Question 6.3: A shaft is rotating at constant angular velocity ω in the be......

The shaft is assumed to be lightly loaded and concentric with the journal (the part of the shaft in contact with the bearing). The control volume selected consists of a unit length of the fluid surrounding the shaft as shown in Figure 6.7. The first law of thermodynamics for the control volume is

{\frac{\delta{\mathcal{Q}}}{d t}}-{\frac{\delta W_{\mathrm{s}}}{d t}}=\int\!\int_{\mathrm{c}.s.}\rho\!\left(e+{\frac{P}{\rho}}\right)(\mathbf{v}\cdot\mathbf{n})\,d A +\frac{\partial}{\partial t}\iiint_{\mathrm{c.v.}}\rho e\,d V+\frac{\delta W_{\mu}}{d t}

From the figure we may observe the following:

1. No fluid crosses the control surface.

2. No shaft work crosses the control surface.

3. The flow is steady.

Thus \delta Q/d t=\delta W_{\mu}/d t=\delta W_{\tau}/d t. The viscous work rate must be determined. In this case all of the viscous work is done to overcome shearing stresses; thus, the viscous work is \textstyle\int\!\int_{{c.s.}}\tau(\mathbf{v}\ \cdot \mathbf{e}_{t})\,d A. At the outer boundary, v=0, and at the inner boundary, \iint_{\mathrm{c.s.}}\tau(\mathbf{v}\,\cdot\,\mathbf{e}_{t})\,d A=-\tau(\omega d/2)A, where e_{t} indicates the sense of the shear stress, \tau , on the surroundings. The resulting sign is consistent with the concept of work being positive when done by a system on its surroundings. Thus,

\frac{\delta Q}{d t}=-\tau\frac{\omega\,d^{2}\pi}{2}

which is the heat transfer rate required to maintain the oil at a constant temperature.

If energy is not removed from the system, then \delta Q/d t=0, and

\frac{\partial}{\partial t}\iiint_{{\mathrm {c.v.}}}e\rho\,d V=-\frac{\delta W_{\mu}}{d t}

As only the internal energy of the oil will increase with respect to time,

{\frac{\partial}{\partial t}}\int\!\!\!\int\!\!\!\int_{\mathrm{c.v.}}e\rho\,d V=\rho\pi\biggl({\frac{D^{2}-d^{2}}{4}}\biggr){\frac{d\mu}{d t}}-{\frac{\delta W_{\mu}}{d t}}=\omega{\frac{d^{2}\pi}{2}}\tau

or, with constant specific heat c

c{\frac{d T}{d t}}={\frac{2\tau\omega\,d^{2}}{\rho(D^{2}-d^{2})}}

where D is the outer bearing diameter.

In this example, the use of the viscous-work term has been illustrated. Note that

1. The viscous-work term involves only quantities on the surface of the control volume.

2. When the velocity on the surface of the control volume is zero, the viscous-work term is zero.