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Chiller Calculations.ipynb

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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Chiller Flow Calculations\n",
+    "\n",
+    "This notebook calculates the flow requirements and flow capacity of cooling the ETL disc section with our chiller.\n",
+    "\n",
+    "\n",
+    "#### Links\n",
+    "\n",
+    "  - Our Chiller: [link](https://www.julabo.com/en-us/products/refrigerated-circulators/refrigerated-heating-circulators/dyneo-dd-1000f)\n",
+    "  - Antifreeze properties: [link](https://www.engineeringtoolbox.com/ethylene-glycol-d_146.html)\n",
+    "   "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Cooling Requirements\n",
+    "\n",
+    "Each full module is estimated to produce 4.53 W of heat. The service hybrid is estimated at 18 W. A half-populated (ie only one side) disc section has 12 modules and 1 service hybrid producing a total of **72.36 Watts** of heat. A fully populated section would have 6 additional modules and 2 additional service hybrids producing 45.18 additional watts of heat. This document will work under the assumption of a half-populated disc section.\n",
+    "\n",
+    "The chiller is operating with a 50/50 mix of water and propylene glycol. The physical properties of interest are:\n",
+    "\n",
+    "  - density: 1.1 g/ml @ -25 $^{\\circ}$C\n",
+    "  - specific heat: ~0.8 J/(g$\\cdot^{\\circ}$C)\n",
+    "  - freezing point: -36.8 $^{\\circ}$C\n",
+    "\n",
+    "From this information, we can calculate the flow rate needed, assuming a specific input and output temperatures of the cooling fluid."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "flow rate: 0.3553392857142858 L/min\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Assume highest possible delta-t \n",
+    "t1 = -36.8  # Celcius\n",
+    "t2 = -20  # Celcius\n",
+    "\n",
+    "dt = t2 - t1\n",
+    "rho = (0.8 * dt) / 1.1  # heat removed per ml\n",
+    "\n",
+    "flow_rate = 72.36 / rho\n",
+    "print(f\"flow rate: {flow_rate*60/1000} L/min\")"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Flow-rate potential\n",
+    "\n",
+    "The flow rate of a liquid through a pipe can be estimated by applying a variant ([ref](http://kb.eng-software.com/eskb/ask-an-engineer/theory-equations-and-calculated-results-questions/relationship-between-pressure-drop-and-flow-rate-in-a-pipeline)) of the Bernoulli equation:\n",
+    "\n",
+    "$$\n",
+    "(P_1 - P_2) = \\rho g \\left[(Z_2 - Z_1) + \\left(\\frac{v_2^2 - v_1^2}{2g}\\right) + H_L\\right]\n",
+    "$$\n",
+    "\n",
+    "Where,\n",
+    "\n",
+    "  - $\\rho$ - fluid density (kg/m$^3$)\n",
+    "  - $g$ - accelleration due to gravity (m/s$^2$)\n",
+    "  - $v_{1,2}$ - velocity at two points in pipe (m/s)\n",
+    "  - $Z_{1,2}$ - elevations at two points in pipe (m)\n",
+    "  - $H_L$ - Head loss, aka energy loss due to friction, vibrations, etc between two points (m)\n",
+    "\n",
+    "For flow through a constant-diameter pipe, the fluid velocities $v_1$ and $v_2$ are equal. Furthermore, if the pipe is horizontal, the two elevations, $Z_1$ and $Z_2$ are also equal. This simplifies the equation to:\n",
+    "\n",
+    "$$\n",
+    "(P_1 - P_2) = \\rho g H_L\n",
+    "$$\n",
+    "\n",
+    "This means that the pressure difference between the ends of the pipe is caused solely by the *head loss*. The head loss can be calculated using the Darcy-Weisbach equation:\n",
+    "\n",
+    "$$\n",
+    "H_L = f \\frac{L}{D} \\frac{v^2}{2g}\n",
+    "$$\n",
+    "\n",
+    "Where,\n",
+    "\n",
+    "  - $f$ - Darcy friction factor (dimensionless)\n",
+    "  - $L$ - Pipe length (m)\n",
+    "  - $D$ - Inside pipe diameter (m)\n",
+    "  - $v$ - fluid velocity (m/s)\n",
+    "\n",
+    "Calculating the *Darcy friction factor* first requires calculating the Reynold's number of the fluid. For this calculation, we will assume that the antifreeze mixture is water-like.\n",
+    "\n",
+    "The Reynolds number is defined as:\n",
+    "\n",
+    "$$\n",
+    "\\mathrm{Re} = \\frac{\\rho u L}{\\mu}\n",
+    "$$\n",
+    "\n",
+    "Where,\n",
+    "\n",
+    "  - $\\rho$ - fluid density (999.7 kg/m$^3$)\n",
+    "  - $u$ - flow speed through pipe (currently unknown but guess 1 m/s - this can be iterated if needed)\n",
+    "  - $\\mu$ - dynamic viscosity (1.308E-3 kg/m$\\cdot$s)\n",
+    "  - $L$ - characteristic linear dimension, diameter of pipe in this case (3E-3 m)\n",
+    "  \n",
+    "Putting in numbers, we get a Reynolds number of 2,293.\n",
+    "\n",
+    "There are many methods to calculating the *darcy friction factor*, some more complicated than others. Let's use the fairly simple [*Halland equation*](https://en.wikipedia.org/wiki/Darcy_friction_factor_formulae#Approximations_of_the_Colebrook_equation) (which may or may not actually be applicable in this situation, see article).\n",
+    "\n",
+    "$$\n",
+    "\\frac{1}{\\sqrt{f}} = -1.8 \\log\\left[\\left(\\frac{\\epsilon/D}{3.7}\\right)^{1.11} + \\frac{6.9}{\\mathrm{Re}}\\right]\n",
+    "$$\n",
+    "\n",
+    "The values $D$ and $\\epsilon$ are the pipe's diameter and \"effective roughness height\", respectively. For these calculations, we will take $D=3$ mm and $\\epsilon=100$ microns. Putting in numbers we get, $\\frac{1}{\\sqrt{f}}=3.3738$, or $f=7.15\n",
+    "E-2$."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Now that we have the *darcy friction factor*, let's rewrite the *head loss* equation to solve for the fluid velocity.\n",
+    "\n",
+    "$$\n",
+    "v^2 = \\frac{2g D}{L f} H_L\n",
+    "$$\n",
+    "\n",
+    "Our pump is rated to supply a pressure of up to 10.2 psi, or 70.33 kPa. Let's assume that the low pressure end of the pipe is at atmospheric pressure, meaning that $\\rho g H_L = 70.33 \\mathrm{kPa}$, giving $H_L=7.17E-3$ m.\n",
+    "\n",
+    "One last bit of information we need is the length of the pipe. Let's use $L=1$ m. Now we have everything we need to calculate the fluid velocity. Putting in numbers yields: $v=0.24$ m/s. Although this is about 25% of our initial velocity estimate, let's push on to the final number. The flow rate is $F=v\\pi\\left(\\frac{D}{2}\\right)^2=1.71E-6$ m$^3$/s, or **0.103 L/min**. This is about 1/3 of what is needed to chill a half-populated disc section."
+   ]
+  }
+ ],
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