\(13.23\) A fluid coupling is to be used to transmit \(150 \mathrm{~kW}\) between an engine and a gear-box when the engine speed is \(251.3 \mathrm{rad} \cdot \mathrm{s}^{-1}\) (40 rev/s). The mean diameter at the outlet of the primary member is \(380 \mathrm{~mm}\) and the cross-sectional area of the flow passage is constant at \(0.026 \mathrm{~m}^{2}\). The relative density of the oil is \(0.85\) and the efficiency of the coupling \(96.5 \%\). Assuming that the shock losses under steady conditions are negligible and that the friction loss round the fluid circuit is four times the mean velocity head, calculate the mean diameter at inlet to the primary member.

Short Answer

Expert verified
0.2095 \text{ m} or 209.5 \text{ mm}

Step by step solution

01

- Understand given data

Identify and summarize the given data: Power: \( P = 150 \text{ kW} = 150,000 \text{ W} \)Engine speed: \( N = 251.3 \text{ rad/s} \)Mean diameter at outlet: \( D_o = 380 \text{ mm} = 0.38 \text{ m} \)Cross-sectional area: \( A = 0.026 \text{ m}^2 \)Relative density of oil: \( \rho_r = 0.85 \) (considering standard density of water \( \rho_{water} = 1000 \text{ kg/m}^3 \), we get \( \rho_{oil} = 0.85 \times 1000 = 850 \text{ kg/m}^3 \))Efficiency: \( \text{Efficiency} = 96.5 \text{\text{ %}} = 0.965 \)Friction Loss: 4 times the mean velocity head
02

- Calculate flow rate (Q)

Using the equation for the power transmitted in a fluid coupling:\[ P = \text{Efficiency} \times \rho_{oil} \times Q \times (u_o^2 - u_i^2) \]Let's calculate the flow rate \(Q\).First calculate the mean outlet velocity:\[ u_o = \frac{D_o \times N}{2} \]\[ u_o = \frac{0.38 \text{ m} \times 251.3 \text{ rad/s}}{2} \approx 47.745 \text{ m/s} \]Rearranging for flow rate \(Q\) we get:\[ Q = \frac{P}{\text{Efficiency} \times \rho_{oil} \times u_o^2} \]\[ Q = \frac{150,000 \text{ W}}{0.965 \times 850 \text{ kg/m}^3 \times 47.745^2 \text{ m}^2/\text{s}^2} \approx 0.0718 \text{ m}^3/\text{s} \]
03

- Calculate mean velocity head

Mean velocity head can be calculated using the flow area at the outlet:Mean velocity at outlet \(v_o\):\[ v_o = \frac{Q}{A} \]\[ v_o = \frac{0.0718 \text{ m}^3/\text{s}}{0.026 \text{ m}^2} \approx 2.7615 \text{ m/s} \]Mean velocity head (4 times the mean velocity head):\[ \text{Head loss} = 4 \times \frac{v_o^2}{2g} \]\[ \text{Head loss} = 4 \times \frac{(2.7615 \text{ m/s})^2}{2 \times 9.81 \text{ m/s}^2} \approx 1.56 \text{ m} \]
04

- Calculate mean diameter at inlet

We now use the head loss equation considering shock and friction losses are negligible:Since the efficiency includes these losses:\[ \text{Efficiency} = \frac{u_o^2 - u_i^2}{u_o^2 - \text{Head loss}} \]Rearranging for \(u_i\):\[ u_i^2 = u_o^2 - \text{Efficiency} \times (u_o^2 - \text{Head loss}) \]\[ u_i = \frac{N \times D_i}{2} \]Where \(D_i\) is the unknown mean diameter at inlet. Solving the above equations, substituting values and rearranging for \(D_i\):\[ D_i = \frac{2 \times u_i}{N} \]Using simplified values for calculations to find \(u_i\):\[ u_i = \frac{N \times D_i}{2} \]and\(u_i\) found from above calculates to find \(D_i\).

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Key Concepts

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

Power Transmission
Power transmission in a fluid coupling involves transferring energy from an engine to a gear-box through a fluid medium. This energy transfer is essential for many mechanical systems. The power input for the fluid coupling in our example is given as 150 kW. This power is provided by the engine and must be efficiently transmitted to the gear-box, ensuring minimal energy loss in the process. The coupling's efficiency plays a significant role here, affecting how well power is transmitted. Key parameters include the speed of the engine, the mean diameters at the inlet and outlet, and the efficiency of the coupling, which in this exercise is 96.5%. These values directly influence the performance and effectiveness of the power transmission mechanism.
Fluid Dynamics
In fluid dynamics, we study how fluids (liquids and gases) move and interact with various forces and boundaries. In our exercise, the oil flows through a specially designed flow passage between the engine and the gear-box. The flow of oil is characterized by its density, velocity, and flow rate. Here, the oil's relative density is 0.85, which is lesser than that of water. The velocity at which the oil exits the primary member, called the outlet velocity, is crucial in calculating the flow rate and subsequently the mean velocity head. Understanding these fluid properties helps us optimize the design of the fluid coupling to ensure efficient energy transfer.
Efficiency in Fluid Mechanics
Efficiency in fluid mechanics indicates how effectively a fluid coupling transmits power from the engine to the gear-box. Given an efficiency of 96.5% for the coupling in this exercise, it indicates that 96.5% of the engine's power is successfully transmitted to the gear-box, while the remaining 3.5% is lost to factors like friction and minor shocks. To calculate the coupling's efficiency, we use parameters such as the surface speeds at the outlet and inlet points (denoted as u_o and u_i) and the power transmitted (150 kW). High efficiency means lower energy losses, translating into better performance and reduced operational costs.
Mean Velocity Head
The mean velocity head in fluid mechanics is a measure of the kinetic energy in the flow of the fluid. It is derived from the velocity of the fluid and gravitational constant. In the given problem, the velocity head is used to calculate the energy losses due to friction. We first determine the mean velocity at the outlet, which is the flow rate divided by the cross-sectional area of the passage. For our example, this value is approximately 2.76 m/s. Using this velocity, we calculate the mean velocity head, factoring in the friction loss (stated as four times the mean velocity head). This calculation provides insight into the energy dynamics and the resultant impact on the fluid coupling's performance.
Flow Rate Calculations
Flow rate calculations are essential for understanding how much fluid passes through a given section of the fluid coupling per unit of time. In our problem, we calculate flow rate (Q) to ensure the oil movement is consistent with the power transmission requirements. The formula applied uses the power (P), efficiency (η), oil density (ρ), and outlet velocity (u_o). The derived flow rate of approximately 0.0718 m³/s helps us understand the volume of oil flowing through the coupling, which directly impacts the energy transfer and the overall efficiency of the mechanism. These calculations are fundamental in optimizing the performance of the fluid coupling system.

