{"id":26,"date":"2022-05-29T17:52:06","date_gmt":"2022-05-29T17:52:06","guid":{"rendered":"https:\/\/web.htk.tlu.ee\/stem\/stem2\/chapter\/soojusjuhtivus\/"},"modified":"2023-11-11T08:21:02","modified_gmt":"2023-11-11T08:21:02","slug":"introduction-to-heat-transfer","status":"publish","type":"chapter","link":"https:\/\/web.htk.tlu.ee\/stem\/stem2\/chapter\/introduction-to-heat-transfer\/","title":{"rendered":"Unit 3: Introduction to heat transfer"},"content":{"raw":"In the following, we give you a short overview of thermal conductivity which is needed to understand <strong>why we need heating of homes<\/strong> (in a cold climate). <strong>What kind of physical processes are taking place? What kind of physical quantities are essential for in-room heating?<\/strong>\r\n\r\nWe need to heat our rooms when the outside air temperature is lower, or even, significantly lower than the recommended temperature. It is clear that there exists a heat transfer from the room to the outside through the walls and ceiling. The heat transfer is going in the opposite direction if the outside air temperature is higher than room temperature. Such kind of heat transfer takes place due to thermal conductivity.\r\n<div class=\"textbox textbox--learning-objectives\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Conduction heat transfer i.e. thermal conduction takes place without bulk motion of the matter.<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<i>Thermal conduction occurs in solid matter, liquids as well as in gases. The only factor for this is the temperature difference in the different parts of the system. Redistribution of the kinetic energy of the particles takes place due to collisions of the molecules in the gases or liquids, or propagation of oscillations through the solid bodies until the temperature in different parts of the system equals. <\/i>\r\n\r\n<em>Thermal conduction can be characterized quantitatively by <strong>heat flux density<\/strong> which is defined as an amount of heat that flows through a unit area per unit time. As we know, the amount of heat per unit time is power. Thus the thermal conduction characterized by power flowing through the (wall) material determines the amount of heat needed for heating the rooms.<\/em>\r\n\r\n<\/div>\r\n<\/div>\r\n<h1>Exercise 3.1<\/h1>\r\nTo make sure you understood this explanation correctly, one small control question.\r\n\r\n&nbsp;\r\n\r\n[h5p id=\"2\"]\r\n\r\nNow you should have some idea of what thermal conductivity is, and what kind of physical quantities can characterize it. Thermal conductivity is quantitatively characterized by the amount of heat <em>Q<\/em> going through the material. In Fig. 3.1 a cuboid is shown. This cuboid can be imagined as a piece of wall with thickness <em>d<\/em> and surface area <em>S<\/em>. As this amount of heat flows through the material, it is often called heat flux. The heat flux per unit time flowing per unit area is called heat flux density (or sometimes just heat flux) <em>q<\/em>.\r\n\r\n<img class=\"alignnone wp-image-156 size-full\" src=\"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-content\/uploads\/sites\/3\/2022\/05\/fig-3-1-1.jpg\" alt=\"\" width=\"370\" height=\"320\" \/>\r\n\r\nIf we talk about the heat flux through the wall of a house (or other certain material), we can use the following formula to calculate the heat flux density <em>q<\/em>:\r\n<div class=\"textbox textbox--learning-objectives\">\r\n<div class=\"textbox__content\">\r\n<p style=\"text-align: center\">[latex]q = (K * S * \\Delta T) \/ d[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p style=\"text-align: right\"><em>(1)<\/em><\/p>\r\nwhere <em>K<\/em> (1 W\/(m\u00d7K)) denotes the material thermal conductivity (coefficient), <em>S<\/em> (1 m<sup>2<\/sup>) is the area of the wall, \u0394T (1 K) is the difference of temperatures on the different sides of the wall, and <em>d<\/em> (1 m) is the thickness of the wall. As we can see from Eq. (1) the heat flux density is given by watts. Thus, indeed, the heat flux density characterizes the power flow through the wall.