9 Calculate the flow rate from this water tank if the 6 in. pipeline has a friction factor of 0.020 and is 50 ft long. Is cavitation to be expected in the pipe entrance

The water tank is missing, so i have attached it.

Complete question is;

Calculate the flow rate from this water tank if the 6 in. pipeline has a friction factor of 0.020 and is 50 ft long. Is cavitation to be expected in the pipe entrance? The water in the tank is 5 ft deep.

V = 30.33 ft/s

Cavitation is not to be expected at the pipe entrance

Explanation:

If we apply the energy equation between the surfaces of the two pools of water, we can find the headloss from;

p1/γ + z1 + V1²/2g = p2/γ + z2 + V2²/2g + h_l

From conditions given, p1, p2, V1 and V2 are all equal to zero.

Thus, we now have;

h_l = z1 - z2

Thus,

h_l = 50 - 0

Δh_l = 50

The headloss can be further expressed from the equation;

Δh_l = h_l,entrance + h_l,pipe + h_l,exit

Now, breaking them down, we have;

h_l,entrance = 0.5(V²/2g)

h_l,pipe = (fL/d) × (V²/2g)

h_l,exit = V²/2g

We are given;

Headloss; Δh_l = 50 ft

friction factor; f = 0.02

Length; L = 50 ft

Diameter; d = 6 inches = 6/12 ft = 0.5 ft

g is acceleration due to gravity = 32.2 ft/s²

Thus;

Δh_l = 0.5(V²/2g) + [(fL/d) × (V²/2g)] + (V²/2g)

Factorizing out V²/2g,we have;

Δh_l = (V²/2g)[0.5 + (fL/d) + 1]

Plugging in the relevant values;

50 = (V²/(2 × 32.2))[0.5 + (0.02 × 50/0.5) + 1]

50 = (V²/64.4)[3.5]

V² = 50 × 64.4/3.5

V² = 920

V = √920

V = 30.33 ft/s

Now, total loss of pressure head is given as;

Total loss of pressure head = h_l,entrance + h_l,exit

Total loss of pressure head = 0.5(V²/2g) + (V²/2g)

Plugging in the relevant values gives;

Total loss of pressure head = (30.33²/(2 × 32.2))(0.5 + 1) = 21.43 ft

Now, due to the fact that the pressure head is 5 ft deep before the water entered the pipe, and it drops by

21.43 ft soon after that, it means the net pressure head is: h = 5 - 21.43 = −16.43 ft just atop the inside part of the

pipe.

The headloss due to friction in the pipe from earlier was seen to be;

h_l,pipe = (fL/d) × (V²/2g)

Thus;

h_l,pipe = (0.02 × 50/0.5) × ((30.33²/(2 × 32.2))

h_l,pipe = 28.57 ft

Converting the net pressure head of −16.43 ft to pressure in psi, from conversion formula, we have;

P_min,gauge = 0.433h(SG) psi

Where;

h is net pressure head

SG is specific gravity

From tables, specific gravity of fresh water is 1

We have h = -16.43

Thus;

P_min,gauge = -16.43 × 0.433 × 1

P_min,gauge = -7.114 psi

Formula for minimum absolute pressure is;

P_min,absol = P_min,gauge + P_atm

Where P_atm is atmospheric pressure in psi and from tables, it has a value of 14.696 psi

Thus;

P_min,absol = -7.114 + 14.696

P_min,absol = 7.582 psi

From online sources relevant to this question, it is gathered that the temperature of the water is at 70°F.

Now, from the table i attached, the vapour pressure in psi at that temperature of 70°F is 0.36341 psi.

Thus is less than the minimum absolute pressure. Thus cavitation will not occur

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explanation: