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Chapter 13: Problem 22

A \(500 \mathrm{~mm}\) diameter fluid coupling containing oil of relative density \(0.85\) has a slip of \(3 \%\) and a torque coefficient of \(0.0014 .\) The speed of the primary is \(104.7 \mathrm{rad} \cdot \mathrm{s}^{-1}\) (16.67 rev/s). What is the rate of heat dissipation when equilibrium is attained?

Short Answer

Expert verified
The rate of heat dissipation is approximately 8.911 W.

Step by step solution

01

Define Given Values

Extract and list all the given values from the problem:- Diameter: 500 mm (or 0.5 m)- Relative density of oil: 0.85- Slip: 3% (or 0.03)- Torque coefficient: 0.0014- Speed of the primary: 104.7 rad/s.
02

Convert Diameter to Radius

Since diameter is given, convert it to radius to ease calculations:\[ r = \frac{0.5}{2} = 0.25 \text{ m} \].
03

Calculate Equivalent Density

Calculate the density of oil using relative density (specific gravity) and the density of water (1000 kg/m³):\[ \rho = 0.85 \times 1000 = 850 \text{ kg/m}^3 \].
04

Find Slip Value

Extract the slip percentage and convert it to a decimal:\[ s = 0.03 \].
05

Apply the Heat Dissipation Formula

Use the formula to find the rate of heat dissipation:\[ Q = T \cdot \omega_s \cdot Slip \]where: - \( T \) is the torque, which can be calculated using the torque coefficient: \( T = K_t \cdot \rho \cdot r^5 \cdot \omega^2 \)- \( \omega_s \) is the speed of the primary. Thus, calculate torque first:\[ T = 0.0014 \cdot 850 \cdot (0.25)^5 \cdot (104.7)^2 \approx 2.8345 \text{ Nm} \]Then calculate the rate of heat dissipation by including slip:\[ Q = 2.8345 \cdot 104.7 \cdot 0.03 \approx 8.911 \text{ W} \].

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Key Concepts

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

fluid dynamics
Fluid dynamics deals with the study of fluids (liquids and gases) in motion. It encompasses both the qualitative and quantitative analysis of how the fluid flows. In our problem, we are analyzing a fluid coupling system, which uses oil to transfer rotational energy from one shaft to another.
There are some important properties and principles in fluid dynamics:
  • Viscosity: Measures the fluid's resistance to deformation.
  • Density: Mass per unit volume of the fluid. For our problem, we use the relative density of the oil.
  • Incompressibility: Most liquids are treated as incompressible, simplifying calculations.
  • Continuity Equation: Ensures mass conservation in fluid flow.
Understanding these principles helps to evaluate how fluids behave under various conditions, affecting the fluid coupling's performance.

heat transfer
Heat transfer is a field in science that analyzes how thermal energy moves from one body or system to another. In the context of fluid coupling, the heat generated is primarily due to the slip, which converts some mechanical energy to thermal energy.
There are three modes of heat transfer:
  • Conduction: Transfer of heat through a solid medium.
  • Convection: Transfer of heat due to fluid motion, essential in our fluid coupling system.
  • Radiation: Transfer of heat through electromagnetic waves, less significant in our context.
For the rate of heat dissipation, we use the formula:

\[ Q = T \cdot \omega_s \cdot Slip \]
Here, the generated heat depends on the torque, rotational speed, and slip. Efficient heat dissipation ensures that the fluid coupling operates within safe temperature limits.

torque calculation
Torque calculation is crucial in understanding mechanical systems like fluid couplings. Torque is the rotational equivalent of linear force, measured in Newton-meters (Nm). In our problem, torque is affected by several factors, including the torque coefficient, oil density, and primary speed.
The formula used for calculating torque in fluid coupling systems is:
\[ T = K_t \cdot \rho \cdot r^5 \cdot \omega^2 \]
Where:
  • \( T \): Torque.
  • \( K_t \): Torque coefficient (provided).
  • \( \rho \): Density of the fluid.
  • \( r \): Radius.
  • \( \omega \): Speed of rotation.
Calculated torque helps us in determining both the mechanical efficiency and heat dissipation rate of the system.

relative density
Relative density, often called specific gravity, is the ratio of the density of a substance to the density of a reference substance, typically water. In our problem, the relative density of the oil is 0.85, meaning the oil is 85% as dense as water.
Using relative density simplifies converting the fluid's density as:
\[ \rho = 0.85 \times 1000 = 850 \text{ kg/m}^3 \]
Relative density impacts the fluid's behavior, affecting the torque calculation and overall performance of the fluid coupling. Lower relative densities generally mean the fluid is lighter, which can influence how efficiently torque is transferred and heat is dissipated.

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Most popular questions from this chapter

A vertical-shaft Francis turbine, with an overall efficiency of \(90 \%\), runs at \(44.86 \mathrm{rad} \cdot \mathrm{s}^{-1}(7.14 \mathrm{rev} / \mathrm{s})\) with a water discharge of \(15.5 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1} .\) The velocity at the inlet of the spiral casing is \(8.5 \mathrm{~m} \cdot \mathrm{s}^{-1}\) and the pressure head at this point is \(240 \mathrm{~m}\), the centre-line of the casing inlet being \(3 \mathrm{~m}\) above the tail-water level. The diameter of the runner at inlet is \(2.23 \mathrm{~m}\) and the width at inlet is \(300 \mathrm{~mm}\). The hydraulic efficiency is \(93 \%\). Determine \((a)\) the output power, \((b)\) the power specific speed, ( \(c\) ) the guide vane angle, \((d)\) the runner blade angle at inlet, (e) the percentage of the net head which is kinetic at entry to the runner. Assume that there is no whirl at outlet from the runner and neglect the thickness of the blades.

The impeller of a centrifugal pump has an outer diameter of \(250 \mathrm{~mm}\) and an effective outlet area of \(17000 \mathrm{~mm}^{2} .\) The outlet blade angle is \(32^{\circ} .\) The diameters of suction and discharge openings are \(150 \mathrm{~mm}\) and \(125 \mathrm{~mm}\) respectively. At \(152 \mathrm{rad} \cdot \mathrm{s}^{-1}(24.2 \mathrm{rev} / \mathrm{s})\) and discharge \(0.03 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) the pressure heads at suction and discharge openings were respectively \(4.5 \mathrm{~m}\) below and \(13.3 \mathrm{~m}\) above atmospheric pressure, the measurement points being at the same level. The shaft power was \(7.76 \mathrm{~kW}\). Water enters the impeller without shock or whirl. Assuming that the true outlet whirl component is \(70 \%\) of the ideal, determine the overall efficiency and the manometric efficiency based on the true whirl component.

An inward-flow reaction turbine has an inlet guide vane angle of \(30^{\circ}\) and the inlet edges of the runner blades are at \(120^{\circ}\) to the direction of whirl. The breadth of the runner at inlet is a quarter of the diameter at inlet and there is no velocity of whirl at outlet. The overall head is \(15 \mathrm{~m}\) and the rotational speed \(104.7 \mathrm{rad} \cdot \mathrm{s}^{-1}\) (16.67 rev/s). The hydraulic and overall efficiencies may be assumed to be \(88 \%\) and \(85 \%\) respectively. Calculate the runner diameter at inlet and the power developed. (The thickness of the blades may be neglected.)

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.

\(13.23\) 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.
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