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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
commitb1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch)
treeab291cffc65280e58ac82470ba63fbcca7805165 /2507/CH5
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Diffstat (limited to '2507/CH5')
-rwxr-xr-x2507/CH5/EX5.1/Ex5_1.sce29
-rwxr-xr-x2507/CH5/EX5.10/Ex5_10.sce18
-rwxr-xr-x2507/CH5/EX5.11/Ex5_11.sce5
-rwxr-xr-x2507/CH5/EX5.2/Ex5_2.sce31
-rwxr-xr-x2507/CH5/EX5.3/Ex5_3.sce33
-rwxr-xr-x2507/CH5/EX5.4/Ex5_4.sce24
-rwxr-xr-x2507/CH5/EX5.5/Ex5_5.sce16
-rwxr-xr-x2507/CH5/EX5.6/Ex5_6.sce16
-rwxr-xr-x2507/CH5/EX5.7/Ex5_7.sce13
-rwxr-xr-x2507/CH5/EX5.8/Ex5_8.sce14
-rwxr-xr-x2507/CH5/EX5.9/Ex5_9.sce10
11 files changed, 209 insertions, 0 deletions
diff --git a/2507/CH5/EX5.1/Ex5_1.sce b/2507/CH5/EX5.1/Ex5_1.sce
new file mode 100755
index 000000000..40ab5be03
--- /dev/null
+++ b/2507/CH5/EX5.1/Ex5_1.sce
@@ -0,0 +1,29 @@
+clc
+clear
+printf("Example 5.1 | Page number 119 \n\n");
+//Find the work output in KJ/kg
+//Given data
+Q = -24; //KJ/kg
+
+p1 = 5e5; //N/m^2
+t1 = 227; //°C
+V1 = 50; //m/s
+v1 = 0.78; //m^3/kg
+
+p2 = 1e5; //N/m^2
+t2 = 57; //°C
+V2 = 100; //m/s
+v2 = 0.97; //m^3/kg
+
+g = 9.81; //m/s^2 //acceleration due to gravity
+
+delta_z = -5; //m
+Cv = 0.7; //KJ/kg
+delta_u = Cv*(t2 - t1); //KJ/kg //change in internal energy //u2-u1
+
+//Solution
+delta_h = delta_u + (p2*v2 - p1*v1)*.001; //KJ/kg //change in enthalpy //h2-h1
+
+W_x = Q - (delta_h + (V2^2 - V1^2)/2*.001 + g*delta_z*.001); //kJ/kg //Work Output
+
+printf("Work output = %.2f KJ/kg",W_x);
diff --git a/2507/CH5/EX5.10/Ex5_10.sce b/2507/CH5/EX5.10/Ex5_10.sce
new file mode 100755
index 000000000..56801bd96
--- /dev/null
+++ b/2507/CH5/EX5.10/Ex5_10.sce
@@ -0,0 +1,18 @@
+clc
+clear
+printf("Example 5.10 | Page number 131 \n\n");
+//Find the mass of air entering and temperature of air in vessel
+
+//Given Data
+m1 = 0.8 //kg //initial mass of air
+p1 = 150 //kPa //initial pressure of air
+T1 = 300 //K //initial temperature of air
+p_p = 600 //kPa //pressure of air in pipe
+T_p = 330 //K // temperature of air in pipe
+
+//Solution
+m2T2 = (p_p/p1)*T1*m1
+m2 = ((0.718*(m2T2/m1-T1))/(331.65)*m1)+m1 //kg //final mass of air
+printf("Mass of air entering in vessel = %.4f kg\n",m2-m1)
+T2 = m2T2/m2 //K //Temperature of air in vessel
+printf("Temperature of air in vessel = %.1f K",T2)
diff --git a/2507/CH5/EX5.11/Ex5_11.sce b/2507/CH5/EX5.11/Ex5_11.sce
new file mode 100755
index 000000000..88a01e5f6
--- /dev/null
+++ b/2507/CH5/EX5.11/Ex5_11.sce
@@ -0,0 +1,5 @@
+clc
+clear
+printf("Example 5.11 | Page number 133 \n\n");
+//This is a theoritical question. Refer textbook for solution.
