P. K. Nag. Engineering Thermodynamics (776119), страница 9
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42 Hr = -249952+(18.7*560) +(70*540) ;3 Hp = 8*( -393522+20288) +9*( -241827+16087)4567+6.25*14171+70*13491;Wcv = 150; // Energy o u t put from e n g i n e i n kWQcv = -205; // Heat t r a n s f e r from e n g i n e i n kWn = ( Wcv - Qcv ) *3600/( Hr - Hp ) ;disp ( ” kg / h ” ,n *114 , ” F u e l c o n s u m p t i o n r a t e i s ” )114Scilab code Exa 16.10 Calculations on burning of liquid octane1 Hr1 = -249952; // For o c t a n e2 Hp1 = Hr1 ;3 // Below v a l u e s a r e c a l c u l a t e d45678910111213using value frontable 16.4T2 = 1000;Hp2 = -1226577T3 = 1200;Hp3 = 46537;T4 = 1100;Hp4 = -595964;Hp = [ Hp2 Hp3 Hp4 ]T = [ T2 T3 T4 ]T1 = interpln ([ Hp ; T ] , Hp1 ) ; // I n t e r p o l a t i o n t of i n d t e m p e r a t u r e a t Hp1disp ( ”K” ,T1 , ” t h e a d e a b a t i c f l a m e t e m p e r a t u r e i s ” )Scilab code Exa 16.11 Calculations on burning of gaseous propane// R e f e r t a b l e 1 6 . 4 f o r v a l u e sT0 = 298;Wrev = -23316 -3*( -394374) -4*( -228583) ;Wrev_ = Wrev /44; // i n kJ / kgHr = -103847;T = 980; // Through t r i a l and e r r o rSr = 270.019+20*205.142+75.2*191.611;Sp = 3*268.194 + 4*231.849 + 15*242.855 +75.2*227.485;9 IE = Sp - Sr ; // I n c r e a s e i n e n t r o p y10 I = T0 *3699.67/44;11 Si = Wrev_ - I ;12345678115disp ( ” kJ / kg ” , Wrev_ , ” R e v e r s i b l e work i s ” )disp ( ” k j / kg mol K” ,Sp - Sr , ” I n c r e a s e i n e n t r o p y d u r i n gcombustion i s ”)14 disp ( ” kJ / kg ” ,I , ” I r r e v e r s i b i l i t y o f t h e p r o c e s s ” )15 disp ( ” kJ / kg ” ,Si , ” A v a i l a b i l i t y o f p r o d u c t s o fcombustion i s ”)1213Scilab code Exa 16.12 Determination of chemical energy of phases of water1 T0 = 298.15; P0 = 1; R = 8.3143;2 xn2 = 0.7567; xo2 = 0.2035; xh2o = 0.0312; xco2 =0.0003;3 // P a r t ( a )4 g_o2 = 0; g_c = 0; g_co2 = -394380;5 A = - g_co2 + R * T0 * log ( xo2 / xco2 ) ;6 disp ( ” kJ / k mol ” ,A , ” The c h e m i c a l e n e r g y o f c a r b o n789101112131415161718is ”)// P a r t ( b )g_h2 = 0; g_h2o_g = -228590;B = g_h2 + g_o2 /2 - g_h2o_g + R * T0 * log ( xo2 ^0.5/ xh2o );disp ( ” kJ / k mol ” ,B , ” The c h e m i c a l e n e r g y o f h y d r o g e ni s ”)// P a r t ( c )g_ch4 = -50790;C = g_ch4 + 2* g_o2 - g_co2 - 2* g_h2o_g + R * T0 * log ((xo2 ^2) /( xco2 * xh2o ) ) ;disp ( ” kJ / k mol ” ,C , ” The c h e m i c a l e n e r g y o f methane i s”)// P a r t ( d )g_co = -137150;D = g_co + g_o2 /2 - g_co2 + R * T0 * log (( xo2 ^0.5) / xco2);disp ( ” kJ / k mol ” ,D , ” The c h e m i c a l e n e r g y o f1161920212223242526272829303132333435Carbonmonoxide i s ” )// P