diff options
Diffstat (limited to '3685/CH19')
48 files changed, 743 insertions, 0 deletions
diff --git a/3685/CH19/EX19.1/Ex19_1.sce b/3685/CH19/EX19.1/Ex19_1.sce new file mode 100644 index 000000000..df2067d27 --- /dev/null +++ b/3685/CH19/EX19.1/Ex19_1.sce @@ -0,0 +1,30 @@ + +clc
+T2 = 488
+T1 = 298
+n = 1.3
+R =8314/44
+rp = (T2/T1)^(n/(n-1))
+
+b = 0.12 // Bore of compressor
+L = 0.15 // Stroke of compressor
+V1 = (%pi/4)*(b)^2*L
+P1 = 120e03 // in kPa
+W = ((n*P1*V1)/(n-1))*(((rp)^((n-1)/n))-1)
+P = (W*1200*0.001)/60
+
+V1_dot = V1*(1200/60)
+m_dot = (P1*V1_dot)/(R*T1)
+
+rp_1 = rp^2
+V2 = (1/rp)^(1/n)*V1
+d = sqrt((V2*4)/(L*%pi))
+printf("\n Example 19.1\n")
+printf("\n Pressure ratio is %f ",rp)
+printf("\n Indicated power is %f kW",P)
+printf("\n Shaft power is %f kW",P/0.8)
+printf("\n Mass flow rate is %f kg/s",m_dot)
+printf("\n Pressure ratio when second stage is added is %f",rp_1)
+printf("\n Volume derived per cycle is V2 %f m^3",V2)
+printf("\n Second stage bore would be %f mm",d*1000)
+//The answers vary due to round off error
diff --git a/3685/CH19/EX19.1/Ex19_1.txt b/3685/CH19/EX19.1/Ex19_1.txt new file mode 100644 index 000000000..8883f78de --- /dev/null +++ b/3685/CH19/EX19.1/Ex19_1.txt @@ -0,0 +1,10 @@ +
+ Example 19.1
+
+ Pressure ratio is 8.476478
+ Indicated power is 11.249010 kW
+ Shaft power is 14.061263 kW
+ Mass flow rate is 0.072307 kg/s
+ Pressure ratio when second stage is added is 71.850672
+ Volume derived per cycle is V2 0.000328 m^3
+ Second stage bore would be 52.744274 mm
\ No newline at end of file diff --git a/3685/CH19/EX19.10/Ex19_10.sce b/3685/CH19/EX19.10/Ex19_10.sce new file mode 100644 index 000000000..c08d0de3f --- /dev/null +++ b/3685/CH19/EX19.10/Ex19_10.sce @@ -0,0 +1,38 @@ +clc
+// Given that
+N = 300 // Speed in RPM
+// Intake condition of compressor
+p1 = 0.98 // Pressure in bar
+T1 = 305 // Temperature in K
+
+p6 = 20// Delivery pressure in bar
+p3 = 5 // Intermediate pressure in bar
+C = .04 // Ratio of clearance volume to the stroke volume
+v = 3 // Volume flow rate of compressure in m^3/min
+p = 1 // pressure in bar
+t = 25 // Temperautre in degree centigrade
+n = 1.3 // Polytropic index
+R = 0.287 // Gas constant in kJ/kgK
+printf("\n Example 19.10\n")
+T = t+273
+r0 = 1+C // Where r0 = v1/vs
+r1 = C*(p3/p1)^(1/n)// Where r1 = v4/vs
+r2=r0-r1//Where r2 is the ratio of volume of air taken at 0.98 bar,305 k and vs
+r3 = r2*(T/T1)*p1/p // Where r3 is the ratio of volume of air taken at free air conditions and vs
+n_vol = r3
+m = p*(1e5)*(v/60)/(R*1000*T)
+T2 = T1*((p3/p1)^((n-1)/n))
+// For perfect intercooling
+T5 = T1
+p5 = p3
+T6 = T5*((p6/p5)^((n-1)/n))
+Wc = (n/(n-1))*m*R*((T2-T1)+(T6-T5))
+m_a_s = m*60/N
+v_fa_s = m_a_s *(R*1000)*T/(p*1e5)
+d = ((v_fa_s/n_vol)*(4/%pi))^(1/3)
+l = d // As given in the question
+P_iso = m*R*T1*(log(p6/p1))
+n_iso = P_iso/Wc
+printf("\n The power required to drive the compressor = %f kW,\n Diameter of cylinder = %f cm, \n Storke of the cylinder = %f cm,\n Isothermal efficiency = %f percent",Wc,d*100,l*100,n_iso*100)
+//The answers given in the book contain calculation error
+
diff --git a/3685/CH19/EX19.10/Ex19_10.txt b/3685/CH19/EX19.10/Ex19_10.txt new file mode 100644 index 000000000..97fd91755 --- /dev/null +++ b/3685/CH19/EX19.10/Ex19_10.txt @@ -0,0 +1,7 @@ +
+ Example 19.10
+
+ The power required to drive the compressor = 18.484703 kW,
+ Diameter of cylinder = 24.539171 cm,
+ Storke of the cylinder = 24.539171 cm,
+ Isothermal efficiency = 83.495502 percent
\ No newline at end of file diff --git a/3685/CH19/EX19.11/Ex19_11.sce b/3685/CH19/EX19.11/Ex19_11.sce new file mode 100644 index 000000000..f87bf6152 --- /dev/null +++ b/3685/CH19/EX19.11/Ex19_11.sce @@ -0,0 +1,21 @@ +clc
+// Given that
+p1 = 1 // Intake pressure of compressor in bar
+T1 = 298 // Intake temperature in K
+p_d = 36 // Delivery pressure in bar
+T2 = 390 // Maximum temperature in any stage in K
+n = 1.3 // Polytropic index
+R = 0.287
+printf("\n Example 19.11\n")
+r = (T2/T1)^(n/(n-1))
+N = ceil(r)
+p2 = (p_d/p1)^(1/N)
+p3 = (p_d/p1)^(2/N)
+p4 = (p_d/p1)^(3/N)
