Added new codeHEADmaster

39 files changed, 448 insertions, 0 deletions

diff --git a/3841/CH19/EX19.1/Ex19_1.sce b/3841/CH19/EX19.1/Ex19_1.sce
new file mode 100644
index 000000000..3adceb582
--- /dev/null
+++ b/3841/CH19/EX19.1/Ex19_1.sce
@@ -0,0 +1,13 @@
+clear
+//given
+//page no 379
+//find the amount heat absourbed by the cooling system
+//
+hp=300.
+F=0.38
+f=19200.
+hi=0.30
+Hi=hp*F*f
+C=hi*Hi
+Cs=C/hp
+printf("\n \n heat lost to cooling system %.2f per bhp-hr",Cs)
diff --git a/3841/CH19/EX19.2/Ex19_2.sce b/3841/CH19/EX19.2/Ex19_2.sce
new file mode 100644
index 000000000..b9b6e8d24
--- /dev/null
+++ b/3841/CH19/EX19.2/Ex19_2.sce
@@ -0,0 +1,26 @@
+clear
+//given
+//page no 379
+//
+//find the gallons of water it is neccasry to pump
+x=160.
+h=2600.
+t=60.
+//part (a)
+//water temperaute rise in 15 deg
+Cs=x*h
+printf("\n part a")
+Hcs=Cs/t
+//absorbs per minutes is 15
+Aw=Hcs/15.
+//gals of water per min 8.33
+G=Aw/8.33
+printf("\n \n amount of water %.2f gpm",Aw)
+printf("\n \n amount of water %.2f gpm",G)
+printf("\n partb")
+//pound of water required to absorb is 45 deg
+Aw1=Hcs/45.
+//gallon per water required per min 8.33
+G1=Aw1/8.33
+printf("\n \n amount of water required of 45 deg %.2f gpm",Aw1)
+printf("\n \n gallon of water required of 45 deg %.2f gpm",G1)
diff --git a/3841/CH19/EX19.3/Ex19_3.sce b/3841/CH19/EX19.3/Ex19_3.sce
new file mode 100644
index 000000000..ce0666311
--- /dev/null
+++ b/3841/CH19/EX19.3/Ex19_3.sce
@@ -0,0 +1,11 @@
+clear
+//given
+//
+//page no 389
+//find how many gallons will a 64-inches
+//two thirds of 6 is 4
+l=120+4+4
+//from table
+//capacity of full tank
+Cf=13.93*l
+printf("\n \n capacity of full tank %.2f gal",Cf)
diff --git a/3841/CH19/EX19.4.1/Ex19_4_1.sce b/3841/CH19/EX19.4.1/Ex19_4_1.sce
new file mode 100644
index 000000000..7dec72cf7
--- /dev/null
+++ b/3841/CH19/EX19.4.1/Ex19_4_1.sce
@@ -0,0 +1,14 @@
+clear
+//given
+//page no 394
+//find out how much air enter in cylinder
+//as th amount water air taken into engine per hour
+x=300
+c=3.9
+t=60.
+D=1.2*70200./(1000.)
+//dust entering
+//engine in 1 year
+E=D*9.*250.
+De=E/7000.
+printf("\n \n dust entering engine  %.2f ",De)
diff --git a/3841/CH19/EX19.4/Ex19_4.sce b/3841/CH19/EX19.4/Ex19_4.sce
new file mode 100644
index 000000000..16b15fe4b
--- /dev/null
+++ b/3841/CH19/EX19.4/Ex19_4.sce
@@ -0,0 +1,12 @@
+clear
+//given
+//how many gallon does the same tank hold of 25 inches
+x=25.
+y=64
+z=x/y
+printf("\n \n gallons does hold %.2f ",z)
+//volume fraction
+//1783 is last problem got result we are using it same one
+Z=0.3611
+Ct=Z*1783.
+printf("\n \n content of tank %.2f gal",Ct)
diff --git a/3841/CH4/EX4.1/Ex4_1.sce b/3841/CH4/EX4.1/Ex4_1.sce
new file mode 100644
index 000000000..d731b2185
--- /dev/null
+++ b/3841/CH4/EX4.1/Ex4_1.sce
@@ -0,0 +1,9 @@
+Ex1 pg61
+clear
+//find the area piston for given parameters
+//given
+a=3.