\r\n\r\nNote: <em>Thermal conductivity<\/em> means the physical phenomena characterizing heat transfer, but often this also means the thermal conductivity\u00a0 coefficient that characterizes the thermal properties of the material.\r\n<h1>Exercise 3.2<\/h1>\r\nLet us study the relation between heat flux density <em>q<\/em> and thermal conductivity <em>K<\/em>. Choose the correct statement.\r\n\r\n&nbsp;\r\n\r\n[h5p id=\"3\"]\r\n<h1>Exercise 3.3<\/h1>\r\nNow you should know, which is the relationship between the heat flux density <em>q<\/em> and thermal conductivity coefficient <em>K<\/em> of the wall material. Choose two wrong statements.\r\n\r\n&nbsp;\r\n\r\n[h5p id=\"4\"]\r\n\r\nBy our everyday experience we can say that a stone conducts heat better than a wood. It means that the heat flux through the stone is greater than through the wood, if other conditions remain the same (see Fig. 3.2).\r\n\r\n<img class=\"alignnone wp-image-157 size-full\" src=\"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-content\/uploads\/sites\/3\/2022\/05\/fig-3-2-1.jpg\" alt=\"\" width=\"678\" height=\"359\" \/>\r\n\r\nUse the <a href=\"https:\/\/lab.concord.org\/embeddable.html#interactives\/energy2d\/htb\/S3A1.json\" target=\"_blank\" rel=\"noopener\">following simulation<\/a> to estimate how many times the thermal conductivities of stone, wood and metal differ.\r\n\r\nYou can check your conclusions by Table 3.2.\r\n\r\n<strong>Table 3.2. Thermal conductivities of selected materials.<\/strong>\r\n<table style=\"border-collapse: collapse;width: 100%\" border=\"0\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 50%\"><strong>Material<\/strong><\/td>\r\n<td style=\"width: 50%\"><strong>Thermal conductivity coefficient <em>K<\/em> (W\/(K\u2219m))<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%\">Brick<\/td>\r\n<td style=\"width: 50%\">0.84<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%\">Wood<\/td>\r\n<td style=\"width: 50%\">0.16<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%\">Glass<\/td>\r\n<td style=\"width: 50%\">0.94<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%\">Rock wool<\/td>\r\n<td style=\"width: 50%\">0.033<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%\">Concrete<\/td>\r\n<td style=\"width: 50%\">1.4-2.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%\">Gyps sheet<\/td>\r\n<td style=\"width: 50%\">0.22<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%\">Copper<\/td>\r\n<td style=\"width: 50%\">384<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h1>Exercise 3.4<\/h1>\r\nTry to estimate the difference of heat fluxes through the brick wall and wooden wall. Choose two correct statements.\r\n\r\n[h5p id=\"16\"]\r\n\r\nIn order to estimate quantitatively how the change of wall material or thickness of material changes the heat flux (density) through the walls, we can use the <a href=\"https:\/\/www.desmos.com\/calculator\/fta3z7mwbd?lang=et\" target=\"_blank\" rel=\"noopener\">program Desmos<\/a>, where the Eq. (1) is entered.\r\n<div class=\"postbox h5p-sidebar\">\r\n<div class=\"h5p-action-bar-settings h5p-panel\"><code><\/code><\/div>\r\n<h1>Exercise 3.5<\/h1>\r\n<div>Try to estimate the difference of heat fluxes through the brick wall and wooden wall.<\/div>\r\n<\/div>\r\n<div><\/div>\r\n<div>\r\n<div class=\"postbox h5p-sidebar\">\r\n<div><\/div>\r\n<div class=\"h5p-action-bar-settings h5p-panel\"><code>[h5p id=\"14\"]<\/code><\/div>\r\n<\/div>\r\n<\/div>\r\n<div><\/div>\r\n<div>In order to estimate quantitatively how the change of wall material or thickness of material changes the heat flux (density) through the walls, we can use the program desmos, where the Eq. (1) is entered.<\/div>\r\n<div><\/div>\r\n&nbsp;\r\n<div><strong><a href=\"https:\/\/www.desmos.com\/calculator\/fta3z7mwbd\" target=\"_blank\" rel=\"noopener\">Go to Desmos.<\/a><\/strong><\/div>\r\n<h1>Problem 3.1<\/h1>\r\n<div>Let us estimate the heat flux density through the one-layer wall with area 30m<sup>2<\/sup> and wall thickness 20 cm.