diff options
Diffstat (limited to '914/CH7')
| -rwxr-xr-x | 914/CH7/EX7.10/ex7_10.sce | 30 | ||||
| -rwxr-xr-x | 914/CH7/EX7.11/ex7_11.sce | 19 | ||||
| -rwxr-xr-x | 914/CH7/EX7.12/ex7_12.sce | 25 | ||||
| -rwxr-xr-x | 914/CH7/EX7.13/ex7_13.sce | 24 | ||||
| -rwxr-xr-x | 914/CH7/EX7.14/ex7_14.sce | 15 | ||||
| -rwxr-xr-x | 914/CH7/EX7.15/ex7_15.sce | 20 | ||||
| -rwxr-xr-x | 914/CH7/EX7.16/ex7_16.sce | 17 | ||||
| -rwxr-xr-x | 914/CH7/EX7.18/ex7_18.sce | 22 | ||||
| -rwxr-xr-x | 914/CH7/EX7.19/ex7_19.sce | 18 | ||||
| -rwxr-xr-x | 914/CH7/EX7.2/ex7_2.sce | 17 | ||||
| -rwxr-xr-x | 914/CH7/EX7.20/ex7_20.sce | 13 | ||||
| -rwxr-xr-x | 914/CH7/EX7.21/ex7_21.sce | 14 | ||||
| -rwxr-xr-x | 914/CH7/EX7.22/ex7_22.sce | 15 | ||||
| -rwxr-xr-x | 914/CH7/EX7.3/ex7_3.sce | 40 | ||||
| -rwxr-xr-x | 914/CH7/EX7.4/ex7_4.sce | 34 | ||||
| -rwxr-xr-x | 914/CH7/EX7.5/ex7_5.sce | 12 | ||||
| -rwxr-xr-x | 914/CH7/EX7.6/ex7_6.sce | 41 | ||||
| -rwxr-xr-x | 914/CH7/EX7.7/ex7_7.sce | 15 | ||||
| -rwxr-xr-x | 914/CH7/EX7.8/ex7_8.sce | 18 |
19 files changed, 409 insertions, 0 deletions
diff --git a/914/CH7/EX7.10/ex7_10.sce b/914/CH7/EX7.10/ex7_10.sce new file mode 100755 index 000000000..eb0a0a1a5 --- /dev/null +++ b/914/CH7/EX7.10/ex7_10.sce @@ -0,0 +1,30 @@ +clc;
+warning("off");
+printf("\n\n example7.10 - pg298");
+// given
+d=0.03; //[m] - diameter
+g=9.784; //[m/sec] - acceleration due to gravity
+deltaz=-1;
+// using the equation (1/2)*(U3^2/alpha3-U1^2/alpha1)+g*deltaz=0
+// assuming
+alpha1=1;
+alpha3=1;
+// also since the diameter of the tank far exceeds the diameter of the hose , the velocity at point 1 must be negligible when compared to the velocity at point 3
+U1=0;
+U3=(-2*g*deltaz+(U1^2)/alpha1)^(1/2);
+p=1000; //[kg/m^3] - density of water
+S3=(%pi/4)*(d)^2
+w=p*U3*S3;
+printf("\n\n the mass flow rate is \n w=%fkg/sec",w);
+// the minimum pressure in the siphon tube is at the point 2. Before the result of 3.13 kg/sec is accepted as the final value, the pressure at point 2 must be calcilated in order to see if the water might boil at this point
+// using deltap=p*((U3^2)/2+g*deltaz)
+deltap=p*((U3^2)/2+g*deltaz);
+p1=1.01325*10^5; //[N/m^2] - is equal to atmospheric pressure
+p2=p1+deltap;
+vp=0.02336*10^5;
+if p2>vp then
+ printf("\n\n the siphon can operate since the pressure at point 2 is greater than the value at which the liquid boils");
+else
+ printf("\n\n the siphon cant operate since the pressuer at point 2 is less than the value at which the liquid boils");
+
+end