+Wc = (N*n*R*T1/(n-1))*((p_d/p1)^((n-1)/(N*n))-1)
+Wc_ = (n/(n-1))*(1*R*T1)*((p_d/p1)^((n-1)/n)- 1)
+T = T1*((p2/p1)^((n-1)/n))
+printf("\n No of stages for min power input = %d,\n Power required = %f kW/kg air,\n The power required for a single stage compressor = %f kW,\n Maximum temperature in any stage = %f K",N,Wc,Wc_,T)
+//The answers given in the book contain round off error
+
+
diff --git a/3685/CH19/EX19.11/Ex19_11.txt b/3685/CH19/EX19.11/Ex19_11.txt new file mode 100644 index 000000000..45764c428 --- /dev/null +++ b/3685/CH19/EX19.11/Ex19_11.txt @@ -0,0 +1,7 @@ +
+ Example 19.11
+
+ No of stages for min power input = 4,
+ Power required = 340.466490 kW/kg air,
+ The power required for a single stage compressor = 476.745441 kW,
+ Maximum temperature in any stage = 366.440061 K
\ No newline at end of file diff --git a/3685/CH19/EX19.12/Ex19_12.sce b/3685/CH19/EX19.12/Ex19_12.sce new file mode 100644 index 000000000..214d10188 --- /dev/null +++ b/3685/CH19/EX19.12/Ex19_12.sce @@ -0,0 +1,21 @@ +clc
+// Given that
+p1 = 700 // Intake pressure of compressor in kPa
+t1 = 38 // Intake temperature in degree centigrade
+c = 0.4 // Ratio of cutoff volume to stroke volume
+p3 = 112 // Back pressure in kPa
+r = 0.85 // Ratio of area of actual indicator diagram to the outlined in the question
+n = 1.3 // Polytropic index
+R = 0.287
+m = 1.25 // Air mass in kg
+printf("\n Example 19.12\n")
+T1 = t1+273
+T2 = T1/((1/c)^(n-1))
+p2 = p1*(c^n)
+V2 = m*R*T2/p2
+v2 = V2/m
+A = R*T1 + R*(T1-T2)/(n-1) - p3*v2
+Io = A*r*m
+printf("\n Indicated output = %f kJ",Io)
+// The answer given in the book vary due to round off error
+
diff --git a/3685/CH19/EX19.12/Ex19_12.txt b/3685/CH19/EX19.12/Ex19_12.txt new file mode 100644 index 000000000..bcf9f1f4a --- /dev/null +++ b/3685/CH19/EX19.12/Ex19_12.txt @@ -0,0 +1,4 @@ +
+ Example 19.12
+
+ Indicated output = 132.877965 kJ
\ No newline at end of file diff --git a/3685/CH19/EX19.13/Ex19_13.sce b/3685/CH19/EX19.13/Ex19_13.sce new file mode 100644 index 000000000..55f1658e6 --- /dev/null +++ b/3685/CH19/EX19.13/Ex19_13.sce @@ -0,0 +1,30 @@ +
+clc
+// Given that
+d = 450 // Bore of low pressure cylinder in mm
+l = 300 // Stroke in mm
+c = 0.05 // Ratio of clearance volume to swept volume
+p1 = 1 // Intake pressure in bar
+t1 = 18 // Intake temperature in degree centigrade
+p4 = 15 // Delivery pressure in bar
+n = 1.3 // Compression and expansion index
+R = 0.29 // Gas constant in kJ/kgK
+printf("\n Example 19.13\n")
+T1 = t1+273
+r = (p4/p1)^(1/3)
+p2 = p1*r
+p3 = p2*r
+Vs = (%pi/4)*((d*1e-3)^2)*(l*1e-3)
+V11 = c*Vs
+V1 = Vs +V11
+V12 = V11*((r)^(1/n))
+Vs_e = V1 - V12
+T3 = T1
+T5 = T3
+T6 = T1*(r^((n-1)/n))
+t6 = T6-273
+V6_7 = (p1/p4)*(T6/T1)*(V1 - V12)
+W = (3*n*R*T1/(n-1))*((p2/p1)^((n-1)/n)-1)
+printf("\n The intermediate pressure are - \n p2 = %f bar,\n p3 = %f bar,\n The effective swept volume = %f m^3,\n Temperature of air delivered per stroke at 15 bar = %f degree centigrade,\n The work done per kg of air = %f kJ",p2,p3,Vs,t6,W)
+// The answers given in the book vary due to round off error
+
diff --git a/3685/CH19/EX19.13/Ex19_13.txt b/3685/CH19/EX19.13/Ex19_13.txt new file mode 100644 index 000000000..acf5520d5 --- /dev/null +++ b/3685/CH19/EX19.13/Ex19_13.txt @@ -0,0 +1,9 @@ +
+ Example 19.13
+
+ The intermediate pressure are -
+ p2 = 2.466212 bar,
+ p3 = 6.082202 bar,
+ The effective sweft volume = 0.047713 m^3,
+ Temperature of air delivered per stroke at 15 bar = 85.394674 degree centigrade,
+ The work done per kg of air = 254.077922 kJ
\ No newline at end of file diff --git a/3685/CH19/EX19.14/Ex19_14.sce b/3685/CH19/EX19.14/Ex19_14.sce new file mode 100644 index 000000000..9affb4883 --- /dev/null +++ b/3685/CH19/EX19.14/Ex19_14.sce @@ -0,0 +1,17 @@ +clc
+// Given that
+p1 = 1.013 // Inlet pressure in bar
+r = 1.5 // Pressure ratio
+Vs = 0.03 // Induce volume of air in m^3/rev
+gama = 1.4
+printf("\n Example 19.14\n")
+p2 = p1*r
+W = (p2-p1)*Vs*100
+pi = (p1+p2)/2
+A_A = (gama/(gama-1))*(p1*Vs)*((pi/p1)^((gama-1)/gama)-1)*100
+Vb = Vs *((p1/pi)^(1/gama))
+A_B = (p2-pi)*Vb*100
+Wr = A_A + A_B
