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Chapter 7: Problem 3

A hose pipe of \(75 \mathrm{~mm}\) bore and length \(450 \mathrm{~m}\) is supplied with water at \(1.4 \mathrm{MPa}\). A nozzle at the outlet end of the pipe is \(3 \mathrm{~m}\) above the level of the inlet end. If the jet from the nozzle is to reach a height of \(35 \mathrm{~m}\) calculate the maximum diameter of the nozzle assuming that \(f=0.01\) and that losses at inlet and in the nozzle are negligible. If the efficiency of the supply pump is \(70 \%\) determine the power required to drive it.

Short Answer

Expert verified
Nozzle diameter: calculate energy, find v h enzime losses some: nozzle size overall time 70%.

Step by step solution

01

Determine Jet Velocity for Desired Height

Using the formula for height of a jet: \[ h = \frac{v^2}{2g} \] where \[ h = 35 \, \text{m}, \ g = 9.81 \, \text{m/s}^2 \]Rearrange the formula to solve for the velocity (v): \[ v = \sqrt{2gh} = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 35 \, \text{m}} \]\[ v = 26.2 \, \text{m/s} \]
02

Calculate Head Losses Due to Friction in Pipe

Frictional head loss in the pipe can be calculated using the Darcy-Weisbach equation:\[ h_f = f \frac{L}{D} \frac{v^2}{2g} \] where \[ f = 0.01, \, L = 450 \, \text{m}, \, D = 0.075 \, \text{m}, \, \ v = ?\] We need to find the velocity in the pipe, which is affected by the nozzle diameter. Let's denote the nozzle diameter as \(d_n\).
03

Apply Continuity Equation to Relate Velocities

Using the continuity equation: \[ A_1 v_1 = A_2 v_2 \] where \[ A_1 = \pi (0.0375)^2, { v_1} = v_{pipe}, \, \ A_2 = \, \pi \left( \frac{d_n}{2} \right)^2 , \, v_2 = 26.2 \, \text{m/s}\] Rearrange to find \(v_{pipe}\): \[v_{pipe} = \frac{ \pi \left( \frac{d_n}{2} \right)^2 v_2 }{ \pi (0.0375)^2} = (0.44 d_n^2)n 26.2 \] Insert the new value into the head loss equation and solve.
04

Determine Jet Outlet Velocity Losses

Combine the energies and solve Reynolds equation: general: Equ = energy balance and minor losses, taking into account height and losses together.
05

Solve Hydraulic Head Loss/Discharge

Final step: solve for energy converted and losses for diameter calculation, nozzle speed gain
06

Power Required for Pump

Determine Power: pump,given efficiency formula use in the context. Solve freq: H(p).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Darcy-Weisbach equation
The Darcy-Weisbach equation is essential in fluid mechanics for calculating the frictional head loss in pipes. It is given by: \[ h_f = f \frac{L}{D} \frac{v^2}{2g} \]Where:
  • \(h_f\) is the frictional head loss
  • \(f\) is the Darcy friction factor, dimensionless
  • \(L\) is the length of the pipe
  • \(D\) is the diameter of the pipe
  • \(v\) is the flow velocity
  • \(g\) is the acceleration due to gravity
In our specific exercise, the length (\(L\)) is 450 m, the pipe diameter (\(D\)) is 0.075 m, and the friction factor (\(f\)) is 0.01. To understand the energy losses due to friction, we first need to calculate the velocity of water in the pipe which will be determined by the diameter of the nozzle.
Jet Velocity Calculation
To see how high a jet of water can reach, we need to calculate its velocity as it exits the nozzle. This can be done using the principle of energy conservation, given by:\[ h = \frac{v^2}{2g} \]Where:
  • \(h\) is the height the jet reaches (35 m)
  • \(g\) is the acceleration due to gravity (9.81 m/s^2)
Solving for the velocity:\[ v = \sqrt{2gh} = \sqrt{2 \times 9.81 \frac{m}{s^2} \times 35 \ m} = 26.2 \ m/s \]This velocity is critical as it determines the energy and momentum of the jet required to reach the given height.
Continuity Equation
The continuity equation helps relate the velocities and cross-sectional areas at different points in the system, ensuring mass conservation. It is represented as: \[ A_1 v_1 = A_2 v_2 \]Where:
  • \(A_1\) and \(A_2\) are the cross-sectional areas of the pipe and nozzle respectively
  • \(v_1\) and \(v_2\) are the flow velocities in the pipe and nozzle
Given the pipe diameter is 0.075 m, calculate the area of the pipe (\(A_1\)):\[ A_1 = \pi \left( \frac{0.075}{2} \right)^2 \]\[ A_2 = \pi \left( \frac{d_n}{2} \right)^2 \]Since the velocity of the jet exiting the nozzle is 26.2 m/s, we can substitute these into the continuity equation to find the velocity in the pipe \(v_{pipe}\) given by: \[ v_{pipe} = \frac{A_2 v_2}{A_1} = (\frac{d_n^2}{0.075^2})26.2 \ m/s \]
Frictional Head Loss
Frictional head loss is important in determining how much pressure is lost as water flows through the pipe. Using the previously mentioned Darcy-Weisbach equation, we substitute the values: \[ h_f = 0.01 \frac{450}{0.075} \frac{v_{pipe}^2}{2 \times 9.81} \]With \(v_{pipe}\) from continuity equation: \[ h_f = 0.01 \frac{450}{0.075} \frac{(\frac{d_n^2}{0.075^2})26.2^2}{2 \times 9.81} \]The total head available would be reduced by frictional head loss, which affects the final velocity and pressure output.
Pump Efficiency
To determine the power required by the pump, we also need to consider its efficiency. Pump efficiency (\(\eta\)) is the ratio of useful hydraulic power delivered to the fluid to the actual power consumed by the pump. It is represented as: \[ \eta = \frac{P_{useful}}{P_{input}} \]Given pump efficiency is 70%, and the useful power can be determined from the pressure and flow rate, the power input (\(P_{input}\)) can be calculated as: \[ P_{useful} = \rho g Q H \]\[ P_{input} = \frac{P_{useful}}{0.7} \]Where:
  • \(\rho\) is water density
  • \(g\) is the gravitational constant
  • \(Q\) is the flow rate
  • \(H\) is the head produced by the pump
Thus, accounting for these values allows us to determine the necessary power to drive the pump effectively.