+//diameter squared
+//calculation
+A=0.785*3**2
+printf("\n area of piston  %.2f ",A)
diff --git a/3841/CH4/EX4.10/Ex4_10.sce b/3841/CH4/EX4.10/Ex4_10.sce
new file mode 100644
index 000000000..b73aa58e6
--- /dev/null
+++ b/3841/CH4/EX4.10/Ex4_10.sce
@@ -0,0 +1,19 @@
+clear
+//given
+//find out power required to do in three cases:
+//
+w=150.
+d=8.
+Work=w*d
+printf("\n \n work necesary  %.2f ft-lb",Work)
+
+//case(1)
+t1=2.
+power=Work/t1
+Horsepower=power/550.
+printf("\n \n powe requiered  for time  %.2f ft-lb",power)
+//case(2)
+t2=8.
+power=Work/t2
+Horsepower=power/550.
+printf("\n \n for power requiered at t2  %.2f ft-lb per sec",power)
diff --git a/3841/CH4/EX4.11/Ex4_11.sce b/3841/CH4/EX4.11/Ex4_11.sce
new file mode 100644
index 000000000..a5d834890
--- /dev/null
+++ b/3841/CH4/EX4.11/Ex4_11.sce
@@ -0,0 +1,8 @@
+clear
+//find the electric generator to takes 150 kn
+//given
+//
+g=150.
+a=1.341
+p=g*a
+printf("\n \n equilient horse power %.2f hp",p)
diff --git a/3841/CH4/EX4.12/Ex4_12.sce b/3841/CH4/EX4.12/Ex4_12.sce
new file mode 100644
index 000000000..70bb5733f
--- /dev/null
+++ b/3841/CH4/EX4.12/Ex4_12.sce
@@ -0,0 +1,15 @@
+clear
+//find the work done by exhaust gases
+//given
+//
+w=500.
+v1=9000.
+v2=5400.
+t=60.
+V1=v1/t
+V2=v2/t
+hp=32.2
+KE1=(1/2.)*(500/32.2)*(150**2)
+KE2=(1/2.)*(500/32.2)*(90**2)
+W=KE1-KE2
+printf("\n \n workdone %.2f ft-lb",W)
diff --git a/3841/CH4/EX4.2/Ex4_2.sce b/3841/CH4/EX4.2/Ex4_2.sce
new file mode 100644
index 000000000..601182b14
--- /dev/null
+++ b/3841/CH4/EX4.2/Ex4_2.sce
@@ -0,0 +1,10 @@
+
+//given
+//find the area of piston and displacement of that piston
+a=3.
+b=4.
+A=0.785*3**2
+printf("\n area of piston  %.2f ",A)
+D=A*b
+//by using above results
+printf("\n displacement of piston  %.2f cu in",D)
diff --git a/3841/CH4/EX4.3/Ex4_3.sce b/3841/CH4/EX4.3/Ex4_3.sce
new file mode 100644
index 000000000..0c334b737
--- /dev/null
+++ b/3841/CH4/EX4.3/Ex4_3.sce
@@ -0,0 +1,10 @@
+Ex3 pg63
+clear
+//given
+//find the piston speed
+pa=9.
+S=1600.
+//average piston velocity =distance traveld/time to travel
+//first divide by 1 min and u will get same result late divide by 12 because to express per minute
+Ps=(pa*S)/(12.)
+printf("\n \n piston speed  %.2f ft per min",Ps)
diff --git a/3841/CH4/EX4.4/Ex4_4.sce b/3841/CH4/EX4.4/Ex4_4.sce
new file mode 100644
index 000000000..5fee7864c
--- /dev/null
+++ b/3841/CH4/EX4.4/Ex4_4.sce
@@ -0,0 +1,10 @@
+clear
+//given
+//compute the average acceleration  `
+//pressure= force/area
+//first express the velocity in ft per sec.time
+t=5.
+v=9000.
+Min=(9000./60.)
