{"id":3164,"date":"2024-08-26T22:18:34","date_gmt":"2024-08-26T22:18:34","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=3164"},"modified":"2024-08-26T22:18:34","modified_gmt":"2024-08-26T22:18:34","slug":"dynamic-systems","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/26\/dynamic-systems\/","title":{"rendered":"Dynamic Systems"},"content":{"rendered":"\n

The following describes how we can arrive at the input-output relationships for systems by representing them by simple models obtained by considering them to be composed of just a few simple basic elements.<\/p>\n\n\n\n

18.10.1 Mechanical Systems<\/h3>\n\n\n\n

Mechanical systems, however complex, have stiffness (or springiness), damping and inertia and can be considered to be composed of basic elements which can be represented by springs, dashpots and masses.<\/p>\n\n\n\n

1. <\/a>Spring<\/strong><\/p>\n\n\n\n

The \u201cspringiness\u201d or \u201cstiffness\u201d of a system can be represented by a spring. For a linear spring (Figure 18.84(a)), the extension\u00a0y<\/em>\u00a0is proportional to the applied extending force\u00a0F<\/em>\u00a0and we have:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

where k<\/em> is a constant termed the stiffness.<\/em><\/a><\/p>\n\n\n\n

2. <\/a>Dashpot<\/strong><\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 18.84<\/strong> Mechanical system building blocks<\/p>\n\n\n\n

The \u201cdamping\u201d of a mechanical system can be represented by a dashpot. This is a piston moving in a viscous medium in a cylinder (Figure 18.84(b)). Movement of the piston inward requires the trapped fluid to flow out past edges of the piston; movement outward requires fluid to flow past the piston and into the enclosed space. The resistive force F which has to be overcome is proportional to the velocity of the piston and hence the rate of change of displacement\u00a0y<\/em>\u00a0with time, i.e.,\u00a0dy\/dt.<\/em>\u00a0Thus:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

where c<\/em> is a constant.<\/p>\n\n\n\n

3. <\/a>Mass<\/strong><\/p>\n\n\n\n

The \u201cinertia\u201d of a system\u2014i.e., how much it resists being accelerated\u2014can be represented by mass. For a mass\u00a0m<\/em>\u00a0(Figure 18.84(c)), the relationship between the applied force\u00a0F<\/em>\u00a0and its acceleration\u00a0a<\/em>\u00a0is given by Newton\u2019s second law as\u00a0F<\/em>=ma.<\/em>\u00a0But acceleration is the rate of change of velocity\u00a0v<\/em>\u00a0with time\u00a0t<\/em>, i.e.,\u00a0a<\/em>=dv\/dt,<\/em>\u00a0and velocity is the rate of change of displacement\u00a0y<\/em>\u00a0with time, i.e.,\u00a0v<\/em>=dy\/dt.<\/em>\u00a0Thus\u00a0a<\/em>=d(dy\/dt)\/dt<\/em>\u00a0and so we can write:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

The following example illustrates how we can arrive at a model for a mechanical system.<\/p>\n\n\n\n

Example 18.5<\/h5>\n\n\n\n

Derive a model for the mechanical system given in\u00a0Figure 18.85(a). The input to the system is the force\u00a0F<\/em>\u00a0and the output is the displacement\u00a0y<\/em>.<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n
\"image\"\/<\/figure>\n\n\n\n

Figure 18.85<\/strong> (a) Mechanical system with mass, damping and stiffness; (b) the free-body diagram for the forces acting on the mass<\/p>\n\n\n\n

Solution<\/h5>\n\n\n\n

To obtain the system model we draw\u00a0free-body diagrams,<\/em>\u00a0these being diagrams of masses showing just the external forces acting on each mass. For the system in\u00a0Figure 18.84(a)\u00a0we have just one mass and so just one free-body diagram and that is shown in\u00a0Figure 18.84(b). As the free-body diagram indicates, the net force acting on the mass is the applied force minus the forces exerted by the spring and by the dashpot:<\/p>\n\n\n\n

net force =\"image\"<\/p>\n\n\n\n

Then applying Newton\u2019s second law, this force must be equal to ma,<\/em> where a<\/em> is the acceleration, and so:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

The relationship between the input F<\/em> to the system and the output y<\/em> is described by the second-order differential equation:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

The term second-order<\/em> is used because the equation includes as its highest derivative d2<\/sup>y\/dt2<\/sup>.<\/em><\/p>\n\n\n\n

18.10.2 Rotational Systems<\/h3>\n\n\n\n

For rotational systems the basic building blocks are a torsion spring, a rotary damper and the moment of inertia (Figure 18.86).<\/p>\n\n\n\n

1. <\/a>Torsional spring<\/strong><\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n
\"image\"\/<\/figure>\n\n\n\n
\"image\"\/<\/figure>\n\n\n\n

