diff options
Diffstat (limited to '2969/CH10')
| -rwxr-xr-x | 2969/CH10/EX10.1/Ex10_1.sce | 22 | ||||
| -rwxr-xr-x | 2969/CH10/EX10.2/Ex10_2.sce | 19 | ||||
| -rwxr-xr-x | 2969/CH10/EX10.3/Ex10_3.sce | 27 | ||||
| -rwxr-xr-x | 2969/CH10/EX10.4/Ex10_4.sce | 36 | ||||
| -rwxr-xr-x | 2969/CH10/EX10.5/Ex10_5.sce | 30 | ||||
| -rwxr-xr-x | 2969/CH10/EX10.6/Ex10_6.sce | 32 |
6 files changed, 166 insertions, 0 deletions
diff --git a/2969/CH10/EX10.1/Ex10_1.sce b/2969/CH10/EX10.1/Ex10_1.sce new file mode 100755 index 000000000..4f2e1112f --- /dev/null +++ b/2969/CH10/EX10.1/Ex10_1.sce @@ -0,0 +1,22 @@ +clc
+clear
+//DATA GIVEN
+V1=1; //volume of air taken in m^3/mim
+p1=1.013; //intake pressure in bar
+T1=15+273; //intake temperature in K
+p2=7; //delivery pressure in bar
+t=1*60; //time in seconds
+//law of compression, pV^1.35=C
+n=1.35;
+R=287;
+
+m=p1*10^5*V1/R/T1; //mass of air delivered in kg/min
+
+//(T2/T1)=(p2/p1)^((n-1)/n);
+T2=T1*(p2/p1)^((n-1)/n); //delivery temp. T2 in K
+
+W=(n)/(n-1)*m*R*(T2-T1)/1000; //indicated work in kJ/min
+
+IP=W/t; //indicated power in kW
+
+printf(' The Indicated power, IP is: %1.2f kW. \n',IP);
diff --git a/2969/CH10/EX10.2/Ex10_2.sce b/2969/CH10/EX10.2/Ex10_2.sce new file mode 100755 index 000000000..eb4a7239e --- /dev/null +++ b/2969/CH10/EX10.2/Ex10_2.sce @@ -0,0 +1,19 @@ +clc
+clear
+//continued from Example 1
+//DATA GIVEN
+V=1; //volume dealt with per min at inlet in m^3/mim
+Vc=1/300; //volume drawn in per cycle, in m^3/cycle
+r=1.5; //stroke to bore ratio
+ETAc=0.85; //mechanical efficiency of the compressor
+ETAmt=0.90; //mechanical efficiency of the motor transmission
+
+//cylinder volume, Vc=(pi/4)D^2*L
+D=[(Vc*4/%pi)/r]^(1/3); //bore in m
+
+//from example 1
+Pi=4.23/ETAc; //power input to the compressor in kW
+MP=Pi/ETAmt; //motor power in kW
+
+printf(' (i) The Cylinder bore,D is: %3.1f mm. \n',(D*1000));
+printf(' (ii) The Motor power is: %1.2f kW. \n',MP);
diff --git a/2969/CH10/EX10.3/Ex10_3.sce b/2969/CH10/EX10.3/Ex10_3.sce new file mode 100755 index 000000000..4f4047596 --- /dev/null +++ b/2969/CH10/EX10.3/Ex10_3.sce @@ -0,0 +1,27 @@ +clc
+clear
+//DATA GIVEN
+T1=20+273; //temperature in K
+p1=1; //pressure in bar
+p2=10; //pressure in bar
+Cv=0.718; //in kJ/kgK
+
+//law of compression, pV^1.2=C
+n=1.2;
+R=0.287; //in kJ/kgK
+
+//(T2/T1)=(p2/p1)^((n-1)/n);
+T2=T1*(p2/p1)^((n-1)/n); //temp. T2 in K
+m=1;
+W=(n)/(n-1)*m*R*T1*[(p2/p1)^((n-1)/n)-1]; //work done per kg of air (kJ/kg of air)
+
+//By the First Law of Thermodynamics
+//heat transferred during compression, Q=W+DU
+//Q=(p1V1-p2V2)/(n-1)+Cv(T2-T1)
+//Q=(T2-T1)*[Cv-R/(n-1)]
+Q=(T2-T1)*[Cv-R/(n-1)];
+
+printf(' (i) The Temperature at the end of compression is: %3.0f K or %3.0f deg. celsius. \n',T2,(T2-273));
+printf(' (ii) The Work done during compression per kg of air is: %3.2f kJ/kg of air. \n',W);
+printf(' The Heat transferred during compression per kg of air is: %2.2f kJ/kg of air. \n',Q);
+printf(' (Negative sign indicates heat REJECTION.) \n');
diff --git a/2969/CH10/EX10.4/Ex10_4.sce b/2969/CH10/EX10.4/Ex10_4.sce new file mode 100755 index 000000000..fb9dd0d51 --- /dev/null +++ b/2969/CH10/EX10.4/Ex10_4.sce @@ -0,0 +1,36 @@ +clc
+clear
+//DATA GIVEN
+p1=1; //suction pressure in bar
+T1=20+273; //suction temperature in K
+p2=6; //discharge pressure in bar
+T2=180+273; //discharge temperature in K
+N=1200; //speed of compressor in R.P.M.
