There is no one answer to this question as posed, so we need to establish some constraints to put this question into a form (or forms) that can be analyzed. Also, you should convert pressures to heads for doing the calculations. That way, all parts of Bernoulli's Equation will have the same units (feet or meters).
Here are some things that must be determined before you can calculate. There may be a few others, but I'm running on inadequate sleep today.
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(a) What is the elevation difference between the main in the road and point of discharge at the downstream end of the copper pipe? This elevation difference affects the maximum flow (see Bernoulli's Equation). It should be obvious that you will get less water out of the copper pipe if its point of discharge is higher than the main compared to if the point of discharge is lower than the main.
(b) What elements along the copper pipe exist that create minor head losses over and above the friction head loss due to the copper pipe itself? Every bend, tee, valve, etc. on the copper pipe adds minor head losses and thus reduces the maximum flow available through the copper pipe.
(c) Does the copper pipe branch and are those branches also drawing water? If so, then the flow into the copper pipe from the main will be more than the flow that comes out of the copper pipe at the point of discharge. This is a more complicated scenario, but it's not all that hard to solve. However, your question implies a homogeneous flow, so we will assume flow through only ONE pipe and no branches, or at least no active branches (i.e. faucets on the branches are closed).
Here are some reasonable assumptions we can make:
(a) The water main in the road is large enough compared to the copper pipe that we can treat it as a constant head reservoir. This means there is no need to investigate what happens upstream of the point of connection to the copper pipe. Assume a stating pressure in the water main (e.g. 50 psig; psig = psi gage), then convert it to a pressure head, then add the pressure head to the water main elevation to establish the hydraulic grade line (HGL).
(b) The point of discharge of the copper pipe is open to the atmosphere, which means that its pressure is about 14.7 psia (psia = psi absolute). This also means its pressure is ZERO psig and its HGL equals its elevation.
(c) Velocity heads will be small enough that we can safely ignore them. In any case, the velocity head at either end of a uniform pipe segment (same diameter and no branches) is identical and these will cancel out.
You need to pick an equation for that part of Bernoulli's Equation that deals with friction losses. For potable water, the two most common choices are Hazen-Williams and Darcy. If you use Hazen-Williams, you need to select a reasonable value for the pipe roughness coefficient, C. If you use Darcy, be prepared to iterate to find the friction factor, f. However, you can take a first cut at Darcy with an assumed f, just assume something that is reasonable for your particular situation. Here, the Moody Diagram is your friend.
Here are two scenarios to look at. The first is simple and idealized; the second is more realistic but more work to solve.
(1) The simplest scenario is a 3/4" copper pipe connected to the main at one end, open to the atmosphere at the other end (e.g. an open faucet or hose bibb), and NO minor losses. In this case, your headloss between the ends of the copper pipe is simply the HGL at the water main minus the HGL at the point of discharge. Insert either Hazen-Williams or Darcy into Bernoulli, rearrange and make substitution as needed (e.g. Q=A*V), then solve for Q.
(2) The more realistic scenario is like the first, but also includes all the minor losses. You will need to come up with minor loss coefficients for each offending bend and fitting (including inlet and outlet minor losses for the ends of the copper pipe), add them up, then fit that total into Bernoulli's Equation. This makes the most sense if you're using Darcy for piping head losses because the forms are similar. Alternatively, if you're using Hazen-Wiliams, you can find tables that provide "equivalent lengths" for bends and fittings (each bend or fitting will produce a minor loss that is equivalent to a length of pipe of the same size). This is a less accurate method, but it's easier to do. To use equivalent lengths, simply add them to the length of the copper pipe and do the calculation in (1) with a longer L.
I hope this helps without giving away the answer.
Fred
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"Is it the only lesson of history that mankind is unteachable?"
--Winston S. Churchill