\ No newline at end of file diff --git a/914/CH7/EX7.11/ex7_11.sce b/914/CH7/EX7.11/ex7_11.sce new file mode 100755 index 000000000..4cd104c97 --- /dev/null +++ b/914/CH7/EX7.11/ex7_11.sce @@ -0,0 +1,19 @@ +clc;
+warning("off");
+printf("\n\n example7.11 - pg300");
+// given
+sp=1.45; // specific gravity of trichloroethylene
+pwater=62.4; //[lb/ft^3] - density of water
+p=sp*pwater;
+d1=1.049; //[inch] - density of pipe at point 1
+d2=0.6; //[inch] - density of pipe at point 2
+d3=1.049; //[inch] - density of pipe at point 3
+// using the formula U1*S1=U2*S2; we get U1=U2*(d2/d1);
+// then using the bernoulli equation deltap/p=(1/2)*(U2^2-U1^2);
+deltap=4.2*(144); //[lb/ft^2] - pressure difference
+U2=((2*(deltap/p)*(1/(1-(d2/d1)^4)))^(1/2))*(32.174)^(1/2);
+// using the formula w=p*U2*S
+w=p*U2*((%pi/4)*(0.6/12)^2);
+w1=w/(2.20462);
+printf("\n\n the mass flow rate is \n w=%flb/sec\n or in SI units \n w=%fkg/sec",w,w1);
+
diff --git a/914/CH7/EX7.12/ex7_12.sce b/914/CH7/EX7.12/ex7_12.sce new file mode 100755 index 000000000..652da6140 --- /dev/null +++ b/914/CH7/EX7.12/ex7_12.sce @@ -0,0 +1,25 @@ +clc;
+warning("off");
+printf("\n\n example7.12 - pg301");
+// given
+Q=50/(7.48*60); //[ft/sec] - volumetric flow rate of water
+d1=1; //[inch] - diameter of pipe
+deltaz=-5; //[ft] - distance between end of pipe and tank
+g=32.1; //[ft/sec] - acceleration due to gravity
+Cp=1; //[Btu/lb*F] - heat capacity of water
+p=62.4; //[lb/ft^3] - density of water
+S1=(%pi/4)*(d1/12)^2;
+U1=Q/S1;
+w=p*Q;
+U2=0;
+gc=32.174;
+// using the formula deltaH=(w/2)*((U2)^2-(U1)^2)+w*g*deltaz
+deltaH=-(w/(2*gc))*((U2)^2-(U1)^2)-w*(g/gc)*deltaz;
+disp(deltaH);
+deltaH=deltaH/778; // converting from ftlb/sec to Btu/sec
+deltaT=deltaH/(w*Cp);
+printf("\n\n The rise in temperature is %fdegF",deltaT);
+
+
+
+
diff --git a/914/CH7/EX7.13/ex7_13.sce b/914/CH7/EX7.13/ex7_13.sce new file mode 100755 index 000000000..08775f3e2 --- /dev/null +++ b/914/CH7/EX7.13/ex7_13.sce @@ -0,0 +1,24 @@ +clc;
+warning("off");
+printf("\n\n example7.13 - pg303");
+// given
+deltaz=30; //[ft] - distance between process and the holding tank
+Q=100; //[gpm] - volumetric flow rate of water
+p1=100; //[psig]
+p2=0; //[psig]
+g=32.1; //[ft/sec] - acceleration due to gravity
+sv=0.0161; //[ft^3/lb] - specific volume of water
+p=1/sv; //[lb/ft^3] - density of water
+e=0.77; // efficiency of centrifugal pump
+deltap=(p1-p2)*(144); //[lbf/ft^2]
+gc=32.174;
+// using the equation deltap/p+g*(deltaz)+Ws=0;
+Wst=-deltap/p-(g/gc)*(deltaz);
+// using the formula for efficiency e=Ws(theoritical)/Ws(actual)
+// therefore
+Wsa=Wst/e;
+// the calulated shaft work is for a unit mass flow rate of water,therfore for given flow rate multiply it by the flow rate
