Quick and dirty schematic:-

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First option if the NVEM will source enough current to switch the relay. There's a reason why it might not, and I'll put this on the bottom of this reply, but try that first. The diode is connected across the coil, it's purpose is to protect the NVEM from back emf. I'll let you google the reason for this rather than lying. It's important.


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The second option introduces a transistor (any general purpose PNP would do with a VCEO rating of 30V, Ic of 50mA, and a hfe (gain) of at least 100. That describes pretty much any general purpose transistor. I'd guess at a 2N3706 but any similar would do. If the output from the NVEM is switched - it goes low - that will drag the base voltage on the transistor significantly below the emitter voltage (24V) and the transistor will conduct C-E, and the relay will actuate. The current through the NVEM will be in the order of 1mA.


The reason why I'm uncertain if the NVEM will source the current for the relay directly is centred around Lies, Damned-Lies, and Data-Sheets.

The EL3H7 opto-isolators internal to the NVEM are rated at 50mA Ic (collector current). That is, beyond 50mA you risk damaging the transistor. But, the actual Ic is a product of the CTR (Current Transfer Ratio) of the device and the forward current through the LED internal to the device. Typically designed around 10mA, could be 20mA, but the CTR is specified as between 40% and 320%. So, guessing a 20mA current flow through the opto-isolator LED, the actual transistor current will be between 40% (4mA) and 320% (64mA), depending on the manufacturing process. So... 4mA won't switch the relay, whereas 64mA would (though the relay coil resistance would limit this to 36mA).

That's why you'd be best doing a bench-test on the relays first.