<\/div>\r\n<div><\/div>\r\n<div><span style=\"background-color: transparent;color: #222222;font-family: monospace;font-size: 1em\">[h5p id=\"15\"]<\/span><\/div>\r\n<div><\/div>\r\n<strong>For discussion:<\/strong> Let us suppose that you have a heating material (or just energy, in general) for 30 days. For how many extra days you have heating material if the temperature difference between inner and outer temperatures is decreased from 20\u00b0C to 18\u00b0C? From 10\u00b0C to 8\u00b0C?","rendered":"<p>In the following, we give you a short overview of thermal conductivity which is needed to understand <strong>why we need heating of homes<\/strong> (in a cold climate). <strong>What kind of physical processes are taking place? What kind of physical quantities are essential for in-room heating?<\/strong><\/p>\n<p>We need to heat our rooms when the outside air temperature is lower, or even, significantly lower than the recommended temperature. It is clear that there exists a heat transfer from the room to the outside through the walls and ceiling. The heat transfer is going in the opposite direction if the outside air temperature is higher than room temperature. Such kind of heat transfer takes place due to thermal conductivity.<\/p>\n<div class=\"textbox textbox--learning-objectives\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Conduction heat transfer i.e. thermal conduction takes place without bulk motion of the matter.<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><i>Thermal conduction occurs in solid matter, liquids as well as in gases. The only factor for this is the temperature difference in the different parts of the system. Redistribution of the kinetic energy of the particles takes place due to collisions of the molecules in the gases or liquids, or propagation of oscillations through the solid bodies until the temperature in different parts of the system equals. <\/i><\/p>\n<p><em>Thermal conduction can be characterized quantitatively by <strong>heat flux density<\/strong> which is defined as an amount of heat that flows through a unit area per unit time. As we know, the amount of heat per unit time is power. Thus the thermal conduction characterized by power flowing through the (wall) material determines the amount of heat needed for heating the rooms.<\/em><\/p>\n<\/div>\n<\/div>\n<h1>Exercise 3.1<\/h1>\n<p>To make sure you understood this explanation correctly, one small control question.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"h5p-2\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-2\" class=\"h5p-iframe\" data-content-id=\"2\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Heat flux density\"><\/iframe><\/div>\n<\/div>\n<p>Now you should have some idea of what thermal conductivity is, and what kind of physical quantities can characterize it. Thermal conductivity is quantitatively characterized by the amount of heat <em>Q<\/em> going through the material. In Fig. 3.1 a cuboid is shown. This cuboid can be imagined as a piece of wall with thickness <em>d<\/em> and surface area <em>S<\/em>. As this amount of heat flows through the material, it is often called heat flux. The heat flux per unit time flowing per unit area is called heat flux density (or sometimes just heat flux) <em>q<\/em>.<\/p>\n<div class=\"wp-nocaption alignnone wp-image-156 size-full\"><img class=\"alignnone wp-image-156 size-full\" src=\"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-content\/uploads\/sites\/3\/2022\/05\/fig-3-1-1.jpg\" alt=\"\" width=\"370\" height=\"320\" srcset=\"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-content\/uploads\/sites\/3\/2022\/05\/fig-3-1-1.jpg 370w, https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-content\/uploads\/sites\/3\/2022\/05\/fig-3-1-1-300x259.jpg 300w, https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-content\/uploads\/sites\/3\/2022\/05\/fig-3-1-1-65x56.jpg 65w, https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-content\/uploads\/sites\/3\/2022\/05\/fig-3-1-1-225x195.jpg 225w, https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-content\/uploads\/sites\/3\/2022\/05\/fig-3-1-1-350x303.jpg 350w\" \/><\/div>\n<p>If we talk about the heat flux through the wall of a house (or other certain material), we can use the following formula to calculate the heat flux density <em>q<\/em>:<\/p>\n<div class=\"textbox