+printf("\n Work input = %f kJ/rev,\n Work input for a vane-type compressor = %f kJ/rev",W,Wr)
+// The answers given in the book vary due to round off error
+
diff --git a/3685/CH19/EX19.14/Ex19_14.txt b/3685/CH19/EX19.14/Ex19_14.txt new file mode 100644 index 000000000..1e8c0f0bb --- /dev/null +++ b/3685/CH19/EX19.14/Ex19_14.txt @@ -0,0 +1,5 @@ +
+ Example 19.14
+
+ Work input = 1.519500 kJ/rev,
+ Work input for a vane-type compressor = 1.348030 kJ/rev
\ No newline at end of file diff --git a/3685/CH19/EX19.15/Ex19_15.sce b/3685/CH19/EX19.15/Ex19_15.sce new file mode 100644 index 000000000..59e407978 --- /dev/null +++ b/3685/CH19/EX19.15/Ex19_15.sce @@ -0,0 +1,20 @@ +clc
+// Given that
+m = 1 // Mass flow rate in kg/s
+r = 2 // Prssure ratio of blower
+t1 = 70 // Inlet temperature in degree centigrade
+p1 = 1 // Inlet pressure in bar
+R = 0.29 // Gas constant in kJ/kgK
+x = 0.7 // Reduction in pressure ratio and intake volume
+gama = 1.4
+printf("\n Example 19.15\n")
+T1 = t1+273
+V = m*R*T1/(p1*100)
+P = V*(p1*r-p1)*100
+p2 = p1*((1/x)^(gama))
+V2 = x*V
+P_ = (gama/(gama-1))*(p1*100*V)*((p2/p1)^((gama-1)/gama)-1) + V2*(p1*r-p2)*100
+
+printf("\n Power required to drive the blower = %f kW,\n Power required = %f kW",P,P_)
+// The answers given in the book vary due to round off error
+
diff --git a/3685/CH19/EX19.15/Ex19_15.txt b/3685/CH19/EX19.15/Ex19_15.txt new file mode 100644 index 000000000..9ab2c2f40 --- /dev/null +++ b/3685/CH19/EX19.15/Ex19_15.txt @@ -0,0 +1,5 @@ +
+ Example 19.15
+
+ Power required to drive the blower = 99.470000 kW,
+ Power required = 77.922089 kW
\ No newline at end of file diff --git a/3685/CH19/EX19.16/Ex19_16.sce b/3685/CH19/EX19.16/Ex19_16.sce new file mode 100644 index 000000000..de2ed5ca3 --- /dev/null +++ b/3685/CH19/EX19.16/Ex19_16.sce @@ -0,0 +1,23 @@ +clc
+// Given that
+r1 = 2.5 // Pressure ratio of compressor for first stage
+r2 = 2.1 // Pressure ratio of compressor for second stage
+m = 5 // Mass flow rate of air in kg/s
+t1 = 10 // Inlet temperature in degree centigrade
+p1 = 1.013 // Inlet pressure in bar
+td = 50 // Temperature drop in intercooler in degree centigreade
+n_iso = .85 // Isentropic efficiency
+cp = 1.005 // Heat capacity of air in kJ/kgK
+x = 0.7 // Reduction in pressure ratio and intake volume
+gama = 1.4 // Ratio of heat capacities for air
+printf("\n Example 19.16\n")
+T1 = t1+273
+T2s = T1*((r1)^((gama-1)/gama))
+T2 = T1 + (T2s-T1)/n_iso
+T3 = T2 - td
+T4s = T3*((r2)^((gama-1)/gama))
+T4 = T3 + (T4s-T3)/n_iso
+P = m*cp*((T2-T1)+(T4-T3))
+printf("\n Actual temperature at the end of first stage = %f K,\n Actual temperature at the end of second stage = %f K,\n The total compressor power = %f kW",T2,T4,P)
+// The answers given in the book vary due to round off error
+
diff --git a/3685/CH19/EX19.16/Ex19_16.txt b/3685/CH19/EX19.16/Ex19_16.txt new file mode 100644 index 000000000..a6aaa13a0 --- /dev/null +++ b/3685/CH19/EX19.16/Ex19_16.txt @@ -0,0 +1,6 @@ +
+ Example 19.16
+
+ Actual temperature at the end of first stage = 382.637049 K,
+ Actual temperature at the end of second stage = 425.041961 K,
+ The total compressor power = 965.010854 kW
\ No newline at end of file diff --git a/3685/CH19/EX19.17/Ex19_17.sce b/3685/CH19/EX19.17/Ex19_17.sce new file mode 100644 index 000000000..e82887aaa --- /dev/null +++ b/3685/CH19/EX19.17/Ex19_17.sce @@ -0,0 +1,23 @@ +clc
+// Given that
+r = 2.5 // Static pressure ratio of supercharger
+p1 = 0.6 // Static inlet pressure in bar
+t1 = 5 // Static inlet temperature in degree centigrade
+A_r = 13 // Air-fuel ratio
+m = 0.04 // The rate of fuel consumed by the engine in kg/s
+gama= 1.39 // For air-fuel mixture
+cp = 1.005 // Heat capacity for air-fuel mixture in kJ/kgk
+n_iso = .84 // Isentropic efficiency of compressor
+v = 120 // Exit velocity from the compressor in m/s
+printf("\n Example 19.17\n")
+T1 = t1+273
+T2s = T1*((r)^((gama-1)/gama))
+T2 = T1 +(T2s-T1)/n_iso
+m_g = m*(A_r+1)
+P = m_g*cp*(T2-T1)
+T02 = T2 + (v^2)/(2*cp*1000)
+t02 = T02-273
+p02 = p1*r*((T02/T2)^(gama/(gama-1)))*100
+printf("\n Power required to drive the compressor = %f kW,\n Stagnatio temperature = %f degree centigrade,\n Stagnation pressure = %f kPa",P,t02,p02)