+Pshaft=6.25; //shaft power in kW
+Ma=1.7; //mass of air delivered in kg/min
+D=0.14; //diameter in m
+L=0.1; //stroke in m
+R=287; //in kJ/kgK
+
+Vd=(%pi/4)*D^2*L*N; //displacement volume for single acting compressor in m^3/min
+FAD=Ma*R*T1/p1/10^5; //m^3/min
+ETAvol=FAD/Vd*100; //actual volumetric efficiency
+
+//(T2/T1)=(p2/p1)^((n-1)/n);
+n=1/[1-(log(T2/T1)/log(p2/p1))]; //index of compression,n
+
+IP=(n)/(n-1)*Ma/60*R/1000*T1*[(p2/p1)^((n-1)/n)-1]; //indicated power in kW
+
+Piso=Ma/60*R/1000*T1*log(p2/p1); //isothermal power
+ETAiso=Piso/IP*100; //isothermal efficiency
+
+ETAmech=IP/Pshaft*100; //mechanical efficiency
+
+ETAovr_iso=Piso/Pshaft*100; //overall isothermal eddiciency
+
+printf(' (i) The actual Volumetric efficiency is: %2.2f percent. \n',ETAvol);
+printf(' (ii) The Indicated Power, IP is: %1.3f KW. \n',IP);
+printf('(iii) The Isothermal efficiency is: %2.2f percent.\n',ETAiso);
+printf(' (iv) The Mechanical efficiency is: %2.1f percent.\n',ETAmech);
+printf(' (v) The Overall isothermal efficiency is: %2.1f percent.\n',ETAovr_iso);
+
diff --git a/2969/CH10/EX10.5/Ex10_5.sce b/2969/CH10/EX10.5/Ex10_5.sce new file mode 100755 index 000000000..b0b8d0d6c --- /dev/null +++ b/2969/CH10/EX10.5/Ex10_5.sce @@ -0,0 +1,30 @@ +//5(b) is as follows:
+clc
+clear
+//DATA GIVEN
+m=6.75; //mass of air in kg/min
+p1=1; //pressure in bar
+T1=21+273; //temp. in K
+p2=1.35; //pressure in bar
+T2=43+273; //temp. in K
+DTcw=3.3; //temp. rise of cooling water in deg. celsius
+Cp=1.003; //Cp for air in kJ/kgK
+//gamma for air=1.4
+g=1.4;
+
+W=m*Cp*(T2-T1); //work in kJ/min
+//If the compression would have been isotropic,
+//T_2=T1*(rp)^[(g-1)/g]
+rp=p2/p1;
+T_2=T1*(rp)^[(g-1)/g];
+Qr=m*Cp*(T_2-T2); //heat rejected to cooling water
+
+Mw=Qr/[4.18*(DTcw)]; //mass of cooling water in kg/min
+
+printf(' (i) The Work is: %3.2f kJ/min. \n',W);
+printf(' (ii) The Mass of cooling water is: %1.2f kg/min. \n',Mw);
+
+//NOTE:
+//in the question compression process is mentioned and p2 is given as 0.35 bar (p2<p1)
+//which is wrong and further p2 is given as 1.35 bar which is allowable
+//so here value of p2 is taken as 1.35 bar.
diff --git a/2969/CH10/EX10.6/Ex10_6.sce b/2969/CH10/EX10.6/Ex10_6.sce new file mode 100755 index 000000000..fd2f4c20f --- /dev/null +++ b/2969/CH10/EX10.6/Ex10_6.sce @@ -0,0 +1,32 @@ +clc
+clear
+//DATA GIVEN
+V1=14; //quantity of air to be delivered, in m^3/mim
+p1=1.013; //intake pressure in bar
+T1=15+273; //intake temperature in K
+p2=7; //delivery pressure in bar
+N=300; //speed of compressor in R.P.M.
+n=1.3; //compression and expansion index
+R=0.287;
+
+//clearance volume, Vc = 0.05 Vs ,Vs=swept volume
+//swept volume Vs=V1-V3=V1-Vc=V1-0.05Vs
+//V1=1.05Vs
+Vpc=V1/N/2; //(V1-V4) volume induced per cycle in m^3
+//V4/V3=(p2/p1)^(1/n)
+c=(p2/p1)^(1/n);
+//V4=c*V3=c*0.05Vs
+//V1-V4=1.05Vs-c*0.05Vs
+Vs=Vpc/(1.05)/(1.05-c*0.05); //volume swept in m^3
+
+//using relation (T2/T1)=(p2/p1)^((n-1)/n);
+T2=T1*(p2/p1)^((n-1)/n); //delivery temp. T2 in K
+
+IP=(n)/(n-1)*p1*10^5*Vpc/100*[(p2/p1)^((n-1)/n)-1]; //indicated power in kW
+
+printf(' (i) The Swept volume of the cylinder, Vs is: %1.4f m^3. \n',Vs);
+printf(' (ii) The delivery temperature, Ts is: %3.0f deg. celsius. \n',(T2-273));
+printf('(iii) The Indicated power, IP is: %2.2f kW. \n',IP);
+
+//NOTE:
+//there is slight variation in answers in textbook due to rounding off of values in book
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