textbox--learning-objectives\">\n<div class=\"textbox__content\">\n<p style=\"text-align: center\">[latex]q = (K * S * \\Delta T) \/ d[\/latex]<\/p>\n<\/div>\n<\/div>\n<p style=\"text-align: right\"><em>(1)<\/em><\/p>\n<p>where <em>K<\/em> (1 W\/(m\u00d7K)) denotes the material thermal conductivity (coefficient), <em>S<\/em> (1 m<sup>2<\/sup>) is the area of the wall, \u0394T (1 K) is the difference of temperatures on the different sides of the wall, and <em>d<\/em> (1 m) is the thickness of the wall. As we can see from Eq. (1) the heat flux density is given by watts. Thus, indeed, the heat flux density characterizes the power flow through the wall.<\/p>\n<p>Note: <em>Thermal conductivity<\/em> means the physical phenomena characterizing heat transfer, but often this also means the thermal conductivity\u00a0 coefficient that characterizes the thermal properties of the material.<\/p>\n<h1>Exercise 3.2<\/h1>\n<p>Let us study the relation between heat flux density <em>q<\/em> and thermal conductivity <em>K<\/em>. Choose the correct statement.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"h5p-3\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-3\" class=\"h5p-iframe\" data-content-id=\"3\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"The heat flux density q and thermal conductivity K\"><\/iframe><\/div>\n<\/div>\n<h1>Exercise 3.3<\/h1>\n<p>Now you should know, which is the relationship between the heat flux density <em>q<\/em> and thermal conductivity coefficient <em>K<\/em> of the wall material. Choose two wrong statements.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"h5p-4\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-4\" class=\"h5p-iframe\" data-content-id=\"4\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"The heat flux density q and thermal conductivity K of the wall material\"><\/iframe><\/div>\n<\/div>\n<p>By our everyday experience we can say that a stone conducts heat better than a wood. It means that the heat flux through the stone is greater than through the wood, if other conditions remain the same (see Fig. 3.2).<\/p>\n<div class=\"wp-nocaption alignnone wp-image-157 size-full\"><img class=\"alignnone wp-image-157 size-full\" src=\"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-content\/uploads\/sites\/3\/2022\/05\/fig-3-2-1.jpg\" alt=\"\" width=\"678\" height=\"359\" srcset=\"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-content\/uploads\/sites\/3\/2022\/05\/fig-3-2-1.jpg 678w, https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-content\/uploads\/sites\/3\/2022\/05\/fig-3-2-1-300x159.jpg 300w, https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-content\/uploads\/sites\/3\/2022\/05\/fig-3-2-1-65x34.jpg 65w, https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-content\/uploads\/sites\/3\/2022\/05\/fig-3-2-1-225x119.jpg 225w, https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-content\/uploads\/sites\/3\/2022\/05\/fig-3-2-1-350x185.jpg 350w\" \/><\/div>\n<p>Use the <a href=\"https:\/\/lab.concord.org\/embeddable.html#interactives\/energy2d\/htb\/S3A1.json\" target=\"_blank\" rel=\"noopener\">following simulation<\/a> to estimate how many times the thermal conductivities of stone, wood and metal differ.<\/p>\n<p>You can check your conclusions by Table 3.2.<\/p>\n<p><strong>Table 3.2. Thermal conductivities of selected materials.<\/strong><\/p>\n<table style=\"border-collapse: collapse;width: 100%\">\n<tbody>\n<tr>\n<td style=\"width: 50%\"><strong>Material<\/strong><\/td>\n<td style=\"width: 50%\"><strong>Thermal conductivity coefficient <em>K<\/em> (W\/(K\u2219m))<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%\">Brick<\/td>\n<td style=\"width: 50%\">0.84<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%\">Wood<\/td>\n<td style=\"width: 50%\">0.16<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%\">Glass<\/td>\n<td style=\"width: 50%\">0.94<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%\">Rock wool<\/td>\n<td style=\"width: 50%\">0.033<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%\">Concrete<\/td>\n<td style=\"width: 50%\">1.4-2.5<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%\">Gyps sheet<\/td>\n<td style=\"width: 50%\">0.22<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%\">Copper<\/td>\n<td style=\"width: 50%\">384<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h1>Exercise 3.4<\/h1>\n<p>Try to estimate the difference of heat fluxes through the brick wall and wooden wall. Choose two correct statements.