Recent content by ttb90

Sorry you are right. The question states the term polytropic expansion which would mean adiabatic in this case. I think polytropic is a general case where n=γ is adiabatic , n=1 is isothermal and n=0 is isobaric. When you say P(V/m)=RT for each corner on the graph, do you mean that because they...

Polyprotic Expansion HELP! Homework Statement Nitrogen at 100°C and specific volume of 0.1846 m^3/kg undergoes a polyprotic expansion process such that pv^(1.2)=const and the final pressure is 100 kPa. A constant pressure compression is then performed to bring the volume back to 0.1846...

I did all the calculations for it and found that Mc=2385.5699 kg/m. I really appreciate your help! Thank you soo much

Read the question again and found that the question asks for the maximum mass of cargo, so since we found 2 masses that satisfy the G'M condition wouldn't that mean that the heavier mass is the answer to Mc. Thoughts? So Mc=11760 kg/m ?

How do I go about showing that? I tried substituting the values into the previous expressions to see if all the conditions are met but couldn't get much out of it to come to a conclusion.. What shall I do?

I tried solving it again and don't think I succeeded. I have attached a scan of my working out so it's easier to read.. Please tell me what I am doing wrong. I think I've solved this like over 10 times and keep finding different answers each time :(

G'M= 5.33/((9327.19+Mc)/1025) - [(12654.38 + 4Mc)/(9327.19+Mc) - ((9327.19+Mc)/8200)] G'M=0 (12654.38+4Mc)/(9327.19+Mc)-(9327.19+Mc)/(8200)=5466.325/(9327.19+Mc) (12654.38+4Mc -5466.325)/(9327.19+Mc)=(9327.19+Mc)/8200 (7188.055+4Mc)/(9327.19+Mc) = (9327.19+Mc)/8200 (9327.19+Mc)^2=58942051+...

Specifically speaking would you say that my moment equation is correct? I think that's the main part of the working out that has to be right as i use the result obtained in the following sections to calculate G'M=0

Definitions: Mb=mass of ballast M= mass of barge (empty)=6327.19kg 0=ƩMa=-(AG' X (M+Mb)+(AG X M) since we know that AG=2m; 0=ƩMa=-(AG' X (M+Mb)+(2 x M) The negative sign is due to the anticlockwise moment M=mass of barge = 6327.19 kg Mb=mass of ballast = 3000 kg Mc=mass of cargo Distance of C...

So I'm finding this last section really hard to start off. This is what I have done so far: Mb=3000kg M= mass of barge (empty)=6327.19kg g=9.81 Mc= mass of cargo ρ=1025kgm^(-3) I struggled to find the moment equation so i started off by doing the following: 1. New draft for the load barge...

Thanks very much for that. There is now a change of scenario and I attached it as figure 2. Question states that 'If the total ballast acting at point A is now increased to 3000 kg/m, what is the maximum mass of cargo per unit length Mc that could be safely carried at position C with the...

SteamKing: This is following I did to find the new position of B. For the height of the center of buoyancy above the bottom of the hull i did the following. Volume of displaced water= V= 4 x h x 1= 4h The weight of the displaced seawater = (6327.19+Mb)g =ρ x g x4h Therefore h=...

Does that mean that when I do find the draft of the ballasted barge we assume that the new position of B is still halfway of the new draft value? Is this assumption correct?

That's how i did mine as well. Took a while to get the answer. I am stuck with this new buoyancy B though, do we assume it stays at the same point after the MB is added to point A. What exactly changes here?

If you look at the attached diagram it shows the height from point A to the new center of gravity to be h.. That's the value I found in the above calculations which is, h=1.5432m