Is pressure really constant?

That's what the problem says

looks like it means the pressure 'difference' between two cross_sections is 7500 pascals.

Meaning $P_2 - P_1 = 7500$ ?

I think the problem here is your (mis)understanding of the wording of the question which you are trying to solve. It is not a problem with understanding Bernouilli's Equation. There is no conceptual difficulty here. So I think the question is "off topic."

@sammygerbil Can I put a picture of the problem and you see for yourself?

If there is a difference in the area of cross-section it means there is a difference in the velocity in the cross-sections (by the equation $A_1v_1 = A_2v_2$), which means there must be a difference in pressure.

Yes, by all means add a picture. I don't think that will change my understanding of the problem. I am not able to vote to close your question, but I will change my down-vote if you convince me that the problem is one of physics rather than language.

Why do you think the pressure is constant (ie P1=P2)?

@sammygerbil I added the picture.

That's it. $A_2$ has a larger pressure than $A_1$ because the velocity of the water would be slower in $A_2$ than $A_1$ because of the larger area. Which means $P_2-P_1 = 7500$

@sammygerbil No. I'm still confused. The question asks for the velocity at cross-section $A_2$. If I try calculating the velocity using $A_1V_1 =A_2V_2$ then I would need to know the area of $A_1$ and $A_2$ or for this question the force at $A_1$ and $A_2$. The only formula I think is valid for this problem is Bernoulli's Equation. What philip said, however, made me think of rearranging Bernoulli's Equation so that $\Delta P$ becomes the subject (since $P_2 - P_1 = \Delta P$).

This is not similar to the hydraulic lift piston problem where you need to know the 'force' at $A_1$ and $A_2$ and assume a constant pressure. The water is flowing, with a velocity related to the area, and because the velocity of water is different at different points in the pipe, there will be a pressure difference. And rightly, we should use the Bernoulli's equation for flowing fluids. Of course in this situation, we could use $A_1v_1 = A_2v_2$ if we know the areas, but is not the case. But we just can't assume a constant pressure when a fluid is flowing at different speeds.

So how can I solve this problem?

$$1/2 \rho v^2_1 = P_2 - P_1 + 1/2 \rho v^2_2$$ $$1/2 \rho v^2_1 = \Delta P + 1/2 \rho v^2_2$$

@philip_0008 using your equation I came up with a negative velocity which doesn't make sense because water is flowing in the positive direction. </br> $$\frac{1}{2} (1000kg/m^3)(3.25m/s) = (7500Pa)+ \frac{1}{2} (1000kg/m^3) v^2_2$$ $$5281.25=7500Pa + 500v^2$$ $$5281.25-7500=500v^2$$ $$-2218.75 = 500v^2$$ $$- \frac{2218.75}{500} = v^2$$ $$\sqrt(v) = -\sqrt \frac{2218.75}{500}$$ $$v = -2.11m/s$$

That's strange. Maybe there's something wrong with the problem.

@AugieJavax98 : Your calculation appears to be correct, apart from taking the square root of a -ve number. $v_2$ is therefore imaginary. Maybe there is more to the question. I note this is part (a). Were there no relevant instructions before or after this part of the question? Otherwise it seems the question is faulty. It happens.

I would guess the $\Delta p$ of 7500 is too high. If you calculate the same problem with lower $\Delta p$ it works. Probably they either chose $\Delta p$ too high or v1 too low. For such a high increase in pressure the corresponding drop in velocity is greater than v1, therefore you would get a negative v2.

Guys a friend of mine showed me how he did this problem and I think it's probably correct. <br/> $$P_1 + \frac{1}{2} \rho v^2_1 = P_2 + \frac{1}{2} \rho V^2_2$$ $$P_2 - P_1 + \frac{1}{2} \rho v^2_1 = \frac{1}{2} \rho v^2_2$$ $$\Delta P =\frac{1}{2} \rho v^2_2 - \frac{1}{2} \rho v^2_1$$ $$\Delta P = \frac{1}{2} \rho (v^2_2 - v^2_1)$$ $$7500Pa = \frac{1}{2} (1000kg/m^3)(v^2_2 - (3.25m/s)^2)$$ $$7500Pa = 500kg/m^3(v^2-10.5625m^2/s^2)$$ $$\frac{7500Pa}{500kg/m^3} = v^2 - 10.5625m^2/s^2$$ $$15m^2/s^2 + 10.5625m^2/s^2 = v^2$$ $$v^2 = 25.5625m^2/s^2$$ $$\sqrt v^2=\sqrt25.5625m^2/s^2$$ $$v=5.06m/s$$

@sammygerbil I also checked the units to make sure that the equation gives me velocity. $$Pa = kg/m^3(v^2-(m/s)^2)$$ $$N/m^2 =kg/m^3(v^2-(m^2/s^2))$$ $$\frac{kgm/s^2/m^2}{kg/m^3} = v^2 - (m^2/s^2)$$ $$\frac{\frac{kgm}{s^2}X\frac{1}{m^2}}{kg/m^3} = v^2 - (m^2/s^2)$$ $$\frac {kg}{s^2m}X\frac{m^3}{kg} = v^2 - (m^2/s^2)$$ $$m^2/s^2 = v^2 - (m^2/s^2)$$ $$v^2 = (m^2/s^2)+(m^2/s^2)$$ $$v^2 = m^2/s^2$$ $$\sqrt v^2 = \sqrt m^2/s^2$$ $$v = m/s$$

@AugieJavax98 : your friend's solution is not correct. There is an error in line 2 : on the left should be $P_1-P_2$ not $P_2-P_1$. I think we have agreed that $P_2>P_1$ which is probably why your friend "fiddled" the equation to get an answer which is not imaginary.

However, I wonder if we are making the error by assuming that $A_2>A_1$ and therefore $P_2>P_1$? That clue comes from the illustration but is not stated in the text. Perhaps the illustration is misleading us. So perhaps we are not allowed to assume that $P_2>P_1$. Perhaps the question intends $A_1>A_2$ and therefore $P_1>P_2$.

the velocity cannot be greater after cross sectional area increased. At least not for an incompressible fluid. @sammy gerbil this would be a good explanation as well