+// The answers given in the book vary due to round off error
+
diff --git a/3685/CH19/EX19.17/Ex19_17.txt b/3685/CH19/EX19.17/Ex19_17.txt new file mode 100644 index 000000000..753b07b80 --- /dev/null +++ b/3685/CH19/EX19.17/Ex19_17.txt @@ -0,0 +1,6 @@ +
+ Example 19.17
+
+ Power required to drive the compressor = 54.603965 kW,
+ Stagnatio temperature = 109.186150 degree centigrade,
+ Stagnation pressure = 160.465578 kPa
\ No newline at end of file diff --git a/3685/CH19/EX19.18/Ex19_18.sce b/3685/CH19/EX19.18/Ex19_18.sce new file mode 100644 index 000000000..725b39fd6 --- /dev/null +++ b/3685/CH19/EX19.18/Ex19_18.sce @@ -0,0 +1,27 @@ +clc
+// Given that
+N = 10000 // Speed in RPM
+V = 1.2 // Volume flow rate of free air in m^3/s
+p1 = 1 // Inlet pressure in bar
+t1 = 27 // Inlet temperature in degree centigrade
+r = 5 // Pressure ratio
+vf = 60 // Velocity flow rate in m/s
+sigma = 0.9 // Slip factor
+n_iso = 0.85 // Isentropic efficiency
+gama = 1.4
+R = 0.287
+cp = 1.005
+printf("\n Example 19.18\n")
+T1 = t1+273
+T2s = T1*((r)^((gama-1)/gama))
+T2 = T1 +(T2s-T1)/n_iso
+m = p1*100*V/(R*288)
+Wc = m*cp*(T2-T1)
+Vb2 = (Wc*1000/(m*sigma))^(1/2)
+D = Vb2*60/(%pi*N)
+Vb1 = Vb2/2
+beta1 = atand(vf/Vb1)
+alpha = atand(vf/(sigma*Vb2))
+printf("\n The temperature of air at outlet = %f degree centigrade,\n Power input = %f kW,\n Diameter of impeller = %f m, \n Blade inlet angle = %d degree,\n Diffuser inlet angle = %f degree ",T2-273,Wc,D,beta1,alpha)
+// The answers given in the book vary due to round off error
+
diff --git a/3685/CH19/EX19.18/Ex19_18.txt b/3685/CH19/EX19.18/Ex19_18.txt new file mode 100644 index 000000000..5cf3a78c8 --- /dev/null +++ b/3685/CH19/EX19.18/Ex19_18.txt @@ -0,0 +1,8 @@ +
+ Example 19.18
+
+ The temperature of air at outlet = 233.053980 degree centigrade,
+ Power input = 300.644961 kW,
+ Diameter of impeller = 0.916123 m,
+ Blade inlet angle = 14 degree,
+ Diffuser inlet angle = 7.912359 degree
\ No newline at end of file diff --git a/3685/CH19/EX19.19/Ex19_19.sce b/3685/CH19/EX19.19/Ex19_19.sce new file mode 100644 index 000000000..6de6a1abf --- /dev/null +++ b/3685/CH19/EX19.19/Ex19_19.sce @@ -0,0 +1,31 @@ +clc
+// Given that
+N = 264 // Speed in RPS
+sigma = 0.91 // Slip factor
+d = 0.482 // Impeller diameter in m
+D = 0.306 // Impeller eye diameter
+D_ = 0.153 // Impeller root eye diameter in m
+vf = 138 // Uniform axial inlet velocity in m/s
+V = 1.2 // Volume flow rate of free air in m^3/s
+m = 9.1 // Air mass flow rate in kg/s
+T1 = 294 // Inlet air stagnation temperature in K
+n_iso = 0.8 // Total head isentropic efficiency
+n_mech = 0.98 // Mechanical efficiency
+gama = 1.4 // Ratio of heat capacities
+cp = 1.006 // Heat capacity in kJ/kgK
+printf("\n Example 19.19\n")
+Wc = m*sigma*(2*%pi*d*N/2)/1000
+P_e = Wc/n_mech
+delta_T = Wc/(m*cp)
+delta_T_ideal = delta_T*n_iso
+T2_i = delta_T_ideal + T1
+r = (T2_i/T1)^(gama/(gama-1)) // Where r = p02/p01
+Vb = 2*%pi*N*D/2
+V_er = (2*%pi*N*D_/2)
+beta1 = atand(vf/Vb)
+beta2 = atand(vf/V_er)
+beta1_ = (beta1 - floor(beta1))*60
+beta2_ = (beta2 - floor(beta2))*60
+printf("\n Total head pressure ratio = %f, \n The required power at input shaft = %f kW,\n Inlet angle at the root = %d degree and %d minute,\n Inlet angle at the tip = %d degree and %d minute",r,P_e,floor(beta1),beta1_,floor(beta2),beta2_)
+// The answers given in the book for total head pressure ratio and required power at input shaft contain calculation error
+
diff --git a/3685/CH19/EX19.19/Ex19_19.txt b/3685/CH19/EX19.19/Ex19_19.txt new file mode 100644 index 000000000..39e7bf658 --- /dev/null +++ b/3685/CH19/EX19.19/Ex19_19.txt @@ -0,0 +1,7 @@ +
+ Example 19.19
+
+ Total head pressure ratio = 1.003448,
+ The required power at input shaft = 3.377984 kW,
+ Inlet angle at the root = 28 degree and 32 minute,
+ Inlet angle at the tip = 47 degree and 24 minute
\ No newline at end of file diff --git a/3685/CH19/EX19.2/Ex19_2.sce b/3685/CH19/EX19.2/Ex19_2.sce new file mode 100644 index 000000000..1c1d37f78 --- /dev/null +++ b/3685/CH19/EX19.2/Ex19_2.sce @@ -0,0 +1,13 @@ +
