{"id":1952,"date":"2024-07-30T21:34:55","date_gmt":"2024-07-30T21:34:55","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=1952"},"modified":"2024-07-30T21:34:56","modified_gmt":"2024-07-30T21:34:56","slug":"combined-heat-transfer","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/07\/30\/combined-heat-transfer\/","title":{"rendered":"Combined Heat Transfer"},"content":{"rendered":"\n

Plane Walls with Convection on Sides<\/h4>\n\n\n\n

There are many cases in practice when different materials are constructed in layers to form a composite wall. This wall may be composed of plaster layer, brick layer, tiles layer, etc. as shown in\u00a0Figure 7.3.<\/a><\/p>\n\n\n\n

In Figure 7.3<\/a> there are three layers A, B, C, of thickness L<\/em>A<\/sub>, L<\/em>B<\/sub>, and L<\/em>C<\/sub>, respectively. The thermal conductivities of the layers are k<\/em>A<\/sub>, k<\/em>B<\/sub>, and k<\/em>C<\/sub>. On one side of the composite wall there is a fluid at temperature t<\/em>\u221e1<\/sub> and heat transfer coefficient from fluid to wall is h<\/em>1<\/sub>; on other side of the composite wall there is fluid at temperature t<\/em>\u221e2<\/sub> and the heat transfer coefficient from fluid to wall is h<\/em>2<\/sub>. Let the temperature of the wall in contact with fluid on one side is t<\/em>1<\/sub> and on other side t<\/em>4<\/sub>. The interface temperatures of the composite wall are t<\/em>2<\/sub> and t<\/em>3<\/sub>. To solve the heat flow problem an analogous of electrical current may be used. The heat flow is caused by temperature difference whereas the current flow is caused by a potential difference. Hence, it is possible to postulate a thermal resistance analogous to an electrical resistance. From Ohm\u2019s law we have, V=IR<\/em> or I=V\/R<\/em>, where V<\/em> is potential difference, I<\/em> is current and R<\/em> is resistance.<\/p>\n\n\n\n

\"Figure<\/figure>\n\n\n\n

Figure 7.3<\/strong> Plane Walls with Convection on Sides<\/p>\n\n\n\n

Thermal resistance, \"equation\" and thermal resistance for a fluid film, <\/em>\"equation\" where Q<\/em> is analogous to I<\/em> and \u0394t<\/em> is analogous to \u0394V<\/em>.<\/p>\n\n\n\n

\"equation\"\/<\/figure>\n\n\n\n

Example 7.2:<\/strong>\u00a0Derive the expression for the heat transfer through series and parallel composite walls as shown in\u00a0Figure 7.4.<\/p>\n\n\n\n

\"Figure<\/figure>\n\n\n\n

Figure 7.4<\/strong> Heat Transfer Through Composite Wall<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

The electrical current analogy of the heat transfer is shown in\u00a0Figure 7.5.<\/p>\n\n\n\n

\"Figure<\/figure>\n\n\n\n

Figure 7.5<\/strong> Electrical Analogy of Heat Transfer for Composite Wall is Shown in Figure 7.4<\/a><\/p>\n\n\n\n

\"equation\"\/<\/figure>\n\n\n\n

Example 7.3:<\/strong> A steel tank of wall thickness 8 mm contains water at 80\u00b0C. Calculate the rate of heat loss per m2<\/sup> of tank surface area when the atmospheric temperature is 20\u00b0C. The thermal conductivity of mild steel is 50 W\/m K, and the heat transfer coefficients for the inside and outside of the tank are 2,500 and 20 W\/m2<\/sup>K, respectively. Calculate also the temperature of the outside surface of the tank.<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

\"equation\"\/<\/figure>\n\n\n\n

Example 7.4:<\/strong>\u00a0A furnace wall consists of 120 mm thick refractory brick and 120 mm thick insulating firebrick separated by an air gap as shown in\u00a0Figure 7.6.\u00a0The outside wall is covered with a 10 mm thickness of plaster. The inner surface of the wall is at 1,000\u00b0C and the room temperature is 20\u00b0C. Calculate the rate at which heat is lost per m2<\/sup>\u00a0of wall surface. The heat transfer coefficient from the outside wall surface to the air on the room is 20 W\/m2<\/sup>K, and the resistance to heat flow of the air gap is 0.15 K\/W. The thermal conductivity of refractory brick, insulating firebrick and plaster are 1.6, 0.3 and 0.14 W\/m K, respectively. Calculate also each interface temperature of the outside of the wall.<\/p>\n\n\n\n