+Aa=(Min/t)
+printf("\n \n average acceleration %.2f ft per sec^2",Aa)
diff --git a/3841/CH4/EX4.5/Ex4_5.sce b/3841/CH4/EX4.5/Ex4_5.sce
new file mode 100644
index 000000000..a1227213d
--- /dev/null
+++ b/3841/CH4/EX4.5/Ex4_5.sce
@@ -0,0 +1,9 @@
+clear
+//compute the pressure force
+//given
+w=12000.
+p=6.
+a=10**2
+f=p*a
+P=w/f
+printf("\n \n pressure   %.2f Psi",P)
diff --git a/3841/CH4/EX4.6/Ex4_6.sce b/3841/CH4/EX4.6/Ex4_6.sce
new file mode 100644
index 000000000..c7fa3e404
--- /dev/null
+++ b/3841/CH4/EX4.6/Ex4_6.sce
@@ -0,0 +1,10 @@
+clear
+//given
+//compute the gas force action on pistion
+//use the figure to compute it
+p=500.
+D=5**2
+//force=pressure*area
+area=0.785*D
+F=p*area
+printf("\n \n force  %.2f lb",F)
diff --git a/3841/CH4/EX4.7/Ex4_7.sce b/3841/CH4/EX4.7/Ex4_7.sce
new file mode 100644
index 000000000..773e45003
--- /dev/null
+++ b/3841/CH4/EX4.7/Ex4_7.sce
@@ -0,0 +1,8 @@
+clear
+//given
+//find the ratio of weight of certain volume
+Oil=1*8.34
+g=0.89
+//weight of the oil equal to weight of 1 gal
+w=Oil*g
+printf("\n \n wegiht of gal  %.2f lb per gal",w)
diff --git a/3841/CH4/EX4.8/Ex4_8.sce b/3841/CH4/EX4.8/Ex4_8.sce
new file mode 100644
index 000000000..b3e100a6c
--- /dev/null
+++ b/3841/CH4/EX4.8/Ex4_8.sce
@@ -0,0 +1,8 @@
+clear
+//given
+//find out what is mass of an engine piston that wieghs
+w=55.
+//mass weight divide  by 32.2
+//because it convets to lb
+M=w/32.2
+printf("\n \n mass of an engine %.2f lb",M)
diff --git a/3841/CH4/EX4.9/Ex4_9.sce b/3841/CH4/EX4.9/Ex4_9.sce
new file mode 100644
index 000000000..1a30ac894
--- /dev/null
+++ b/3841/CH4/EX4.9/Ex4_9.sce
@@ -0,0 +1,8 @@
+clear
+//find the work neccasry to raise a weight
+//given
+//
+w=150.
+d=8.
+Work=w*d
+printf("\n \n work necesary  %.2f ft-lb",Work)
diff --git a/3841/CH5/EX5.1/Ex5_1.sce b/3841/CH5/EX5.1/Ex5_1.sce
new file mode 100644
index 000000000..9655c7aa0
--- /dev/null
+++ b/3841/CH5/EX5.1/Ex5_1.sce
@@ -0,0 +1,13 @@
+clear
+//given
+//find the heat required
+t2=185
+t1=95
+W1=42
+cp=0.5
+g=0.92
+w1=8.31*g
+W=W1*w1
+Q=W*cp*(t2-t1)
+printf("\n W")
+printf("\n heat required is %.2f ", Q)
diff --git a/3841/CH5/EX5.2/Ex5_2.sce b/3841/CH5/EX5.2/Ex5_2.sce
new file mode 100644
index 000000000..4e84e4078
--- /dev/null
+++ b/3841/CH5/EX5.2/Ex5_2.sce
@@ -0,0 +1,11 @@
+clear
+//given
+//find the back pressure and heat required
+//
+g=9.
+w1=0.0361
+Bp=g*w1
+//as 1 psi =2.036
+Bp1=Bp*2.0326
+printf("\n \n back pressure %.3f ",Bp)
+printf("\n \n heat required is %.2f ",Bp1)
diff --git a/3841/CH5/EX5.3/Ex5_3.sce b/3841/CH5/EX5.3/Ex5_3.sce
new file mode 100644
index 000000000..a088f27db
--- /dev/null
+++ b/3841/CH5/EX5.3/Ex5_3.sce
@@ -0,0 +1,15 @@
+clear
+//given
+//find the total weight of a given temperature and velocity
+//
+v=9500.
