diff options
Diffstat (limited to 'sample_notebooks/kumargugloth/Chapter1.ipynb')
| -rw-r--r--[-rwxr-xr-x] | sample_notebooks/kumargugloth/Chapter1.ipynb | 290 |
1 files changed, 36 insertions, 254 deletions
diff --git a/sample_notebooks/kumargugloth/Chapter1.ipynb b/sample_notebooks/kumargugloth/Chapter1.ipynb index df9ba4d0..fdfb0cb9 100755..100644 --- a/sample_notebooks/kumargugloth/Chapter1.ipynb +++ b/sample_notebooks/kumargugloth/Chapter1.ipynb @@ -1,7 +1,7 @@ {
"metadata": {
"name": "",
- "signature": "sha256:84e452258bd05b64c16351467c4970051f4494cb47d7a832df03bdce07abddb8"
+ "signature": "sha256:281275d36b0e16d144d1212530d5ebac420ea6bfd258dbfe43c04ce417d0dbbc"
},
"nbformat": 3,
"nbformat_minor": 0,
@@ -13,7 +13,7 @@ "level": 1,
"metadata": {},
"source": [
- "Chapter1-Introduction"
+ "Chapter1-Atomic Weight "
]
},
{
@@ -21,34 +21,23 @@ "level": 2,
"metadata": {},
"source": [
- "Ex1-pg9"
+ "Ex1-pg12"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
- "\n",
"import math\n",
- " #determine\n",
- "##This numerical is Ex 1_1E,page 9.\n",
- "Pso=20.5\n",
- "Psc=20.5*550##converting hp to fps system\n",
- "Qo=385.\n",
- "Qc=385./449.##converting gpm to ft^3/s\n",
- "E=0.83\n",
- "dp=E*Psc/(Qc*144.)\n",
- "print\"%s %.2f %s \"%('The pressure rise is ',dp,' psi')\n",
- "print(\"After rounding off,pressure rise is 75.8 psi\")\n",
- "dpr=75.8\n",
- "dHw=75.8*144/62.4##62.4 is accelaration due to gravity in fps system\n",
- "print\"%s %.2f %s \"%(' The head of water is ',dHw,' ft of water')\n",
- "print(\"After rounding off the value of head of water the answer is 175 ft of water.\")\n",
- "dhwr=175##rounded off value of head of water\n",
- "sg=0.72##specific gravity of oil\n",
- "dHo=dhwr/sg\n",
- "print\"%s %.2f %s \"%(' The head of oil is ',dHo,' ft of oil')\n",
- "print(\"After rounding off the value of head of oil the answer is 243 ft of oil.\")\n"
+ "##Intitalisation of variables\n",
+ "#calculate the Molecular weight of carbon dioxide\n",
+ "dco= 1.9635 ##gms/lit\n",
+ "do= 1.4277 ##gms/lit\n",
+ "mo= 32. ##gms\n",
+ "##CALCULATIONS\n",
+ "mwt= dco*mo/do\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'% ('Molecular weight of carbon dioxide = ',mwt,'')\n"
],
"language": "python",
"metadata": {},
@@ -57,125 +46,7 @@ "output_type": "stream",
"stream": "stdout",
"text": [
- "The pressure rise is 75.79 psi \n",
- "After rounding off,pressure rise is 75.8 psi\n",
- " The head of water is 174.92 ft of water \n",
- "After rounding off the value of head of water the answer is 175 ft of water.\n",
- " The head of oil is 243.06 ft of oil \n",
- "After rounding off the value of head of oil the answer is 243 ft of oil.\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2-pg10"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math \n",
- "#determine\n",
- "##This numerical is Ex 1_1S,page 10.\n",
- "E=0.83##efficiency\n",
- "Ps=15300.\n",
- "Q=87.4\n",
- "Qs=87.4/3600.##flow rate in meter cube per sec\n",
- "rho=998.\n",
- "g=9.81\n",
- "sg=0.72\n",
- "dp=E*Ps/Qs\n",
- "print\"%s %.2f %s \"%('\\n The change in pressure (dp)is ',dp,'')\n",
- "dpr=523000##rounded value of dp\n",
- "print(\"The rounded off value of dp is 523kPa.\")\n",
- "dHw=dpr/(rho*g)\n",
- "print\"%s %.2f %s \"%(' dHw is equal to ',dHw,' m of water')\n",
- "print(\"The rounded off value of dHw is 53.4 m of water.\")\n",