+w=(Q*p)/(7.48*60);
+Wsactual=Wsa*w;
+power=-Wsactual/(778*0.7070);
+printf("\n\n the required horsepower is %fhp",power);
diff --git a/914/CH7/EX7.14/ex7_14.sce b/914/CH7/EX7.14/ex7_14.sce new file mode 100755 index 000000000..e438dc824 --- /dev/null +++ b/914/CH7/EX7.14/ex7_14.sce @@ -0,0 +1,15 @@ +clc;
+warning("off");
+printf("\n\n example7.14 - pg304");
+// given
+p1=5; //[atm] - initial pressure
+p2=0.75; //[atm] - final pressure after expansion through turbine
+T=450; //[K] - temperature
+y=1.4; // cp/cv for nitrogen
+// using the equation Ws=-(y/(y-1))*(p1/density1)*((p2/p1)^((y-1)/y)-1)
+R=8314; // gas constant
+p1bydensity=R*T;
+Ws=-(y/(y-1))*(p1bydensity)*((p2/p1)^((y-1)/y)-1);
+printf("\n\n the shaft work of the gas as it expands through the turbine and transmits its molecular energy to the rotating blades is \n Ws=%eJ/kmol",Ws);
+
+
diff --git a/914/CH7/EX7.15/ex7_15.sce b/914/CH7/EX7.15/ex7_15.sce new file mode 100755 index 000000000..a08d5cc82 --- /dev/null +++ b/914/CH7/EX7.15/ex7_15.sce @@ -0,0 +1,20 @@ +clc;
+warning("off");
+printf("\n\n example 7.15 - pg311");
+// given
+T=273.15+25; //[K] - temperature
+R=8.314; //[kPa*m^3/kmol*K] - gas constant
+p=101.325; //[kPa] - pressure
+M=29; // molecular weight of gas
+pa=(p*M)/(R*T);
+sg=13.45; // specific gravity
+pm=sg*1000;
+g=9.807; //[m/sec^2] - acceleration due to gravity
+deltaz=15/100; //[m]
+// using the equation p2-p1=deltap=(pm-pa)*g*deltaz
+deltap=-(pm-pa)*g*deltaz;
+printf("\n\n the pressure drop is %eN/m^2",deltap);
+printf("\n the minus sign means the upstream pressure p1 is greater than p2, i.e ther is a pressure drop.");
+
+
+
diff --git a/914/CH7/EX7.16/ex7_16.sce b/914/CH7/EX7.16/ex7_16.sce new file mode 100755 index 000000000..d2bdeec15 --- /dev/null +++ b/914/CH7/EX7.16/ex7_16.sce @@ -0,0 +1,17 @@ +clc;
+warning("off");
+printf("\n\n example7.16 - pg312");
+// given
+T=536.67; //[degR]; - temperature
+R=10.73; //[(lbf/in^2*ft^3)*lb*mol^-1*degR] - gas constant
+p=14.696; //[lbf/in^2];
+g=9.807*3.2808; //[ft/sec^2] - acceleration due to gravity
+M=29; // molecular weight of gas
+pa=(p*M)/(R*T);
+sg=13.45; // specific gravity
+pm=sg*62.4;
+deltaz=15/(2.54*12); //[ft]
+gc=32.174;
+// using the equation p2-p1=deltap=(pm-pa)*g*deltaz
+deltap=(pm-pa)*(g/gc)*deltaz;
+printf("\n\n the pressure drop is %flbf/ft^2",deltap);
diff --git a/914/CH7/EX7.18/ex7_18.sce b/914/CH7/EX7.18/ex7_18.sce new file mode 100755 index 000000000..4c574e2fe --- /dev/null +++ b/914/CH7/EX7.18/ex7_18.sce @@ -0,0 +1,22 @@ +clc;
+warning("off");
+printf("\n\n example7.18 - pg315");
