05.04.2016 Views

Modern Engineering Thermodynamics

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

19.5 Thermomechanical Coupling 779<br />

Pressure p 1 Pressure p 2<br />

Temperature T 2<br />

Temperature T 1<br />

Capillary tube, porous plug,<br />

or permeable membrane<br />

FIGURE 19.9<br />

The thermomechanical effect.<br />

In Example 19.4, the pressure difference and the temperature difference have opposite signs. However, the<br />

osmotic heat conductivity (k o ) can be negative for certain substances, and this produces pressure and temperature<br />

differences with the same algebraic sign.<br />

Figure 19.9 shows two rigid insulated vessels connected by a capillary tube (or a tube filled with a porous substance,<br />

or one having a permeable membrane). If we construct this system so that there is either (1) zero heat<br />

transfer between the vessels, (2) zero temperature difference between the vessels, or (3) zero pressure difference<br />

between the vessels, then the following phenomena occur.<br />

19.5.1 Zero Heat Transfer<br />

When the vessels are at different pressures and temperatures but no heat transfer occurs between the vessels<br />

(J Q = 0), Eq. (19.49) requires that a temperature gradient exists along the connecting tube of the form<br />

<br />

<br />

<br />

dT<br />

dx = − k o dp<br />

k t + μk 2 o /ðTk pÞ dx<br />

Then, from Eq. (19.50), the following mass flux occurs through the connecting tube:<br />

<br />

<br />

J JQ M =0 = ρk o<br />

Tk t + μk 2 o /k − ρk <br />

<br />

p dp<br />

p μ dx<br />

which vanishes only under the improbable circumstance of Tk t = 0.<br />

(19.54)<br />

19.5.2 Zero Temperature Gradient<br />

When both vessels are at different pressures but equal temperatures, T = constant and dT/dx = 0. Then, Eqs.<br />

(19.49) and (19.50) give<br />

and<br />

So that, using Eq. (19.50), we can write<br />

<br />

J Q<br />

J M<br />

<br />

<br />

J T<br />

dp<br />

Q =constant<br />

= −k o<br />

dx<br />

<br />

J T M =constant<br />

= − ρk <br />

p dp<br />

μ dx<br />

T<br />

=constant<br />

= μk o<br />

ρk p<br />

<br />

= − T dp<br />

JM<br />

ρ dx<br />

=0<br />

At this point, we define the isothermal energy transport rate _Q i that is “induced” by the thermomechanical mass<br />

flow rate _m with 14<br />

<br />

J Q T Q<br />

= _ i<br />

=constant<br />

_m = μk o<br />

(19.55)<br />

ρk p<br />

J M<br />

14 Notice that we appear to have a heat transfer here that is not due to a temperature difference. It is because such thermal energy<br />

transports exist that we chose earlier to define heat transfer in general terms as a nonwork, nonmass flow energy transport.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!