#Variable declaration M_dot1=20 #rate of mass to be cooled M_dot2=25 #rate of cooling water Cp=4.18e3 #Heat capacity T1=360 #Initial temp. T2=340 #Final temp. theta_1=300 #Temperature of cooing water entering U=2e3 #Overall heat transfer coefficient #Calculation import sympy import math Q=M_dot1*Cp*(T1-T2) #Heat load x=sympy.Symbol('x') theta_2=sympy.solve(Q-(M_dot2*Cp*(x-300))) theta_m1=((T1-theta_2)-(T2-theta_1))/(math.log((T1-theta_2)/(T2-theta_1))) A1=Q/(U*theta_m1) theta_m2=((T1-theta_1)-(T2-theta_2))/(math.log((T1-theta_1)/(T2-theta_2))) A2=Q/(U*theta_m2) #Result print"(a).In counter flow,The surface area required",round(A1,2),"m**2" print"(b).In cocurrent flow,The surface area required",round(A2,2),"m**2"
(a).In counter flow,The surface area required 19.92 m**2 (b).In cocurrent flow,The surface area required 21.28 m**2
#Variable declaration dx=0.5 #Thickness of wall T1=400 #Temperartue of inner surface T2=300 #Temperature of outer surface K=0.7 #Thermal conductivity A=1 #Area of heat transfer #Calculation #From equation 9.12: Q=K*A*(T1-T2)/dx #Result print"The heat loss per square metre of surface =",Q,"W/m**2"
The heat loss per square metre of surface = 140.0 W/m**2
#Variable declaration dx1=0.20 #thickness of firebrick dx2=0.10 #thickness of insulating brick dx3=0.20 #thickness of building brick k1=1.4 #Thermal conductivity of firebrick k2=0.21 #Thermal conductivity of insulating brick k3=0.7 #Thermal conductivity of building brick T1=1200 #Temperature at junction 1 T4=330 #Temperature at junction 4 #Calculation Q=(T1-T4)/((dx1/k1)+(dx2/k2)+(dx3/k3)) #The ratio (Temperature drop over firebrick)/(Total temperature drop) R=(dx1/k1)/((dx1/k1)+(dx2/k2)+(dx3/k3)) #Temperature drop over firebrick dT=(T1-T4)*R T2=(T1-dT) #Result print"\n Heat loss per unit area =%d"%Q,"W/m**2" print"\n Temperature drop over firebrick =%d"%dT,"K" print"\n The temperature at the firebrick-insulating brick interface =",round(T2),"K"
Heat loss per unit area =961 W/m**2 Temperature drop over firebrick =137 K The temperature at the firebrick-insulating brick interface = 1063.0 K
#Variable declaration T=295 #initial temperature of surfaces T2f=375 #Final temperature of far surface dT1=900 #Temperature of near face raised #Calculation import sympy R=(T2f-T)/(2*(dT1-T)) #ratio of theta to twice of theta dash x=sympy.Symbol('x') t=sympy.solve((1.30**2*x)-346**2) #Result print"Time taken to rise from 295 to 375 K =",round(t/3600,1),"h"
Time taken to rise from 295 to 375 K = 19.7 h
#Variable declaration T=295 #initial temperature of surfaces T2f=375 #Final temperature of far surface dT1=900 #Temperature of near face raised DH=4.2e-7 #Thermal diffusivity #The development of the temperature profile is shown in Figure 9.12 #The problem will be solved by taking relatively large intervals for dx. #Choosing dx = 50 mm, the construction shown in Figure 9.12 dx=50e-3 #Because the second face is perfectly insulated, the temperature gradient must # be zero at this point. #It is seen that the temperature is #less than 375 K after time 23dt and greater than 375 K after time 25dt #Thus: #t=24*dt #from equation 9.43 dt=dx**2/(2*DH) t=24*dt #Result print"\n The time taken to rise from 295 to 375 K =",round(t/3600.0,1),"h"
The time taken to rise from 295 to 375 K = 19.8 h