+printf("This is a theoritical question. Refer textbook for solution.")
diff --git a/2507/CH5/EX5.2/Ex5_2.sce b/2507/CH5/EX5.2/Ex5_2.sce
new file mode 100755
index 000000000..0afc9a1f8
--- /dev/null
+++ b/2507/CH5/EX5.2/Ex5_2.sce
@@ -0,0 +1,31 @@
+clc
+clear
+printf("Example 5.2 | Page number 120 \n\n");
+//Part (a) Find the change in enthalpy across turbine, if inlet velocity is negligible
+//Part (b) Find the change in enthalpy across turbine, if inlet velocity is 60m/s
+
+//Given Data
+m = 5000/3600 // kg/s // flow rate
+W_x = 550 // KJ/s //power developed by turbine
+Q = 0 //Heat loss is negligible
+
+//Solution
+//Part (a)
+printf("Part(a)\n\n")
+V1 = 0 // m/s //inlet velocity
+V2 = 360 // m/s //exit velocity
+g = 9.81 // m/s^2
+delta_z = 0 //m //z2-z1
+
+delta_h = ((Q-W_x)/m)-g*delta_z*.001-((V2^2-V1^2)/2000) //KJ/Kg //change in enthalpy
+printf("Change in enthalpy = %.2f KJ/kg\n",delta_h)
+
+//Part (a)
+printf("\nPart(b)\n\n")
+V1 = 60 // m/s //inlet velocity
+V2 = 360 // m/s //exit velocity
+g = 9.81 // m/s^2
+delta_z = 0 //m //z2-z1
+
+delta_h = ((Q-W_x)/m)-g*delta_z*.001-((V2^2-V1^2)/2000) //KJ/Kg //change in enthalpy
+printf("Change in enthalpy = %.2f KJ/kg\n",delta_h)
diff --git a/2507/CH5/EX5.3/Ex5_3.sce b/2507/CH5/EX5.3/Ex5_3.sce
new file mode 100755
index 000000000..7dd360ecb
--- /dev/null
+++ b/2507/CH5/EX5.3/Ex5_3.sce
@@ -0,0 +1,33 @@
+clc
+clear
+printf("Example 5.3 | Page number 122 \n\n");
+//Part(a) Determine the temperature of air at inlet to the turbine
+//PArt(b) Determine power developed by turbine
+
+//Given Data
+mA = 0.8 // kg/s //flow rate of stream A
+pA = 15e5 // N/m^2 //Pressure of stream A
+tA = 250 //°C //temperature of stream A
+
+
+mB = 0.5 // kg/s //flow rate of stream B
+pB = 15e5 // N/m^2 //Pressure of stream B
+tB = 200 //°C //temperature of stream B
+
+Q = 0 //No heat loss
+
+p1 = 10e5 // N/m^2 //pressure supply of chamber
+t2 = 30 //°C //exhaust air temperature from turbine
+
+Cv = 0.718 // KJ/kgK //heat capacity at constant volume
+Cp = 1 // KJ/kgK //heat capacity at constant pressure
+
+//solution
+//Part(a)
+printf("Part (a)\n")
+t1 = ((mA*Cp*tA)+(mB*Cp*tB))/((mA*Cp)+(mB*Cp)) // °C //the temperature of air at inlet to the turbine
+printf("The temperature of air at inlet to the turbine = %.2f °C\n",t1);
+//Part(b)
+printf("\nPart (b)\n")
+WT = -1*(mA+mB)*Cp*(t2-t1) // °kW //power developed by turbine
+printf("Power developed by turbine = %.2f kW",WT);
diff --git a/2507/CH5/EX5.4/Ex5_4.sce b/2507/CH5/EX5.4/Ex5_4.sce
new file mode 100755
index 000000000..58d505d56
--- /dev/null
+++ b/2507/CH5/EX5.4/Ex5_4.sce
@@ -0,0 +1,24 @@
+clc
+clear
+printf("Example 5.4 | Page number 123 \n\n");
+//Find inlet and exit velocities
+
+//Given Data
+d1 = 0.15 //m //inlet diameter
+m = 4000/3600 // kg/s //flow rate
+v1 = 0.285 //m^3/kg //specific volume at entry
+d2 = 0.25 //m //exit diameter
+v2 = 15 // m^3/kg //specific volume at exit
+
+//Solution
+
+A1 = %pi*d1^2/4 //m^2 //inlet cross sectional area
+A2 = %pi*d2^2/4 // m^2 // exit cross sectional area
+printf("Inlet cross sectional area (A1)= %.5f m^2\n",A1);
+printf("Exit cross sectional area (A2)= %.5f m^2\n",A2);
+
+V1 = m*v1/A1 //m/s //inlet velocity
+V2 = m*v2/A2 //m/s //exit velocity
+
+printf("\nInlet velocity = %.1f m/s",V1);
+printf("\nExit velocity = %.1f m/s",V2);
diff --git a/2507/CH5/EX5.5/Ex5_5.sce b/2507/CH5/EX5.5/Ex5_5.sce