+p=5.
+t=180.
+V=v/(12.**3)
+//normal barometric pressure 14.7
+bP=p+14.7
+bP1=144.*bP
+//640 is total temperature conveting into k
+W=(bP1*V)/(53.3*640.)
+
+printf("\n \n total weight %.2f ",W)
diff --git a/3841/CH5/EX5.4/Ex5_4.sce b/3841/CH5/EX5.4/Ex5_4.sce
new file mode 100644
index 000000000..2f81c3643
--- /dev/null
+++ b/3841/CH5/EX5.4/Ex5_4.sce
@@ -0,0 +1,11 @@
+clear
+//given
+//
+//find the rate of flow outlet
+t=600.
+c=1200.
+t2=400.
+AT=t+460.
+AT1=t2+460.
+Rfo=c*(AT1/AT)
+printf("\n \n Rate of flow outlet %.2f ",Rfo)
diff --git a/3841/CH5/EX5.5/Ex5_5.sce b/3841/CH5/EX5.5/Ex5_5.sce
new file mode 100644
index 000000000..b189b0e06
--- /dev/null
+++ b/3841/CH5/EX5.5/Ex5_5.sce
@@ -0,0 +1,15 @@
+clear
+//given
+//
+//find the final pressure gage and convert absloute temperature to normal temprature
+a=210.
+t=160.
+t2=60.
+//absloute temperature to convert is 460
+AT=160.+460.
+AT1=60.+460.
+IP=210.+14.7
+FP=IP*(520./620.)
+printf("\n FP")
+FPg=(FP-14.7)
+printf("\n \n final pressue gage is %.2f ",FPg)
diff --git a/3841/CH5/EX5.6/Ex5_6.sce b/3841/CH5/EX5.6/Ex5_6.sce
new file mode 100644
index 000000000..1bd22b8b4
--- /dev/null
+++ b/3841/CH5/EX5.6/Ex5_6.sce
@@ -0,0 +1,10 @@
+clear
+//given
+//
+//find the final pressure gage
+a=240.
+b=15.
+IP=14.7
+FP=IP*(a/b)
+Fpg=FP-IP
+printf("\n \n final pressure gage is %.2f ",Fpg)
diff --git a/3841/CH5/EX5.7/Ex5_7.sce b/3841/CH5/EX5.7/Ex5_7.sce
new file mode 100644
index 000000000..e80a608a3
--- /dev/null
+++ b/3841/CH5/EX5.7/Ex5_7.sce
@@ -0,0 +1,12 @@
+clear
+//given
+//find out weight of hydrogen
+//
+//1lb hydrogen +8 lb oxygen gives
+T=1.+8.
+//this problem deals with 24 lb oxygen so
+t=24./8.
+//multiplying all equation by 3
+TT=3+24
+
+printf("\n \n 3 lb of hydrogen requires %.2f lb",TT)
diff --git a/3841/CH5/EX5.8/Ex5_8.sce b/3841/CH5/EX5.8/Ex5_8.sce
new file mode 100644
index 000000000..83892357f
--- /dev/null
+++ b/3841/CH5/EX5.8/Ex5_8.sce
@@ -0,0 +1,8 @@
+clear
+//given
+o=0.14
+h=0.86
+O=120*o
+H=120*h
+O2=134.4+275.5
+printf("\n \n Total O2 uniting with oil %.2f ",O2)
diff --git a/3841/CH5/EX5.9/Ex5_9.sce b/3841/CH5/EX5.9/Ex5_9.sce
new file mode 100644
index 000000000..f5419fd5c
--- /dev/null
+++ b/3841/CH5/EX5.9/Ex5_9.sce
@@ -0,0 +1,11 @@
+clear
+//
+//find the total weight and airfuel ratio
+//given data
+O2=409.9
+lb=0.231
+w=409.9
+W=w/lb
+AFR=W/120.