#Variable declaration d=25e-3 #Diameter of copper sphere l=25e-3 #Side length of a copper cube h=75 #External heat transfer coefficient rho_cu=8950 #Density of copper at mean temperature Cp=0.38e3 #Heat capacity of copper at mean temperature k=385 # Thermal conductivity of copper at mean temperature Tf=923 #Temperature of the furnace Ta=368 #Temperature at which they are annealed t=5*60 # time taken #Calculation import math import sympy V_Ae_S=(d/6.0) #V/Ae tor the sphere V_Ae_C=(l/6.0) #V/Ae tor the cube Bi=h*(V_Ae_S)/k #The use of a lumped capacity method is therefore justified tao=rho_cu*Cp*V_Ae_S/h x=sympy.Symbol('x') T=sympy.solve(((x-Ta)/(Tf-Ta))-math.exp(-t/tao)) #Result print"Temperature of the sphere and of the cube at the end of 5 minutes =%d"%(T-273),"C"
Temperature of the sphere and of the cube at the end of 5 minutes =208 C
#Variable declaration k=2.5 #Thermal conductivity DH=2e-7 #Thermal diffusivity of the surrounding fluid h=100 #External heat transfer coefficient To=293 #Initial Temperature T_dash=373 #Oven Temperture Tc=353 #temperature throughout the whole of the sheet reaches a minimum l=10e-3 #thickness of sheet L=l/2 #Calculation #For the given process, the Biot number Bi=h*L/k Bi_1=1/Bi lim_val=(T_dash-Tc)/(T_dash-To) #From Figure 9.17, the Fourier number Fo=7.7 t=Fo*L**2/DH #Result print"The minimum time for which the sheet must be heated =%d"%(t/60.0),"min"
The minimum time for which the sheet must be heated =16 min
#Variable declaration l=5 #Length of the channel of uranium reactor Q=.25e6 #Heat release from uranium reactor k=33 #Thermal conductivity of the uranium #Calculation import math Q_m=Q/l #Heat release rate #Thus, from equation 9.52: dT=Q_m/(4*math.pi*k) #Result print"The temperature difference between the surface and the centre of the uranium element =",round(dT),"K"
The temperature difference between the surface and the centre of the uranium element = 121.0 K
#Variable declaration Cp=2380 #specific heat capacity of nitrobenzene k=0.15 Meu=0.70e-3 #Viscosity of nitrobenzene d_i=15e-3 #internal diameter of tube d_o=19e-3 #external diameter of the tube d_s=0.44 #shell diameter b_s=0.150 #baffle spacing p=0.025 #pitch c=0.006 #clearance #(i)Tube side coefficient h_i=1000 #based on inside area #Calculation import math h_io=1000*d_i/d_o #based on outside area #(ii) Shell side coefficient. A=d_s*b_s*c/p #Area for flow G_s_=4/A #Taking Meu/Meu_s=1 in equation 9.91 d_e=4*((25e-3**2-(math.pi*d_o**2/4))/(math.pi/d_o)) h_o=0.36*k/d_e*(d_e*G_s_/Meu)**0.55*(Cp*Meu/k)**0.33 #(iii) Overall coefficient #The math.logarithmic mean temperature difference is given by: Tm=(((400.0-345.0)-(315.0-305.0))/math.log((400.0-345.0)/(315.0-305.0))) #The corrected mean temperature difference is Tm_c=Tm*0.8 Q=4*Cp*(400.0-315.0) #The surface area of each tube A_t=0.0598 U_o=Q/(2*166*5*A_t*Tm_c) #(iv) Scale resistance. R_d=(1/U_o)-(1.0/750.0)-(1.0/1000.0) #Result print"\n Value of scale resistance that could be allowed =",round(R_d,5),"m**2 K/W"
Value of scale resistance that could be allowed = 0.00026 m**2 K/W
