{"id":1402,"date":"2020-10-30T23:30:00","date_gmt":"2020-10-31T03:30:00","guid":{"rendered":"http:\/\/underthehood.blogwyrm.com\/?p=1402"},"modified":"2022-07-28T06:28:23","modified_gmt":"2022-07-28T10:28:23","slug":"elementary-compressible-flow-part-3","status":"publish","type":"post","link":"https:\/\/underthehood.blogwyrm.com\/?p=1402","title":{"rendered":"Elementary Compressible Flow \u2013 Part 3"},"content":{"rendered":"\n<p>In this installment we derive a central result in compressible fluid flow showing the inner working of the converging-diverging nozzle.\u00a0 The key relations are the jump conditions are the mass conservation equation<\/p>\n\n\n\n<p>\\[ \\rho_1 V_1 A_1 = \\rho_2 V_2 A_2 \\;&nbsp; \\]<\/p>\n\n\n\n<p>and the energy equation<\/p>\n\n\n\n<p>\\[ h_1 + \\frac{V_1^2}{2} = h_2 + \\frac{V_2^2}{2} \\; .\\]<\/p>\n\n\n\n<p>As a reminder, the enthalpy is defined in terms of the specific internal energy $e$ by $h = e + P\/\\rho$.&nbsp; Correspondingly, the first law of thermodynamics (as usually expressed in terms of the specific internal energy)<\/p>\n\n\n\n<p>\\[ de = T ds + \\frac{P}{\\rho^2} d\\rho \\]<\/p>\n\n\n\n<p>becomes<\/p>\n\n\n\n<p>\\[ dh = T ds + \\frac{dP}{\\rho} \\; .\\]<\/p>\n\n\n\n<p>The final set of equations come from the specific fluid that is flowing.&nbsp; For simplicity, we will deal with compressible gas flow and will take the gas to be ideal.&nbsp; The equation of state for an ideal gas is<\/p>\n\n\n\n<p>\\[ \\frac{P}{\\rho} = RT \\]<\/p>\n\n\n\n<p>with the usual relations between the heat capacity at constant pressure $c_P$ and the heat capacity at constant volume $c_V$ of<\/p>\n\n\n\n<p>\\[ c_P \u2013 c_V = R \\;\u00a0 \\]<\/p>\n\n\n\n<p>and<\/p>\n\n\n\n<p>\\[ c_P\/c_V = \\gamma \\; , \\]<\/p>\n\n\n\n<p>where $\\gamma$ is a constant usually taken to have the value of 1.4 for ordinary air.<\/p>\n\n\n\n<p>For an ideal gas, the specific enthalpy (like the specific energy) is strictly a function of temperature given by<\/p>\n\n\n\n<p>\\[ dh = c_P dT \\; , \\]<\/p>\n\n\n\n<p>where we will often assume that $c_P$ is constant over a wide range of conditions; an assumption that allows us to integrate to $h = c_P T$ immediately.&nbsp; In other words, we assume the gas is a calorically perfect ideal gas \u2013 this is actually a reasonably good assumption given that <a href=\"https:\/\/www.engineeringtoolbox.com\/nitrogen-d_977.html\">$c_P$ for nitrogen<\/a> varies only about 1.5\n\n\n\n<p>The speed of sound in the ideal gas is also only a function of temperature given by<\/p>\n\n\n\n<p>\\[ c = \\sqrt{ \\gamma R T } \\; .\\]<\/p>\n\n\n\n<p>The final assumption is that the flow is isentropic, which is a good assumption since the flow happens quickly so that a negligible amount of heat flows into the system.&nbsp; Under this assumption, the equation of state can be substituted into the first law to yield<\/p>\n\n\n\n<p>\\[ c_P dT = \\frac{dP}{\\rho} \\; .\\]<\/p>\n\n\n\n<p>The next step is to eliminating the density by way of the ideal gas law to get an equation in terms of pressure and temperature<\/p>\n\n\n\n<p>\\[ c_P dT = \\frac{dP}{P\/RT} \\; . \\]<\/p>\n\n\n\n<p>Re-arranging gives a simple set of differential equations<\/p>\n\n\n\n<p>\\[ \\frac{dT}{T} = \\frac{R}{c_P} \\frac{dP}{P} \\; ,\\]<\/p>\n\n\n\n<p>which integrates to<\/p>\n\n\n\n<p>\\[ \\frac{T_1}{T_2} = \\left( \\frac{P_1}{P_2} \\right)^{\\frac{\\gamma-1}{\\gamma}} \\; ,\\]<\/p>\n\n\n\n<p>where $R = c_P &#8211; c_V$ and $c_P\/c_V = \\gamma$.