A pipe with diameter $7.5 \,\rm{cm}$ is connected from the bottom of an oil tank to an empty tank. The empty tank is $83$ meters above sea level. The bottom of the oil tank is $93$ meters above sea level. The oil depth/height of the oil tank is $28$ meters. Assume Bernoulli's principle applies for the oil stream and calculate the velocity of the oil exiting the pipe at the empty tank.
I've been trying to solve the above problem using Bernoulli's principle $$p_2+\rho gh_2+\frac12\rho v^2_2=p_1+\rho gh_1+\frac12\rho v^2_1$$ but have been unable to. I also tried using the continuity equation $A_2v_2=A_1v_1$ and found that since the pipe's diameter is constant the area would be the same and the velocity would therefore be $v_1=v_2$ but I was not able to solve the problem using it.
How am I supposed to go about solving this when no velocities are given and the diameter of the pipe is constant?
If the density is needed to answer the question I assume it's $850 \, \rm{Kg} \, \rm{m}^{-3}$ since that's what's listed for oil in a density table in the book. The correct answer is supposed to be (approximately) $27 \,\rm{m}\, \rm{s}^{-1}$.
Assume $p_1=p_2$, and assume at the top of the oil tank the kinetic energy of the oil is zero, then you're left with
$$\frac{1}{2}\rho v^2 = \rho g \Delta h$$
or
$$v = \sqrt{2 g h} \approx \sqrt{2 \cdot 9.81 \cdot 38} \frac{m}{s} = 27.3\frac{m}{s}$$
(Technically, you could assume that $p_1, p_2$ corresponds to the atmospheric pressure at the top of the oil tank and bottom of the empty tank, but that differential is negligible compared to the pressure differential from the oil.)