+clc
+c = 0.05 // Clearance volume
+p1 = 96 // Inlet pressure in bar
+p2 = 725 // Outlet pressure in bar
+pa = 101.3 // Atmospheric pressure
+Ta = 292 // Atmospheric temperature in kelvin
+T1 = 305 // Inlet temperature in Kelvin
+n = 1.3 // polytropic index
+printf("\n Example 19.2 \n ")
+n_v = (1+c-c*((p2/p1)^(1/n)))*(p1/pa)*(Ta/T1)
+printf("\n Volumetric efficiency of system is %f percent", n_v*100)
+// Answer is not mentioned in book
diff --git a/3685/CH19/EX19.2/Ex19_2.txt b/3685/CH19/EX19.2/Ex19_2.txt new file mode 100644 index 000000000..6c1a227ac --- /dev/null +++ b/3685/CH19/EX19.2/Ex19_2.txt @@ -0,0 +1,4 @@ +
+ Example 19.2
+
+ Volumetric efficiency of system is 73.779396 percent
\ No newline at end of file diff --git a/3685/CH19/EX19.20/Ex19_20.sce b/3685/CH19/EX19.20/Ex19_20.sce new file mode 100644 index 000000000..eda9554e2 --- /dev/null +++ b/3685/CH19/EX19.20/Ex19_20.sce @@ -0,0 +1,26 @@ +clc
+// Given that
+N = 16000 // Speed in RPM
+t1 = 17 // Intake temperture of gas in degree centigrade
+rp = 4 // Pressure ratio
+sigma = 0.85// Slip factor
+n_iso = 0.82 // Isentropic efficiency
+alpha_wirl = 20 // Pre-wirl angle in degree
+d1 = 200 // Mean diameter of impeller eye in mm
+V1 = 120 //Absolute air velocity in m/s
+gama = 1.4 // Ratio of heat capacities
+cp = 1.005 // Heat capacity in kJ/kgK
+printf("\n Example 19.20\n")
+T1 = t1 + 273
+T2s = T1*((rp)^((gama-1)/gama))
+delta_Ts = T2s-1
+delta_T = delta_Ts/n_iso
+Wc = 1 *cp*delta_T
+Vb1 = (%pi*d1*(1e-3)*N)/60
+Vw1 = V1*sind(alpha_wirl)
+Vb2 = 459.78 // By solving quadratic equation 172.81e3=0.85*Vb2^2-167.55*41.05
+d2 = Vb2*60/(%pi*N)
+
+printf("\n Impeller tip diameter = %f mm",d2*1000)
+// The answer given in the book varies due to round off error
+
diff --git a/3685/CH19/EX19.20/Ex19_20.txt b/3685/CH19/EX19.20/Ex19_20.txt new file mode 100644 index 000000000..f264e027f --- /dev/null +++ b/3685/CH19/EX19.20/Ex19_20.txt @@ -0,0 +1,4 @@ +
+ Example 19.20
+
+ Impeller tip diameter = 548.821948 mm
\ No newline at end of file diff --git a/3685/CH19/EX19.21/Ex19_21.sce b/3685/CH19/EX19.21/Ex19_21.sce new file mode 100644 index 000000000..b8bd2ee94 --- /dev/null +++ b/3685/CH19/EX19.21/Ex19_21.sce @@ -0,0 +1,23 @@ +clc
+// Given that
+m = 2.5 // Mass flow rate in kg/s
+p1 = 1 // Inlet pressure in bar
+T1 = 300 // Inlet temperature in bar
+n_s = 0.88 // Stage efficiency
+Wc = 600 // Power input in kW
+delta_t = 21 // Temperature rise in first stage in degree centigrade
+gama = 1.4 // Ratio of heat capacities
+cp = 1.005 // Heat capacity in kJ/kgK
+printf("\n Example 19.21\n")
+x = n_s*gama/(gama-1)// Where x = (n/(n-1))
+T = Wc/(m*cp)+T1
+p = p1*((T/T1)^(x))
+T2 = T1 + n_s*delta_t
+r = ((T2/T1)^(gama/(gama-1)))// Where r = p2/p1
+N = log(p/p1)/log(r)
+N_ = ceil(N)
+Ts = T1*(p/p1)^((gama-1)/gama)
+n_inter = (Ts-T1)/(T-T1)
+printf("\n The delivery pressure = %f bar,\n The no of stages = %d,\n The internal efficiency = %f ",p,N_,n_inter)
+
+
diff --git a/3685/CH19/EX19.21/Ex19_21.txt b/3685/CH19/EX19.21/Ex19_21.txt new file mode 100644 index 000000000..cfcda31a1 --- /dev/null +++ b/3685/CH19/EX19.21/Ex19_21.txt @@ -0,0 +1,6 @@ +
+ Example 19.21
+
+ The delivery pressure = 6.071253 bar,
+ The no of stages = 9,
+ The internal efficiency = 0.846898
\ No newline at end of file diff --git a/3685/CH19/EX19.22/Ex19_22.sce b/3685/CH19/EX19.22/Ex19_22.sce new file mode 100644 index 000000000..a86de067f --- /dev/null +++ b/3685/CH19/EX19.22/Ex19_22.sce @@ -0,0 +1,26 @@ +clc
+// Given that
+D = 0.5 // Mean diameter of impeller in m
+N = 15000 // Speed in RPM
+Vf = 230 // Velocity of flow in m/s
+p1 = 1 // Inlet pressure in bar
+T1 = 300 // Inlet temperature in K
+Vw1 = 80 // Velocity of whirl at inlet in m/s
+n_s = 0.88 // Stage efficiency
+rp = 1.5 // Pressure ratio
+gama = 1.4
+cp = 1.0005
+printf("\n Example 19.22\n")
+Vb = (%pi*D*N/60)
+Ts = T1*((rp)^((gama-1)/gama))
+T = T1 + (Ts-T1)/n_s
+Wc = cp*(T-T1)
+Vw2 = Vw1 + (Wc*1000)/(Vb)
+beta1 = atand(Vf/(Vb-Vw1))
+beta2 = atand(Vf/(Vb-Vw2))
+theta = beta2-beta1
+R = 1-((Vw1+Vw2)/(2*Vb))
+
