Water at 20^◦ C flows from an 2-m-diameter open tank through a 3-cm opening on the side of the tank (close to the bottom of the tank) as illustrated in Figure EP 5.20. The elevation of the water in the tank at point 1 is 20 m above the centerline of the opening at point 2. The empty tank weights 3

Question

Water at [latex] 20^{\circ} C[/latex] flows from an 2-m-diameter open tank through a 3-cm opening on the side of the tank (close to the bottom of the tank) as illustrated in Figure EP 5.20. The elevation of the water in the tank at point 1 is 20 m above the centerline of the opening at point 2. The empty tank weights 3 N, and assume a static coefficient of friction between the tank and the floor, μ of 0.44. (a) Determine the ideal velocity and flowrate of the free jet as it leaves the opening at point 2. (b) Determine the anchoring force required to keep the tank from moving to the left due to the free jet acting to the right. (c) Determine the friction force, [latex] F_{f} [/latex] acting between the tank and the floor due to the weight of the tank and the water. (d) State if the friction force acting between the tank and the floor,[latex] F_{f} [/latex] is sufficient to keep tank from moving to the left.

Accepted Answer

(a) In order to determine the ideal velocity of the free jetat point 2, the Bernoulli equation is applied between points 1 and 2. Because the cross-sectional area of the tank at point 1 is much larger than the cross-sectional area of the opening, the velocity at point 1 is assumed to be zero. Assuming the datum is at point 2 yields the derivation and application of Torricelli’s theorem as follows:

[latex]Z_{1}: = 20 m[/latex] [latex]Z_{2}: = 0 m[/latex] [latex] P_{1}: = 0 N/m^2 [/latex] [latex] P_{2}: = 0 N/m^2 [/latex] [latex]V_{1}: = 0 \frac{m}{sec} [/latex]

[latex] ho : = 998 \frac{kg}{m^{3}} [/latex] [latex]g: = 9.81 \frac{m}{sec^{2}} [/latex] [latex] \gamma : = ho .g = 9.79 imes 10^{3} \frac{kg}{m^{2}s^{2}} [/latex]

Guess value: [latex]V_{2}: = 1 \frac{m}{sec} [/latex]

Given

[latex] \frac{P_{1}}{\gamma } + Z_{1} + \frac{V^{2}_{1}}{2. g} =\frac{P_{2}}{\gamma } + Z_{2} + \frac{V^{2}_{2}}{2. g}[/latex]

[latex]V_{2}: = Find (V_{2}) = 19.809 \frac{m}{s} [/latex]

Alternatively, one may directly apply Torricelli’s theorem as follows:

[latex]h : = Z_{1} = 20 m [/latex] [latex]V_{2}: =\sqrt{2.g.h} = 19.809 \frac{m}{s} [/latex]

Thus, Torricelli’s theorem, which is a special case of the Bernoulli equation, specifically illustrates the conversion of potential energy stored in the height of the fluid in the tank, h to kinetic energy stored in the ideal velocity of the free jet of fluid at the opening in the side of the tank. Furthermore, in order to determine the flowrate, the continuity equation is applied at point 2 as follows:

[latex]D_{2}: = 3 cm [/latex] [latex]A_{2}: = \frac{\pi .D^{2}_{2}}{4} = 7.069 imes 10^{-4} m^{2} [/latex]
[latex] Q_{2}: = V_{2}.A_{2} = 0.014 \frac{m^{3}}{s} [/latex]

(b) In order to determine the anchoring force required to keep the tank from moving to the left due to the free jet acting to the right, the integral form of the momentum equation is applied in the x direction for the control volume illustrated in Figure EP 5.20. The flow enters at control face 1 and leaves at control face 2. The viscous forces are ignored because of the assumption of ideal flow, there is no pressure force component because the open tank flow is open to the atmosphere ([latex] p_{1} = p_{2} = 0 N/m^{2} [/latex] ), and there is no gravitational force component along the x-axis because the flow at point 2 is assumed to be horizontal.

[latex] \sum{F_{x}} = R_{x} = ( ho Q_{2}v_{2x}) - ( ho Q_{1}v_{1x}) [/latex]

[latex] V_{1x}: = V_{1} = 0 [/latex] [latex] V_{2x}: = V_{2} = 19.809 \frac{m}{s} [/latex] [latex] Q: = Q_{2} = 0.014 \frac{m^{3}}{s} [/latex]

Guess value: [latex] R_{x}: = 1000 N [/latex]

Given

[latex] R_{x}: = ho .Q (V_{2x} - V_{1x} ) [/latex]

[latex] R_{x}: = Find (R_{x}) = 276.816 N [/latex]

Thus, the anchoring force required to keep the tank from moving to the left, [latex] R_{x} [/latex] acts to the right as assumed and is illustrated in Figure EP 5.20. (c) The friction force, [latex] F_{f}[/latex] = μW acting between the tank and the floor due to the weight of the tank and the water is computed as follows:

[latex] \mu : = 0.44 [/latex] [latex] W_{tank}: = 3 N [/latex] [latex] D_{tank}: = 2 m [/latex]

[latex]A_{tank}: = \frac{\pi .D^{2}_{tank}}{4} = 3.142 m^{2} [/latex] [latex]V_{water}: = A_{tank}.h = 62.832 m^{3} [/latex]

[latex] W_{water}: = \gamma .V_{water} = 6.151 imes 10^{5} N [/latex] [latex] W: = W_{tank} + W_{water} = 6.152 imes 10^{5} N [/latex]

[latex] F_{f}: = \mu .W = 2.707 imes 10^{5} N [/latex]

(d) Thus, the friction force acting between the tank and the floor, [latex] F_{f}[/latex] is (more than) sufficient to keep tank from moving to the left.

Question 5.20: Water at 20^◦ C flows from an 2-m-diameter open tank through...

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