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Most popular questions from this chapter

A trailer pump is to supply a hose \(40 \mathrm{~m}\) long and fitted with a \(50 \mathrm{~mm}\) diameter nozzle capable of throwing a jet of water to a height \(40 \mathrm{~m}\) above the pump level. If the power lost in friction in the hose is not to exceed \(15 \%\) of the available hydraulic power, determine the diameter of hose required. Friction in the nozzle may be neglected and \(f\) for the hose assumed to be in the range \(0.007-0.01\).

Between the connecting flanges of two pipes \(A\) and \(B\) is bolted a plate containing a sharp-edged orifice \(C\) for which \(C_{\mathrm{c}}=0.62\). The pipes and the orifice are coaxial and the diameters of \(A\), \(B\) and \(C\) are respectively \(150 \mathrm{~mm}, 200 \mathrm{~mm}\) and \(100 \mathrm{~mm}\). Water flows from \(A\) into \(B\) at the rate of \(42.5 \mathrm{~L} \cdot \mathrm{s}^{-1}\). Neglecting shear stresses at boundaries, determine ( \(a\) ) the difference of static head between sections in \(A\) and \(B\) at which the velocity is uniform, (b) the power dissipated.

Two reservoirs are joined by a sharp-ended flexible pipe \(100 \mathrm{~mm}\) diameter and \(36 \mathrm{~m}\) long. The ends of the pipe differ in level by \(4 \mathrm{~m}\); the surface level in the upper reservoir is \(1.8 \mathrm{~m}\) above the pipe inlet while that in the lower reservoir is \(1.2 \mathrm{~m}\) above the pipe outlet. At a position \(7.5 \mathrm{~m}\) horizontally from the upper reservoir the pipe is required to pass over a barrier. Assuming that the pipe is straight between its inlet and the barrier and that \(f=0.01\) determine the greatest height to which the pipe may rise at the barrier if the absolute pressure in the pipe is not to be less than \(40 \mathrm{kPa}\). Additional losses at bends may be neglected. (Take atmospheric pressure \(=101.3 \mathrm{kPa}\).)

A large tank with vertical sides is divided by a vertical partition into two sections \(A\) and \(B\), with plan areas of \(1.5 \mathrm{~m}^{2}\) and \(7.5 \mathrm{~m}^{2}\) respectively. The partition contains a \(25 \mathrm{~mm}\) diameter orifice \(\left(C_{d}=0.6\right)\) at a height of \(300 \mathrm{~mm}\) above the base. Initially section \(A\) contains water to a depth of \(2.15 \mathrm{~m}\) and section \(B\) contains water to a depth of \(950 \mathrm{~mm}\). Calculate the time required for the water levels to equalize after the orifice is opened.

Calculate the power required to pump sulphuric acid (dynamic viscosity \(0.04 \mathrm{~Pa} \cdot \mathrm{s}\), relative density \(1.83\) ) at \(45 \mathrm{~L} \cdot \mathrm{s}^{-1}\) from a supply tank through a glass-lined \(150 \mathrm{~mm}\) diameter pipe, \(18 \mathrm{~m}\) long, into a storage tank. The liquid level in the storage tank is \(6 \mathrm{~m}\) above that in the supply tank. For laminar flow \(f=16 / \mathrm{Re}\); for turbulent flow \(f=0.0014\left(1+100 \mathrm{Re}^{-1 / 3}\right)\) if \(\mathrm{Re}<10^{7}\). Take all losses into account.
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