<\/p>\n\n\n\n<p>When the analogous steps are followed with $P$ eliminated in favor of $\\rho$, one arrives at<\/p>\n\n\n\n<p>\\[ \\frac{T_1}{T_2} = \\left( \\frac{\\rho_1}{\\rho_2} \\right)^{\\gamma \u2013 1} \\; . \\]<\/p>\n\n\n\n<p>The value of this latter equation is that it allows the elimination of temperature (simply equation both forms of $T_1\/T_2$ and the emergence of the polytropic equation<\/p>\n\n\n\n<p>\\[ \\frac{P_1}{P_2} = \\left( \\frac{\\rho_1}{\\rho_2} \\right)^{\\gamma} \\; \\]<\/p>\n\n\n\n<p>relating pressure and density for an isentropic flow.<\/p>\n\n\n\n<p>With these assumptions and corresponding relations in hand, we now use them to relate the conditions from the inlet of the gas from a reservoir (e.g. a fuel tank) to the outlet of the nozzle.&nbsp; The rocket engine will be the prototype for this situation.&nbsp; We need to be able to relate the thermodynamics at each point in the flow to the thermodynamics in the reservoir.&nbsp; Since the gas in the reservoir is not moving with respect to the nozzle we introduce the concept of stagnation, in which the flow is envisioned to have come to a stop and the only energy present being stored as internal energy.&nbsp; The third jump condition relating enthalpy and bulk flow velocity at different points will be important.<\/p>\n\n\n\n<p>Since there is a qualitative change between subsonic and supersonic flow (see last post\u2019s discussion of <a href=\"http:\/\/underthehood.blogwyrm.com\/?p=1393\">how the area is related to Mach number<\/a>), we will need to stitch together the two flows (stagnant-to-sonic with sonic-to-supersonic) at the critical point where $M = 1$.&nbsp; Variables associated with this point are called critical and are decorated with a star (e.g. $A^*, P^*$ and so on).&nbsp; In addition, we will decorate all variables associated with the reservoir where the flow is stagnant with a 0 subscript (e.g. $A_0, P_0$ and so on).<\/p>\n\n\n\n<p>Let\u2019s start on the subsonic side by relating the thermodynamics in the tank to the corresponding variables in the flow as it moves in the converging portion of the nozzle.&nbsp; Since the flow is stagnant in the tank, we find that the stagnation enthalpy $h_0$ relates to the flow anywhere else in the nozzle by<\/p>\n\n\n\n<p>\\[ h_0 = h_1 + \\frac{V_1^2}{2} \\; . \\]<\/p>\n\n\n\n<p>Since the gas is calorically perfect, the enthalpy can be eliminated in favor of the temperatures to arrive at<\/p>\n\n\n\n<p>\\[ c_P T_0 = c_P T_1 + \\frac{V_1^2}{2} \\; .\\]<\/p>\n\n\n\n<p>Dividing by $c_P T_1$ and using the fact that $T_1 = {c_1}^2\/\\gamma R$ gives<\/p>\n\n\n\n<p>\\[ \\frac{T_0}{T_1} = 1 + \\frac{V_1^2}{2 c_P T_1} = 1 + \\frac{V_1^2}{2 c_P ({c_1}^2\/\\gamma R)} \\; .\\]<\/p>\n\n\n\n<p>In this last relation, the ratio of the flow velocity to the speed of sound is the Mach number.