+// given
+at=0.049; //[in^2] - cross sectional area of the manometer tubing
+aw=15.5; //[in^2] - cross sectional area of the well
+g=32.174; //[ft/sec^2] - acceleration due to gravity
+gc=32.174;
+sg=13.45; //[ specific garvity of mercury
+p=62.4; //[lb/ft^3] - density of water;
+pm=sg*p;
+deltaz_waterleg=45.2213;
+// using the equation A(well)*deltaz(well)=A(tube)*deltaz(tube)
+deltazt=70; //[cm]
+deltazw=deltazt*(at/aw);
+deltaz=deltazt+deltazw;
+deltap_Hg=-pm*(g/gc)*(deltaz/(2.54*12));
+disp(deltap_Hg);
+deltazw=45.2213; //[cm]
+deltap_tap=deltap_Hg+p*(g/gc)*(deltazw/(12*2.54));
+printf("\n\n deltap_tap=%f lbf/ft^2",deltap_tap);
+printf("\ndeltap is negative and therefore p1 is greater than p2");
diff --git a/914/CH7/EX7.19/ex7_19.sce b/914/CH7/EX7.19/ex7_19.sce new file mode 100755 index 000000000..fb01e42e0 --- /dev/null +++ b/914/CH7/EX7.19/ex7_19.sce @@ -0,0 +1,18 @@ +clc;
+warning("off");
+printf("\n\n example7_19 - pg317");
+// given
+p=749/760; //[atm]
+T=21+273.15; //[K]
+R=82.06; //[atm*cm^3/K] - gas constant
+v=(R*T)/p; //[cm^3/mole] - molar volume
+M=29; //[g/mole] - molecular weight
+pair=M/v;
+m_air=53.32; //[g]
+m_h2o=50.22; //[g]
+ph2o=0.998; //[g/cm^3] - density of water
+V=(m_air-m_h2o)/(ph2o-pair); //[cm^3]
+density=m_air/V;
+printf("\n\n The density of coin is \n density=%fg/cm^3",density);
+printf("\n\n Consulting a handbook it is seen that this result is correct density for gold");
+
diff --git a/914/CH7/EX7.2/ex7_2.sce b/914/CH7/EX7.2/ex7_2.sce new file mode 100755 index 000000000..98d3fd3eb --- /dev/null +++ b/914/CH7/EX7.2/ex7_2.sce @@ -0,0 +1,17 @@ +clc;
+warning("off");
+printf("\n\n example7.2 - pg273");
+// given
+id=4; //[m] - inside diameter
+h=2; //[m] - water level
+ro=0.03; //[m] - radius of exit hole
+rt=id/2; //[m] - inside radius
+g=9.80665; //[m/sec^2] - gravitational acceleration
+// using the equation dh/h^(1/2)=-((ro^2)/(rt^2))*(2*g)^(1/2)dt and integrating between h=2 and h=1
+t1=integrate('(1/h^(1/2))*(1/(-((ro^2)/(rt^2))*(2*g)^(1/2)))','h',2,1);
+printf("\n\n Time required to remove half of the contents of the tank is \n t=%fsec=%fmin",t1,t1/60);
+//integrating between h=2 and h=0
+t2=integrate('(1/h^(1/2))*(1/(-((ro^2)/(rt^2))*(2*g)^(1/2)))','h',2,0);
+printf("\n\n Time required to empty the tank fully is \n t=%fsec=%fmin",t2,t2/60);
+
+
diff --git a/914/CH7/EX7.20/ex7_20.sce b/914/CH7/EX7.20/ex7_20.sce new file mode 100755 index 000000000..bf2ad597d --- /dev/null +++ b/914/CH7/EX7.20/ex7_20.sce @@ -0,0 +1,13 @@ +clc;
+warning("off");
+printf("\n\n example7.20 - pg318");
+// given
+P=749/760; //[atm] - pressure
