{"id":245,"date":"2015-08-11T16:01:39","date_gmt":"2015-08-11T21:01:39","guid":{"rendered":"http:\/\/charles-oneill.com\/blog\/?p=245"},"modified":"2024-06-14T17:04:39","modified_gmt":"2024-06-14T22:04:39","slug":"moist-air-density","status":"publish","type":"post","link":"https:\/\/charles-oneill.com\/blog\/moist-air-density\/","title":{"rendered":"Moist Air Density"},"content":{"rendered":"\n<hr class=\"wp-block-separator has-css-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Calculator:<\/h2>\n\n\n\n<script type=\"text\/javascript\">\nfunction updateOutput() { \n\t\t\/\/get form\n\tvar form = document.getElementById(\"calc\");\n\t\t\/\/get output\n\tvar out = form.elements[\"rho\"];\n        var daout = form.elements[\"da\"];\n\t\t\/\/get two numbers\n\tvar P = parseFloat(form.elements[\"p\"].value);\n\tvar T = parseFloat(form.elements[\"T\"].value)+459.67;\n\tvar psi = parseFloat(form.elements[\"psi\"].value)\/100;\n\t\t\/\/get operator\n\tvar Md = 28.97;\n\tvar Mw = 18.00;\n\tvar Rbar = 1545;\n\tvar ps = 0.08865*Math.exp( ( -0.002369*(T-8375.65)*(T-491.67)) \/ (T-28.818));\n\tout.value = (P*Md + psi*ps*(Mw-Md))\/Rbar\/T\/12.0\/32.174*12*12*12;\n        daout.value = \"\";\n}\nwindow.onload = updateOutput;\n<\/script>\n<form id=\"calc\" oninput=\"updateOutput()\">\n<table>\n       <tbody><tr>\n         <tr><td><b> Inputs<\/b><\/td><td><\/td><\/tr>\n         <tr><td>Pressure [psi]<\/td>\n\t\t <td><input name=\"p\" value=\"14.69\"><\/td>\n\t\t<\/tr>\n\t\t<tr>\n         <td>Temperature [F]<\/td>\n\t\t  <td><input name=\"T\" value=\"59\"><\/td>\n\t\t<\/tr><tr>\n         <td>Relative Humidity [%]<\/td>\n\t\t  <td><input name=\"psi\" value=\"0\"><\/td>\n       <\/tr>\n         <tr><td><b> Calculated <\/b><\/td><\/tr><td><\/td>\n         <tr>\n         <td> Density [slug\/ft<sup>3<\/sup>]<\/td>\n\t\t <td> <input name=\"rho\"> <\/td>\n       <\/tr>\n        <tr>\n         <td> Density Alt [ft]<\/td>\n\t\t <td> <input name=\"da\"> <\/td>\n       <\/tr>\n      <\/tbody><\/table>\n <\/form>\n\n\n\n<hr class=\"wp-block-separator has-css-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Theory:<\/h2>\n\n\n<p><strong>Warning: Don&#8217;t make this fundamental error in determining the density of moist air. <\/strong><\/p>\n<p>Recently, I&#8217;ve been preparing a graduate level course on Wing and Airfoil Theory. One lecture delves into modeling the real atmosphere including humid, wet, moist air. Specifically, I wanted to calculate the density of humid air. Visit my <a href=\"https:\/\/charles-oneill.com\/blog\/atmosphere\/\">Atmosphere<\/a> page for more information and tools.<\/p>\n<p><strong>[Update 1st March 2019] A moist air density calculator is at the top of this page.<\/strong><\/p>\n<p>My first thought was to grab a psychometric chart from my wife&#8217;s ASHRAE handbook which is available in pdf format at: <a href=\"http:\/\/charles-oneill.com\/blog\/wp-content\/uploads\/2015\/08\/ASHRAE-Chart.pdf\">ASHRAE Psychometric Chart No. 1<\/a><a href=\"http:\/\/charles-oneill.com\/blog\/wp-content\/uploads\/2015\/08\/ashrae.png\"><img loading=\"lazy\" decoding=\"async\" class=\" wp-image-246 size-medium aligncenter\" src=\"http:\/\/charles-oneill.com\/blog\/wp-content\/uploads\/2015\/08\/ashrae-300x233.png\" alt=\"ASHRAE Psychometric Chart  No. 1\" width=\"300\" height=\"233\" srcset=\"https:\/\/charles-oneill.com\/blog\/wp-content\/uploads\/2015\/08\/ashrae-300x233.png 300w, https:\/\/charles-oneill.com\/blog\/wp-content\/uploads\/2015\/08\/ashrae-800x621.png 800w, https:\/\/charles-oneill.com\/blog\/wp-content\/uploads\/2015\/08\/ashrae.png 955w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a>The chart plots lines of density, dry-bulb temperature, wet-bulb temperature, enthalpy, and relative humidity. As an example, at 90 F and 90% relative humidity, the chart indicates a volume per cubic foot of dry air of approximately 14.48. This gives a density of 0.00214 slugs per cubic foot.<\/p>\n<p>However, this density is <strong>not<\/strong> the density of humid air. I didn&#8217;t catch the mistake until verifying my own independent calculating routine.<\/p>\n<p>\\(\\) My derivation started with an ideal gas approximation of dry air and water vapor. The total pressure \\(p\\) is the summation of partial pressures for dry air \\(p_d\\) and water vapor \\(p_w\\).