#Variable declaration G=15.0 #Mass flow rate of benzene d_s=1.0 #Internal diameter of Heat Exchanger l=5.0 #Length of tubes od=19e-3 #Outer diameter of tubes C=6e-3 #Clearance l_b=0.25 #Baffle spacing Meu=.5e-3 Y=25e-3 #dimension of square pitch N=19.0 #no. of Baffles #Calculation import math As=d_s*l_b*C/Y #Cross-flow area G_dash_s=G/As #Mass flow d_e=4*(Y**2-(math.pi*od**2.0/4.0))/(math.pi*od) #Equivalent Diameter Re=G_dash_s*d_e/Meu #From Figure 9.29: f_dash=0.280 rho_b=881 #density of benzene DPf=f_dash*G_dash_s**2*(N+1)*d_s/(2*rho_b*d_e) #Result print"The pressure drop over the tube bundle =%d"%DPf,"N/m**2=",round(DPf/(rho_b*9.81)),"m of Benzene" print"NOTE:Approx value of pressure drop is given in book"
The pressure drop over the tube bundle =8680 N/m**2= 1.0 m of Benzene NOTE:Approx value of pressure drop is given in book
#Variable declaration d=0.15 #Diameter of pipe Ts=400 #Surface temperature Ta=294 #Air temperture k=0.0310 #Thermal conductivity ---Table 6, Appendix A1 #Calculation X=36/k**4 #From Equation 9.102: GrPr=X*(Ts-Ta)*d**3 #From Table 9.5: n=0.25 C_dd=1.32 #Thus, in Equation 9.104: h=C_dd*(Ts-Ta)**n*d**(3*n-1) #result print"\n The heat transfer coefficient =",round(h,2),"W/m**2 K"
The heat transfer coefficient = 6.81 W/m**2 K
#Variable declaration lamda=1e-6 #Wavelength E_l_b=1e9 #Emissive power at given lambda C2=1.439e-2 C1=3.742e-16 #Calculation import math T=C2/lamda/math.log(C1/(E_l_b*lamda**5)) #With an error of +2 per cent, the correct value is given by: E_l_b_n=(100-2)*E_l_b/100 #In equation 9.108:1 T_n=C2/lamda/math.log(C1/(E_l_b_n*lamda**5)) #Result print"\n The temperature of surface =%d"%T,"K" print"\n The temperature of surface with +2 per cent error=",round(T_n),"K"
The temperature of surface =1121 K The temperature of surface with +2 per cent error= 1120.0 K
#Variable declaration d=10e-3 #Diameter of carbide elements l=0.5 #Length of carbide elements Ts=1750 #Maximun surface temperature of carbide P=500e3 #Thermal power output required sigma=5.67e-8 #Calculation Eb=sigma*Ts**4 A=math.pi*d*l P1=Eb*A #Power dissipated by one element n=P/P1 #Number of elements required #Result print"Number of elements required =%d"%round(n)
Number of elements required =60
#Variable declaration A=10.0 #Area of the surface P_r=1000e3 #Power radiated T1=1500.0 #First Temperature T2=1600.0 #Second Temperatue sigma=5.67e-8 #Calculation E=P_r/A #The emissive Power #From equation 9,118: e=E/(sigma*T1**4) E2=e*sigma*T2**4 #Result print"\n Emissivity when T=1500 K =",round(e,3) print"\n The Emissive power when T=1600 K =",round(E2*1e-2),"kW"
Emissivity when T=1500 K = 0.348 The Emissive power when T=1600 K = 1295.0 kW
import math #Variable declaration A1=2.0 #Area of rectangle(Surface 1) A2=math.pi*1**2.0/4.0 #Area of disc (Surface 2) T1=1500.0 #Temperature of Surface 1 T2=750.0 #Temperature of Surface 2 F12=0.25 #View factor sigma=5.67e-8 #Calculation #From equation 9. 1 26: F21=A1*F12/A2 Q12=sigma*A1*F12*(T1**4-T2**4) #Result print"View factor, F12 =",round(F21,3) print"The net radiation transfer =",round(Q12*1e-3,1),"kW" print"NOTE:Calculation mistake in book"
View factor, F12 = 0.637 The net radiation transfer = 134.6 kW NOTE:Calculation mistake in book
#Variable declaration X=4.0 #width of horizontal plate and length vertical plate Y=6.0 #length of horizontal plate Z=3.0 #height of verical plate #Calculation W=Y/X H=Z/X A1=Z*X #Area of plate 1 A2=X*Y #Area of plate 2 F12=0.12 #From equation 9.126: F21=A1*F12/A2 #For the two spheres r1=1.0 #Diameter of sphere 1 r2=2.0 #Diameter of sphere 2 F12b=1.0 F21b=(r1/r2)**2 F22b=1.0-F21b #Result print"For vertical plate" print"View Factor, F12=",F12 print"View Factor, F21=",F21 print"\nFor sphere:" print"View Factor, F21=",F21b print"View Factor, F22=",F22b