a r t ( e )g_ch3oh = -166240;E = g_ch3oh + 1.5* g_o2 - g_co2 - 2* g_h2o_g + R * T0 *log (( xo2 ^1.5) /( xco2 *( xh2o ^2) ) )disp ( ” kJ / k mol ” ,E , ” The c h e m i c a l e n e r g y o f m e t h a n o li s ”)// P a r t ( f )F = R * T0 * log (1/ xn2 ) ;disp ( ” kJ / k mol ” ,F , ” The c h e m i c a l e n e r g y o f n i t r o g e ni s ”)// P a r t ( g )G = R * T0 * log (1/ xo2 ) ;disp ( ” kJ / k mol ” ,G , ” The c h e m i c a l e n e r g y o f Oxygen i s ”)// P a r t ( h )H = R * T0 * log (1/ xco2 ) ;disp ( ” kJ / k mol ” ,H , ” The c h e m i c a l e n e r g y o fcarbondioxide i s ”)// P a r t ( i )g_h2o_l = -237180;I = g_h2o_l - g_h2o_g + R * T0 * log (1/ xh2o ) ;disp ( ” kJ / k mol ” ,I , ” The c h e m i c a l e n e r g y o f w a t e r i s ” )Scilab code Exa 16.13 Calculation on burning of liquid octane1 b = 8/(0.114+0.029) ; // By c a r b o n b a l a n c e2 C = 18/2; // By h y d r o g e n b a l a n c e3 a = b *0.114 + ( b /2) *0.029 + b *0.016 + C /2 ; // By4567oxygen b a l a n c eWcv = 1; // Power d e v e l o p e d by e n g i n e i n kWn_fuel = (0.57*1) /(3600*114.22) ;Qcv = Wcv - n_fuel *3845872; // 5 .
3 3disp ( ”kW” ,Qcv , ” The r a t e o f h e a t t r a n s f e r from t h eengine i s ”)1178 // P a r t ( b )9 ach = 5407843; // c h e m i c a l e n e r g y o f l i q u i d o c t a n e10 n2 = Wcv /( n_fuel * ach ) ;11 disp ( ”%” , n2 *100 , ” The s e c o n d law e f f i c i e n c y i s ” )118Chapter 17Compressible fluid flowScilab code Exa 17.1 Calculation s on flow of air through a duct1 T0 = 37+273; P = 40; g = 1.4;2 function [ x ] = speed (a ,b , f )3N = 100;4eps = 1e -5;5if (( f ( a ) * f ( b ) ) >0) then6error ( ’ no r o o t p o s s i b l e f ( a ) ∗ f ( b ) >0 ’ ) ;7abort ;8end ;9if ( abs ( f ( a ) ) < eps ) then10error ( ’ s o l u t i o n a t a ’ ) ;11abort ;12end13if ( abs ( f ( b ) ) < eps ) then14error ( ’ s o l u t i o n a t b ’ ) ;15abort ;16end17while (N >0)18c = ( a + b ) /219if ( abs ( f ( c ) ) < eps ) then20x = c ;21x;11922232425262728293031323334353637383940414243444546return ;end ;if (( f ( a ) * f ( c ) ) <0 ) thenb = c ;elsea = c ;endN = N -1;enderror ( ’ no c o n v e r g e n c e ’ ) ;abort ;endfunctiondeff ( ’ [ y ]= p ( x ) ’ ,[ ’ y = x ˆ4 + ( 5 ∗ ( x ˆ 2 ) ) − 3 .
2 2 5 ’ ])x = speed (0.5 ,1 , p ) ;M = x ; // Mach numberg = 1.4; // gammaR = 0.287;T = T0 /(1+(( g -1) /2) * M ^2) ;c = sqrt ( g * R * T *1000) ;V = c*M;P0 = P *(( T0 / T ) ^( g /( g -1) ) ) ;disp (M , ”Mach number i s ” )disp ( ”m/ s ” ,V , ” V e l o c i t y i s ” )disp ( ” kPa ” ,P0 , ” S t a g n a t i o n p r e s s u r e i s ” )Scilab code Exa 17.2 Calculations on canonical air diffuser1234567P1 = 0.18 e03 ; // i n KpaR = 0.287; T1 = 310; P0 = 0.1 e03 ;A1 = 0.11; V1 = 