+printf("\n Fluid deflection angle = %f degree,\n Power input = %f kJ/kg,\n The degree of reaction = %d percent",theta,Wc,R*100)
+// The answers given in the book vary because of round off error
+
diff --git a/3685/CH19/EX19.22/Ex19_22.txt b/3685/CH19/EX19.22/Ex19_22.txt new file mode 100644 index 000000000..3de4477f0 --- /dev/null +++ b/3685/CH19/EX19.22/Ex19_22.txt @@ -0,0 +1,6 @@ +
+ Example 19.22
+
+ Fluid deflection angle = 11.812325 degree,
+ Power input = 41.892843 kJ/kg,
+ The degree of reaction = 66 percent
\ No newline at end of file diff --git a/3685/CH19/EX19.23/Ex19_23.sce b/3685/CH19/EX19.23/Ex19_23.sce new file mode 100644 index 000000000..516b21a55 --- /dev/null +++ b/3685/CH19/EX19.23/Ex19_23.sce @@ -0,0 +1,21 @@ +clc
+// Given that
+v = 5 // Volume flow rate in m^3/s
+d = 1 // Mean impeller diameter in m
+D = 0.6 // Hub diameter in m
+N = 600 // Rotational speed in RPM
+h = 35 // Theoratical head in mm
+rho = 1.2 // Density of air in kg/m^3
+rho_w = 1000 // Density of water in kg/m^3
+printf("\n Example 19.23\n")
+Vf = v*4/(%pi*(d^2 - D^2))
+Vb = (%pi*d*N/60)
+Vb_ = (%pi*D*N/60)
+H = h/rho
+Vw2 = H*9.81/(Vb)
+Vw2_ = H*9.81/(Vb_)
+beta_tip = atand(Vf/(Vb-Vw2))
+beta_hub = atand(Vf/(Vb_-Vw2_))
+printf("\n Blade angle at the tip = %f degree,\n Blade angle at the hub = %f degree",beta_tip,beta_hub)
+// The answers given in the book vary because of round off error
+
diff --git a/3685/CH19/EX19.23/Ex19_23.txt b/3685/CH19/EX19.23/Ex19_23.txt new file mode 100644 index 000000000..55e843789 --- /dev/null +++ b/3685/CH19/EX19.23/Ex19_23.txt @@ -0,0 +1,5 @@ +
+ Example 19.23
+
+ Blade angle at the tip = 24.031920 degree,
+ Blade angle at the hub = 69.747752 degree
\ No newline at end of file diff --git a/3685/CH19/EX19.24/Ex19_24.sce b/3685/CH19/EX19.24/Ex19_24.sce new file mode 100644 index 000000000..e1007ffe5 --- /dev/null +++ b/3685/CH19/EX19.24/Ex19_24.sce @@ -0,0 +1,32 @@ +clc
+// Given that
+N0 = 9000 // Rotational speed in RPM
+Q = 6 // Volume flow rate in m^3/s
+p1 = 1 // Initial pressure in bar
+t1 = 25 // Initial temperature in degree centigrade
+p2 = 2.2 // Compressed pressure in bar
+n = 1.33 // Compression index
+Vf = 75 // Velocity of flow in m/s
+beta1 = 30 // Blade angle at inlet in degree
+beta2 = 55 // Blade angle at outlet in degree
+d = 0.75 // Diameter of impeller in m
+cp = 1.005
+printf("\n Example 19.24\n")
+T1 = t1+273
+T2 = T1*(p2/p1)^((n-1)/n)
+Wc = cp*(T2-T1)
+x = Wc // Where x = Vw2*Vb2
+y = Vf/tand(beta2)// Where y = Vb2-Vw2(Equation 1)
+z = (y^2 +4*x*1000)^(0.5) // Where z = Vw2+Vb2(Equation 2)
+// By solving Equation 1 and Equation 2
+Vb2 = (y+z)/2
+Vw2 = ((z-y)/2)
+N = Vb2*60/(%pi*d)
+Vb1 = Vf/tand(beta1)
+D1 = Vb1*60/(%pi*N)
+b1 = Q/(%pi*D1*Vf)
+Q_ = Q* (1/p2)*(T2/T1)
+b2 = Q_/(%pi*d*Vf)
+printf("\n Speed of impeller = %f RPM,\n Impeller width at inlet = %f cm,\n Impeller width at outlet = %f cm,",N,b1*100,b2*100)
+// The answers given in the book vary because of round off error
+
diff --git a/3685/CH19/EX19.24/Ex19_24.txt b/3685/CH19/EX19.24/Ex19_24.txt new file mode 100644 index 000000000..870155fd7 --- /dev/null +++ b/3685/CH19/EX19.24/Ex19_24.txt @@ -0,0 +1,6 @@ +
+ Example 19.24
+
+ Speed of impeller = 7181.025316 RPM,
+ Impeller width at inlet = 7.370608 cm,
+ Impeller width at outlet = 1.876801 cm,
\ No newline at end of file diff --git a/3685/CH19/EX19.3/Ex19_3.sce b/3685/CH19/EX19.3/Ex19_3.sce new file mode 100644 index 000000000..284d30c8d --- /dev/null +++ b/3685/CH19/EX19.3/Ex19_3.sce @@ -0,0 +1,36 @@ +clc
+P1 = 101.3e03
+P4 = P1 // in Pa
+P2 = 8*P1
+P3 = P2
+T1 = 288
+Vs = 2000
+V3 = 100
+Vc = V3
+V1 = Vs + Vc
+n = 1.25
+R = 287
+V4 = ((P3/P4)^(1/n))*V3
+W = ((n*P1*(V1-V4)*1e-06)/(n-1))*(((P2/P1)^((n-1)/n))-1)
+P = (W*800*0.001)/60
+
+m = (P1*(V1-V4)*1e-06)/(R*T1)
+m_dot = m*800
+
+FAD = (V1-V4)*1e-06*800
+
+Wt = P1*(V1-V4)*1e-06*log(P2/P1)
+n_isothermal = (Wt*800*0.001)/(P*60)
+
+Pi = P/0.85
+n_v =100*(V1-V4)/Vs
+printf("\n Example 19.3\n")
+printf("\n Indicated poer is %f kW",P)
+printf("\n Volumetric efficiency is %f percent",n_v)
+printf("\n Mass flow rate is %f kg/min",m_dot)
+printf("\n Free air delivery is %f m^3/min",FAD)
