{"id":5385,"date":"2026-09-15T00:27:25","date_gmt":"2026-09-14T22:27:25","guid":{"rendered":"https:\/\/ghetool.eu\/?post_type=course&#038;p=5385"},"modified":"2026-09-18T12:35:13","modified_gmt":"2026-09-18T10:35:13","slug":"deel-7-antwoorden","status":"publish","type":"course","link":"https:\/\/ghetool.eu\/nl_nl\/course\/part-7-answers\/","title":{"rendered":"Deel 7: Antwoorden"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">In this chapter, we will provide you with the answers to the question at the end of each chapter of the fifth part or the course.<\/p>\n\n\n\n<div class=\"note\">To get as much out of this design course, we highly suggest you try to solve these questions first for yourself before looking at the solution here.<\/div>\n\n\n\n<div class=\"note\">Please note that, since geothermal borefield design is a rather complicated task, there is sometimes no definitive answer. The solutions we propose here are our interpretation of the questions, but this does not necessarily mean that other solutions would not be valid.<\/div>\n\n\n\n\n\n\n\n\n\n<iframe title=\"Deel 7: Antwoorden\" width=\"800\" height=\"600\" src=\"https:\/\/www.youtube.com\/embed\/nF2fqnLwlQM?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n\n\n\n\n\n\n\n\n\n<h2 class=\"wp-block-heading\">Question 1.1<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><i>(Go to the <a href=\"https:\/\/ghetool.eu\/course\/average-inlet-outlet-temperatures\">original question<\/a>)<\/i><\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">Why is it not possible to work with the inlet or outlet temperatures when a constant, measured effective borehole thermal resistance is used?<\/p>\n<\/blockquote>\n\n\n\n<p class=\"wp-block-paragraph\">Internally, the effective borehole thermal resistance is used to calculate the average fluid temperature based on the borehole wall temperature and the required extraction or injection power. This step in the process is feasible when working with either a constant, manually entered borehole resistance or one that is dynamically calculated in GHEtool.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">However, when converting the average fluid temperature to the inlet or outlet temperature, information about the flow rate is required, as this determines the temperature difference between the inlet and outlet temperatures. This information is not available when the borehole resistance is entered directly, and therefore the inlet and outlet temperatures cannot be used in this case.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Question 1.2<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><i>(Go to the <a href=\"https:\/\/ghetool.eu\/course\/average-inlet-outlet-temperatures\">original question<\/a>)<\/i><\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">Since the temperatures depend on the fluid temperature type selected (inlet, outlet or average), how does this affect the Reynolds number and turbulence in your simulation? Does this depend on the type of fluid temperature you are working with?<\/p>\n<\/blockquote>\n\n\n\n<p class=\"wp-block-paragraph\">The Reynolds number, and hence also the turbulence, is independent of the choice of fluid temperature <strong>when the design stays the same<\/strong>. Switching from average to inlet or outlet temperatures does indeed change how the fluid temperatures are reported, but the actual underlying borehole resistance and average fluid temperature remain the same.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Question 1.3<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><i>(Go to the <a href=\"https:\/\/ghetool.eu\/course\/average-inlet-outlet-temperatures\">original question<\/a>)<\/i><\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">The last simulation was carried out with a variable flow rate based on a constant temperature difference. What would change if a constant flow rate were used?<a href=\"https:\/\/ghetool.eu\/course\/introduction-part-7\/\"><\/a><\/p>\n<\/blockquote>\n\n\n\n<p class=\"wp-block-paragraph\">When a constant temperature difference is used, the conversion from the average to the inlet and outlet fluid temperatures is direct. When a constant flow rate is used, two main things will happen:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Due to the constant flow rate, the borehole resistance will be more constant throughout the simulation period (as discussed back in <a href=\"https:\/\/ghetool.eu\/course\/variable-flow-rates\/\">Part 3.3<\/a>).