<\/p>\n<p>$$\\rho = \\frac{p_d}{R_d T_d}+\\frac{p_v}{R_v T_v}$$<\/p>\n<p>Since the temperatures are identical, the density is:<\/p>\n<p>$$\\rho = \\frac{p_d M_d + p_v M_v}{\\bar R T}$$<\/p>\n<p>Given that the relative humidity relates the partial pressure of water vapor to the saturation pressure of water pressure \\(p_v = \\phi p_s \\), the density formula is rearranged to<\/p>\n<p>$$\\rho = \\frac{p M_d + \\phi p_s \\left(\u00a0 M_v &#8211; M_d \\right)}{\\bar R T}$$<\/p>\n<p>Since the molecular mass weight of dry air is \\( 28.97 \\frac{lbm}{lbmol} \\) while water vapor is lower at \\( 18.00 \\frac{lbm}{lbmol} \\), the density of wet air is <strong>always<\/strong> lower than dry air.<\/p>\n<p>The reduction in magnitude depends on the saturation pressure of water vapor. For the region of interest to aerodynamics, the Arden-Buck equation provides a curve fit approximation to the partial pressure of water vapor. Converted to standard units, the partial pressure in psi given a temperature in R is:<\/p>\n<p>$$p_s(T) = 0.08865 \\exp\\left(\\frac{-0.002369 (T-8375.65)(T-491.67)}{T-28.818}\\right)$$<\/p>\n<p>Combining the density and partial pressure formulas gives an alternative estimate of the density of air at 90 F and 90% humidity of 0.002203 slugs per cubic foot.<\/p>\n<p>So what gives? Who is correct? Why is the gold-standard ASHRAE chart suggesting a lower density? The key words are: volume per cubic foot of <strong>dry air<\/strong>. <a href=\"http:\/\/charles-oneill.com\/blog\/wp-content\/uploads\/2015\/08\/ashrae-dryair.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-253 size-medium\" src=\"http:\/\/charles-oneill.com\/blog\/wp-content\/uploads\/2015\/08\/ashrae-dryair-300x205.png\" alt=\"ashrae-dryair\" width=\"300\" height=\"205\" srcset=\"https:\/\/charles-oneill.com\/blog\/wp-content\/uploads\/2015\/08\/ashrae-dryair-300x205.png 300w, https:\/\/charles-oneill.com\/blog\/wp-content\/uploads\/2015\/08\/ashrae-dryair.png 607w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><\/p>\n<p>The ASHRAE chart gives the density of only the air and NOT the water vapor. The ASHRAE chart clearly states that only the dry air component is considered.<\/p>\n<p>Adding in the water vapor requires finding the humidity ratio.<\/p>\n<p><a href=\"http:\/\/charles-oneill.com\/blog\/wp-content\/uploads\/2015\/08\/ashrae-ratio.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-254\" src=\"http:\/\/charles-oneill.com\/blog\/wp-content\/uploads\/2015\/08\/ashrae-ratio.png\" alt=\"ashrae-ratio\" width=\"495\" height=\"45\" srcset=\"https:\/\/charles-oneill.com\/blog\/wp-content\/uploads\/2015\/08\/ashrae-ratio.png 495w, https:\/\/charles-oneill.com\/blog\/wp-content\/uploads\/2015\/08\/ashrae-ratio-300x27.png 300w\" sizes=\"(max-width: 495px) 100vw, 495px\" \/><\/a><\/p>\n<p>For the 90 F and 90% case, the humidity ratio is 0.028. Each 100 pounds of dry air has 2.8 pounds of water. Correcting the previous ASHRAE density gives:<\/p>\n<p>$$ \\rho = (1+0.028) 0.00214 \\frac{slugs}{ft^3} $$ which is 0.002207 slugs per cubic foot.<\/p>\n<p>Mystery solved.<\/p>\n<p><strong>Warning<\/strong>: There are websites out there inadvertently giving the wrong density of wet air. One example is <a href=\"http:\/\/denysschen.com\/denysschen\/catalogue\/density.aspx\">denysschen.com.<\/a> A single sentence would correct the confusion. On the other hand, the site at <a href=\"http:\/\/hvac-calculator.com\/wet_air_density.php\">hvac-calculator.com<\/a> appears to give the correct value.<\/p>\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calculator: Inputs Pressure [psi] Temperature [F] Relative Humidity [%] Calculated Density [slug\/ft3] Density Alt [ft] Theory: Warning: Don&#8217;t make this fundamental error in determining the density of moist air. Recently, I&#8217;ve been preparing a graduate level course on Wing and Airfoil Theory. One lecture delves into modeling the real atmosphere including humid, wet, moist air. [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"_links":{"self":[{"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/posts\/245"}],"collection":[{"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/comments?post=245"}],"version-history":[{"count":23,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/posts\/245\/revisions"}],"predecessor-version":[{"id":1605,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/posts\/245\/revisions\/1605"}],"wp:attachment":[{"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/media?parent=245"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/categories?post=245"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/charles-oneill.com\/blog\/wp-json\/wp\/v2\/tags?post=245"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}