new file mode 100755
index 000000000..06f910160
--- /dev/null
+++ b/2507/CH5/EX5.5/Ex5_5.sce
@@ -0,0 +1,16 @@
+clc
+clear
+printf("Example 5.5 | Page number 125 \n\n");
+//Find air temperature after throttling
+
+//Given Data
+p1 = 10//bar //inlet pressure
+t1 = 300 //°C //inlet temperature
+
+p2 = 0.1 //bar //exit pressure
+Cp = 1 //kJ/kgK // heat capacity at constant pressure
+//Solution
+//Adiabatic process
+delta_h = 0 //change in enthalpy
+t2 = delta_h/Cp + t1
+printf("Temperature of air after throttling = %.1f °C",t1)
diff --git a/2507/CH5/EX5.6/Ex5_6.sce b/2507/CH5/EX5.6/Ex5_6.sce
new file mode 100755
index 000000000..a8f7e7820
--- /dev/null
+++ b/2507/CH5/EX5.6/Ex5_6.sce
@@ -0,0 +1,16 @@
+clc
+clear
+printf("Example 5.6 | Page number 126 \n\n");
+// Calculate work input to compressor
+
+//Given Data
+p1 = 1e5 // N/m^2 //inlet pressure
+v1 = 0.08 //m^3/kg // specific volume at inlet
+p2 = 7e5 // N/m^2 //exit pressure
+v2 = 0.016 // m^3/kg //specific volume at exit
+u1 = 48 // kJ/kg // internal energy at inlet
+u2 = 200 // kJ/kg // internal energy at exit
+Q = -120 // kJ/kg // heat loss
+//Solution
+Wc = ((u2 - u1) + (p2*v2 - p1*v1)*.001 - Q)*-1 // kJ/kg // work input to compressor
+printf("Work input to compressor (Wc) = %.1f kJ/kg",Wc)
diff --git a/2507/CH5/EX5.7/Ex5_7.sce b/2507/CH5/EX5.7/Ex5_7.sce
new file mode 100755
index 000000000..ee2e20e3f
--- /dev/null
+++ b/2507/CH5/EX5.7/Ex5_7.sce
@@ -0,0 +1,13 @@
+clc
+clear
+printf("Example 5.7 | Page number 128 \n\n");
+//Find mass flow rate of cooling water
+mh = 9.45 // kg/s // flow rate of steam
+h_h2 = 140 // kJ/kg // enthalpy of condensate
+h_h1 = 2570 // kJ/kg // inlet enthalpy of steam
+t1 = 25 // °C //inlet temperature of cooling water
+t2 = 36 // °C //exit temperature of cooling water
+c = 4.189 // kJ/kg deg // specific heat of water
+//Solution
+mc = -1*(mh*(h_h2-h_h1))/(c*(t2-t1)) // kg/s //mass flow rate of cooling water
+printf("Mass flow rate of cooling water = %.2f kg/s",mc)
diff --git a/2507/CH5/EX5.8/Ex5_8.sce b/2507/CH5/EX5.8/Ex5_8.sce
new file mode 100755
index 000000000..dd8510634
--- /dev/null
+++ b/2507/CH5/EX5.8/Ex5_8.sce
@@ -0,0 +1,14 @@
+clc
+clear
+printf("Example 5.8 | Page number 129 \n\n");
+//Redo example 5.7 for heat loss 10% of heat transferred
+mh = 9.45 // kg/s // flow rate of steam
+h_h2 = 140 // kJ/kg // enthalpy of condensate
+h_h1 = 2570 // kJ/kg // inlet enthalpy of steam
+t1 = 25 // °C //inlet temperature of cooling water
+t2 = 36 // °C //exit temperature of cooling water
+c = 4.189 // kJ/kg deg // specific heat of water
+fractionalheatloss = 0.1 // fractional heat loss
+//Solution
+mc = -1*((1-fractionalheatloss)*mh*(h_h2-h_h1))/(c*(t2-t1)) // kg/s //mass flow rate of cooling water
+printf("Mass flow rate of cooling water = %.1f kg/s",mc)
diff --git a/2507/CH5/EX5.9/Ex5_9.sce b/2507/CH5/EX5.9/Ex5_9.sce
new file mode 100755
index 000000000..fe56e70e1
--- /dev/null
+++ b/2507/CH5/EX5.9/Ex5_9.sce
@@ -0,0 +1,10 @@
+clc
+clear
+printf("Example 5.9 | Page number 130 \n\n");
+//Find inlet air temperature
+V1 = 300 //m/s //inlet air velocity
+t2 = 100 //°C //exit air temperature
+V2 = 15 //m/s //exit air velocity
+//Solution
+t1 = t2 + .001*(V2^2 - V1^2)/2 // °C //inlet air temperature
+printf("Inlet air temperature = %.1f °C",t1)
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