<\/li>\n\n\n\n<li>The temperature difference between the inlet and outlet will now vary from month to month or hour to hour. Since $\\dot{m}$ in the formula $\\dot{Q}=\\dot{m}C_p\\Delta T$ is now constant, this results in a variable $\\Delta T$.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">The actual consequences for your design therefore depend on what this constant flow rate is. When the flow rate is set equal to the maximum flow rate when working with a constant temperature difference, the average fluid temperature will stay the same at the most critical point (since the borehole resistance would be identical), and so will the inlet and outlet temperatures.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">For all other time steps, the flow rate will now be higher, leading to both a fluid temperature that is closer to the borehole wall temperature (since the borehole resistance will be better) and inlet and outlet temperatures that will be closer to the average fluid temperature due to the higher flow rate. Overall, your fluid temperatures will be more optimistic.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">When the constant flow rate is set to a value that is lower than the maximum flow rate when working with a variable one, the critical results will be worse due to the lower flow rate, and so will the results for all the time steps where this constant flow rate is lower than would have been the case with a temperature dependent flow rate.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Question 2.1<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><i>(Go to the <a href=\"https:\/\/ghetool.eu\/course\/calculate-required-borehole-depth\">original question<\/a>)<\/i><\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">There is another way to overcome the gradient error without adding extra boreholes to the system or improving the borehole thermal resistance. Can you figure out what it is?<\/p>\n<\/blockquote>\n\n\n\n<p class=\"wp-block-paragraph\">The gradient error occurs when the increase in undisturbed ground temperature is larger than the decrease in the difference between the fluid temperature and the borehole wall temperature. If we were able to reduce the effect of the thermal gradient on the design, we might overcome the error.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The trick lies in understanding that the undisturbed ground temperature is linked to the <strong>borehole depth<\/strong>, whereas the temperature difference between the borehole wall and the fluid is linked to the <strong>borehole length<\/strong>. In most cases, both are (apart from the buried depth) rather similar, but in the case of tilted boreholes, they are not.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">When a borehole is tilted, the borehole length increases faster than the borehole depth, reducing the relative importance of the ground thermal gradient. By selecting a sufficiently large tilt for the boreholes, the gradient error can be overcome.<\/p>\n\n\n\n<div class=\"caution\">When working with tilted boreholes, there are a couple of important aspects:<ul><li>The drilling angle should be technically feasible, so always discuss with your drilling contractor which technique they are using and what angle they can handle. For non guided drilling, this is typically around 20\u00b0 with respect to a purely vertical borehole.<\/li><li>With angled boreholes, you need to be extra careful that your boreholes stay within your property boundaries.<\/li><\/ul><\/div>\n\n\n\n<h2 class=\"wp-block-heading\">Question 2.2<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><i>(Go to the <a href=\"https:\/\/ghetool.eu\/course\/calculate-required-borehole-depth\">original question<\/a>)<\/i><\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">In the example above, it was not possible to stay within the temperature limits with four boreholes, as indicated by the gradient error. Based on your answer to the previous question, can you modify the design so that it works with four boreholes? Use the case with constant ground properties, and do not change the load or the borehole thermal resistance.