+T=21+273.15; //[K] - temperature
+poak=38*(1/62.4); //[g/cm^3] - density of oak
+pbrass=534/62.4; //[g/cm^3] - density of brass
+m_brass=6.7348; //[g]
+pair=0.001184; //[g/cm^3] - density of air
+// using the formula m_oak=m_brass*((1-(pair/pbrass))/(1-(pair/poak)))
+m_oak=m_brass*((1-(pair/pbrass))/(1-(pair/poak)));
+printf("\n\n m_oak=%fg",m_oak);
diff --git a/914/CH7/EX7.21/ex7_21.sce b/914/CH7/EX7.21/ex7_21.sce new file mode 100755 index 000000000..c07fdb8ce --- /dev/null +++ b/914/CH7/EX7.21/ex7_21.sce @@ -0,0 +1,14 @@ +clc;
+warning("off");
+printf("\n\n example7.21 - pg320");
+// given
+T=545.67; //[degR] - temperature
+R=1545; //[Torr*ft^3/degR*mole] - gas constant
+M=29; //[g/mole] - molecular weight
+g=9.807; //[m/sec^2] - acceleration due to gravity
+gc=9.807;
+po=760; //[Torr] - pressure
+deltaz=50; //[ft]
+// using the equation p=po*exp(-(g/gc)*M*(deltaz/R*T))
+p=po*%e^(-(g/gc)*M*(deltaz/(R*T)));
+printf("\n\n p=%fTorr\n Thus, the pressure decrease for an elevation of 50ft is very small",p);
diff --git a/914/CH7/EX7.22/ex7_22.sce b/914/CH7/EX7.22/ex7_22.sce new file mode 100755 index 000000000..58cd4271f --- /dev/null +++ b/914/CH7/EX7.22/ex7_22.sce @@ -0,0 +1,15 @@ +clc;
+warning("off");
+printf("\n\n example7.22 - pg321");
+// given
+To=545.67; //[degR] - air temperature at beach level
+betaa=-0.00357; //[degR/ft] - constant
+R=1545; //[Torr*ft^3/degR*mole] - gas constant
+M=29;
+deltaz=25000; //[ft]
+// using the equation ln(p/po)=((M)/(R*betaa))*ln(To/(To+betaa*deltaz)
+p=po*exp(((M)/(R*betaa))*log(To/(To+betaa*deltaz)));
+printf("\n\n p=%fTorr",p);
+// using the equation T=To+betaa*deltaz
+T=To+betaa*deltaz;
+printf("\n\n T=%fdegR",T);
diff --git a/914/CH7/EX7.3/ex7_3.sce b/914/CH7/EX7.3/ex7_3.sce new file mode 100755 index 000000000..3001e6e85 --- /dev/null +++ b/914/CH7/EX7.3/ex7_3.sce @@ -0,0 +1,40 @@ +clc;
+warning("off");
+printf("\n\n example7.3 - pg274");
+// given
+// composition of fuel gas
+nH2=24;
+nN2=0.5;
+nCO=5.9;
+nH2S=1.5;
+nC2H4=0.1;
+nC2H6=1;
+nCH4=64;
+nCO2=3.0;
+// calculating the theoritical amount of O2 required
+nO2theoreq=12+2.95+2.25+0.30+3.50+128;
+// since fuel gas is burned with 40% excess O2,then O2 required is
+nO2req=1.4*nO2theoreq;
+nair=nO2req/0.21; // as amount of O2 in air is 21%
+nN2air=nair*(0.79); // as amount of N2 in air is 79%
+nN2=nN2+nN2air;
+nO2=nO2req-nO2theoreq;
+nH2O=24+0+0.2+3.0+128;
+nCO2formed=72.1;
+nCO2=nCO2+nCO2formed;
+nSO2=1.5;
+ntotal=nSO2+nCO2+nO2+nN2+nH2O;
+mpSO2=(nSO2/ntotal)*100;
+mpCO2=(nCO2/ntotal)*100;
+mpO2=(nO2/ntotal)*100;
+mpN2=(nN2/ntotal)*100;
+mpH2O=(nH2O/ntotal)*100;
