Recent content by Maiq

You just have to match up the values. If you have a phase it will be in the saturated tables, high pressures in compressed liquid tables, and high temperatures in superheated vapor tables. If you can't find the numbers on the tables then you have to use interpolation.

No need for the quadratic formula. look at your Forces in the direction. You made a mistake.

α is angular acceleration. you used the angular velocity, 605rpm (63rad/s) to find the torque. Your KE equation is correct but you used the wrong value for angular velocity. The angular velocity was given.

Lets see your work then.

Also I can tell the quality is not going to be 100%. You can use the volume and mass to find the specific volume. Than use that specific volume and the specific volumes from the table for state 1 and 2 to find the quality.

You already found the answer to a. Look at what you put for c. You also already found the final pressure.

Yes it is positive. If you look at the FBD you can see that all the forces besides BC are in the negative direction. So what would that mean for BC?

Sorry about that. I was using the tables in my thermo book that list Cp and Cv. I worked out c using those Cp and Cv values and the specific gas constant I mentioned earlier and got 3.46. I kinda assumed that maybe you had those tables also.

^ It looks like its just mgh=(1/2)mΔv^{2}

The specific gas constant is just R divided by the gas's molar mass. I guess the m is for CO2's molar mass which is 44. Does dividing 37 by 44 give you the right Cp?

For R you need to use the specific gas constant for CO2. Which is 0.1889 \frac{kj}{kg K}

Ok so it does work actually but you would have to use sin(\frac{dθ}{dt})=\frac{y}{x}. Since the angle you start with will be the same as the angle at the end, \frac{dθ}{dt}=0 therefore sin(\frac{dθ}{dt})=0. So you end up with 0=\frac{vsin(θ)t-(1/2)gt^{2}}{vcos(θ)t}=tan(θ)-\frac{gt}{2vcos(θ)}...

Never mind, this equation doesn't seem to work. What I was trying to do was use sin(θ)=y/x. I now realize that the θ in this sin(θ) changes with time, while the angle θ is constant in the position equations.

If you divided your vertical position by your horizontal position what would you have? (Hint: its a trig function)

Actually I think the position would just be 5 \hat{r} since at any point its radial position would be 5. In my dynamics class we also used the equation \vec{r}=r \hat{e}_{r}. \hat{e}_{θ} was only used with velocity and acceleration equations because those would change with θ.