Bernoulli's Principle: Venturi Pipe

[sirclash]

1. Where did i go wrong, please explain.
Homework Statement
The Horizontal Venturi pipe has a cross sectional area of 50cm^2 at the wider portions(A_1) and 25cm^2 at the constriction(A_2). Air density of 1.29 kg/(m^3) is flowing in the pipe.
there's a attachment of a picture to reference.
A)Write down Bernoulli's equation for the streamline of the wider part to the streamline of the constricted part.
B)Derive an expression for the speed of the air, V_2 (Constricted), in terms of the cross sectional area A_1 and A_2, The pressure difference (P_1 - P_2), and the density of air (roh).
C) The difference in height, h, of the columns in the U-shaped tube is 15 cm. Find the pressure difference given the density of mercury is 13600 kg/(m^3)
D) Calculate V_2

Homework Equations

P_1 + (roh*V_1^2)/2 + roh*g*y_1= P_2 + (roh*V_2^2)/2 + roh*g*y_2

The Attempt at a Solution

A) P_1 + (roh*V_1^2)/2 = P_2 + (roh*V_2^2)/2
B) I rearranged the the answer and used V_1=(A_2*V_2)/(A_1) to get,
V_2= (2(P_1 - P_2))/(roh(1-(A_2)/(A_1))^2) is this right??
C) P_1 - P_2 = (roh of mercury)*g*h i get P_1 - P_2 = 1999.2 pascals?
D) I got 1033.17 m/s and that can be right unless i just converted wrong.

 

[amy andrews]

Just checking, did you convert the centimeter values to meters? That's the first thing to look at... if so, you'll have to give us more detailed calculations so we can see where you went wrong, because 1033.17 m/s is indeed an absurd answer.

 

[Matdehaast]

Dunno man but for part B I get something different. See if it helps... What I get for part B is:

V_2 = (A_1*V_1)/A_2

Therefore V_2= SQRT of 2*A_1(P_1-P_2)/(Rho)(1-(A_1)^2)

Peace
Mat