+printf("\n\n gas N2 O2 H2O CO2 SO2");
+printf("\n\n moles %f %f %f %f %f",nN2,nO2,nH2O,nCO2,nSO2);
+printf("\n\n mole percent %f %f %f %f %f",mpN2,mpO2,mpH2O,mpCO2,mpSO2);
+
+
+
+
+
+
diff --git a/914/CH7/EX7.4/ex7_4.sce b/914/CH7/EX7.4/ex7_4.sce new file mode 100755 index 000000000..3e8a75af9 --- /dev/null +++ b/914/CH7/EX7.4/ex7_4.sce @@ -0,0 +1,34 @@ +clc;
+warning("off");
+printf("\n\n example7.4 - pg280");
+// given
+id=6; //[inch] - inlet diameter
+od=4; //[inch] - outlet diameter
+Q=10; //[ft^3/sec] - water flow rate
+alpha2=%pi/3; //[radians] - angle of reduction of elbow
+alpha1=0;
+p1=100; //[psi] - absolute inlet pressure
+p2=29; //[psi] - absolute outlet pressure
+S1=(%pi*((id/12)^2))/4;
+S2=(%pi*((od/12)^2))/4;
+U1=Q/S1;
+U2=Q/S2;
+mu=6.72*10^-4; //[lb*ft^-1*sec^-1]
+p=62.4; //[lb/ft^3]
+Nrei=((id/12)*U1*p)/(mu);
+disp(Nrei,"Nre(inlet)=");
+Nreo=((od/12)*U2*p)/(mu);
+disp(Nreo,"Nre(outlet)=");
+// thus
+b=1;
+w1=p*Q; //[lb/sec] - mass flow rate
+w2=w1;
+gc=32.174;
+// using the equation (w/gc)*((U1)*(cos(alpha1))-(U2)*(cos(alpha2)))+p1*S1*cos(alpha1)-p2*S2*cos(alpha2)+Fextx=0;
+Fextx=-(w1/gc)*((U1)*(cos(alpha1))-(U2)*(cos(alpha2)))-p1*144*S1*cos(alpha1)+p2*144*S2*cos(alpha2);
+disp(Fextx,"Fext,x=");
+Fexty=-(w1/gc)*((U1)*(sin(alpha1))-(U2)*(sin(alpha2)))-p1*144*S1*sin(alpha1)+p2*144*S2*sin(alpha2);
+disp(Fexty,"Fext,y=");
+printf("\n\n the forces Fxt,x and Fext,y are the forces exerted on the fluid by the elbow.Fext,x acts to the left and Fext,y acts in the positive y direction.Note that the elbow is horizantal,and gravity acts in the z direction");
+
+
diff --git a/914/CH7/EX7.5/ex7_5.sce b/914/CH7/EX7.5/ex7_5.sce new file mode 100755 index 000000000..3a9a937c6 --- /dev/null +++ b/914/CH7/EX7.5/ex7_5.sce @@ -0,0 +1,12 @@ +clc;
+warning("off");
+printf("\n\n example7.5 - pg 282");
+// given
+Fextx=-2522; //[lb] - force in x direction
+Fexty=2240; //[lb] - force in y direction
+// the force exerted by the elbow on the fluid is the resolution of Fext,x and Fext,y , therefore
+Fext=((Fextx)^2+(Fexty)^2)^(1/2);
+alpha=180+(atan(Fexty/Fextx))*(180/%pi);
+printf("\n\n the force has a magnitude of %flb and a direction of %f from the positive x direction(in the second quadrant",Fext,alpha);
+
+
diff --git a/914/CH7/EX7.6/ex7_6.sce b/914/CH7/EX7.6/ex7_6.sce new file mode 100755 index 000000000..cd6b07980 --- /dev/null +++ b/914/CH7/EX7.6/ex7_6.sce @@ -0,0 +1,41 @@ +clc;
+warning("off");
+printf("\n\n example7.6 - pg283");
+// given
+id=6; //[inch] - inlet diameter
+od=4; //[inch] - outlet diameter
+Q=10; //[ft^3/sec] - water flow rate
+alpha2=%pi/3; //[radians] - angle of reduction of elbow
