Question: If I were wanting to drive i.e. industrial relays which require 24v from a Microchip PIC what would I need to look at in terms of a general transistor circuit (and it's pitfalls when driving such a component).

Answer: 
I'm sorry, but I think Matt's answer is not a good one at all.

The MOSFET in his schematic is a *P-channel*, not an N-channel. The diode doesn't offer any protection for the FET; it may be destroyed together with the FET. Besides it's a 20 V diode, so even if it would protect against the induction voltage the 24 V supply may already kill it. The 7406 is superfluous, besides its maximum voltage is 30 V, not 40 V, and that 30 V is **Absolute Maximum Ratings**, not for continuous use. The circuit will also draw an unnecessary 5 mA with the relay on, and 10 mA no less with the relay off. Also the 100 Ω resistor doesn't "dampen oscillations".

What you need is a **logic level gate** MOSFET. You're using a PIC, which probably will have a supply voltage of minimum 3.3 V. Let us know if the voltage is lower. A logic gate FET will switch on with a 3.3 V gate voltage, so the PIC can drive it directly. No 7406 needed. 

A relay typically needs less than 500 mW, at 24 V that would be 20 mA, but this is an industrial relay, and will probably need more. Let's be generous and say it needs 1 A (that's 24 W!). If we can find a FET with an \$R\_{DS(ON)}\$ of less than 350 mΩ we'll be able to use an SMD; these are much cheaper than PTH parts. At the high 1 A it will dissipate 350 mW. What else? Power supply is 24 V, so let's take a maximum \$V\_{DS}\$ of minimum 40 V. One FET which fulfills these requirements is the [BUK98150](http://www.nxp.com/documents/data_sheet/BUK98150-55A.pdf):

Max. \$V\_{DS}\$ 55 V  

Max. \$I\_D\$ 5 A  

Max. \$R\_{DS(ON)}\$ < 200 mΩ @ 3.3 V  

Max. \$V\_{GS(th)}\$ 2 V

Looks good. The BUK98150 will sink 2 A at 2.6 V gate voltage.

![enter image description here](https://i.stack.imgur.com/DSVkd.png)

This graph shows an \$R\_{DS(ON)}\$ of 175 mΩ @ 3 V and 2 A, for 1 A it will be less. Then dissipated power will be 175 mW, which the SOT-223 package can handle easily. The 175 mV drop is negligible. 

![enter image description here](https://i.stack.imgur.com/ZjVyT.png)

This is the circuit. Contrary to Matt's it only consumes 0.1 mW. I've kept his 100 Ω resistor, which limits the short current spikes when switching; a microcontroller doesn't like capacitive loads much. The 100 kΩ ensures that the gate won't float if the PIC's I/O would be switched to input accidentally. 

As you can see the diode goes over the relay, not the FET. You can use a Schottky diode here. [This one](http://www.nxp.com/documents/data_sheet/PMEG4010EH_EJ_ET.pdf) has a maximum reverse voltage of 40 V.
I personally would follow stevenvh's answer only I would move R2 to the pic side of R1. The reason for this is R1 and R2 are acting as a voltage divider and therefore you are not getting the full pic voltage on to the gate pin of the MOSFET. by moving R2 to the other side of R1 you'll get the full voltage at the gate pin. 
Admittedly the voltage loss will be very small as the resistor values chosen here will only loose 0.003v @ 3.3v so it wouldn't really make a difference in this example but with different resistor values it could make a difference so out of principle I would always put R2 on the other side of R1