267;w = ( P1 /( R * T1 ) ) * A1 * V1 ;g = 1.4;c1 = sqrt ( g * R * T1 *1000) ;M1 = V1 / c1 ;1208 A1A_ = 1.0570; // A1/A∗ A∗ = A9 P1P01 = 0.68207;10 T1T01 = 0.89644;11 F1F_ = 1.0284;12 A2A1 = 0.44/0.11 ; // A2A1 = A2/A113 A2A_ = A2A1 * A1A_ ;14 M2 = 0.135; P2P02 = 0.987; T2T02 = 0.996; F2F_ =1516171819202122232425263.46;P2P1 = P2P02 / P1P01 ;T2T1 = T2T02 / T1T01 ;F2F1 = F2F_ / F1F_ ;P2 = P2P1 * P1 ;T2 = T2T1 * T1 ;A2 = A2A1 * A1 ;F1 = P1 * A1 *(1+ g * M1 ^2) ;F2 = F2F1 * F1 ;Tint = F2 - F1 ;Text = P0 *( A2 - A1 ) ;NT = Tint - Text ;disp ( ”kN” ,NT , ” Net t h r u s t i s ” )Scilab code Exa 17.3 Calculations on air flow through convergent divergent nozzle1234567891011M2 = 2.197; P2P0 = 0.0939; T2T0 = 0.5089;P0 = 1000; T0 = 360; g = 1.4; R = 0.287;P2 = P2P0 * P0 ;T2 = T2T0 * T0 ;c2 = sqrt ( g * R * T2 *1000) ;V2 = c2 * M2 ;// f o r a i rP_P0 = 0.528; T_T0 = 0.833; // T == T∗P_ = P_P0 * P0 ; T_ = T_T0 * T0 ;rho_ = P_ /( R * T_ ) ;V_ = sqrt ( g * R * T_ *1000) ;12112 At = 500 e -06; // t h r o a t a r e a13 w = At * V_ * rho_ ;14 disp ( ”When d i v e r g e n t s e c t i o n a c t a s a n o z z l e ” )15 disp ( ” kg / s ” ,w , ”Maximum f l o w r a t e o f a i r i s ” )16 disp ( ”K” ,T2 , ” S t a t i c t e m p e r a t u r e i s ” )17 disp ( ” kPa ” ,P2 , ” S t a t i c P r e s s u r e i s ” )18 disp ( ”m/ s ” ,V2 , ” V e l o c i t y a t t h e e x i t from t h e n o z z l e19202122232425262728293031i s ”)// P a r t ( b )Mb = 0.308;P2P0b = 0.936;T2T0b = 0.9812;P2b = P2P0b * P0 ;T2b = T2T0b * T0 ;c2b = sqrt ( g * R * T2b *1000) ;V2b = c2b * Mb ;disp ( ”When d i v e r g e n t s e c t i o n a c tdisp ( ” kg / s ” ,w , ”Maximum f l o w r a t edisp ( ”K” ,T2b , ” S t a t i c t e m p e r a t u r edisp ( ” kPa ” ,P2b , ” S t a t i c P r e s s u r edisp ( ”m/ s ” ,V2b , ” V e l o c i t y a t t h ei s ”)as a d i f f u s e r ”)of a i r i s ”)i s ”)i s ”)e x i t from t h e n o z z l eScilab code Exa 17.4 Calculations on pitot tube immersed in a supersonicflow123456789Px = 16 kPa ; Poy = 70 kPa ;Mx = 1.735; Pyx = 3.34; // Pyx = Py/Pxrho_yx = 2.25;Tyx = 1.483; Poyox = 0.84; My = 0.631;Tox = 573; Toy = Tox ;Tx = Tox /(1+(( g -1) /2) * Mx ^2) ;Ty = Tyx * Tx ;Pox = Poy / Poyox ;// From t a b l e12210 Mx = 1.735;11 disp ( Mx , ”Mach number o f t h e t u n n e li s ”)Scilab code Exa 17.5 Calculations on a CD nozzle operating at off designcondition12345678910111213141516171819202122Ax = 18.75; A_ = 12.50; // A = A∗AA_ = 1.5; // A/A∗Mx = 1.86; Pxox = 0.159; R = 0.287;Pox = 0.21 e03 ; // i n kPaPx = Pxox * Pox ;// from t h e g a s t a b l e on n o r m a l s h o c kMx = 1.86; My = 0.604; Pyx = 3.87; Poyx = 4.95;Poyox = 0.786;Py = Pyx * Px ;Poy = Poyx * Px ;My = 0.604;Ay_ = 1.183;A2 = 25; Ay = 