For vertical plate View Factor, F12= 0.12 View Factor, F21= 0.06 For sphere: View Factor, F21= 0.25 View Factor, F22= 0.75
#Variable declaration ri_u=0.2 # Inner radius of the upper ring ro_u=0.3 # Outer radius of the upper ring ri_l=0.3 # Inner radius of the lower ring ro_l=0.4 # Outer radius of the lower ring F12_34=0.4 F12_4=0.22 F1_34=0.55 F14=0.30 #Calculation A12_A2=ro_l**2/(ro_l**2-ri_l**2) A1_A2=ro_u**2/(ro_l**2-ri_l**2) F23=((A12_A2)*(F12_34-F12_4))+((A1_A2)*(F1_34-F14)) #Result print"\n F23 =",round(F23,2)
F23 = 0.73
#Variable declaration d=1 #Diameter of plate r1=0.5 r4=r1 #Radius of the imaginary disc sealing the hemisphere L=r1 #The distance between the plate and the bottom of the dome #Calculation import math A1=math.pi*d**2/4.0 #Area of the plate A2=2*math.pi*d**2/4.0 #Area of the underside of the Hemisphere A4=math.pi*r4**2/4.0 #Area of an imaginary disc sealing the hemisphere and parallel #to the plate T1=750 #Temperature of the plate T2=1200 #Temperature of hemispherical cone T3=290 #Temperature of the surroundings sigma=5.67e-8 R1=r1/L R4=r4/L S=1+(1+R4**2)/(R1**2) F14=0.5*(S-(S**2-4*(r4/r1)**2)**0.5) F12=F14 F13=1-F12 Q1=sigma*A1*F12*(T2**4-T1**4)+sigma*A1*F13*(T3**4-T1**4) #Result print"The net rate of heat transfer by radiation to the plate =",round(Q1*1e-3,1),"kW"
The net rate of heat transfer by radiation to the plate = 21.4 kW
#Variable declaration d=2 #Diameter of the cylinder h=1 #Depth of insulated cylinder T1=1500 T2=373 #From Figure 9.40ii, with i = 1, j = 2 r1=1 r2=1 L=1 #Calculation A1=math.pi*d**2.0/4.0 #Radiant heater surface A2=A1 #Under-Surface of the vessel A_R=math.pi*d*h #The view factor may also be obtained from Figure 9.39ii as follows: R1=r1/L R2=r2/L S=1+(1+R2**2)/(R1**2) F12=0.5*(S-(S**2-4*(r2/r1)**2)**0.5) sigma=5.67e-8 #Using the summation rule #F11=0 F1R=1-F12 F2R=F1R Q2=(A1*F12+((1/(A1*F1R)+(1/(A2*F2R))))**-1)*sigma*(T1**4-T2**4) #If the surroundings without insulation are surface 3 at T3=290 F23=F2R #from equation 9.135 Q2_d=sigma*A1*F12*(T1**4-T2**4)+sigma*A2*F23*(T3**4-T2**4) red=(Q2-Q2_d)/Q2*100 #Percentage Reduction #Result print"The rate of radiant heat transfer to the vessel =%d"%(Q2*1e-3),"kW" print"If the insulation were removed ,The rate of radiant heat transfer to the vessel=",round(Q2_d*1e-3),"kW" print"Reduction percentage =",round(red),"%"
The rate of radiant heat transfer to the vessel =620 kW If the insulation were removed ,The rate of radiant heat transfer to the vessel= 342.0 kW Reduction percentage = 45.0 %
#Variable declaration e=0.75 #Emissivity of grey surface r=1-e #reflectivity of surface Ts=400 #Temperature of surface T_amb=295 sigma=5.67e-8 q1=3e3 #Rate of radiation arriving at grey surface #Calculation #From equation 9.118 Eb=sigma*Ts**4 #From equation 9.138 qo=e*Eb+r*q1 #From equation 9.140 Q_A=e/r*(Eb-qo) q=Q_A #For convective heat transfer from the surface qc=-1*q hc=qc/(Ts-T_amb) #Result print"\n Radiosity =%d"%round(qo),"W/m**2" print"\n The net rate of radiation trasfer = %d"%q,"W/m**2" print"\n Coefficient of heat transfer =",round(hc,1),"W/m**2 K"
Radiosity =1839 W/m**2 The net rate of radiation trasfer = -1161 W/m**2 Coefficient of heat transfer = 11.1 W/m**2 K