- "dHwr=53.4##rounded off value of dHw\n",
- "print(\"Thus we can determine head of oil.\")\n",
- "dHoil=dHwr/sg\n",
- "print\"%s %.2f %s \"%(' dHoil is given by ',dHoil,' m of oil')\n",
- "print(\"The rounded off value of dHoil is 74.2 m of oil.\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- " The change in pressure (dp)is 523070.94 \n",
- "The rounded off value of dp is 523kPa.\n",
- " dHw is equal to 53.42 m of water \n",
- "The rounded off value of dHw is 53.4 m of water.\n",
- "Thus we can determine head of oil.\n",
- " dHoil is given by 74.17 m of oil \n",
- "The rounded off value of dHoil is 74.2 m of oil.\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3-pg10"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine\n",
- "##This numerical is Ex 1_2E,page 10.\n",
- "Q=12000.\n",
- "A=3.5\n",
- "rho_a=0.0762\n",
- "E=0.85\n",
- "r=2.5##resistance of duct system\n",
- "V=Q/(60.*A)\n",
- "print\"%s %.2f %s \"%('The air flow velocity at discharge is ',V,' ft/s')\n",
- "KE=(rho_a*(V**2))/(32.2*2)\n",
- "print\"%s %.2f %s \"%('\\n The product is ',KE,' lb/ft^2')\n",
- "##PE=KE\n",
- "Hv=KE/62.4\n",
- "print\"%s %.2f %s \"%('\\n The dynamic head is ',Hv,' ft')\n",
- "print(\"The value of dynamic head in inches of water is 0.74.\")\n",
- "Hvi=0.74##Head in inches\n",
- "Ht=r+Hvi\n",
- "print\"%s %.2f %s \"%('\\n The total head is ',Ht,' inches of water')\n",
- "p_tot=Ht*62.4\n",
- "Ps=Q*p_tot/(60.*12.*E)\n",
- "print\"%s %.2f %s \"%('\\n The shaft power is ',Ps,' ft-lb/s')\n",
- "print(\"The shaft power is 7.2 hp.\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The air flow velocity at discharge is 57.14 ft/s \n",
- "\n",
- " The product is 3.86 lb/ft^2 \n",
- "\n",
- " The dynamic head is 0.06 ft \n",
- "The value of dynamic head in inches of water is 0.74.\n",
- "\n",
- " The total head is 3.24 inches of water \n",
- "\n",
- " The shaft power is 3964.24 ft-lb/s \n",
- "The shaft power is 7.2 hp.\n"
+ "Molecular weight of carbon dioxide = 44.01 \n"
]
}
],
@@ -186,91 +57,23 @@ "level": 2,
"metadata": {},
"source": [
- "Ex4-pg11"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "##This numerical is Ex 1_2S,page 11.\n",
- "Q=340.\n",
- "A=0.325\n",
- "V=Q/(60.*A)\n",
- "print\"%s %.2f %s \"%('The air flow velocity at discharge is ',V,' m/s')\n",
- "rho_a=1.22\n",
- "Vr=17.4\n",
- "Hd=(rho_a*(Vr**2))/2.\n",
- "print\"%s %.2f %s \"%('\\n The dynamic pressure head is ',Hd,' Pa')\n",
- "Hdr=184.7##rounded off value of Hd\n",
- "rho_w=998.##density of water=rhow\n",
- "g=9.81\n",
- "H=0.0635\n",
- "dp=rho_w*g*H##static pressure head\n",
- "print\"%s %.2f %s \"%('\\n The static pressure head is ',dp,' Pa')\n",
- "dpr=621.7\n",
- "p_tot=Hdr+dpr\n",
- "print\"%s %.2f %s \"%('\\n The total pressure head is ',p_tot,' Pa')\n",
- "p_tot=806.4\n",
- "E=0.85##efficiency\n",
- "Ps=Q*p_tot/(60*E)\n",
- "print\"%s %.2f %s \"%('\\n The shaft power is',Ps, 'W')\n",
- "print(\"The shaft power is 5.376 kW.\")\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The air flow velocity at discharge is 17.44 m/s \n",
- "\n",
- " The dynamic pressure head is 184.68 Pa \n",
- "\n",
- " The static pressure head is 621.69 Pa \n",
- "\n",
- " The total pressure head is 806.40 Pa \n",
- "\n",
- " The shaft power is 5376.00 W \n",
- "The shaft power is 5.376 kW.\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5-pg11"
