- Variant 1. Theory
- Variant 2. Calculation formulas
- Variant 2a. Calculation formulas ORIGINAL
- Variant 3. Num. methods
- Variant 4. Online web form
APPENDIXES
Flow Measurement
Variant 2. Standart
Calculation formulas
The water tank shown in Fig. A-FM.02, is shaped like a cylinder. The diameter of the tank is 1 meter. At the bottom of the tank there is a circular pointed hole with a diameter of 0.01 meter. The discharge coefficient of the opening is 0.60. If the tank is filled to a depth of 1 meter, as shown in Fig. A-FM.02, how long will it take to empty it?
Formulas:
\[\sigma\cdot v(h)\cdot \Delta t=-S(h)\Delta h\]
\[\Delta t\to 0\]
\[-S(h)\frac{dh}{dt} = \sigma v(h)\]
\[v(h)=\sigma \mu \sqrt{2gh}\]
\[dt=\frac{-S(h)}{\sigma \mu \sqrt{2gh}}dh\]
\[t=\frac{1}{\sigma \mu \sqrt{2gh}}\int_{h}^{H}\frac{S(h)}{\sqrt{h}}dh\]
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Problem A-1.
How long will it take to empty it?
The water tank shown in Fig. A-FM.02, is shaped like a cylinder. The diameter of the tank is 1 meter. At the bottom of the tank there is a circular pointed hole with a diameter of 0.01 meter. The discharge coefficient of the opening is 0.60. If the tank is filled to a depth of 1 meter, as shown in Fig. A-FM.02, how long will it take to empty it?
\[H=1 m,\mu=0.62 , r=0.015 m, \rho = 1000 \frac{kg}{m^{3}}, g=9.81 \frac{m}{s^{2}} \]
Counting
\[\sigma=\pi \cdot r ^{2}=3.1415 \cdot 0.015^{2} = 7.06858335 \cdot 10^{-4} m^{2}\]
\[S(h)=\pi \cdot h=3.1415 \cdot h m^{2}\]
\[t=\frac{1}{\sigma \mu \sqrt{2gh}}\int_{h}^{H}\frac{S(h)}{\sqrt{h}}dh\]
\[t=\frac{1}{\sigma \mu \sqrt{2gh}}\int_{h}^{H}\frac{S(h)}{\sqrt{h}}dh\]
\[h_D(H,h)=\frac{2\cdot h^{2}-3\cdot H\cdot h}{3\cdot h-6\cdot H} [m]\]
\[h_D(H,h)=\frac{2\cdot 7^{2}-3\cdot 7\cdot 7}{3\cdot 7-6\cdot 7} = 2.33 m \]
Variant 1. Standart
Facts:
- \[P(\rho ,g,H,h,B)=3\cdot 9.81\cdot \left ( 7\cdot 7-\frac{7^{2}}{2} \right )\cdot 1000 = 721035 N \]
- \[h_D(H,h)=\frac{2\cdot 7^{2}-3\cdot 7\cdot 7}{3\cdot 7-6\cdot 7} = 2.33 m \]