Need to run a 75mmx30mm 12v .24a blower for 10 sec
It is not so clear how familiar you are with circuits,
 but the following should give you a good head start:

How it works:

The \$R_4\$ represents your load, and upon closing \$SW2\$ the lower side switch \$M_1\$'s gate is pulled up to \$7.5V\$ through the reference voltage provided by the zener diode. At this point your blower is up and running. 

Note that the push-pull button must be held long enough to charge the capacitor \$C_T\$, which should not be longer than a few miliseconds (\$\approx 15ms\$). So, you shouldn't expect any problems with that

Since \$SW2\$ will be open and block the current flow back to the supply, the capacitor \$C_T\$ is forced to discharge through \$R_T\$. And the time required to turn off the transistor \$M_1\$ and consequently the load is given approximately by:

$$t=-R_T\cdot C_T\ \ln(\frac{V_{th}}{V_Z})\approx-1M\Omega\cdot 4.7\mu F\ \ln(\frac{2V}{7.5V})=6.2s$$







simulate this circuit &ndash; Schematic created using CircuitLab