Figure 18.86<\/strong> Rotational system elements: (a) torsional spring or elastic twisting of a shaft; (b) rotational dashpot; (c) moment of inertia<\/p>\n\n\n\n

The \u201cspringiness\u201d or \u201cstiffness\u201d of a rotational spring is represented by a torsional spring. For a torsional spring, the angle \u03b8 rotated is proportional to the torque T<\/em>:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

where k<\/em> is a measure of the stiffness of the spring.<\/p>\n\n\n\n

2. <\/a>Rotational dashpot<\/strong><\/p>\n\n\n\n

The damping inherent in rotational motion is represented by a rotational dashpot. For a rotational dashpot, i.e., effectively a disk rotating in a fluid, the resistive torque T<\/em> is proportional to the angular velocity \u03c9 and thus:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

where c<\/em> is the damping constant.<\/a><\/p>\n\n\n\n

3. <\/a>Inertia<\/strong><\/p>\n\n\n\n

The inertia of a rotational system is represented by the moment of inertia of a mass. A torque T<\/em> applied to a mass with a moment of inertia I results in an angular acceleration a and thus, since angular acceleration is the rate of change of angular velocity \u03c9 with time, i.e., d<\/em>\u03c9\/dt,<\/em> and angular velocity \u03c9 is the rate of change of angle with time, i.e., d<\/em>\u03b8\/dt<\/em>, then the angular acceleration is d(<\/em>d\u03b8\/dt)\/dt<\/em> and so:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

The following example illustrates how we can arrive at a model for a rotational system.<\/p>\n\n\n\n

Example 18.6<\/h5>\n\n\n\n

Develop a model for the system shown in\u00a0Figure 18.87(a)\u00a0of the rotation of a disk as a result of twisting a shaft.<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n
\"image\"\/<\/figure>\n\n\n\n

Figure 18.87<\/strong> Example<\/p>\n\n\n\n

Solution<\/h5>\n\n\n\n

Figure 18.87(b)\u00a0shows the free-body diagram for the system. The torques acting on the disk are the applied torque\u00a0T<\/em>, the spring torque\u00a0k<\/em>\u03b8 and the damping torque\u00a0c<\/em>\u03c9. Hence:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

We thus have the second-order differential equation relating the input of the torque to the output of the angle of twist:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

18.10.3 Electrical Systems<\/h3>\n\n\n\n

The basic elements of electrical systems are the resistor, inductor and capacitor (Figure 18.88<\/a>).<\/p>\n\n\n\n

1. <\/a>Resistor<\/strong><\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 18.88<\/strong> Electrical system building blocks<\/p>\n\n\n\n

For a resistor,<\/em> resistance R<\/em>, the potential difference v<\/em> across it when there is a current i<\/em> through it is given by:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

2. <\/a>Inductor<\/strong><\/p>\n\n\n\n

For an inductor,<\/em> inductance L<\/em>, the potential difference v<\/em> across it at any instant depends on the rate of change of current i<\/em> and is:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

3. <\/a>Capacitor<\/strong><\/p>\n\n\n\n

For a capacitor,<\/em> the potential difference v<\/em> across it depends on the charge q<\/em> on the capacitor plates with v<\/em>=q\/C,<\/em> where C<\/em> is the capacitance. Thus:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n
\"image\"\/<\/figure>\n\n\n\n

Since current i<\/em> is the rate of movement of charge:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

and so we can write:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

To develop the models for electrical circuits we use Kirchhoff\u2019s laws. These can be stated as:<\/p>\n\n\n\n

1. <\/a>Kirchhoff\u2019s current law<\/strong><\/p>\n\n\n\n

The total current flowing into any circuit junction is equal to the total current leaving that junction, i.e., the algebraic sum of the currents at a junction is zero.<\/p>\n\n\n\n

2. <\/a>Kirchhoff\u2019s voltage law<\/strong><\/p>\n\n\n\n

In a closed circuit path, termed a loop, the algebraic sum of the voltages across the elements that make up the loop is zero. This is the same as saying that for a loop containing a source of e.m.f., the sum of the potential drops across each circuit element is equal to the sum of the applied e.m.f.\u2019s, provided we take account of their directions.<\/p>\n\n\n\n

The following examples illustrate the development of models for electrical systems.<\/p>\n\n\n\n

Example 18.7<\/h5>\n\n\n\n

Develop a model for the electrical system described by the circuit shown in\u00a0Figure 18.89. The input is the voltage\u00a0v<\/em>\u00a0when the switch is closed and the output is the voltage\u00a0vc<\/em>\u00a0across the capacitor.<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 18.89<\/strong> Electrical system with resistance and capacitance<\/p>\n\n\n\n