&nbsp; Dropping the \u20181\u2019 subscript and using the relations between $c_P$, $c_V$, and $R$, we arrive at what we will call the stagnation temperature relation<\/p>\n\n\n\n<p>\\[ \\frac{T_0}{T} = 1 + \\frac{(\\gamma-1)}{2} M^2 \\;. \\]<\/p>\n\n\n\n<p>Using the relationship between the temperature and pressure derived above gives the corresponding stagnation pressure equation<\/p>\n\n\n\n<p>\\[ \\frac{P_0}{P} = \\left[ 1 + \\frac{(\\gamma-1)}{2} M^2 \\right]^{\\frac{\\gamma}{\\gamma &#8211; 1}} \\; \\]<\/p>\n\n\n\n<p>Return to the mass flow equation and, first eliminate the density using the equation of state and multiply by $c\/c$, then express the sound speed in terms of the temperature to get<\/p>\n\n\n\n<p>\\[ {\\dot m} = \\rho A V = \\frac{P}{R T} A \\frac{V}{c} c = P A M \\frac{c}{RT} = P A M \\sqrt{ \\frac{\\gamma}{R T} } \\; .\\]<\/p>\n\n\n\n<p>Next, solve the stagnation pressure equation for $P$&nbsp;<\/p>\n\n\n\n<p>\\[ P = P_0 \\left( \\frac{T_0}{T} \\right)^{\\frac{\\gamma}{1 &#8211; \\gamma}} \\]<\/p>\n\n\n\n<p>and then substitute into the mass flow equation<\/p>\n\n\n\n<p>\\[ {\\dot m} = P_0 \\left( \\frac{T_0}{T} \\right)^{\\frac{\\gamma}{1 &#8211; \\gamma}} A M \\sqrt{ \\frac{\\gamma}{R T}} \\left( \\frac{T_0}{T_0} \\right)^{1\/2} \\\\ = P_0 \\sqrt{\\frac{\\gamma}{R T_0}} \\left(\\frac{T_0}{T}\\right)^{\\frac{\\gamma}{1-\\gamma}+\\frac{1}{2}} A M \\; .\\]<\/p>\n\n\n\n<p>Simplifying and using the stagnation temperature equation gives<\/p>\n\n\n\n<p>\\[ {\\dot m} = P_0 \\sqrt{\\frac{\\gamma}{RT_0}} A M \\left[ 1 + \\frac{\\gamma-1}{2} M^2 \\right]^{\\frac{\\gamma+1}{2(1-\\gamma)}} \\; .\\]<\/p>\n\n\n\n<p>This combination is constant no matter where in the flow it is evaluated.&nbsp; Equate the expression evaluated at the critical point to the expression anywhere else in the flow to get<\/p>\n\n\n\n<p>\\[ P_0 \\sqrt{\\frac{\\gamma}{RT_0}} A^* \\left[ 1 + \\frac{\\gamma-1}{2} \\right]^{\\frac{\\gamma+1}{2(1-\\gamma)}} = P_0 \\sqrt{\\frac{\\gamma}{RT_0}} A M \\left[ 1 + \\frac{\\gamma-1}{2} M^2 \\right]^{\\frac{\\gamma+1}{2(1-\\gamma)}} \\; .\\]<\/p>\n\n\n\n<p>Eliminating the terms common to both sides and rearranging we can express the cross-sectional area at any point in the nozzle to the point where the flow goes critical as<\/p>\n\n\n\n<p>\\[ \\frac{A}{A*} = \\frac{1}{M} \\left(1+\\frac{\\gamma-1}{2}\\right)^{\\frac{\\gamma+1}{2(1-\\gamma)}} \\left[ 1+ \\frac{\\gamma-1}{2} M^2 \\right]^{\\frac{\\gamma+1}{2(\\gamma-1)}} \\; .\\]<\/p>\n\n\n\n<p>This is the well-known relation that forms the basis for designing and analyzing the nozzle flow.&nbsp; It will be the basis for our final post on compressible flow next month.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>In this installment we derive a central result in compressible fluid flow showing the inner working of the converging-diverging nozzle.\u00a0 The key relations are the jump conditions are the mass&#8230; <a class=\"read-more-button\" href=\"https:\/\/underthehood.blogwyrm.com\/?p=1402\">Read more &gt;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-1402","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/posts\/1402","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1402"}],"version-history":[{"count":26,"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/posts\/1402\/revisions"}],"predecessor-version":[{"id":1814,"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=\/wp\/v2\/posts\/1402\/revisions\/1814"}],"wp:attachment":[{"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1402"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1402"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/underthehood.blogwyrm.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1402"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}