Confusions with common mode voltage in a single ended system
One way to calculate the common mode voltage is very simple:

$$V_{cm} = \frac{V_+ + V_-}{2}$$

From this formula you can find the common-mode voltage in any configuration where you can work out the voltage at the inverting and non-inverting inputs.

For example,



Here, the common-mode voltage is \$\frac{V_\rm{cm} + V_\rm{d}}{2}\$ (\$V_\rm{cm}\$ being the value of the source in the diagram, not the actual common-mode voltage) because the inverting input is fixed at 0 V. If the amplifier "OA" is actually an op-amp, the result is likely that the output is driven to one or the other power supply rails (or as close as that op-amp is capable of driving it), because there is no negative feedback to reduce the circuit gain.


  So the above circuit [your first diagram] can be modelled as below [your second diagram]??


The two diagrams are not equivalent. The first one shows no means of establishing the common-mode voltage, so the internal bias networks of the amplifier will drive it to some value. But they may do so only very weakly, and that the common mode voltage might even drift around as a few electrons are blown onto or off of the circuit by passing breezes (depending on the exact design of the op-amp).

Your second diagram shows a voltage source establishing a common mode voltage with a low impedance, which is probably a much better way of driving an op-amp circuit.