Solution<\/h5>\n\n\n\n

Using Kirchhoff\u2019s voltage law gives:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

and, since VR<\/sub><\/em>=Ri<\/em> and i<\/em>=C(dvc<\/sub>\/dt)<\/em> we obtain the equation:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

The relationship between an input v<\/em> and the output vc<\/sub><\/em> is a first order differential equation. The term first-order<\/em> is used because it includes as its highest derivative dvC<\/sub>\/dt.<\/em><\/p>\n\n\n\n

Example 18.8<\/h5>\n\n\n\n

Develop a model for the circuit shown in\u00a0Figure 18.90\u00a0when we have an input voltage\u00a0v<\/em>\u00a0when the switch is closed and take an output as the voltage\u00a0v<\/em>c<\/sub>\u00a0across the capacitor.<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 18.90<\/strong> Electrical system with resistance, inductance and capacitance<\/p>\n\n\n\n

Solution<\/h5>\n\n\n\n

Applying Kirchhoff\u2019s voltage law gives:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

and so:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Since i<\/em>=C<\/em>(dvC<\/sub><\/em>\/dt<\/em>), then di<\/em>\/dt<\/em>=C<\/em>(d<\/em>2<\/sup>vC<\/sub><\/em>\/dt<\/em>2<\/sup>) and thus, we can write:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

The relationship between an input v<\/em> and output vc<\/sub><\/em> is described by a second-order differential equation.<\/p>\n\n\n\n

18.10.4 Thermal Systems<\/h3>\n\n\n\n

Thermal systems have two basic building blocks, resistance and capacitance (Figure 18.91).<\/p>\n\n\n\n

1. <\/a>Thermal resistance<\/strong><\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n
\"image\"\/<\/figure>\n\n\n\n

Figure 18.91<\/strong> (a) Thermal resistance; (b) thermal capacitance<\/p>\n\n\n\n

The thermal resistance\u00a0R<\/em>\u00a0is the resistance offered to the rate of flow of heat\u00a0q<\/em>\u00a0(Figure 18.91(a)) and is defined by:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

where T<\/em>1<\/sub>– is the temperature difference through which the heat flows.<\/p>\n\n\n\n

For heat conduction through a solid we have the rate of flow of heat proportional to the cross-sectional area and the temperature gradient. Thus for two points at temperatures T<\/em>1<\/sub> and T<\/em>2<\/sub> and a distance L<\/em> apart:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

with k<\/em> being the thermal conductivity. Thus with this mode of heat transfer, the thermal resistance R<\/em> is L\/Ak.<\/em> For heat transfer by convection between two points, Newton\u2019s law of cooling gives:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

<\/a>where (T<\/em>2<\/sub>–T<\/em>1<\/sub>) is the temperature difference, h<\/em> the coefficient of heat transfer and A<\/em> the surface area across which the temperature difference is. The thermal resistance with this mode of heat transfer is thus 1\/Ah.<\/em><\/p>\n\n\n\n

2. <\/a>Thermal capacitance<\/strong><\/p>\n\n\n\n

The thermal capacitance (Figure 18.91(b)) is a measure of the store of internal energy in a system. If the rate of flow of heat into a system is\u00a0q<\/em>1<\/sub>, and the rate of flow out\u00a0q<\/em>2<\/sub>\u00a0then the rate of change of internal energy of the system is\u00a0q<\/em>1-<\/sub>q<\/em>2<\/sub>. An increase in internal energy can result in a change in temperature:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

where m<\/em> is the mass and c<\/em> the specific heat capacity. Thus the rate of change of internal energy is equal to mc<\/em> times the rate of change of temperature. Hence:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

This equation can be written as:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

where the capacitance C<\/em>=mc.<\/em><\/p>\n\n\n\n

The following example illustrates the development of models for thermal systems.<\/p>\n\n\n\n

Example 18.9<\/h5>\n\n\n\n

Develop a model for the simple thermal system of a thermometer at temperature\u00a0T<\/em>\u00a0being used to measure the temperature of a liquid when it suddenly changes to the higher temperature of\u00a0TL<\/sub><\/em>\u00a0(Figure 18.92).<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 18.92<\/strong> Example<\/p>\n\n\n\n

Solution<\/h5>\n\n\n\n

When the temperature changes there is heat flow q<\/em> from the liquid to the thermometer. The thermal resistance to heat flow from the liquid to the thermometer is:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Since there is only a net flow of heat from the liquid to the thermometer the thermal capacitance of the thermometer is:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Substituting for q<\/em> gives:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

which, when rearranged gives:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

This is a first-order differential equation.<\/p>\n\n\n\n

Example 18.10<\/h5>\n\n\n\n

Determine a model for the temperature of a room (Figure 18.93) containing a heater which supplies heat at the rate\u00a0q<\/em>1<\/sub>\u00a0and the room loses heat at the rate\u00a0q<\/em>2.<\/sub><\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 18.93<\/strong> Example<\/p>\n\n\n\n

