Question
The manometer fluid in Fig. P3.120 is mercury. Estimate the volume flow in the tube if the flowing fluid is(a) gasoline and $(b)$ nitrogen, at $20^{\circ} \mathrm{C}$ and 1 atm.
Step 1
The flowing fluid is gasoline. We will use Bernoulli's equation for this, which is $P_1/\rho g + V_1^2/2g = P_2/\rho g + V_2^2/2g$. In this case, $V_2 = 0$. Therefore, Bernoulli's equation becomes $V_1^2/2g - (P_2 - P_1)/\rho g = \Delta h Show more…
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As shown in Fig. 1.6, a manometer is attached to a tank of gas in which the pressure is $104.0 \mathrm{kPa}$. The manometer liquid is mercury, with a density of $13.59 \mathrm{~g} / \mathrm{cm}^{3}$. If $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$ and the atmospheric pressure is $101.33 \mathrm{kPa}$, calculate (a) the difference in mercury levels in the manometer, in $\mathrm{cm}$. (b) the gage pressure of the gas, in $\mathrm{kPa}$.
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In Fig. P2.11, pressure gage $A$ reads $1.5 \mathrm{kPa}$ (gage). The fluids are at $20^{\circ} \mathrm{C}$. Determine the elevations $z$, in meters, of the liquid levels in the open piezometer tubes $B$ and $C$.
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