<\/p>\n<\/blockquote>\n\n\n\n<p class=\"wp-block-paragraph\">When the boreholes are placed in a square with 6 m borehole to borehole spacing and the boreholes are tilted 35\u00b0 outwards (as in the graph below), it is feasible to stay within the temperature limits with 4 boreholes of 154 m borehole length. The corresponding depth is 127.22 m, with an average undisturbed ground temperature of 11.84\u00b0C.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img fetchpriority=\"high\" decoding=\"async\" width=\"358\" height=\"309\" src=\"https:\/\/ghetool.eu\/wp-content\/uploads\/2026\/09\/Grid-boreholes.png\" alt=\"\" class=\"wp-image-5389\" srcset=\"https:\/\/ghetool.eu\/wp-content\/uploads\/2026\/09\/Grid-boreholes.png 358w, https:\/\/ghetool.eu\/wp-content\/uploads\/2026\/09\/Grid-boreholes-300x259.png 300w, https:\/\/ghetool.eu\/wp-content\/uploads\/2026\/09\/Grid-boreholes-14x12.png 14w\" sizes=\"(max-width: 358px) 100vw, 358px\" \/><figcaption class=\"wp-element-caption\">Borehole coordinates.<\/figcaption><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Question 2.3<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><i>(Go to the <a href=\"https:\/\/ghetool.eu\/course\/calculate-required-borehole-depth\">original question<\/a>)<\/i><\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">If the inlet fluid temperature had been used instead of the average fluid temperature to calculate the required depth, how would the results have changed?<\/p>\n<\/blockquote>\n\n\n\n<p class=\"wp-block-paragraph\">The type of fluid temperature changes how critical the temperature thresholds are. When switching from the average to the inlet fluid temperature, all fluid temperatures during extraction will be colder and those during injection will be higher. Therefore, the required drilling depth will increase from 363 m to 413 m. When the outlet temperatures are selected, the temperatures will be more optimistic, and therefore the drilling depth can be reduced.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Question 3.1<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><i>(Go to the <a href=\"https:\/\/ghetool.eu\/course\/calculate-required-borefield-size-and-depth\">original question<\/a>)<\/i><\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">Can you explain why, in the first example, optimising for the minimum number of boreholes and the minimum total borehole length yielded the same result?<\/p>\n<\/blockquote>\n\n\n\n<p class=\"wp-block-paragraph\">In both cases, the borefield was limited by the minimum fluid temperature limit (as seen in the figure below). When this is the case, drilling deeper is the best solution, since it both increases the undisturbed ground temperature and brings the fluid temperature closer to the borehole wall temperature due to the lower specific heat extraction rate. Therefore, the solutions with the minimum number of boreholes and the minimum total borehole length overlap in this case.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img decoding=\"async\" width=\"1024\" height=\"361\" src=\"https:\/\/ghetool.eu\/wp-content\/uploads\/2026\/08\/Temp-req-size-min-limit-1-1024x361.png\" alt=\"Temperature profile for the borefield configuration with the minimum total borehole length.\" class=\"wp-image-5307\" srcset=\"https:\/\/ghetool.eu\/wp-content\/uploads\/2026\/08\/Temp-req-size-min-limit-1-1024x361.png 1024w, https:\/\/ghetool.eu\/wp-content\/uploads\/2026\/08\/Temp-req-size-min-limit-1-300x106.png 300w, https:\/\/ghetool.eu\/wp-content\/uploads\/2026\/08\/Temp-req-size-min-limit-1-768x270.png 768w, https:\/\/ghetool.eu\/wp-content\/uploads\/2026\/08\/Temp-req-size-min-limit-1-18x6.png 18w, https:\/\/ghetool.eu\/wp-content\/uploads\/2026\/08\/Temp-req-size-min-limit-1.png 1420w\" sizes=\"(max-width: 1024px) 100vw, 1024px\" \/><figcaption class=\"wp-element-caption\">Temperature profile for the borefield configuration with the minimum total borehole length.<\/figcaption><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">When the borefield is limited by the maximum temperature limit, the situation is different, since having more and shallower boreholes is typically the way to reduce the total borehole length, except when the fluid regime switches from turbulent to laminar, as discussed in <a href=\"https:\/\/ghetool.eu\/course\/cope-with-imbalance\/\">Part 5.4<\/a>.