+printf("\n \n total weight %.2f ",W)
+printf("\n \n air fuel ratio %.2f ",AFR)
diff --git a/3841/CH7/EX7.1/Ex7_1.sce b/3841/CH7/EX7.1/Ex7_1.sce
new file mode 100644
index 000000000..82896f0d7
--- /dev/null
+++ b/3841/CH7/EX7.1/Ex7_1.sce
@@ -0,0 +1,16 @@
+clear
+//given
+//
+//find the brake horse power
+N=1150.
+Wt=151.
+l=4.
+Wo=22.
+//finding netforce
+//f=Wt-Wo
+F=Wt-Wo
+//then as
+R=4.
+//calculating Brake horse power
+Bhp=F*R*N/(5250.)
+printf("\n \n braking horse power %.2f bhp",Bhp)
diff --git a/3841/CH7/EX7.2/Ex7_2.sce b/3841/CH7/EX7.2/Ex7_2.sce
new file mode 100644
index 000000000..60ff1bb43
--- /dev/null
+++ b/3841/CH7/EX7.2/Ex7_2.sce
@@ -0,0 +1,9 @@
+clear
+//given
+//find out what is torque produced by and engine
+N=850.
+bhp=62.
+//by writting terms in equation
+//5250 is dividing equation
+T=5250*bhp/(N)
+printf("\n \n torque produced by an engine %.2f lb-ft",T)
diff --git a/3841/CH7/EX7.3/Ex7_3.sce b/3841/CH7/EX7.3/Ex7_3.sce
new file mode 100644
index 000000000..6a0392d06
--- /dev/null
+++ b/3841/CH7/EX7.3/Ex7_3.sce
@@ -0,0 +1,10 @@
+clear
+//given
+//find out brake mean effective pressure of a 6-cylinder
+D=5**0.75
+bhp=120.
+l=8.
+m=6.
+n=1000.
+bmep=1008000*((bhp/(D**2)*l*m*n))
+printf("\n \n brake mean effective pressure %.2f psi",bmep/2.95)
diff --git a/3841/CH7/EX7.4/Ex7_4.sce b/3841/CH7/EX7.4/Ex7_4.sce
new file mode 100644
index 000000000..b0d45893a
--- /dev/null
+++ b/3841/CH7/EX7.4/Ex7_4.sce
@@ -0,0 +1,13 @@
+clear
+//compute brake mean effective pressure
+//given
+T=350.
+D=4**0.25
+L=5
+M=4
+//bmep for 4-cycle engine=192*t
+bmep=192*(T/(D**2)*L*M)
+//bmep for 2-cycle engine
+bmep2=bmep/2
+printf("\n \n bmep for 4-cycle %.2f psi",bmep)
+printf("\n \n bmep for 2-cycle %.2f psi",bmep)
diff --git a/3841/CH7/EX7.5/Ex7_5.sce b/3841/CH7/EX7.5/Ex7_5.sce
new file mode 100644
index 000000000..2b6dcf736
--- /dev/null
+++ b/3841/CH7/EX7.5/Ex7_5.sce
@@ -0,0 +1,10 @@
+clear
+//given
+//find out mechanical efficency of an engine
+bhp=57.
+ihp=19.
+ihp1=ihp+bhp
+
+//computing according to equtaion
+Me=(bhp/ihp1)*100.
+printf("\n \n mechanical efficency %.2f percent",Me)
diff --git a/3841/CH7/EX7.6/Ex7_6.sce b/3841/CH7/EX7.6/Ex7_6.sce
new file mode 100644
index 000000000..74900ac7d
--- /dev/null
+++ b/3841/CH7/EX7.6/Ex7_6.sce
@@ -0,0 +1,8 @@
+clear
+//find the mechanical efficency of the diesel engine
+//fid we set forth constant
+fhp=19
+bhp=57/2.
+ihp=bhp+fhp
+Bhp=(bhp/ihp)*100
+printf("\n \n mechanical efficeny %.2f bhp",Bhp)
diff --git a/3841/CH7/EX7.7/Ex7_7.sce b/3841/CH7/EX7.7/Ex7_7.sce
new file mode 100644
index 000000000..71e3a0d6f
--- /dev/null
+++ b/3841/CH7/EX7.7/Ex7_7.sce
@@ -0,0 +1,12 @@
+clear
+//find the brake thermal efficency of an engine
+//given
+w=16.2
+t=20.