18.75;A2_ = ( A2 / Ay ) * Ay_ ;// From i s e n t r o p i c t a b l eM2 = 0.402;P2oy = 0.895;P2 = P2oy * Poy ;syx = -R * log ( Poy / Pox ) ; // sy−s xdisp ( M2 , ” E x i t mach number i s M2” )disp ( ” kPa ” ,P2 , ” E x i t p r e s s u r e i s ” )disp ( ” kPa ” ,Pox - Poy , ” E x i t S t a g n a t i o n p r e s s u r e i s ” )disp ( ” kJ / kg K” ,syx , ” Entropy i n c r e a s e i s ” )Scilab code Exa 17.6 Calculations on expansion of air through a convergent nozzle1231234567891011g = 1.4; R = 0.287; d = 1.4; // d e lP0 = 1.4; // i n b a rT0 = 280; T1 = T0 ;cp = 1.005; A2 = 0.0013P_ = P0 /(( g +1) /2) ^( d /( d -1) ) ; // P = P∗P1 = P0 ; Pb = 1; P2 = Pb ;T2 = T1 *( P2 / P1 ) ^(( d -1) / d ) ;V2 = sqrt (2* cp *( T1 - T2 ) *1000) ;m_dot = ( A2 * V2 * P2 *100) /( R * T2 ) ;disp ( ” kg / s ” , m_dot , ” Mass f l o w r a t e i s ” )disp ( ” The mass f l o w r a t e can be i n c r e a s e d by r a i s i n gthe supply p r e s s u r e ”)Scilab code Exa 17.7 Calculations on an ideal gas undergoing a normalshock123456789101112Mx = 1.8; Pyx = 3.6133;Px = 0.5; Tx = 280; Ty = 429;Py = Pyx * Px ; cp = 1.005;disp ( ” b a r ” ,Py , ” P r e s s u r e Py i s ” )Pxox = 0.17404;Pox = Px / Pxox ;disp ( ” b a r ” ,Pox , ” S t a g n a t i o n p r e s s u r e i s ” )Txox = 0.60680;Tox = Tx / Txox ;disp ( ”K” ,Tox , ” S t a g n a t i o n t e m p e r a t u r e i s ” )sysx = cp * log ( Ty / Tx ) -R * log ( Py / Px ) ;disp ( ” kJ / kg K” ,syx , ” The c h a n g e i n s p e c i f i c e n t r o p yi s ”)124Chapter 18Gas compressorsScilab code Exa 18.1 Calculations on a single reciprocating compressor1234567891011121314151617181920T2 = 488; T1 = 298; n = 1.3; R =8314/44;rp = ( T2 / T1 ) ^( n /( n -1) ) ;disp ( rp , ” P r e s s u r e r a t i o i s ” )b = 0.12; // Bore o f c o m p r e s s o rL = 0.15; // S t r o k e o f c o m p r e s s o rV1 = ( %pi /4) *( b ) ^2* L ;P1 = 120 e03 ; // i n kPaW = (( n * P1 * V1 ) /( n -1) ) *((( rp ) ^(( n -1) / n ) ) -1) ;P = ( W *1200*0.001) /60 ;disp ( ”kW” ,P , ” I n d i c a t e d power i s ” )disp ( ”kW” ,P /0.8 , ” S h a f t power i s ” )V1_dot = V1 *(1200/60) ;m_dot = ( P1 * V1_dot ) /( R * T1 ) ;disp ( ” kg / s ” , m_dot , ” Mass f l o w r a t e i s ” )rp_1 = rp ^2;disp ( rp_1 , ” P r e s s u r e r a t i o when s e c o n d s t a g e i s addedi s ”)V2 = (1/ rp ) ^(1/ n ) * V1 ;disp ( ”m3” ,V2 , ” Volume d e r i v e d p e r c y c l e i s V2” )d = sqrt (( V2 *4) /( L * %pi ) ) ;disp ( ”mm” ,d *1000 , ” S e c o n d s t a g e b o r e would be ” )125Scilab code Exa 18.2 Calculations on a single reciprocating air compressor123456789101112131415161718192021P1 = 101.3 e03 ; P4 = P1 ; // i n PaP2 = 8* P1 ; P3 = P2 ;T1 = 288; Vs = 2000;V3 = 100; Vc = V3 ;V1 = Vs + Vc ;n = 1.25; R = 287;V4 = (( P3 / P4 ) ^(1/ n ) ) * V3 ;W = (( n * P1 *( V1 - V4 ) *1 e -06) /( n -1) ) *((( P2 / P1 ) ^(( n -1) / n )) -1) ;P = ( W *800*0.001) /60 ;disp ( ”kW” ,P , ” I n d i c a t e d p o e r i s ” )disp ( ”%” ,100*( V1 - V4 ) / Vs , ” V o l u m e t r i c e f f i c i e n c y i s ” )m = ( P1 *( V1 - V4 ) *1 e -06) /( R * T1 ) ;m_dot = m *800;disp ( ” kg / min ” , m_dot , ” Mass f l o w r a t e i s ” )FAD = ( V1 - V4 ) *1 e -06*800;disp ( ”m3/ min ” ,FAD , ” F r e e a i r d e l i v e r y i s ” )Wt = P1 *( V1 - V4 ) *1 e -06* log ( P2 / P1 ) ;n_isothermal = ( Wt *800*0.001) /( P *60) ;disp ( ”%” ,100* n_isothermal , ” I s o t h e r m a l e f f i c i e n c y i s ”)Pi = P /0.85;disp ( ”kW” ,Pi , ” I n p u t power i s ” )Scilab code Exa 18.3 Calculations on a two stage air compressor with perfect intercoolings1 P1 = 1; P3 = 9;1262345678P2 = sqrt ( P1 * P3 ) ;T1 = 300; cp = 1.005;R = 0.287; n = 1.3;W = ((2* n * R * T1 ) /( n -1) ) *(( P2 / P1 ) ^(( n -1) / n ) -1) ;T2 = T1 *( P2 / P1 ) ^(( n -1) / n ) ;H = cp *( T2 - T1 ) ;disp ( ” kJ / kg ” ,H , ” Heat r e j e c t e d t o t h e i n t e r c o o l e r)is ”Scilab code Exa 18.4 Calculations on a single acting two stage air compressors1 P1 = 1.013; P4 = 80;2 P2 = sqrt ( P1 * P4 ) ;3 V_dot = 4/60; // i n m3/ s4 n = 1.25;5 n_mech = 0.75;6 W_dot = ((2* n ) /( n -1) ) *(( P1 *100* V_dot ) / n_mech ) *(( P2 /78910111213141516P1 ) ^(( n -1) / n ) -1) ;N = 250;L = (3*60) /(2* N ) ; // S t r o k e l e n g t h o f p i s t o n i n mVlp = 4/ N ;n_vol = 0.8;Dlp = sqrt (( Vlp *4) /( n_vol * L * %pi ) ) ;Dhp = Dlp * sqrt ( P1 / P2 ) ;disp ( ”kW” , W_dot , ”Minimum power r e q u i r e d i s ” )disp ( ”cm” ,L *100 , ” S t r o k e o f t h e c o m p r e s s o r i s ” )disp ( ”cm” , Dhp *100 , ” Bore o f h i g h p r e s s u r e c o m p r e s s u r ei s ”)disp ( ”cm” , Dlp *100 , ” Bore o f l o p r e s s u r e c o m p r e s s u r ei s ”)Scilab code Exa 18.5 Determination of out put power of an air engine127V12 = 0.4; // V12 = V1/V2T1 = 38+273; n = 1.3; P3 = 112; // back p r e s s u r em = 1.25; R = 0.287;T2 = (( V12 ) ^( n -1) ) * T1 ;P1 = 700; // i n kPaP2 = P1 *( V12 ) ^ n ;V2 = ( m * R * T2 ) / P2 ;v2 = V2 / m ;A = R * T1 + R *( T1 - T2 ) /( n -1) - P3 * v2 ; // Area o fi n d i c a t o r diagram10 IO = A *0.85* m ;11 disp ( ” kJ ” ,IO , ” I n d i c a t e d o u t p u t i s ” )123456789Scilab code Exa 18.6 Calculations on a three stage acting reciprocatingair compressor123456789101112131415161718P1 = 1; P41 = 15; // P41 = P4/P1P21 = ( P41 ) ^(1/3) ;P2 = P21 * P1 ; n = 1.3; R = 0.287;P3 = P21 * P2 ;P11 = P2 ; P12 = P1 ;b = 0.45; s = 0.3; // Bore and s t r o k e o f c y l i n d e rVs = ( %pi /4) * b ^2* s ; // Swept volume o f t h