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Question 3.2<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><i>(Go to the <a href=\"https:\/\/ghetool.eu\/course\/calculate-required-borefield-size-and-depth\">original question<\/a>)<\/i><\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">How would the results change if the simulation were carried out using the inlet fluid temperature instead of the average fluid temperature?<\/p>\n<\/blockquote>\n\n\n\n<p class=\"wp-block-paragraph\">temperature makes the coldest temperatures even colder. This means that the total borehole length will increase.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Question 4.1<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><i>(Go to the <a href=\"https:\/\/ghetool.eu\/course\/thermal-response-test-trt\">original question<\/a>)<\/i><\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">What happens when the undisturbed ground temperature is 12.7\u00b0C instead of 14.7\u00b0C?<\/p>\n<\/blockquote>\n\n\n\n<p class=\"wp-block-paragraph\">In order to understand the importance of the undisturbed ground temperature in the TRT analysis, let us recall that the whole analysis is based on a linear correlation:$$\\overline{T}_f-T_0=k\\cdot \\ln(t)+m$$In this equation, the thermal conductivity is solely determined by the slope $k$ of this line, whereas the vertical offset $m$ and the slope $k$ are used for the effective borehole resistance. Looking at the equation for the borehole resistance below, having a lower undisturbed ground temperature will increase the $\\overline{T}_f(t)-T_0$ term and therefore increase the borehole resistance from 0.0623 m\u00b7K\/W to 0.1025 m\u00b7K\/W.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$R_b^*=\\frac{H}{\\dot{Q}}\\cdot(\\overline{T}_f(t)-T_0)-\\frac{1}{4\\pi \\lambda}\\cdot \\left[ ln(t)+ln \\left(\\frac{4\\alpha}{r_0^2}\\right)-0.5772\\right]$$<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Question 4.2<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><i>(Go to the <a href=\"https:\/\/ghetool.eu\/course\/thermal-response-test-trt\">original question<\/a>)<\/i><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The answer to this question is rather similar to the one before. Looking at the equation to calculate the ground thermal conductivity $\\lambda$ from the slope $k$:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$R_b^*=\\frac{H}{\\dot{Q}}\\cdot(\\overline{T}_f(t)-T_0)-\\frac{1}{4\\pi \\lambda}\\cdot \\left[\\ln(t)+\\ln \\left(\\frac{4\\alpha}{r_0^2}\\right)-0.5772\\right]$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">it is clear that the volumetric heat capacity plays no role here. Therefore, the ground thermal conductivity is not dependent on this estimate.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The effective borehole thermal resistance, however, does depend on the volumetric heat capacity $C_v$ estimate via the ground thermal diffusivity $\\alpha=\\lambda\/C_v$. Changing the volumetric heat capacity will therefore only change the borehole resistance.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n","protected":false},"excerpt":{"rendered":"<p>In dit hoofdstuk vind je de antwoorden op de vragen uit de verschillende hoofdstukken van deel 7.<\/p>","protected":false},"template":"","section":[126],"chapter":[144],"authors":[39],"class_list":["post-5385","course","type-course","status-publish","hentry","section-chapter-5","chapter-part-7","authors-wouter-peere"],"_links":{"self":[{"href":"https:\/\/ghetool.eu\/nl_nl\/wp-json\/wp\/v2\/course\/5385","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ghetool.eu\/nl_nl\/wp-json\/wp\/v2\/course"}],"about":[{"href":"https:\/\/ghetool.eu\/nl_nl\/wp-json\/wp\/v2\/types\/course"}],"wp:attachment":[{"href":"https:\/\/ghetool.eu\/nl_nl\/wp-json\/wp\/v2\/media?parent=5385"}],"wp:term":[{"taxonomy":"section","embeddable":true,"href":"https:\/\/ghetool.eu\/nl_nl\/wp-json\/wp\/v2\/section?post=5385"},{"taxonomy":"chapter","embeddable":true,"href":"https:\/\/ghetool.eu\/nl_nl\/wp-json\/wp\/v2\/chapter?post=5385"},{"taxonomy":"authors","embeddable":true,"href":"https:\/\/ghetool.eu\/nl_nl\/wp-json\/wp\/v2\/authors?post=5385"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}