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The "wye' fitting shown in the figure below lies in a horizontal plane. The fitting splits the inlet flow into two equal parts. At section 1, the water velocity is 12 ft/s and the pressure is 20 psig. The temperature of water is 60°F. Assume inviscid flow. (7201 P2 D2 D2 = 3 in. Qi Di = 6 in. D3 D3 = 4 in. P3 14201 Calculate the water pressure at sections 2 and 3. psig P2 = i P3 psig

User SeaSpell
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Final answer:

The water pressure at sections 2 and 3 is both 20 psig.

Step-by-step explanation:

In order to calculate the water pressure at sections 2 and 3, we need to apply the principle of conservation of mass and Bernoulli's equation. At section 1, the water velocity is given as 12 ft/s and the pressure is 20 psig. Since the fitting splits the inlet flow into two equal parts, each part will have a velocity of 12 ft/s and the same area as the inlet.

Using the conservation of mass, we can calculate the area at section 2 and 3. Let's assume the area at section 1 is A1, and the areas at sections 2 and 3 are A2 and A3, respectively. Since the flow is split equally, A2 = A3 = A1/2.

Now, using Bernoulli's equation that states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume is constant along a streamline, we can set up equations for sections 1, 2, and 3. At section 1, the equation can be written as:

20 psig + (1/2) * (12 ft/s)^2 + 0 = P2 + (1/2) * (12 ft/s)^2 + 0

Since the fitting is in a horizontal plane, the height terms cancel out. Rearranging the equation, we get:

P2 = 20 psig

Similarly, at section 3, the equation can be written as:

20 psig + (1/2) * (12 ft/s)^2 + 0 = P3 + (1/2) * (12 ft/s)^2 + 0

Rearranging the equation, we get:

P3 = 20 psig

User Jalo
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