# Recent content by dlacombe13

1. ### Very simple question about calculating weight

Okay thanks. It just strikes me as odd that the formula has mass, but yet you end up with weight. It's also crazy since this means the weight of the truss is only 10lbs and yet it holds up a 10,000lb load...But then again this is just a 2D model of what would really be a 3D structure.
2. ### Very simple question about calculating weight

Homework Statement Calculate the weight of each member of the truss structure. Homework Equations density = mass/volume The Attempt at a Solution This is part of a lab that we are doing in my mechanics of materials course. I have designed a structure, and I must calculate the weight of each...
3. ### Thermodynamics -- Internal Energy of Water

Oh yes, oops. So then ΔU = 2093.5 kJ And then: V = m/ρ = 100kg / 1000kg/m3 ==> V = 0.1m3 ΔH = ΔU + VΔP = 2093.5 kN*m + (0.1m3)*(900 kN/m2) =2183.5 kN*m ==> ΔH = 2183.5 kJ
4. ### Thermodynamics -- Internal Energy of Water

Oh okay I see now, so C = 4.187 kJ/kgK. So I found that ΔU = (100kg) * (4/187 kg/kgK) * (278.15K) = 116,461.41 kJ. Now I am ready to use that in the equation for ΔH = ΔU + VΔP. So in this case, volume does not change? And if so, would I have to find volume by using V = m/ρ ?
5. ### Thermodynamics -- Internal Energy of Water

Right, I was originally thinking that. So is C for water the same in all of its physical states?
6. ### Thermodynamics -- Internal Energy of Water

OK thanks, that's what I thought. So my tables that I am provided start at P = 25 bar. I have been doing a lot of searching on the internet, since some of my other problems involve compressed liquid. I found something called Compressed Liquid Approximation. It states that: u(T,P) ≅ uf(T) v(T,P)...
7. ### Thermodynamics -- Internal Energy of Water

Homework Statement Water is initially at P = 1 bar and T = 20°C. 100kg of water is pumped to a higher pressure at which P = 10 bar and T = 25°C. Find ΔU and ΔH Homework Equations H = m*h du = c*dT dh = c*dT + v*dP The Attempt at a Solution So far I have looked in my table and found that at P...
8. ### Internal Energy Thermodynamics

I am confused because I do not understand which value to choose for Cp if it changes with volume. Even if I were to integrate I would have: ∫dU = m∫Cp dT But how can I do that? The Cp is really throwing me off, and I have no idea how to use it in this problem. The same goes for Cv, since that...
9. ### Internal Energy Thermodynamics

No I wouldn't think that mass would change. I went ahead and calculated the volume of the second state, although I'm not sure if it was needed. Would I then have to use dU = mcvdT ? And since it cv is not constant, would I need to integrate both sides of this equation?
10. ### Internal Energy Thermodynamics

Okay so I am familiar with the PV=nRT formula. In our class, we usually use PV=mRT, and use the specific R for the gas in question. Would my first step be to calculate the mass of CO2 in state 1 using this formula?
11. ### Internal Energy Thermodynamics

Homework Statement CO2 is at P=3atm, T = 295K and V=1.2m3. It is isobarically heated to T = 500K. Find ΔU and ΔH Homework Equations dU = cpdT The Attempt at a Solution I am having a hard time in general in this class. I understand that in this problem, ΔP = 0. Does this mean that there must...
12. ### Thermodynamics - Need help with inclined plane problem

Oh okay that makes sense, thank you both for your help. I appreciate it!
13. ### Thermodynamics - Need help with inclined plane problem

Right, so the equation is the same, and we get the same answer. So why exactly are you guys pointing out my equation and saying it is dimensionally inconsistent?. What does that mean?
14. ### Thermodynamics - Need help with inclined plane problem

Oops, should be: Qout = 352.77MJ - 105.83MJ = 246.94MJ And that equation came from my notes. Why is it plus and not minus?
15. ### Thermodynamics - Need help with inclined plane problem

So Qin = Wout/0.30 = 3.53GJ dE/dt = Qin - Qout - Wout 0 = 3.53GJ - Qout - 105.83MJ Qout = 3.53GJ - 105.83MJ = 3.42GJ