the sulfur in the oal is onverted to by the reation S + O. In this reation, kilograms of sulfur will reat to form 64 kilograms of. The mass ratio of produed to sulfur reated is. With this assumption, the annual average emission rate for this problem an be omputed as follows. w x10 kg oal 0.0 kg S 64 kg year w S S year kg oal kg S 65 days m.54kg/s day s 9. Calulate the emission rate of per heat input (g/gj) for oal with a heating value of 0 J/kg and a sulfur ontent of % by weight. In this problem, we are given the sulfur ontent, w S 0.0, and the heat of ombustion of the, 0 J/kg. We want to find the ratio of emissions to the heat input rate of the,. We an do this by using the relation between sulfur and emissions derived in the solution to problem 9.1, and the usual relationship that the heat

8 input is the mass flow rate of the times its heat of ombustion. This gives the following result. w Sm S w S Substituting the problem data gives. w m S S kg oal 1000 g oal 1000 J 0.0 g S 64g 0 J kg oal GJ g oal g S 1.g/GJ 9.4 The ozone data in the table represent hourly values on a high pollution day in Los Angeles. Plot the diurnal ozone profile. Determine the arithmeti average and geometri average ozone onentration for that day. Determine the maximum eight-hour average onentration and ompare it with the new ambient ozone standard of 80 ppb by volume. S The first two olumns in the table below are taken from the problem statement. The third olumn shows the results of omputing the eight-hour averages. The data from the problem statement are plotted in the figure below to give the diurnal ozone profile.

9 Ozone (ppb Hour vol) Eight-ho ur average

10 The daily average onentrations, both the arithmeti mean and the geometri mean are omputed, respetively. This gives the following results. Daily arithmeti mean 116 ppb vol Daily geometri mean 5 ppb vol In data with wide variations between the lowest and highest data points, the geometri mean gives greater weight to the lower data (as ompared to the arithmeti mean.) The individual eight-hour means in the table above were omputed. The maximum eight-hour is shown below. aximum daily eight-hour mean 6 ppb vol The maximum eight-hour value is muh greater than the new 80 ppb NAAS for ozone.