I've implemented above circuit and it was successful. The given circuit needs 12VDC to run properly.
I want to run this circuit on 5-6VDC or 9VDC whichever is possible. I know I have to change resistors but not confirmed which to use.
Also if I use 9V battery only to activate 6V relay for 8hrs daily, for how many days will that battery work?
All you really need to change is the relay - pick one with an appropriate coil voltage to match your supply.
The resistor values on the opamp inputs all stay the same since they're used in voltage divider networks off the supply - they're 'ratiometric'.
The only resistor you might consider changing is R2 - you could reduce that one a little in proportion to the supply voltage.
So for a 6V supply I'd consider changing R2 to a 5k6 or 4k7.
The LM358 Opamp in this circuit is an example of a "single supply" opamp and its input common-mode voltage range includes ground.
A different Opamp (like the dreaded 741) may not have this feature and so you wouldn't be able to simply reduce the supply voltage and trust in the ratiometric nature of the inputs.
Other than the relay this circuit will run on anything from 3V to 32V.
R2 is the only other component that will need to change. 10K is way too high anyway, make it like 1K.
In order to change it you need to use a different relay, and since you are running it off a battery you are probably better off with a solid-state relay that can tolerate a larger drive voltage range.
As for battery life... it really depends on the battery, but there is not much you can do to improve it with this circuit other than what I have already mentioned and increasing the size of the pot.
BTW: This circuit needs some hysteresis or it will turn itself off again when the light turns on.
I'll give you something to get started. LDR is a resistor that goes from very high value (100k-1M Ohm) when dark and a low value when light is on.
The LDR is forming a voltage divider with R1, the result of which goes to one of the inputs of the OPAMP U1, which is being used as a comparator. The other input of this very comparator is tied to a potentiometer. So the comparison of the voltages resulting of the voltage divider and the potentiometer is what sets wether the Relay should act or not.
Knowing this you should evaluate:
Regarding to the battery: duration of the battery is dependant on the current that is drawn from it. It is not possible to answer that question with the information you provided.
You have 4 components dependent upon VDD:
1) OpAmp/comparator will work at 6 volts? check it
2) base drive to bipolar is adequate to saturate? you need the standard operating current of the (new 5 volt relay you must chose), and add 10% for winding wire tolerance, and reduce the bipolar beta by 90% to get deep into the saturation region, to avoid time*power product overheating of the bipolar (SOA safe operating area); thus you compute R = 4v/Ib_to_saturate; this assumes your opamp/comparator serves as good comparator without 1Meg resistor from VIn+ to Vout
3) a relay for 5 volts
4) replace the 1n4148 with diode able to handle (one-time-pulse rating) the 200 or 500 milliamp of you new relay