+printf("\n Isothermal efficiency is %f percent",100*n_isothermal)
+printf("\n Input power is %f kW",Pi)
+
+//The answers vary due to round off error
+
diff --git a/3685/CH19/EX19.3/Ex19_3.txt b/3685/CH19/EX19.3/Ex19_3.txt new file mode 100644 index 000000000..301360c08 --- /dev/null +++ b/3685/CH19/EX19.3/Ex19_3.txt @@ -0,0 +1,9 @@ +
+ Example 19.3
+
+ Indicated poer is 5.475656 kW
+ Volumetric efficiency is 78.609842 percent
+ Mass flow rate is 1.541459 kg/min
+ Free air delivery is 1.257757 m^3/min
+ Isothermal efficiency is 80.642806 percent
+ Input power is 6.441949 kW
\ No newline at end of file diff --git a/3685/CH19/EX19.4/Ex19_4.sce b/3685/CH19/EX19.4/Ex19_4.sce new file mode 100644 index 000000000..d9f366b71 --- /dev/null +++ b/3685/CH19/EX19.4/Ex19_4.sce @@ -0,0 +1,26 @@ +clc
+// Given that
+m = 3 // Mass flow rate in kg/min
+p1 = 1 // Initial pressure in bar
+T1 = 300 // Initial temperature in K
+p3 = 6 // Pressure after compression in bar
+p5 = 15 // Maximum pressure in bar
+N = 300 // Rpm of compressure
+n = 1.3 // Index of compression and expansion
+r = 1.5 // Stroke to bore ratio
+R = 287 // Gas constant of air
+t = 15 // Temperature in degree centigrade
+printf("\n Example 19.4\n")
+T = t+273
+Wc = (n/(n-1))*(m/60)*(R*(1e-3)*T1)*(((p3/p1)^((n-1)/n))-1)
+r1 = (p5/p1)^(1/n)// Where r1 = V1/Vc
+r2 = r1-1 // Where r2 = Vs/Vc
+r3 = (p3/p1)^(1/n)
+n_vol = (r1-r3)*(T/T1)/r2
+V = m*R*T/(2*(1e5)*N)
+Vs = V/n_vol
+d = (Vs*4/(%pi*r))^(1/3)
+l = r*d
+printf("\n Power input is %f kW, \n Volumetric efficiency is %f percent, \n Bore of the cylinder is %f m, \n Stroke of the cylinder is %f m",Wc,n_vol*100,d,l)
+//The answers vary due to round off error
+
diff --git a/3685/CH19/EX19.4/Ex19_4.txt b/3685/CH19/EX19.4/Ex19_4.txt new file mode 100644 index 000000000..160c83f58 --- /dev/null +++ b/3685/CH19/EX19.4/Ex19_4.txt @@ -0,0 +1,7 @@ +
+ Example 19.4
+
+ Power input is 9.552761 kW,
+ Volumetric efficiency is 55.465731 percent,
+ Bore of the cylinder is 0.184932 m,
+ Stroke of the cylinder is 0.277398 m
\ No newline at end of file diff --git a/3685/CH19/EX19.5/Ex19_5.sce b/3685/CH19/EX19.5/Ex19_5.sce new file mode 100644 index 000000000..d11090b9f --- /dev/null +++ b/3685/CH19/EX19.5/Ex19_5.sce @@ -0,0 +1,29 @@ +clc
+// Given that
+d = 15 // Diameter in cm
+l = 18 // Stroke in cm
+C = 0.04 // Ratio of clearance volume and sweft volume
+p1 = 1 // Pressure in bar
+t1 = 25 // Temperature in degree centigrade
+p2 = 8// Pressure in bar
+N = 1200 // Rpm of compressure
+W = 18 // Actual power input in kW
+m = 4 // Mass flow rate in kg/min
+R = 0.287
+printf("\n Example 19.5\n")
+T1 = t1+273
+v = R*T1/(p1*100)
+V = m*v
+Vs = (%pi/4)*((d*(1e-2))^2)*(l*1e-2)*N
+n_vol = V/Vs
+n = (log(p2/p1))/(log((1+C-n_vol)/C))
+// The value of n given in the example is wrong
+n = 1.573
+T2 = T1*(p2/p1)^((n-1)/n)
+Wc = (n/(n-1))*(m*R/60)*(T2-T1)
+n_mech = Wc/W
+W_isothermal = m*R*T1*log(p2/p1)/60
+n_iso = W_isothermal/W
+printf("\n Power required to drive the unit is %f kW,\n Isothermal efficiency is %f percent,\n Mechanical efficiency is %f percent",Wc,n_iso*100,n_mech*100)
+//The answers vary due to round off error
+
diff --git a/3685/CH19/EX19.5/Ex19_5.txt b/3685/CH19/EX19.5/Ex19_5.txt new file mode 100644 index 000000000..b115ad4a3 --- /dev/null +++ b/3685/CH19/EX19.5/Ex19_5.txt @@ -0,0 +1,6 @@ +
+ Example 19.5
+
+ Power required to drive the unit is 17.732605 kW,
+ Isothermal efficiency is 65.869006 percent,
+ Mechanical efficiency is 98.514474 percent
\ No newline at end of file diff --git a/3685/CH19/EX19.6/Ex19_6.sce b/3685/CH19/EX19.6/Ex19_6.sce new file mode 100644 index 000000000..4072365e8 --- /dev/null +++ b/3685/CH19/EX19.6/Ex19_6.sce @@ -0,0 +1,25 @@ +clc
+// Given that
+d = 40 // Diameter in cm
+l = 50 // Stroke in cm
+D = 5 // Piston rod diameter in cm
+C = 0.04 // Ratio of clearance volume and sweft volume
+p1 = 1 // Pressure in bar
+t1 = 15 // Temperature in degree centigrade
+p2 = 7.5// Pressure in bar
+N = 300 // Rpm of compressure
+n_vol = 0.8 // Volumetric efficiency
+n_mech = 0.95 // Mechanical efficiency
+n_iso = .7 // Isothermal efficiency
+R = 0.287
+printf("\n Example 19.6\n")
