Answer:
F1=177.88 Newtons
Explanation:
Let's start with the Bernoulli's equation:
[tex]P_{1} + \frac{1}{2}\beta V_{1} ^{2} + \beta gh_{1} =P_{2} + \frac{1}{2}\beta V_{2} ^{2} + \beta gh_{2}[/tex]
Where:
P is pressure, V is Velocity, g is gravity, h is height and β is density (for water β=1000 kg/m3); at the points 1 and 2 respectively.
From the Bernoulli's equation and assuming that h is constant and P2 is zero (from the data), we have:
[tex]P_{1} + \frac{1}{2}\beta V_{1} ^{2} = \frac{1}{2}\beta V_{2} ^{2}[/tex]
As we know, P1 must be equal to [tex]\frac{F_{1} }{A_{1}}[/tex], so, replacing P1 in the equation, we have:
[tex]P_{1} = \frac{F_{1}}{A_{1}} = \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2})[/tex]
And
[tex]F_{1} = {A_{1}} ( \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2}))[/tex]
Now, let's find the velocity to replace the values on the expression:
We can express the flow in function of velocity and area as [tex]Q = V A[/tex], where Q is flow, V is velocity and A is area. As the same, we can write this: [tex]Q_{1} = V_{1} A_{1}\\Q_{2} = V_{2} A_{2}[/tex]. In the last two equations, let's clear Velocities.
[tex]V_{1} = \frac{Q_{1}}{A_{1}}\\V_{2} = \frac{Q_{2}}{A_{2}}[/tex]
and replacing V1 and V2 on the last equation resulting from Bernoulli's (the one that has the force on it):
[tex]F_{1} = {A_{1}} ( \frac{1}{2}\beta((\frac{Q_{2}}{A_{2}})^{2} - (\frac{Q_{1}}{A_{1}})^{2}))[/tex]
First, we have to consider that from a mass balance, the flow is the same, so Q1=Q2, what changes, is the velocity. Knowing this, let's write the areas, diameters, density and flow on International Units System (S.I.), because the exercise is asking us the answer in Newtons.
[tex]D_{1}=1.5 inches=0.0381 mts\\D_{2}=1 inches=0.0254 mts\\A_{1}=\frac{\pi D_{1}^{2} }{4}=0.00114mts^{2}\\A_{2}=\frac{\pi D_{2}^{2} }{4}=0.000507mts^{2}\\\beta=1000 kgs/m^{3}\\Q=200gpm=0.01mts^{3}/seg[/tex]
Replacing the respective values in this last expression, we obtain:
F1 = 177.88 N
