A water-powered backup sump pump uses tap water at a pressure of \(3.00 \mathrm{~atm}\left(p_{1}=3 p_{\mathrm{atm}}=\right.\) \(3.03 \cdot 10^{5} \mathrm{~Pa}\) ) to pump water out of a well, as shown in the figure \(\left(p_{\text {well }}=p_{\text {ttm }}\right)\). This system allows water to be pumped out of a basement sump well when the electric pump stops working during an electrical power outage. Using water to pump water may sound strange at first, but these pumps are quite efficient, typically pumping out \(2.00 \mathrm{~L}\) of well water for every \(1.00 \mathrm{~L}\) of pressurized tap water. The supply water moves to the right in a large pipe with cross-sectional area \(A_{1}\) at a speed \(v_{1}=2.05 \mathrm{~m} / \mathrm{s}\). The water then flows into a pipe of smaller diameter with a cross-sectional area that is ten times smaller \(\left(A_{2}=A_{1} / 10\right)\). a) What is the speed \(v_{2}\) of the water in the smaller pipe, with area \(A_{2} ?\) b) What is the pressure \(p_{2}\) of the water in the smaller pipe, with area \(A_{2} ?\) c) The pump is designed so that the vertical pipe, with cross-sectional area \(A_{3}\), that leads to the well water also has a pressure of \(p_{2}\) at its top. What is the maximum height, \(h,\) of the column of water that the pump can support (and therefore act on ) in the vertical pipe?

Step by step solution

01

Find the speed of water in the smaller pipe (v2)

The equation of continuity states that the product of the cross-sectional area and the speed of the fluid remains constant in a pipe. Given that the cross-sectional area of the smaller pipe is 10 times smaller than that of the larger pipe (\(A_2 = A_1/10\)), we have: $$A_1v_1 = A_2v_2$$ Solve for \(v_2\): $$v_2 = \frac{A_1v_1}{A_2}$$ We also know \(A_2 = A_1/10\). Therefore, we can write: $$v_2 = \frac{A_1v_1}{A_1/10} = 10v_1$$ Plug in the given value for \(v_1 = 2.05~m/s\): $$v_2 = 10(2.05~m/s)$$ $$v_2 = 20.5~m/s$$

02

Determine the pressure in the smaller pipe (p2)

We can use Bernoulli's theorem to find the pressure in the smaller pipe. For the larger pipe, we have: $$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$$ Now, we need to solve for \(p_2\): $$p_2 = p_1 + \frac{1}{2}\rho v_1^2 - \frac{1}{2}\rho v_2^2$$ We are given that \(p_1 = 3\;\text{atm} = 3.03\times10^5~\mathrm{Pa}\), we have calculated \(v_1 = 2.05\;\text{m/s}\) and \(v_2 = 20.5\;\text{m/s}\). The density of water can be considered as \(\rho = 1000~\mathrm{kg/m^3}\). Plug in the values: $$p_2 = 3.03\times10^5~\mathrm{Pa} + \frac{1}{2}(1000~\mathrm{kg/m^3})(2.05~\mathrm{m/s})^2 - \frac{1}{2}(1000~\mathrm{kg/m^3})(20.5~\mathrm{m/s})^2$$ $$p_2 = 3.03\times10^5~\mathrm{Pa} - 2.09\times10^5~\mathrm{Pa}$$ $$p_2 = 9.4\times10^4~\mathrm{Pa}$$

03

Find the maximum height of the water column (h) supported by the pump

We are given that the vertical pipe has a pressure of \(p_2\) at its top. To find the maximum height it can support, we can use the concept of hydrostatic pressure: $$\Delta p = p_2 - p_\text{well} = \rho gh$$ Where \(\Delta p\) is the pressure difference, and \(g\) is the acceleration due to gravity (\(9.81~\mathrm{m/s^2}\)). In this case, the pressure difference is equal to the pressure at the top of the vertical pipe (\(p_2\)). Solve for \(h\): $$h = \frac{\Delta p}{\rho g}$$ $$h = \frac{9.4\times10^4\;\text{Pa}}{1000\;\text{kg/m}^3 \times 9.81\;\text{m/s}^2}$$ $$h = 9.58~\mathrm{m}$$ The speed of the water in the smaller pipe is \(20.5\;\mathrm{m/s}\), the pressure in the smaller pipe is \(9.4\times10^4\;\mathrm{Pa}\), and the maximum height the pump can support is \(9.58~\mathrm{m}\).

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