What is the temperature of water at the end of the pipe? I need to solve the following problem for a practical application.
Water, initially at temperature $T_0$, flows with a constant speed $v$ through a PVC tube of length $l$ and diameter $D$. The temperature of the tube is constant on its whole inner surface $T_t$. What is the temperature of the water $T_1$ at the end of the tube?
UPDATE: Real-world application I had in mind. (It may sound weird, you have been warned!)
Four hours a week my girlfriend is released from her mother duties and I'm taking 100% responsibility for our three-month-old daughter. This involves feeding her breast milk stored in the fridge. Unfortunately, I face two challenges:


*

*Arrival of hunger is very unpredictable. It may happen anywhere between 2 and 3 hours.  This makes it really difficult to pre-heat the right amount of milk, as milk that is warmed up but not consumed immediately is to be treated as wasted.

*My daughter is not the most patient when it comes to hunger. To re-iterate, every second the milk is delayed she cries as if her existence was threatened. As her parent, this is uttermost painful and stressful.
So I thought of the following solution: Buy a 10-meter PVC pipe of 3mm inner section (outer section is 5mm). It would start in the breast milk container in the fridge, pass through a giant bowl of water at 40°C,  potentially spiraling several times inside it, and end 1-meter lower where I sit in the sofa. In essence, this would be a gravity-fed pipe for which initial calculations revealed a speed $v=0.39m/s$ and a discharge rate of $167mL/min$, if we crudely assume that breast milk has similar properties to water.
Not sure this was tried before, but it seems promising to me.
The flow-rate is sufficiently high to feed an infant in real-time. The only remaining question was whether the milk warms up fast enough to reach infant consumption temperature. Fortunately, if we plug the numbers in the answer given by Gert, then spiraling 1 meter through the water tank would already bring the milk to reasonable consumption temperature.
 A: Real world assumptions:


*

*No temperature gradients over the cross-section of the pipe.

*Plug flow (turbulent flow).

*Water heat capacity $c_p$ and density $\rho$ are temperature invariant.


Consider an infinitesimal mass element $dm$ at temperature $T(x)$ travelling down the pipe at speed $v$. We apply Newton's law of cooling to it:

$$\frac{dQ}{dt}=hdA\big(T(x)-T_t\big)$$
Where $Q$ is heat energy, $h$ is the heat transfer coefficient (water to PVC) and $dA$ the surface area of the mass element.
$$dA=\pi D dx$$
$$\frac{dQ}{dt}=h\pi D\big(T(x)-T_t\big)dx$$
When the mass element's temperature has dropped by $dT(<0)$, then:
$$dQ=-c_pdmdT$$
Divide by $dt$:
$$\frac{dQ}{dt}=-c_p\dot{m}dT$$
Where: $\frac{dm}{dt}=\dot{m}$ is the mass throughput of the water. So we get:
$$h\pi D\big(T(x)-T_t\big)dx=-c_p\dot{m}dT$$
For plug flow:
$$\dot{m}=\frac{\pi D^2}{4}\rho v$$
With $\rho$ the density of the water. Slightly reworked we get:
$$-\frac{4h}{c_p\rho vD}dx=\frac{dT}{T(x)-T_t}$$
Lets call:
$$\alpha=\frac{4h}{c_p\rho vD}$$
Now integrate between relevant boundaries:
$$\int_0^l-\alpha dx=\int_{T_0}^{T_1}\frac{dT}{T(x)-T_t}$$
$$-\alpha l=\ln\frac{T_1-T_t}{T_0-T_t}$$
$$\Large{T_1=T_t+(T_0-T_t)e^{-\alpha l}}$$
Note that for $l \to +\infty \implies T_1 \to T_t$.

Note also that due to the simplifying real world assumptions this is a 'quick and easy' solution. For laminar flow with temperature gradients $\frac{\partial dT}{\partial y} \neq 0$ and $\frac{\partial dT}{\partial z} \neq 0$, a quasi-2D Fourier heat PDE would have to be solved, which is mathematically far more demanding. Despite the simpler solution's limitations, this is the model widely used by engineers for the design of e.g. tubular heat-exchangers.

The OP's choice of setting $T_t$ as the constant temperature of the inside wall of the tube is a little awkward. Normally we would set $T_t=\text{constant}$ for the cooling fluid surrounding the PVC tube. That then requires the calculation of a thermal resistance $U$ of the pipe, as shown here. In the calculation above, $U$ would replace $k$.