Solution<\/h5>\n\n\n\n

We will assume that the air in the room is at a uniform temperature T<\/em>. If the air and furniture in the room have a combined thermal capacity C<\/em>, since the energy rate to heat the room is q<\/em>1<\/sub>–q<\/em>2<\/sub>, we have:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

If the temperature inside the room is T<\/em> and that outside the room T<\/em>0<\/sub> then<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

where R<\/em> is the thermal resistance of the walls. Substituting for q<\/em>2<\/sub> gives:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Hence:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

This is a first-order differential equation.<\/p>\n\n\n\n

18.10.5 Hydraulic Systems<\/h3>\n\n\n\n

For a fluid system the three building blocks are resistance, capacitance and inertance; these are the equivalents of electrical resistance, capacitance and inductance.\u00a0The equivalent of electrical current is the volumetric rate of flow and of potential difference is pressure difference. Hydraulic fluid systems are assumed to involve an incompressible liquid; pneumatic systems, however, involve compressible gases and consequently there will be density changes when the pressure changes. Here we will just consider the simpler case of hydraulic systems.\u00a0Figure 18.94\u00a0shows the basic form of building blocks for hydraulic systems.<\/p>\n\n\n\n

1. <\/a>Hydraulic resistance<\/strong><\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n
\"image\"\/<\/figure>\n\n\n\n
\"image\"\/<\/figure>\n\n\n\n

Figure 18.94<\/strong> Hydraulic building blocks<\/p>\n\n\n\n

Hydraulic resistance\u00a0R<\/em>\u00a0is the resistance to flow which occurs when a liquid flows from one diameter pipe to another (Figure 18.94(a)) and is defined as being given by the hydraulic equivalent of Ohm\u2019s law:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

2. <\/a>Hydraulic capacitance<\/strong><\/p>\n\n\n\n

Hydraulic capacitance\u00a0C<\/em>\u00a0is the term used to describe energy storage where the hydraulic liquid is stored in the form of potential energy (Figure 18.94(b)). The rate of change of volume\u00a0V<\/em>\u00a0of liquid stored is equal to the difference between the volumetric rate at which liquid enters the container\u00a0q<\/em>1<\/sub>\u00a0and the rate at which it leaves\u00a0q<\/em>2<\/sub>, i.e.,<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

But V<\/em>=Ah<\/em> and so:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

The pressure difference between the input and output is:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Hence, substituting for h<\/em> gives:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

The hydraulic capacitance C<\/em> is defined as:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

and thus, we can write:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

3. <\/a>Hydraulic inertance<\/strong><\/p>\n\n\n\n

Hydraulic inertance is the equivalent of inductance in electrical systems. To accelerate a fluid a net force is required and this is provided by the pressure difference (Figure 18.93(c)). Thus:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

where a<\/em> is the acceleration and so the rate of change of velocity v<\/em>. The mass of fluid being accelerated is m<\/em>=ALp<\/em> and the rate of flow q<\/em>=Av<\/em> and so:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n
\"image\"\/<\/figure>\n\n\n\n

where the inertance I<\/em> is given by I<\/em>=Lp\/A.<\/em><\/a><\/p>\n\n\n\n

The following example illustrates the development of a model for a hydraulic system.<\/p>\n\n\n\n

Example 18.11<\/h5>\n\n\n\n

Develop a model for the hydraulic system shown in\u00a0Figure 18.95\u00a0where there is a liquid entering a container at one rate\u00a0q<\/em>1<\/sub>\u00a0and leaving through a valve at another rate\u00a0q<\/em>2<\/sub>.<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Figure 18.95<\/strong> Example<\/p>\n\n\n\n

Solution<\/h5>\n\n\n\n

We can neglect the inertance since flow rates can be assumed to change only very slowly. For the capacitance term we have:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

For the resistance of the valve we have:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Thus, substituting for q2<\/sub>, and recognizing that the pressure difference is hpg,<\/em> gives:<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n
\"image\"\/<\/figure>\n\n\n\n

This is a first-order differential equation.<\/p>\n","protected":false},"excerpt":{"rendered":"

The following describes how we can arrive at the input-output relationships for systems by representing them by simple models obtained by considering them to be composed of just a few simple basic elements. 18.10.1 Mechanical Systems Mechanical systems, however complex, have stiffness (or springiness), damping and inertia and can be considered to be composed of […]<\/p>\n","protected":false},"author":1,"featured_media":3165,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[423],"tags":[],"class_list":["post-3164","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-control-and-instrumentation-systems"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/08\/technique.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3164","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/comments?post=3164"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3164\/revisions"}],"predecessor-version":[{"id":3166,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/3164\/revisions\/3166"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/3165"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=3164"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=3164"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=3164"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}