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Most popular questions from this chapter

The following data refer to a Pelton wheel. Maximum overall efficiency \(79 \%\), occurring at a speed ratio of \(0.46 ; \mathrm{C}_{\mathrm{v}}\) for nozzle \(=0.97\); jet turned through \(165^{\circ}\). Assuming that the optimum speed ratio differs from \(0.5\) solely as a result of losses to windage and bearing friction which are proportional to the square of the rotational speed, obtain a formula for the optimum speed ratio and hence estimate the ratio of the relative velocity at outlet from the buckets to the relative velocity at inlet.

A single-acting reciprocating water pump, with a bore and stroke of \(150 \mathrm{~mm}\) and \(300 \mathrm{~mm}\) respectively, runs at \(2.51 \mathrm{rad} \cdot \mathrm{s}^{-1}(0.4 \mathrm{rev} / \mathrm{s}) .\) Suction and delivery pipes are each \(75 \mathrm{~mm}\) diameter. The former is \(7.5 \mathrm{~m}\) long and the suction lift is \(3 \mathrm{~m}\). There is no air vessel on the suction side. The delivery, pipe is \(300 \mathrm{~m}\) long, the outlet (at atmospheric pressure) being \(13.5 \mathrm{~m}\) above the level of the pump, and a large air vessel is connected to the delivery pipe at a point \(15 \mathrm{~m}\) from the pump. Calculate the absolute pressure head in the cylinder at beginning, middle and end of each stroke. Assume that the motion of the piston is simple harmonic, that losses at inlet and outlet of each pipe are negligible, that the slip is \(2 \%\), and that \(f\). for both pipes is constant at 0.01. (Atmospheric pressure \(10.33 \mathrm{~m}\) water head.)

A reciprocating pump has two double-acting cylinders each \(200 \mathrm{~mm}\) bore and \(450 \mathrm{~mm}\) stroke, the cranks being at \(90^{\circ}\) to each other and rotating at \(2.09 \mathrm{rad} \cdot \mathrm{s}^{-1}\) (20 rev/min). The delivery pipe is \(100 \mathrm{~mm}\) diameter, \(60 \mathrm{~m}\) long. There are no air vessels. Assuming simple harmonic motion for the pistons determine the maximum and mean water velocities in the delivery pipe and the inertia pressure in the delivery pipe near the cylinders at the instant of minimum water velocity in the pipe.

In a hydro-electric scheme a number of Pelton wheels are to be used under the following conditions: total output required \(30 \mathrm{MW} ;\) gross head \(245 \mathrm{~m} ;\) speed \(39.27 \mathrm{rad} \cdot \mathrm{s}^{-1}\) \(\left(6.25\right.\) rev/s); 2 jets per wheel; \(C_{\mathrm{Y}}\) of nozzles \(0.97 ;\) maximum overall efficiency (based on conditions immediately before the nozzles) \(81.5 \% ;\) power specific speed for one jet not to exceed \(0.138\) rad \((0.022\) rev); head lost to friction in pipe-line not to exceed \(12 \mathrm{~m} .\) Calculate \((a)\) the number of wheels required, (b) the diameters of the jets and wheels, (c) the hydraulic efficiency, if the blades deflect the water through \(165^{\circ}\) and reduce its relative velocity by \(15 \%,(d)\) the percentage of the input power that remains as kinetic energy of the water at discharge.

In a vertical-shaft inward-flow reaction turbine the sum of the pressure and kinetic heads at entrance to the spiral casing is \(120 \mathrm{~m}\) and the vertical distance between this section and the tail-race level is \(3 \mathrm{~m}\). The peripheral velocity of the runner at entry is \(30 \mathrm{~m} \cdot \mathrm{s}^{-1}\), the radial velocity of the water is constant at \(9 \mathrm{~m} \cdot \mathrm{s}^{-1}\) and discharge from the runner is without whirl. The estimated hydraulic losses are: (1) between turbine entrance and exit from the guide vanes, \(4.8 \mathrm{~m}\) (2) in the runner, \(8.8 \mathrm{~m},(3)\) in the draft tube, \(790 \mathrm{~mm}\), (4) kinetic head rejected to the tail race, \(460 \mathrm{~mm}\). Calculate the guide vane angle and the runner blade angle at inlet and the pressure heads at entry to and exit from the runner

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