  # sciencer

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## Everything posted by sciencer

1. Understood, thank you. Is the correct terminology 'linear change', whilst the other phenomenon is 'exponential change'? Seems strange because the inference of the term 'exponential' is that the same end point is obtained by rapidly increasing/decreasing rate of change. A polynomial factor would achieve the exponential change, but no polynomial was applied in this example?
2. > > A quantity can decrease in multiple ways in a 10 year period. > > It can decrease linearly. That means you lose 89/10 (or 8.9) in year 1. And again 8.9 in year 2. And again 8.9 in year 3. > > It can also be an exponential decay. You can lose for example 50% every year. Starting with 100, you lose 50 in year 1. You have 50 left, and therefore you lose only 25 in year 2. And only 12.5 in year 3. > > What you just did is that you assumed that the decrease (-89/10) is the linear value. But then you applied it to the exponential decay. > A linear change is the requested assumption. Thanks for the explanation of the exponential application, which was not understood and hence presumably the source of the error. If the change of -8.9% is applied to the original quantity, the result in year 10 becomes 34, still wrong. If anyone knows the relevant areas of maths that should be researched (i.e. don't know what search criterion to use!), it would be appreciated to know; thank you. > > Since this sounds an awful lot like homework, I will not give you the answer. > > Not sure what constitutes homework so cannot comment.
3. If a quantity of 100 in year 1 changes to 11 in year 10, the change would be -89 %. But when this change is converted to an annual value (-89/10) and then applied to each year, year 2 would calculate as 91.1. This is repeated such that according to the annual value, year 10 quantity is 39, which is a big difference from 11! What is the correct way to apply an annual change, please?
4. OK, so for your example your term 'midpoint' would be the median change to go from 0 to 24 over the period? The median function in the spreadsheet does give the expected answer.
5. 0=2-(1+x) -> -2=-(1+x) -> -2=-1-x -> -2+1=-x -> -1=-x -> x=1. Can't get 2! Thanks.
6. If 0=2-(1+x), isn't the correct answer x=2? A simple question, before the author of the book is notified of the print error
7. Understood, but that detracts from the original question: what is the theory behind this spreadsheet function? Seems like a sum of a geometric series but I want a 'mean', perhaps analogous to a geometric mean?
8. Yes but such software explains how to use the function and it is the theory that is not known. Why is it not possible to simply subtract the initial rate from the final rate and divide by the total time to obtain a simple average annual change, i.e. (6-3)/4=0.75? The power function result is approximately 0.15.
9. A spreadsheet (e.g Libreoffice Calc) has a power function for the example scenario: annual change (%) is 3,4,5,6 for years 1 to 4 respectively. The power function is: power= ((a4/a1),1/5)-1 where a1,a2,a3,a4 correspond to the values 3,4,5,6. Can someone explain the theory behind this function please? Thanks.
10. Please, what is the function to create a rate depending upon a collection of other rates, some of which may be zero? Using a spreadsheet, functions harmonic mean can be applied to a range of rates (percentages), but this function fails when any rate is zero (which is required). Any advice of an alternative function to harmonic mean please?
11. Thank you. For this data set representing measurement of temperature as a function of time, is it correct that the data would be independent (i.e. Markov process) when the time interval is regular, for example a measurement made every 10 seconds?
12. Variance is a measure of the variation in a data set, defined as Σ(y-mean(y))^2/(n-1). Is there a statistical term for the following calculation: data set time data 1 10 2 11 3 -22 4 34 5 -45 6 56 7 67 8 -78 variation between adjacent data v=(datum(t+1)-datum(t))^2 square root of v to obtain positive values of variation=v^(1/2) Σv^(1/2) median(v^(1/2))
13. The objective is to calculate energy output of fuel when stoichiometry is not known, therefore enthalpies with mole values were used, instead of enthalpies of formation of reactants and products. The equation is: biomass (CxHyOz)(s) + xO2(g) -> xCO2(g) + yH2O(l) so a simple empirical formula could be x=4, y=6, z=3. The enthalpies units are J/hr, because of the fuel feed rate. The data was tab separated so a quick copy into Calc will show better the units for the values, e.g. heat capacity Cp (J/K/mol) CO2(g)=37 H2O(l)=75 N(g)=29 O(g)=29
14. I am seeking comment on the method of the following calculations to determine the energy output of combustion of fuel. Any errors in the methodolgy would be appreciated. The method is using total enthalpies with mole quantities The ultimate analysis of the fuel is used to obtain weight values of the composition: C 45 (% weight) 243 (mol) 0 (ΔHf kJ/mol) H 6 433 (including air, water) 218 N 5 23 473 O 35 155 (including air, water) 249 water 5 - ash 4 - CO2 188 (mol/hr) -394 (ΔHf kJ/mol) 37 (Cp J/K/mol) H2O 227 -286 75 N2 3120 - 29 O 620 - 29 fuel feed 5 kg/hr air supply 1500 l/min combustion temperature 1070 K From dQ/dt=Σ H(outlets)-Σ H(inlets) C: 0 H: 433*218 N: 23*473 O: 155*249 total enthalpy: 1.44 x10^5 CO2: 188*(-394+((1070*37)-(298*37))) H20: 227*(-286+((1070*75)-(298*75))) N2: 3120*(0+((1070*29)-(298*29))) O2: 620*(0+((1070*29)-(298*29))) total enthalpy: 1.02 x10^8 energy balance = 1.02 x10^8 J/hr = 28 kW
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