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In series would usually be best, with two caveats: the setup has to be such that the upstream meter does not affect the downstream meter, either systematically or by adding noise. if the total flow is near the full scale of the sensor, and the sensor type is such that there is a "sweet spot" in accuracy or noise over the range The first of the ...


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Obviously putting it in series is better than the other two options. Putting in series, means that If you put two (perfectly calibrated) flow meters in series, then you will be able to make inferences on the actual value based on the calibration uncertainty, the reading uncertainty, and the sampling statistics. Problems with the other setups I'll start with ...


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I've had a go at calculated the pressure loss along the pipe, can someone verify this? A 5MW Francis turbine has been chosen using this; So in generator mode, the rated discharge will be: $$P_g=\rho \times g \times h \times \eta_g \times Q_g$$ Rearranging for Q while assuming generator efficiency is 0.9 gives: $$Q_g = \frac{P_g}{\rho \times g \times h \times ...


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You can't pump water into the ocean and gain the potential energy of it again like this. In this example, if you pump water out, the tank fills with water anyway because it's really just a balloon. If you somehow had a cave you could pump out then it's a simple head calc, using the total pipe equivalent length. Basically, putting this in the ocean instead ...


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The key principle is that, when expressed in terms of the non-dimensional head coefficient $K_H := gH/\left(D\omega\right)^2$ and the non-dimensional flow coefficient $K_Q := Q/\left(D^3\omega\right)$, there is a single head-flow characteristic for the whole family of geometrically similar pumps. That is to say, when expressed in terms of those quantities, ...


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We know $Q = AV$, and $A = \pi$$D^2/4$, let's rearrange the terms and get to the root "D": $Q_1^2 = (A_1V_1)^2 = (\pi$$D_1^2V_1/4)^2$ = $(70-H_1)/1.41 * 10^{-3}$ $D_1^2 = (4/V_1\pi)\sqrt{(70 - H_1)/1.41 * 10^{-3})}$ $D_1 = \sqrt{(4/V_1\pi)\sqrt{(70 - H_1)/1.41 * 10^{-3})}} $ You can do the same for $D_2$ and get $D_1/D_2$. (First to check my ...


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The symbol $H_{\mathsf{C}}$ here doesn't represent a head loss, it represents an "absolute" head at point $\mathsf{C}$. Assuming that the fluid velocity in the region vertically above $\mathsf{C}$ is negligible, so that the pressure variation in that region is hydrostatic, that absolute head is equal to the vertical co-ordinate of the free surface ...


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