Pressure difference between two points in a tapered pipe

[I_Try_Math]

Homework Statement

The pipe in the figure is transporting oil (density 850 kg/m3). The velocity at point 1
is 0.5m/s, but at point 2 it is 1.0m/s. Calculate the difference in height in the two open thin tubes

Relevant Equations

Bernoulli's equation

Trying to find an equation for the pressure difference between point 1 and 2. Not sure if my overall reasoning is incorrect, or I introduced a sign error somehow. The equation I come up with implies that the difference is a negative number, but that can't be true if the area of pipe at point 1 is larger than it is at point 2, as it in the given diagram?

Here's my work:

$l_1$ and $l_2$ are intended to mean the distance from points 1 and 2 to where the fluid meets the air.

$P_1 + \rho g l_1 = P_{atm}$

$P_2 + \rho g l_2 = P_{atm}$

$P_1 + \rho g l_1 = P_2 + \rho g l_2$

$P_1 - P_2 + \rho g l_1 = \rho g l_2$

$P_1 - P_2 = \rho g l_2 - \rho g l_1$

$P_1 - P_2 = \rho g(l_2 - l_1)$

$P_1 - P_2 = \rho g(-h)$

 

[Lnewqban]

The height difference should not have any sign.
The lesser height reached by its liquid column is the effect caused by the lower internal static pressure in cross-section 2.