# Formulating the control volume for a mechanical system

1. Aug 8, 2014

### bugatti79

Folks,

I am having difficulty correctly representing a mechanical system within a correct "control volume at an instant" in order to identify the various energy balance terms given below

$\displaystyle \dot E_{st}=\frac{d E_{st}}{dt}=\dot E_{in} - \dot E_{out}+ \dot E_g$ (1)

that correlates to this derived expression

$\displaystyle m*c_p*\frac{dT}{dt}=q=P_{loss}-hA(T_s-T_{amb})$ (2) where q is Watts. The last term is the general term for convection and radiation.

We have the measured power loss $P_{loss},dt, T_s,T_{amb}$ from test. Then we approximate $m, h, A,h$ in order to predict $dT$ which was done successfully.

However, despite all this, i would like to know how the above expression (2) was derived from first principles, ie from (1) in the first place.

Ie, is $P_{loss}=\dot E_g$ the energy generated?

I can write out my interpretation and post it as a picture if anyone is interested in correcting me where i have gone wrong...
thanks
bugatti

2. Aug 8, 2014

### 256bits

Eq(1) states that the energy within the control volume equals the energy coming in minus the energy going out plus internal thermal energy generation.
Eg would be a chemical reaction( either exothermic or endothermic), electrical resistance, ...
Ein , Eout would be the work or heat that enters or leaves the control volume.

Does anything in Eq(2) match up knowing that?

3. Aug 8, 2014

### Staff: Mentor

Yes, $P_{loss}=\dot E_g$ and $\dot E_{out}=hA(T_s-T_{amb})$

Chet

4. Aug 9, 2014

### bugatti79

For this axle we know that the main form of heat dissipation is free convection (there will be some conduction to the ground) and thus $\dot E_{out}=hA(T_s-T_{amb})$

Its not clear to me whether $P_{loss}=\dot E_g$ or $P_{loss}=\dot E_{in}$ (The heat comes from the friction between the gears and the churning of the oil)

Also I dont know where $\displaystyle mc_p\frac{dT}{dt}$ fits in in eqn 1.... To me, this term is the steady flow thermal energy equation which is normally used for "open systems with flowing fluid" but we do not have fluid flowing across our "closed system" boundary??...

5. Aug 9, 2014

### bugatti79

So we either have
$\dot E_{in}-\dot E_{out}=0$ plus the mcp(dT/dt) term or

$\dot E_{g}-\dot E_{out}=0$ plus the mcp(dT/dt) term.....???

6. Aug 9, 2014

### bugatti79

I believe it is actually

$\dot E_{in}-\dot E_{out}=\dot E_{st}$ where power loss, convection and mcpdT/dt are the 1, 2 and 3rd terms respectively.

Thanks guys

7. Aug 9, 2014

### Staff: Mentor

It depends on what you choose for the boundary of your control volume. Does the control volume include the gears and oil? If so, $P_{loss}=\dot E_g$.
No. This term is the rate of accumulation of heat within the control volume, and corresponds to your dEst/dt. This is not heat carried by flow into or out of the control volume. That would be covered by Edot,in. Edot,in includes heat by fluid flow into the control volume, heat conducted into the control volume through its boundary, and work done on the boundary of the control volume by the surroundings.

For more details, presenting the material in a more precise manner, see Bird, Stewart, and Lightfoot, Transport Phenomena.

Chet

8. Aug 9, 2014

### bugatti79

Yes, there is gears and oil. Ok So it is

$\dot E_g -\dot E_{out}= \dot E_{st}$
thanks