<\/p>\n<div id=\"h5p-16\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-16\" class=\"h5p-iframe\" data-content-id=\"16\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Let us continue the study of the equation (1).\"><\/iframe><\/div>\n<\/div>\n<p>In order to estimate quantitatively how the change of wall material or thickness of material changes the heat flux (density) through the walls, we can use the <a href=\"https:\/\/www.desmos.com\/calculator\/fta3z7mwbd?lang=et\" target=\"_blank\" rel=\"noopener\">program Desmos<\/a>, where the Eq. (1) is entered.<\/p>\n<div class=\"postbox h5p-sidebar\">\n<div class=\"h5p-action-bar-settings h5p-panel\"><code><\/code><\/div>\n<h1>Exercise 3.5<\/h1>\n<div>Try to estimate the difference of heat fluxes through the brick wall and wooden wall.<\/div>\n<\/div>\n<div><\/div>\n<div>\n<div class=\"postbox h5p-sidebar\">\n<div><\/div>\n<div class=\"h5p-action-bar-settings h5p-panel\"><code><\/p>\n<div id=\"h5p-14\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-14\" class=\"h5p-iframe\" data-content-id=\"14\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Estimate the difference of heat fluxes through the brick wall and wooden wall\"><\/iframe><\/div>\n<\/div>\n<p><\/code><\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div>In order to estimate quantitatively how the change of wall material or thickness of material changes the heat flux (density) through the walls, we can use the program desmos, where the Eq. (1) is entered.<\/div>\n<div><\/div>\n<p>&nbsp;<\/p>\n<div><strong><a href=\"https:\/\/www.desmos.com\/calculator\/fta3z7mwbd\" target=\"_blank\" rel=\"noopener\">Go to Desmos.<\/a><\/strong><\/div>\n<h1>Problem 3.1<\/h1>\n<div>Let us estimate the heat flux density through the one-layer wall with area 30m<sup>2<\/sup> and wall thickness 20 cm.<\/div>\n<div><\/div>\n<div><span style=\"background-color: transparent;color: #222222;font-family: monospace;font-size: 1em\"><\/p>\n<div id=\"h5p-15\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-15\" class=\"h5p-iframe\" data-content-id=\"15\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Problem 3.1\"><\/iframe><\/div>\n<\/div>\n<p><\/span><\/div>\n<div><\/div>\n<p><strong>For discussion:<\/strong> Let us suppose that you have a heating material (or just energy, in general) for 30 days. For how many extra days you have heating material if the temperature difference between inner and outer temperatures is decreased from 20\u00b0C to 18\u00b0C? From 10\u00b0C to 8\u00b0C?<\/p>\n","protected":false},"author":2,"menu_order":4,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[50],"contributor":[],"license":[],"part":73,"_links":{"self":[{"href":"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-json\/pressbooks\/v2\/chapters\/26"}],"collection":[{"href":"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-json\/wp\/v2\/users\/2"}],"version-history":[{"count":24,"href":"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-json\/pressbooks\/v2\/chapters\/26\/revisions"}],"predecessor-version":[{"id":318,"href":"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-json\/pressbooks\/v2\/chapters\/26\/revisions\/318"}],"part":[{"href":"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-json\/pressbooks\/v2\/parts\/73"}],"metadata":[{"href":"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-json\/pressbooks\/v2\/chapters\/26\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-json\/wp\/v2\/media?parent=26"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-json\/pressbooks\/v2\/chapter-type?post=26"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-json\/wp\/v2\/contributor?post=26"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/web.htk.tlu.ee\/stem\/stem2\/wp-json\/wp\/v2\/license?post=26"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}