The length of a steel beam increases by 0.78 mm when its temperature is raised from 22 degrees C to 35 degrees C. What is the length of the beam at 22 degrees C (in meters)?I used: L = (0.78 mm)/[(9/5)(.00000645 F)(13)] = 5.17 meters but Mastering Physics said Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures. I'm confused because this is how we learned this kind of problem in class, so if anybody knows what I did wrong, feel free to correct my errors! Thanks
? got me is that the whole problem or is there more?
the welded joints,...are then sturdy steel, and conducts warmth very very right this moment, so develop from warmth often is the comparable because of fact the plate steel around it. you may desire to slope the backside and have a condensate seize gadget, steam secure practices stress alleviation valve, sized for the quantity and pressures, totally lag the vessel, adhere to ALL secure practices specs., (of your codes), which comprise the welding NDT. in the tank layout enable for the thermal develop of the vessel. cheers.
It seems you are trying to convert celsius to Fahrenheit, Why? There is no need. Delta L= alpha (initial length) (delta celcius) what you need is an alpha, it should be given in degrees celsius to the negative first (steel's alpha happens to be 0.000036 1/celsius). You also need the length of the beam at 35 degrees C. This answer (delta L) needs to be added (if heat is rising) or subtracted (if heat is dropping) to the original length.
we may use the formula ΔL=α(Lo)(Tf-To) where Lf is the length of steel at temp. Tf Lo is length at To α is the coefficient of linear expansion of steel which is 11x10^-6/°C ΔL=0.78mm (the change in length) Lo=? ΔL=α(Lo)(Tf-To) Lo=ΔL/[α(Tf-To)] Lo=(0.78x10^-3m)/[(11x10^-6/°C)(35°C-2...? Lo=5.455m answer