+Vs = (%pi/4)*((d*(1e-2))^2)*(l*(1e-2))
+Vs_ = (%pi/4)*(((d*(1e-2))^2)-(D*(1e-2))^2)*(l*1e-2)
+Vs_min = (Vs+Vs_)*2*N
+V1 = Vs_min*n_vol
+W_iso = p1*V1*(log(p2/p1))
+Win = W_iso/n_iso
+Wc = Win/n_mech
+printf("\n Power required to drive the compressure is %f kW",Wc)
+//The answers vary due to round off error
+
diff --git a/3685/CH19/EX19.6/Ex19_6.txt b/3685/CH19/EX19.6/Ex19_6.txt new file mode 100644 index 000000000..3036ee2b7 --- /dev/null +++ b/3685/CH19/EX19.6/Ex19_6.txt @@ -0,0 +1,4 @@ +
+ Example 19.6
+
+ Power required to drive the compressure is 181.333212 kW
\ No newline at end of file diff --git a/3685/CH19/EX19.7/Ex19_7.sce b/3685/CH19/EX19.7/Ex19_7.sce new file mode 100644 index 000000000..c1bdb8f2a --- /dev/null +++ b/3685/CH19/EX19.7/Ex19_7.sce @@ -0,0 +1,16 @@ +clc
+// Given that
+p1 = 1 // Pressure in bar
+t1 = 27 // Temperature in degree centigrade
+n = 1.3 // Index of the compression process
+p3 = 9// Pressure in bar
+R = 0.287
+printf("\n Example 19.7\n")
+T1 = t1+273
+p2 = sqrt(p1*p3)
+Wc = ((2*n*R*T1)/(n-1))*(((p2/p1)^((n-1)/n))-1)
+T2 = T1*((p2/p1)^((n-1)/n))
+H = 1.005*(T2-T1)
+printf("\n Minimum work done is %f kJ/kg,\n Heat rejected to intercooler is %f kJ/kg",Wc,H)
+//The answers vary due to round off error
+
diff --git a/3685/CH19/EX19.7/Ex19_7.txt b/3685/CH19/EX19.7/Ex19_7.txt new file mode 100644 index 000000000..dbce463ae --- /dev/null +++ b/3685/CH19/EX19.7/Ex19_7.txt @@ -0,0 +1,5 @@ +
+ Example 19.7
+
+ Minimum work done is 215.324046 kJ/kg,
+ Heat rejected to intercooler is 87.001072 kJ/kg
\ No newline at end of file diff --git a/3685/CH19/EX19.8/Ex19_8.sce b/3685/CH19/EX19.8/Ex19_8.sce new file mode 100644 index 000000000..9ab250626 --- /dev/null +++ b/3685/CH19/EX19.8/Ex19_8.sce @@ -0,0 +1,22 @@ +clc
+// Given that
+V = 4 // Volume flow rate in m^3/min
+p1 = 1.013 // Pressure in bar
+t1 = 15 // Temperature in degree centigrade
+N = 250 // Speed in RPM
+p4 = 80// Delivery pressure in bar
+v = 3 //Speed of piston in m/sec
+n_mech = .75 // Mechanical efficiency
+n_vol = .8 // Volumetric efficiency
+n = 1.25 // Polytropic index
+printf("\n Example 19.8\n")
+T1 = t1+273
+p2 = sqrt(p1*p4)
+W = (2*n/(n-1))*(p1*100/n_mech)*(V/60)*((p2/p1)^((n-1)/n) - 1)
+L = v*60/(N*2)
+Vs = V/N
+D_LP = sqrt(Vs*V/(%pi*L*n_vol))
+D_HP = D_LP*sqrt(p1/p2)
+printf("\n Minimum power required by the compressure is %f kW,\n Bore of the compressure in low pressure side is %f cm,\n Bore of the compressure in high pressure side is %f cm,\n Stroke of the compressure is %d cm",W,D_LP*100,D_HP*100,L*100)
+//The answers vary due to round off error
+
diff --git a/3685/CH19/EX19.8/Ex19_8.txt b/3685/CH19/EX19.8/Ex19_8.txt new file mode 100644 index 000000000..5a15c6210 --- /dev/null +++ b/3685/CH19/EX19.8/Ex19_8.txt @@ -0,0 +1,7 @@ +
+ Example 19.8
+
+ Minimum power required by the compressure is 49.337005 kW,
+ Bore of the compressure in low pressure side is 26.596152 cm,
+ Bore of the compressure in high pressure side is 8.921722 cm,
+ Stroke of the compressure is 36 cm
\ No newline at end of file diff --git a/3685/CH19/EX19.9/Ex19_9.sce b/3685/CH19/EX19.9/Ex19_9.sce new file mode 100644 index 000000000..af7e86b49 --- /dev/null +++ b/3685/CH19/EX19.9/Ex19_9.sce @@ -0,0 +1,19 @@ +clc
+// Given that
+p1 = 1 // Pressure in bar
+T1 = 300 // Temperature in K
+p4 = 9// Compressed pressure in bar
+n = 1.3 // Polytropic index
+R = 0.287 // Gas constant in kJ/kgK
+cp = 1.042 // Heat capapcity in kJ/kgK
+printf("\n Example 19.9\n")
+p2 = sqrt(p1*p4)
+T2 =T1*((p2/p1)^((n-1)/n))
+Wc = (2*n/(n-1))*R*1*(T2-T1)
+Wc_ = Wc/2
+Q = 1*cp*(T2-T1)
+Q_ = cp*(T1-T2)+Wc_
+H = Q+2*Q_
+printf("\n Compressor work = %f kJ/kg,\n Total heat transfer to the surrounding = %f kJ/kg",Wc_,H)
+//The answers given in the book contain calculation error
+
diff --git a/3685/CH19/EX19.9/Ex19_9.txt b/3685/CH19/EX19.9/Ex19_9.txt new file mode 100644 index 000000000..1503053e4 --- /dev/null +++ b/3685/CH19/EX19.9/Ex19_9.txt @@ -0,0 +1,5 @@ +
+ Example 19.9
+
+ Compressor work = 107.662023 kJ/kg,
+ Total heat transfer to the surrounding = 125.119950 kJ/kg
\ No newline at end of file |