Recent content by DrVirz

Hi all, Just doing some thermo study and am stuck on a question. I am not sure where to start this Q as normally I am given a property at the exit..? Any help is appreciated.

Well that's great news! For some reason I thought my solution seemed to simple! Thanks for clearing that up!

Hi all, I've attempted this question however my solution seems too simple..? I have taken the point 'A' (bottom right corner of beam) as the point with max compressive stress, I am not sure if this is correct. I have just assumed this by observation.

Oh! My bad! Been a long day. I will adjust to MPa or N/mm^2

Hi all, I was just wondering if someone could confirm that my answer to the below question is correct. The point where the question asks you to find the shear stress is at 2m (n). However, I don't believe this value needs to be included in the calculation anywhere?

This clears up a lot! Thanks for that. I will go and do the rest of the question when I get home later today.

This is where I am getting confused. Because the beam has 2 hinges, would I split the beam into 3 parts. Then at each hinge I would have a force in the y direction (and x however this is neglected). So for section 1 at B I would have a positive y force, for section 2 at B I would have a negative...

Hi all, Having trouble with the following question, trying to work out the FBD and forces. I am fine with working out the max tensile and compressive forces however I can't get that far as the two hinges are throwing me off. As I understand, the moment at the hinges should be 0, is this...

Ah! That makes much more sense to me now. Was a confusing question! Thanks for your help!

Sorry but this has confused me. In the first part I multiplied the evap. specific vol by 0.2? And in the last part i multiplied the total mass by 0.2?

Ok, so. I find the specific volume of the mixture (0.001044+0.2(1.67185)) = 0.335414 m3/kg Then, the total mass of the mixture (0.2/0.335414) = 0.59627 kg Therefore the mass of the water should be 0.2 x 0.59627 = 0.1192 kg Which is still not correct :S

Yes I am using steam tables. So the mass of the water at State 1 is simply Volume/Specific Volume... 0.04/0.001043= 38.351kg Mass of the sat. vapor at state 1 is 0.16/1.694 = 0.0945 What I would then do is find the mass of the sat. vapor when the tank is full @ 500 kPa, i.e 0.2/37380 ... Then...

Hi all, I have another thermo q I am stuck on. You can see the question and my attempted answer below. Part 1 is fine, Part 2 i don't know where to start, and part 3 I thought was correct however the answer is wrong! I think I am doing something wrong with the U2 part of internal energy, just...

Yes I did see that in Delta's post. I am just saying, naturally, I refer to the tables for my data values and when I didn't come across a value for water it threw me off. I thought I had done something wrong. That's what happens when you have been studying all day I guess, time to take a rest.

Sorry in my previous post i meant to say CANT find the Cv/Cp value for water in any of my tables.