Double_Interpolation

# Double_Interpolation - ([email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */@30 2747.8 kJ kg kJ a kg...

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With the example that we had in section, we need to find enthalpy, h, at T = 450C and P = 30MPa: From the temperature tables for superheated water, we have: T h @ 28MPa h @ 32MPa 440C 2812.6kJ/kg 2688kJ/kg 480C 3028.5kJ/kg 2949.2kJ/kg Obviously, we don't have the value for enthalpy at the specified T, and P. Assuming the relationships between h, T and P are linear, what we need to do is three separate interpolations. (Don’t get tricked by the name, “double-interpolation”.) First, we need to deal with the first row, T = 440C. Find the h @ 30MPa: (2812.6 2688.0) (28.0 32.0) (28.0 30.0)
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Unformatted text preview: (2812.6 @30 ) @30 2747.8 kJ kg kJ a kg kJ a kg MPa MPa h MPa h MPa--=--= Then, routinely, deal with the 2 nd row, T = 480C, the same way as above: (3028.5 2949.2) (28.0 32.0) (28.0 30.0) (3028.5 @30 ) @30 2988.85 kJ kg kJ b kg kJ b kg MPa MPa h MPa h MPa--=--= Now, life has become so much easier, and we have the following: T h @ 30MPa 440C 2747.8 480C 2988.85 Do interpolation once again to find enthalpy at T=450C: (2747.8 2988.5) (440 480) (440 450) (2747.8 @30 ) @30 2808.1 kJ o kg o kJ kg kJ kg C C h MPa h MPa--=--= Bingo! That’s all we need!...
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## This note was uploaded on 03/15/2010 for the course ENGRD 221 taught by Professor Staff during the Fall '08 term at Cornell.

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