\"Figure<\/figure>\n\n\n\n

Figure 7.6<\/strong> Composite Walls<\/p>\n\n\n\n

<\/a>Solution:<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

\"equation\"\/<\/figure>\n\n\n\n
\"equation\"\/<\/figure>\n\n\n\n

Heat Transfer Through Hollow Cylinder<\/h4>\n\n\n\n

Consider a cylinder of internal radius r<\/em>1<\/sub> and external radius r<\/em>2<\/sub> as shown in Figure 7.7.<\/a> Let the inside and outside temperatures be t<\/em>1<\/sub> and t<\/em>2<\/sub>, respectively. Consider the heat flow through a small element of thickness dr<\/em> at any radius r<\/em>, where the temperature is t<\/em>. Let the thermal conductivity of the material be k<\/em>, temperature of fluid flow inside the cylinder be t<\/em>f1<\/sub>, heat transfer coefficient be h<\/em>f1<\/sub>, temperature of fluid flow outside the cylinder be t<\/em>f2<\/sub> and heat transfer coefficient be h<\/em>f2<\/sub>. The heat transfer through the elemental area of length l<\/em> can be given by,<\/p>\n\n\n\n

\"Figure<\/figure>\n\n\n\n

Figure 7.7<\/strong> Heat Transfer Through Cylindrical Wall<\/p>\n\n\n\n

\"equation\"\/<\/figure>\n\n\n\n

Integrating between the inside and outside surfaces,<\/p>\n\n\n\n

\"equation\"\/<\/figure>\n\n\n\n

If we put mean area A<\/em>m<\/sub> and mean radius r<\/em>m<\/sub> in the above equation as<\/p>\n\n\n\n

\"equation\"\/<\/figure>\n\n\n\n

In the case of a composite cylinder the most convenient approach is again that of the electrical analogy; by using\"equation\" where x<\/em> is thickness of a layer and A<\/em>m<\/sub> is the logarithmic mean area for that layer.<\/p>\n\n\n\n

The film of fluid on the inside and outside surfaces can be treated as<\/p>\n\n\n\n

\"equation\"\/<\/figure>\n\n\n\n

Example 7.5:<\/strong> A steel pipe of 80 mm bore and 8 mm wall thickness, carrying steam at 250\u00b0C, is insulated with 36 mm of a moulded high temperature diatomaceous earth covering. This covering is in turn insulated with 50 mm of asbestos felt. If the atmospheric temperature is 20\u00b0C, calculate the rate at which heat is lost by the steam per metre length of pipe. The heat transfer coefficients for the inside and outside surfaces are 525 and 27 W\/m2<\/sup>K, respectively, and the thermal conductivity of steel, diatomaceous earth and asbestos felt are 55, 0.1 and 0.08 W\/m K, respectively. Calculate also the temperature of the outside surface.<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Given: r<\/em>i<\/sub> = 40mm; r<\/em>1<\/sub> = 48mm; r<\/em>2<\/sub> = 84mm; r<\/em>0<\/sub> = 134mm; h<\/em>1<\/sub> = 525 W\/m2<\/sup>K; m2<\/sup>K<\/p>\n\n\n\n

\"equation\"\/<\/figure>\n\n\n\n

Now resistance of per unit length of pipe<\/p>\n\n\n\n

\"equation\"\/<\/figure>\n","protected":false},"excerpt":{"rendered":"

Plane Walls with Convection on Sides There are many cases in practice when different materials are constructed in layers to form a composite wall. This wall may be composed of plaster layer, brick layer, tiles layer, etc. as shown in\u00a0Figure 7.3. In Figure 7.3 there are three layers A, B, C, of thickness LA, LB, and LC, respectively. The thermal […]<\/p>\n","protected":false},"author":1,"featured_media":1953,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[338],"tags":[],"class_list":["post-1952","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-heat-transfer"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/07\/download-11.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/1952","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=1952"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/1952\/revisions"}],"predecessor-version":[{"id":1954,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/1952\/revisions\/1954"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/1953"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=1952"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=1952"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=1952"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}