Recent content by marsten

They apply at 25 °C and 1 atm. I'll ask him if it's necessary to take into consideration. The calculation of the flue gas enthalpy is just a small part of a bigger project I'm working on. Thank you so much for your help Chet! Marcus

I don't know. But maybe you're right. I found a table and the enthalpy of formation is : ##h_{f_{CO_{2}}}=-394\frac{kJ}{kmol}## ##h_{f_{H_{2}O}}=-242\frac{kJ}{kmol}## Should I just subtract this number from the enthalpy that I originally used, like this : ##h_{fg}=\frac{8\%\cdot...

It's normal cubic meter.Ok, so I have the volume of each component of the flue gas : ##V_{CO_{2}}=0,59\frac{m_{n}^{3}}{m_{n}^{3}fuel}## ##V_{H_{2}O}=0,55\frac{m_{n}^{3}}{m_{n}^{3}fuel}## ##V_{N_{2}}=5,09\frac{m_{n}^{3}}{m_{n}^{3}fuel}## ##V_{O_{2}}=0,81\frac{m_{n}^{3}}{m_{n}^{3}fuel}## And thus...

Thank you for your answer. I will try to make things clearer with units. The h represents the enthalphy. I think you are more familiar with using i. The unit in this case is ##\frac{kJ}{m_{n}^{3}}## The V represents the volume of each flue gas component. The unit is...

Hi 1. Homework Statement I'm trying to calculate the enthalphy of flue gas leaving a gas turbine, in kJ/kg. I just want to confirm that I'm using a correct method. Homework Equations hfg = (VCO2⋅ hCO2 + VH2O⋅ hH2O + VN2⋅ hN2 + VO2⋅ hO2) / (VCO2 + VH2O + VN2 + VO2) ρfg = (p⋅ M) / (R⋅ Tfg)...

Yep!

Nope, not the correct answer. Missed one thing. It's \mu_k=0,43 Thanks again!

The final answer will be \mu_k=0,14 Thank you very much!

Ok, so it will be something like this? s=\frac{a_{x2}t_2^2}{2}=\frac{g(sin\alpha)\tau^2}{2\cdot2^2}=\frac{g(sin\alpha)\tau^2}{8} ⇒ \tau=\sqrt{\frac{8s}{g(sin\alpha)}}

The different accelerations that I have come up with so far is: Acceleration with friction: a_{x1}=g(sin\alpha-\mu_kcos\alpha) Acceleration without friction: a_{x2}=g(sin\alpha) And the given times: Time with friction: t_1=\tau Time without friction: t_2=\frac{1}{2}\tau And by using the...

So as long as I put \mu_k=0 I will always get the same result (4,91 m/s^2). Could this acceleration help me anywhere? Or is it not to any use for me?

Sorry. The acceleration that I first calculated must be the acceleration without friction and without toboggan.

Ahh, should I use 2m? Using the same equation gives me: 2mgsin\alpha-\mu_k2mgcos\alpha=ma_x ⇔ a_x=2g(sin\alpha-\mu_kcos\alpha) ⇒ a_x=2*9,81*sin(30) ⇒ a_x=9,81 m/s^2

Some equation to find the distance? Like: s=v_0t+½a_xt^2=½a_xt^2

Homework Statement A girl, with the mass m, slides down an icy slope at the time t1=τ. The slope is inclined at 30° below horizontal. The coefficient of kinetic friction between the girl and icy slope is μk. The girl notices that she can slide down the slope significantly faster if she sits on...