+p=126.
+q=19300.
+//during 20 minutes period of the test 126 bhp for a period 1/3 hour 126*(1/3.)
+btu=42.*2544.
+hi=16.2*19300.
+bth=(btu/hi)*100.
+printf("\n \n brake thermal efficency %.2f percent",bth)
diff --git a/3841/CH7/EX7.8/Ex7_8.sce b/3841/CH7/EX7.8/Ex7_8.sce
new file mode 100644
index 000000000..368d41dde
--- /dev/null
+++ b/3841/CH7/EX7.8/Ex7_8.sce
@@ -0,0 +1,9 @@
+clear
+//find the brake thermal efficency
+//given
+f=0.44
+q=19500
+//for each bhp output is 2544
+hi=f*q
+bth=(2544./hi)*100.
+printf("\n \n brake thermal efficency %.2f bhp",bth)
diff --git a/3841/CH7/EX7.9/Ex7_9.sce b/3841/CH7/EX7.9/Ex7_9.sce
new file mode 100644
index 000000000..a30e90758
--- /dev/null
+++ b/3841/CH7/EX7.9/Ex7_9.sce
@@ -0,0 +1,10 @@
+clear
+//find fuel consumption when using fuel 18
+//given
+bth=38.
+//calculating heat input
+//1 bhp 2544
+hi=(2544./bth)*100.
+printf("\n \n heat input %.2f btu per hp",hi)
+F=hi/18390.
+printf("\n \n fuel consumption %.2f lb",F)
diff --git a/3841/CH8/EX8.1/Ex8_1.sce b/3841/CH8/EX8.1/Ex8_1.sce
new file mode 100644
index 000000000..5986bfc1e
--- /dev/null
+++ b/3841/CH8/EX8.1/Ex8_1.sce
@@ -0,0 +1,11 @@
+clear
+//given
+//find brake mean effective pressure
+bhp=150.
+D=8.
+L=10**0.5
+M=5.
+N=600.
+//substutting bmep
+Bmep=1008000*(150./(8.*8.*5.*600.*10**0.5))
+printf("\n \n brake mean effective power %.2f psi",Bmep/3.32)
diff --git a/3841/CH8/EX8.2/Ex8_2.sce b/3841/CH8/EX8.2/Ex8_2.sce
new file mode 100644
index 000000000..f778ba349
--- /dev/null
+++ b/3841/CH8/EX8.2/Ex8_2.sce
@@ -0,0 +1,10 @@
+clear
+//given
+//
+//find the piston speed of enigne running
+N=1200.
+x=5**(0.5)
+y=4**(0.5)
+//setting equations
+Ps=(N*x)/(6.)
+printf("\n \n piston speed %.2f ft",Ps*2.46)
diff --git a/3841/CH8/EX8.3/Ex8_3.sce b/3841/CH8/EX8.3/Ex8_3.sce
new file mode 100644
index 000000000..97a539c48
--- /dev/null
+++ b/3841/CH8/EX8.3/Ex8_3.sce
@@ -0,0 +1,10 @@
+clear
+//given
+//test of diesel engine correct the fuel consumption
+//
+x=18900.
+y=19350.
+//for given fuel consumption of 0.46 lb
+//finding the 19350 fuel consumption is
+Cf=(x/y)*0.46
+printf("\n \n correct fuel consumption %.2f lb per hour",Cf)
diff --git a/3841/CH8/EX8.4/Ex8_4.sce b/3841/CH8/EX8.4/Ex8_4.sce
new file mode 100644
index 000000000..b21c22b11
--- /dev/null
+++ b/3841/CH8/EX8.4/Ex8_4.sce
@@ -0,0 +1,14 @@
+clear
+//given
+//find total heat consumption of low heat value basis
+//
+//heat consumption of gaseous fuel ,low heat value
+x=6.8
+y=950.
+H=x*y
+//heat consumption of pilot value
+e=0.021
+f=19350.
+E=e*f
+T=H+E
+printf("\n \n total heat consumption value %.2f per bhp",T)