- #1
Saladsamurai
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I am trying to follow along with the text's derivation of the Bernoulli equation and I am a little confused with a particular line. The images show an elemental streamtube control volume (CV) of variable area A(s) and length ds.
We assume that though the properties may change with time and position 's' they remain uniform over the cross section 'A.'
So treating as a 1-dimensional inlet/outlet CV we can write conservation of mass as:
\frac{d}{d\,t}(\int_{CV}\rho d\,V) +\dot{m}_{out} - \dot{m}_{in} = 0 \approx \frac{\partial{\rho}}{\partial{t}}d\,V + d\,\dot{m}
Okay. I am missing somethings here:
(1) I am assuming that they set \dot{m}_{in} =\dot{m}
and
\dot{m}_{out} =\dot{m} + d\,\dot{m}
so
\dot{m}_{out} - \dot{m}_{in} =d\,\dot{m}.
Does that sound right?
(2) So the thing that is really bothering me is this. Where did the integral go? AND why did d/d\,t turn into \partial/\partial{t}
I keep getting jammed up on these total derivatives turning into partials. What does it all mean?
thanks
We assume that though the properties may change with time and position 's' they remain uniform over the cross section 'A.'
So treating as a 1-dimensional inlet/outlet CV we can write conservation of mass as:
\frac{d}{d\,t}(\int_{CV}\rho d\,V) +\dot{m}_{out} - \dot{m}_{in} = 0 \approx \frac{\partial{\rho}}{\partial{t}}d\,V + d\,\dot{m}
Okay. I am missing somethings here:
(1) I am assuming that they set \dot{m}_{in} =\dot{m}
and
\dot{m}_{out} =\dot{m} + d\,\dot{m}
so
\dot{m}_{out} - \dot{m}_{in} =d\,\dot{m}.
Does that sound right?
(2) So the thing that is really bothering me is this. Where did the integral go? AND why did d/d\,t turn into \partial/\partial{t}
I keep getting jammed up on these total derivatives turning into partials. What does it all mean?
thanks
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