e c y l i n d e rV11 = 0.05* Vs ; // C l e a r a n c e volumeV1 = V11 + Vs ;V12 = V11 *( P11 / P12 ) ^(1/ n ) ;disp ( ”m3” ,V1 - V12 , ” E f f e c t i v e s w e p t volume o f t h e LPc y l i n d e r i s ”)T1 = 291; T3 = T1 ; T5 = T1 ;P43 = P21 ; // P4/P3T6 = T5 *( P43 ) ^(( n -1) / n ) ;disp ( ”K” ,T6 , ” D e l i v e r y t e m p e r a t u r e i s ” )P4 = 15; // D e l i v e r y p r e s s u r eV6_7 = ( P1 / P4 ) *( T6 / T1 ) *( V1 - V12 ) ; // V6−V7disp ( ”m3” , V6_7 , ” Volume o f t h e a i r d e l i v e r e d ” )12819 W = ((3* n * R * T1 ) /( n -1) ) *(( P21 ) ^(( n -1) / n ) -1) ;20 disp ( ” kJ ” ,W , ”Work done p e r kg o f t h e a i r i s ” )Scilab code Exa 18.7 Determining the work input for a vane type compressor1234567891011P1 = 1.013;P2 = 1.5* P1 ;Vs = 0.03; Va = Vs ;WD = ( P2 - P1 ) * Vs *100;Pi = ( P1 + P2 ) /2;g = 1.4;Aa = (( g * P1 *100* Vs ) /( g -1) ) *(( Pi / P1 ) ^(( g -1) / g ) -1) ;Vb = Va *( P1 / Pi ) ^(1/ g ) ;Ab = Vb *( P2 - Pi ) *100;WR = Aa + Ab ;disp ( ” kJ / r e v ” ,WR , ”Work r e q u i r e d i s ” )Scilab code Exa 18.8 Determination of power required to drive the rootsblower123456789101112// For B l o w e rm_dot = 1; R = 0.287; T1 = 343;P1 = 100; P2 = 2* P1 ; g = 1.4;V_dot = ( m_dot * R * T1 ) / P1 ;PRb = V_dot *( P2 - P1 ) ;disp ( ”kW” ,PRb , ” Power r e q u i r e d by t h e b l o w e r i s ” )// For van c o m p r e s s o rP1v = 1; V21 = 0.7 // V2/V1P2v = P1v *(1/ V12 ) ^ g ;V2_dot = 0.7;V1_dot = 0.7* V_dot ;P3v = 2;12913 PRv = (( g * P1v *100* V_dot ) /( g -1) ) *(( P2v / P1v ) ^(( g -1) / g )14-1) + V1_dot *100*( P3v - P2v ) ;disp ( ”kW” ,PRv , ” Power R e q u i r e d by van c o m p r e s s o r i s ” )Scilab code Exa 18.9 Calculations on a gas turbine utilizing a two stagecentrifugal compressor12345678910T1 = 283; P21 = 2.5; // P2/P1P32 = 2.1; // P3/P2ns = 0.85; ma = 5; cp = 1.005;T2s = T1 *( P21 ) ^(( g -1) / g ) ;T2 = T1 + ( T2s - T1 ) / ns ;T3 = T2 -50;T4s = T3 *( P32 ) ^(( g -1) / g ) ;T4 = T3 + ( T4s - T3 ) / ns ;P = ma * cp *(( T2 - T1 ) +( T4 - T3 ) ) ;disp ( ”kW” ,P , ” T o t a l c o m p r e s s o r power i s ” )Scilab code Exa 18.10 Calculations on a rotatry compressorT1 = 278; P21 = 2.5; // P2/P1cp = 1.005; ns = 0.84; V2 = 120;T2s = T1 *( P21 ) ^(( g -1) / g ) ;T2 = T1 + ( T2s - T1 ) / ns ;mg = 0.04*(13+1) ;P = mg * cp *( T2 - T1 ) ;T02 = T2 + V2 ^2/(2* cp *1000) ;P1 = 0.6;P2 = P21 *0.6;P02 = P2 *( T02 / T2 ) ^( g /( g -1) ) ;disp ( ”kW” ,P , ” Power r e q u i r e d t o d r i v e t h e c o m p r e s s o ri s ”)12 disp ( ”K” ,T02 , ” S t a g n a t i o n t e m p e r a t u r e i s ” )123456789101113013disp ( ” b a r ” ,P02 , ” S t a g n a t i o n p r e s s u r e i s ” )131.
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