Avariable, q, is reported to have a value of 2.360 x 10 kg .m²/h. a. write a dimensional equation for q', the equivalent variable value expressed in u. s. customary units, using seconds as the unit for time. b. estimate q' without using a calculator, following the procedure outlined in section 2.5b. (show your calculations.) then determine q with a calculator, expressing your answer in both scientific and decimal notation and making sure it has the correct number of significant figures. 2.5b validating results every problem you will ever have to solve-in this and other courses and in your professional career-will involve two critical questions: (1) how do i get a solution? (2) when i get one, how do i know it's right? most of this book is devoted to question 1-that is, to methods of solving problems that arise in the design and analysis of chemical processes. however, question 2 is equally important, and serious problems can arise when it is not asked. all successful engineers get into the habit of asking it whenever they solve a problem and they develop a wide variety of strategies for answering it. among approaches you can use to validate a quantitative problem solution are back-substitution, order-of- magnitude estimation, and the test of reasonableness • back-substitution is straightforward: after you solve a set of equations, substitute your solution back into the equations and make sure it works. • order-of-magnitude estimation means coming up with a crude and easy-to-obtain approximation of the answer to a problem and making sure that the more exact solution comes reasonably close to it. • applying the test of reasonableness means verifying that the solution makes sense. if, for example, a calculated velocity of water flowing in a pipe is faster than the speed of light or the calculated temperature in a chemical reactor is higher than the interior temperature of the sun, you should suspect that a mistake has been made somewhere. the procedure for checking an arithmetic calculation by order-of-magnitude estimation is as follows: 1. substitute simple integers for all numerical quantities, using powers of 10 (scientific notation) for very small and very large numbers. 27.36 + 20 or 30 (whichever makes the subsequent arithmetic easier) 63,472 + 6 x 104 0.002887 + 3 x 10-3 2. do the resulting arithmetic by hand, continuing to round off intermediate answers. (36,720) (0.0624) 0.000478 (4 x 10^)(5 x 10-2) - = 4 x 10(4-2+4) = 4 x 106 5 x 10-4 the correct solution (obtained using a calculator) is 4.78 x 106. if you obtain this solution, since it is of the same magnitude as the estimate, you can be reasonably confident that you haven't made a gross error in the calculation. 3. if a number is added to a second, much smaller, number, drop the second number in the approximation. = 0.25 4.13 + 0.04762 the calculator solution is 0.239.