+ "Ex2-pg13"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
- "#determine \n",
"import math\n",
- "##This numerical is Ex 1_3E,page 11.\n",
- "H=295.##net head in ft\n",
- "Q=148.##water flow rate\n",
- "n=1800.##rpm\n",
- "E=0.87##efficiency\n",
- "a=62.4##product of density and accelaration due to gravity\n",
- "omega=(n*2.*math.pi)/60.\n",
- "dp=a*H\n",
- "print\"%s %.2f %s \"%('The pressure is ',dp,' lb/ft^2')\n",
- "Ps=E*Q*dp\n",
- "print\"%s %.2f %s \"%('\\n Output power is equal to ',Ps,' lb-ft/s')\n",
- "print(\"The output output power can also be written as 2.37*10^6 lb-ft/s\")\n",
- "print(\"Output power in terms of horsepower is given by 4309hp.\")\n",
- "Psr=2370000##rounded off value of Ps\n",
- "Torque=Psr/omega\n",
- "print\"%s %.2f %s \"%(' The output torque is ',Torque,' lb-ft.')\n",
- "print(\"The output torque can also be written as 12.57*10^3 lb-ft\")\n",
- "\n"
+ "##Intitalisation of variables\n",
+ "#calculate the atomic weight of lead\n",
+ "shl= 0.031 ##cal deg^-1 g^-1\n",
+ "ewlc= 103.605 ##gms\n",
+ "n= 2.\n",
+ "##CALCULATIONS\n",
+ "aw= n*ewlc\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'% ('Atomic weight of lead = ',aw,' gms')\n"
],
"language": "python",
"metadata": {},
@@ -279,49 +82,33 @@ "output_type": "stream",
"stream": "stdout",
"text": [
- "The pressure is 18408.00 lb/ft^2 \n",
- "\n",
- " Output power is equal to 2370214.08 lb-ft/s \n",
- "The output output power can also be written as 2.37*10^6 lb-ft/s\n",
- "Output power in terms of horsepower is given by 4309hp.\n",
- " The output torque is 12573.24 lb-ft. \n",
- "The output torque can also be written as 12.57*10^3 lb-ft\n"
+ "Atomic weight of lead = 207.21 gms\n"
]
}
],
- "prompt_number": 6
+ "prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
- "Ex6-pg12"
+ "Ex3-pg13"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
- "#determine c\n",
"import math\n",
- "##This numerical is Ex 1_3S,page 12.\n",
- "H=90.\n",
- "Q=4.2##water flow rate(in m^3/s)\n",
- "n=1800.\n",
- "E=0.87##efficiency\n",
- "rho=998.\n",
- "g=9.81\n",
- "omega=(n*2.*math.pi)/60.\n",
- "dp=rho*g*H\n",
- "print\"%s %.2f %s \"%('The pressure is ',dp,' N/m^2')\n",
- "Ps=E*Q*dp\n",
- "print\"%s %.2f %s \"%('\\n Output power is equal to ',Ps,' N-m/s')\n",
- "print(\"After rounding off the value of output power is 3220 kW.\")\n",
- "Psr=3220000.##rounded off value of Ps\n",
- "Torque=Psr/omega\n",
- "print\"%s %.2f %s \"%(' The output torque is ',Torque,' N-m.')\n",
- "print(\"After rounding off the output torque comes out to be 17.1*10^3 N-m.\")\n"
+ "##Intitalisation of variables\n",
+ "\n",
+ "ewt= 17.337 ##gms\n",
+ "n=3.\n",
+ "##CALCULATIONS\n",
+ "aw= ewt*n\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'% ('Atomic weight of chromium = ',aw,' gms')\n"
],
"language": "python",
"metadata": {},
@@ -330,16 +117,11 @@ "output_type": "stream",
"stream": "stdout",
"text": [
- "The pressure is 881134.20 N/m^2 \n",
- "\n",
- " Output power is equal to 3219664.37 N-m/s \n",
- "After rounding off the value of output power is 3220 kW.\n",
- " The output torque is 17082.63 N-m. \n",
- "After rounding off the output torque comes out to be 17.1*10^3 N-m.\n"
+ "Atomic weight of chromium = 52.01 gms\n"
]
}
],
- "prompt_number": 7
+ "prompt_number": 2
}
],
"metadata": {}
|