+alpha1=0;
+p1=100; //[psi] - absolute inlet pressure
+p2=29; //[psi] - absolute outlet pressure
+patm=14.7; //[psi] - atmospheric pressure
+p1gauge=p1-patm;
+p2gauge=p2-patm;
+S1=(%pi*((id/12)^2))/4;
+S2=(%pi*((od/12)^2))/4;
+U1=Q/S1;
+U2=Q/S2;
+p=62.4; //[lb/ft^3]
+b=1;
+w1=p*Q; //[lb/sec] - mass flow rate
+w2=w1;
+gc=32.174;
+// using the equation Fpress=p1gauge*S1-p2gauge*S2*cos(alpha2);
+Fpressx=p1gauge*144*S1-p2gauge*144*S2*cos(alpha2);
+Fpressy=p1gauge*144*S1*sin(alpha1)-p2gauge*144*S2*sin(alpha2);
+wdeltaUx=(w1/gc)*((U2)*(cos(alpha2))-(U1)*(cos(alpha1)));
+wdeltaUy=(w1/gc)*((U2)*(sin(alpha2))-(U1)*(sin(alpha1)));
+Fextx=wdeltaUx-Fpressx;
+Fexty=wdeltaUy-Fpressy;
+Fext=((Fextx)^2+(Fexty)^2)^(1/2);
+alpha=180+(atan(Fexty/Fextx))*(180/%pi);
+printf("\n\n The force has a magnitude of %flb and a direction of %f from the positive x direction(in the second quadrant",Fext,alpha);
+printf("\n\n Also there is a force on the elbow in the z direction owing to the weight of the elbow plus the weight of the fluid inside");
+
+
+
+
+
+
+
diff --git a/914/CH7/EX7.7/ex7_7.sce b/914/CH7/EX7.7/ex7_7.sce new file mode 100755 index 000000000..c89951ff7 --- /dev/null +++ b/914/CH7/EX7.7/ex7_7.sce @@ -0,0 +1,15 @@ +clc;
+warning("off");
+printf("\n\n example7.7 - pg293");
+// given
+Uo=1; //[m/sec]
+// using Ux/Uo=y/yo
+// assuming any particular value of yo will not change the answer,therefore
+yo=1;
+Uxavg=integrate('(Uo*y)/yo','y',0,yo);
+Ux3avg=integrate('((Uo*y)/yo)^3','y',0,yo);
+// using the formula alpha=(Uxavg)^3/Ux3avg
+alpha=(Uxavg)^3/Ux3avg;
+disp(alpha,"alpha=");
+printf("\n\n Note that the kinetic correction factor alpha has the same final value for laminar pipe flow as it has for laminar flow between parallel plates.");
+
diff --git a/914/CH7/EX7.8/ex7_8.sce b/914/CH7/EX7.8/ex7_8.sce new file mode 100755 index 000000000..95f469c47 --- /dev/null +++ b/914/CH7/EX7.8/ex7_8.sce @@ -0,0 +1,18 @@ +clc;
+warning("off");
+printf("\n\n example7.8 - pg293");
+// given
+Q=0.03; //[m^3/sec] - volumetric flow rate
+id=7; //[cm] - inside diameter
+deltaz=-7; //[m] - length of pipe
+T1=25; //[degC] - lowere side temperature
+T2=45; //[degC] - higher side temperature
+g=9.81; //[m/sec^2] - acceleration due to gravity
+deltaP=4*10^4; //[N/m^2] - pressure loss due to friction
+p=1000; //[kg/m^3] - density of water
+w=Q*p;
+C=4184; //[J/kg*K) - heat capacity of water
+deltaH=w*C*(T2-T1);
+// using the formula Qh=deltaH+w*g*deltaz
+Qh=deltaH+w*g*deltaz;
+printf("\n\n the duty on heat exchanger is \n Q=%6eJ/sec",Qh);
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