diff options
Diffstat (limited to '2657/CH1')
-rwxr-xr-x | 2657/CH1/EX1.1/Ex1_1.sce | 15 | ||||
-rwxr-xr-x | 2657/CH1/EX1.2/Ex1_2.sce | 12 | ||||
-rwxr-xr-x | 2657/CH1/EX1.3/Ex1_3.sce | 17 | ||||
-rwxr-xr-x | 2657/CH1/EX1.4/Ex1_4.sce | 26 | ||||
-rwxr-xr-x | 2657/CH1/EX1.5/Ex1_5.sce | 20 | ||||
-rwxr-xr-x | 2657/CH1/EX1.6/Ex1_6.sce | 20 | ||||
-rwxr-xr-x | 2657/CH1/EX1.7/Ex1_7.sce | 24 |
7 files changed, 134 insertions, 0 deletions
diff --git a/2657/CH1/EX1.1/Ex1_1.sce b/2657/CH1/EX1.1/Ex1_1.sce new file mode 100755 index 000000000..0b5f05eaa --- /dev/null +++ b/2657/CH1/EX1.1/Ex1_1.sce @@ -0,0 +1,15 @@ +//Calculation of cubic capacity and clearance volume
+clc,clear
+//Given:
+n=4 //Number of cylinders
+d=68/10 //Bore in cm
+l=75/10 //Stroke in cm
+r=8 //Compression ratio
+//Solution:
+V_s=(%pi/4)*d^2*l //Swept volume of one cylinder in cm^3
+cubic_capacity=n*V_s //Cubic capacity in cm^3
+//Since, r = (V_c + V_s)/V_c
+V_c=V_s/(r-1) //Clearance volume in cm^3
+//Results:
+printf("\n The cubic capacity of the engine = %.1f cm^3",cubic_capacity)
+printf("\n The clearance volume of a cylinder, V_c = %.1f cm^3\n\n",V_c)
diff --git a/2657/CH1/EX1.2/Ex1_2.sce b/2657/CH1/EX1.2/Ex1_2.sce new file mode 100755 index 000000000..6ed257156 --- /dev/null +++ b/2657/CH1/EX1.2/Ex1_2.sce @@ -0,0 +1,12 @@ +//Calculation of brake power and friction power
+clc,clear
+//Given:
+ip=10 //Indicated power in kW
+eta_m=80 //Mechanical efficiency in percent
+//Solution:
+//Since, eta_m = bp/ip
+bp=(eta_m/100)*ip //Brake power in kW
+fp=ip-bp //Friction power in kW
+//Results:
+printf("\n The brake power delivered, bp = %d kW\n",bp)
+printf(" The friction power, fp = %d kW\n\n",fp)
diff --git a/2657/CH1/EX1.3/Ex1_3.sce b/2657/CH1/EX1.3/Ex1_3.sce new file mode 100755 index 000000000..9afce8cca --- /dev/null +++ b/2657/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,17 @@ +//Calculation of mechanical efficiency
+clc,clear
+//Given:
+bp=100 //Brake power at full load in kW
+fp=25 //Frictional power in kW (printing error)
+//Solution:
+eta_m=bp/(bp+fp) //Mechanical efficiency at full load
+//(a)At half load
+bp=bp/2 //Brake power at half load in kW
+eta_m1=bp/(bp+fp) //Mechanical efficiency at half load
+//(b)At quarter load
+bp=bp/2 //Brake power at quarter load in kW
+eta_m2=bp/(bp+fp) //Mechanical efficiency at quarter load
+//Results:
+printf("\n The mechanical efficiency at full load, eta_m = %d percent",eta_m*100)
+printf("\n The mechanical efficiency,\n (a)At half load, eta_m = %.1f percent\n (b)At quarter load, eta_m = %d percent\n\n",eta_m1*100,eta_m2*100)
+//Data in the book is printed wrong
diff --git a/2657/CH1/EX1.4/Ex1_4.sce b/2657/CH1/EX1.4/Ex1_4.sce new file mode 100755 index 000000000..f23f858f5 --- /dev/null +++ b/2657/CH1/EX1.4/Ex1_4.sce @@ -0,0 +1,26 @@ +//Calculations on four stroke petrol engine
+clc,clear
+//Given:
+bp=35 //Brake power in kW
+eta_m=80 //Mechanical efficiency in percent
+bsfc=0.4 //Brake specific fuel consumption in kg/kWh
+A_F=14/1 //Air-fuel ratio
+CV=43000 //Calorific value in kJ/kg
+//Solution:
+//(a)
+ip=bp*100/eta_m //Indicated power in kW
+//(b)
+fp=ip-bp //Frictional power in kW
+//(c)
+//Since, 1 kWh = 3600 kJ
+eta_bt=1/(bsfc*CV/3600) //Brake thermal efficiency
+//(d)
+eta_it=eta_bt/eta_m*100 //Indicated thermal efficiency
+//(e)
+m_f=bsfc*bp //Fuel consumption in kg/hr
+//(f)
+m_a=A_F*m_f //Air consumption in kg/hr
+//Results:
+printf("\n (a)The indicated power, ip = %.2f kW\n (b)The friction power, fp = %.2f kW",ip,fp)
+printf("\n (c)The brake thermal efficiency, eta_bt = %.1f percent\n (d)The indicated thermal efficiency, eta_it = %.1f percent",eta_bt*100,eta_it*100)
+printf("\n (e)The fuel consumption per hour, m_f = %.1f kg/hr\n (f)The air consumption per hour, m_a = %d kg/hr\n\n",m_f,m_a)
diff --git a/2657/CH1/EX1.5/Ex1_5.sce b/2657/CH1/EX1.5/Ex1_5.sce new file mode 100755 index 000000000..2a372f95d --- /dev/null +++ b/2657/CH1/EX1.5/Ex1_5.sce @@ -0,0 +1,20 @@ +//Calculations on SI engine
+clc,clear
+//Given:
+F_A=0.07/1 //Fuel-air ratio
+bp=75 //Brake power in kW
+eta_bt=20 //Brake thermal efficiency in percent
+rho_a=1.2 //Density of air in kg/m^3
+rho_f=4*rho_a //Density of fuel vapour in kg/m^3
+CV=43700 //Calorific value of fuel in kJ/kg
+//Solution:
+m_f=bp*3600/(eta_bt*CV/100) //Fuel consumption in kg/hr
+m_a=m_f/F_A //Air consumption in kg/hr
+V_a=m_a/rho_a //Volume of air in m^3/hr
+V_f=m_f/rho_f //Volume of fuel in m^3/hr
+V_mixture=V_f+V_a //Mixture volume in m^3/hr
+//Results:
+printf("\n The air consumption, m_a = %.1f kg/hr",m_a)
+printf("\n The volume of air required, V_a = %.1f m^3/hr",V_a)
+printf("\n The volume of mixture required = %.1f m^3/hr\n\n",V_mixture) //(printing error)
+//Answer in the book is printed wrong
diff --git a/2657/CH1/EX1.6/Ex1_6.sce b/2657/CH1/EX1.6/Ex1_6.sce new file mode 100755 index 000000000..2b5f46d4a --- /dev/null +++ b/2657/CH1/EX1.6/Ex1_6.sce @@ -0,0 +1,20 @@ +//Calculations on diesel engine
+clc,clear
+//Given:
+bp=5 //Brake power in kW
+eta_it=30 //Indicated thermal efficiency in percent
+eta_m=75 //Mechanical efficiency in percent (printing error)
+//Solution:
+ip=bp*100/eta_m //Indicated power in kW
+CV=42000 //Calorific value of diesel(fuel) in kJ/kg
+m_f=ip*3600/(eta_it*CV/100) //Fuel consumption in kg/hr
+//Density of diesel(fuel) = 0.87 kg/l
+rho_f=0.87 //Density of fuel in kg/l
+V_f=m_f/rho_f //Fuel consumption in l/hr
+isfc=m_f/ip //Indicated specific fuel consumption in kg/kWh
+bsfc=m_f/bp //Brake specific fuel consumption in kg/kWh
+//Results:
+printf("\n The fuel consumption of engine, m_f in,\n (a)kg/hr = %.3f kg/hr\n (b)litres/hr = %.2f l/hr",m_f,V_f)
+printf("\n\n (c)Indicated specific fuel consumption, isfc = %.3f kg/kWh",isfc)
+printf("\n (d)Brake specific fuel consumption, bsfc = %.3f kg/kWh\n\n",bsfc)
+//Data in the book is printed wrong
diff --git a/2657/CH1/EX1.7/Ex1_7.sce b/2657/CH1/EX1.7/Ex1_7.sce new file mode 100755 index 000000000..fda20d8dd --- /dev/null +++ b/2657/CH1/EX1.7/Ex1_7.sce @@ -0,0 +1,24 @@ +//Calculations on two stroke CI engine
+clc,clear
+//Given:
+bp=5000 //Brake power in kW
+fp=1000 //Friction power in kW
+m_f=2300 //Fuel consumption in kg/hr
+A_F=20/1 //Air-fuel ratio
+CV=42000 //Calorific value of fuel in kJ/kg
+//Solution:
+//(a)
+ip=bp+fp //Indicated power in kW
+//(b)
+eta_m=bp/ip //Mechanical efficiency
+//(c)
+m_a=A_F*m_f //Air consumption in kg/hr
+//(d)
+eta_it=ip*3600/(m_f*CV) //Indicated thermal efficiency
+//(e)
+eta_bt=eta_it*eta_m //Brake thermal efficiency
+//Results:
+printf("\n (a)The indicated power, ip = %d kW",ip)
+printf("\n (b)The mechanical efficiency, eta_m = %d percent",eta_m*100)
+printf("\n (c)The air consumption, m_a = %d kg/hr",m_a)
+printf("\n (d)The indicated thermal efficiency, eta_it = %.1f percent\n (e)The brake thermal efficiency, eta_bt = %.1f percent\n\n",eta_it*100,eta_bt*100)
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