#Variable declaration sigma=5.67e-8 T=[1000,500,300] #tempertaure of surfaces A=[1.07,1.07,0.628] #Array of area of surfaces e=[0.75,0.50,1.0] #Array of emissivity of the surfaces r=[0.250,0.50] # Array of radius of two surfaces L=0.2 #distance between two discs X=*3 Y=*3 R=*2 #Calculation from scipy.optimize import fsolve for i in range(0,2): X[i]=A[i]/r[i] Y[i]=A[i]*e[i]/r[i] R[i]=r[i]/L F11=0 F22=0 S=1+(1+R**2)/(R**2) F12=0.5*(S-(S**2-4*(r/(2*r))**2)**0.5) A1_F11=0 A2_F22=0 A1_F12=A*F12 A1_F13=A-(A*F11+A*F12) #for surface 2: A2_F21=A1_F12 A2_F23=A1_F13 #for surface 3: #By reciprocity rule A3_F31=A1_F13 A3_F32=A2_F23 A3_F33=A-(A3_F31+A3_F32) #From equation 9.112: E_b=*3 for i in range(0,3): E_b[i]=sigma*T[i]**4/1000.0 #since surface 3 is a black body q_o3=E_b #From equations 9.157 and 9.158: #we get f=[0,0] def F(p): x,y=p f=(A1_F11-A/r)*x+A2_F21*y+A3_F31*q_o3+E_b*A*e/r f=(A1_F12*x)+((A2_F22-A/r)*y)+E_b*A*e/r return(f,f) q_o=fsolve(F,(0,0)) #From equation 9.140: Q1=(A*e/r)*(E_b-q_o) Q2=(A*e/r)*(E_b-q_o) #Result print"\n Power input to the heater ,Q1 =",round(Q1,2),"kW" print"\n The rate of heat transfer to the plate ,Q2 =",round(Q2,2),"kW"
Power input to the heater ,Q1 = 36.19 kW The rate of heat transfer to the plate ,Q2 = -14.44 kW
#Variable declaration d=0.5 #diameter of chamber l=2.0 #Length of chamber e=0.5 #Emissivity T_s=750.0 #Temperature at which the chamber is maintained P=150e3 T_g=1250.0 sigma=5.67e-8 P_c=0.1*P P_w=P_c #Calculation import math V=math.pi/4.0*d**2*l #Volume of the chamber A_s=(2*math.pi/4.0*d**2)+(math.pi*d*l) #total surface are of chamber L_e=3.6*(V/A_s) C_w=1.4 e_w1=C_w*0.075 C_c1=1.2 e_c1=(C_c1*0.037) A=(P_w+P_c)*L_e B=P_c/(P_c+P_w) De=0.001 e_g=e_w1+0.044-De e_w2=(0.12*C_w) a_w=e_w2*(T_g/T_s)**0.65 C_c2=1.02 e_c2=(0.08*C_c2) a_c=e_c2*(T_g/T_s)**0.65 a_g=a_w+a_c #If the surrounding surface is black, then: Q=sigma*A_s*(e_g*T_g**4-a_g*T_s**4) #For grey walls, the correction factor allowing for multiple reflection of #incident radiation is: C_g=0.5/(1.0-(1.0-0.326)*(1.0-0.5)) Q_w=(Q*C_g) #Result print"\n Radiation to the walls if the surface is black =%.1f"%(Q*1e-3),"kW" print"\n Net radiation to the walls =%.2f"%(Q_w*1e-3),"kW"
Radiation to the walls if the surface is black =50.3 kW Net radiation to the walls =37.97 kW
#Variable declaration: def mole(w,m): n = w/m return(n) def partial(n1): p = 308*(n1/total) return(p) w_steam = 0.57 #mass flow rate of steam entering in [kg/sec] w_CO2 = 0.20 #mass flow rate of CO2 entering in [kg/sec] m_water = 18 #molecular mass of water in kg m_CO2 = 44 #molecular mass of CO2 in kg n_steam = mole(w_steam,m_water) #number of moles in kmol n_CO2 = mole(w_CO2,m_CO2) #number of moles in kmol #Calculation import math total = n_steam + n_CO2 p_steam = partial(n_steam) p_CO2 = partial(n_CO2) mean_mol = (0.57 + 0.20)/total #mean molecular weight of the mixture in kg/kmol outlet_steam = 11.7 #partial pressure of water in kN/m**2 outlet_CO2 = 308 - outlet_steam #partial pressure of water in kN/m**2 n_s = n_CO2*outlet_steam/outlet_CO2 steam_condensed = n_steam - n_s p_steam_401K = 252.2 #[kN/m**2] p_CO2_401K = 308 - 252.2 #[kN/m**2] steam_remaining = 0.0045*p_steam_401K/p_CO2_401K s_c = n_steam - steam_remaining #[kmol] Heat_cond = s_c*18*(2180 + 1.93*(404-401)) #[kW] Heat_uncondensed_steam = 0.0203*18*1.93*(404-401) #[kW] Heat_CO2 = 0.020*0.92*(404-401) total_heat = Heat_cond + Heat_uncondensed_steam + Heat_CO2 flow_water = 1407.3/(4.187*(319-300)) #[kg/sec] hi = 6.36 #[kW/m**2 K]Based on flow velocity of 1425 kg/m**2 sec ho = 5.25 #[kW/m**2 K]Based on outside area Cp = (0.20*0.92 + 0.57*1.93)/0.77 #[kJ/kg K] k_mean = 0.025 #[kW/m K] a = 0.0411 #[m**2] mass_velocity = (0.20+0.57)/0.0411 #[kg/m**2 sec] hg = 107 #[W/m**2 K] at Re = 29,800 at equivalent diameter = 0.024m u_pD = 0.62 #(u/pD)**0.67 = 0.62 Cpu_k = 1.01 #(Cp*u/k)**0.67 Psf = (122.6 - 38)/math.log(122.6/38) kG = hg*(Cpu_k)/(1000*Cp*Psf*u_pD) #Result print"\n Assuming no scale resistance, the overall coefficient =",round(1407.3/(34.8*74.2),3) ,"W/m** K" print"\n Actual coeficient =",round(1407.3/(53.9*74.2),3) ,"kW/m**2 K" print"\n Dirt factor =",round((0.545-0.352)/(0.545*0.352),2),"K/kW"
Assuming no scale resistance, the overall coefficient = 0.545 W/m** K Actual coeficient = 0.352 kW/m**2 K Dirt factor = 1.01 K/kW
#Variable declaration d_v=1.0 #diameter of the vessel L=0.3 #diameter of propeller agitator N=2.5 #rotating speed of propeller agitator T=310.0 #Temperature G=0.5 #circulation speed of cooling water d_o=25e-3 #outer diameter of stainless steel coil d=22e-3 #inner diameter of stainless steel coil d_w=(d_o+d)/2 d_c=0.8 #diameter of helix T_m=290.0 #mean temperature k1=0.59 Meu1=1.08e-3 C_p1=4.18e3 x_w=1.5e-3 #Calculation h_i=(k1/d)*(1+3.5*(d/d_c))*0.023*(d*1315.0/Meu1)**0.8*(C_p1*Meu1/k1)**0.4 #The external film coefficient is given by equation 9.204: C_p2=1.88e3 #Specefic heat capacity Meu2=6.5e-3 #viscosity k2=0.40 rho=1666 Meu_s=8.6e-3 h_o=0.87*(C_p2*Meu2/k2)**(1.0/3.0)*(L**2*N*rho/Meu2)**0.62*(Meu2/Meu_s)**0.14*k2/d_v k_w=15.9 R_o=0.0004 R_i=0.0002 U_o=((1/h_o)+(x_w*d_o/(k_w*d_w))+(d_o/(h_i*d))+(R_o)+(R_i*d_o/d))**-1 #Result print"The overall coeffecient of heat transfer =",round(U_o),"W/m**2.K"
The overall coeffecient of heat transfer = 498.0 W/m**2.K
#Variable declaration C_p=4e3 #Calculation import sympy from scipy.integrate import quad x=sympy.Symbol('x') T_max=sympy.solve((600*0.5)*(393-x)-(10*6)*(x-293)) #solving the equation finally we get def f(T): return(11111*(1/(376.3-T))) t1=quad(f,293.0,353.0) #The steam is turned off for 7200 s and during this time a heat balance gives: #on solving as given in book we get T=346.9 #The time taken to reheat the liquid to 353 K is then given by: t2=quad(f,T,353) #Result print"\n\n Maximum temperature to which it can be heated =",round(T_max,1),"K" print"\n Time taken to heat the liquid from 293 K to 353 K =%d"%t1,"s" print"\n Time taken to reheat the liquid to 353 K =%d"%round(t2),"s"
Maximum temperature to which it can be heated = 376.3 K Time taken to heat the liquid from 293 K to 353 K =14155 s Time taken to reheat the liquid to 353 K =2584 s
#Variable declaration #As in Example 9.1, the heat load = 1672 kW Q=1672 #With reference to Figure 9.71: T1=360 T2=340 theta1=300 #Temperature of cooling water entering theta2=316 #Calculation X=(theta2-theta1)/(T1-theta1) Y=(T1-T2)/(theta2-theta1) #from Figure 9.58 F=0.97 theta_m=41.9 #and hence: A=Q/(2*F*theta_m) #the heat transfer area #Result print"\n The heat transfer area is =",round(A,1),"m**2"
The heat transfer area is = 20.6 m**2
#Variable declaration #As in Example 9.1, the heat load = 1672 kW Q=1672.0 #With reference to Figure 9.71: T1=360.0 T2=340.0 theta1=300.0 #Temperature of cooling water entering theta2=316.0 F_theta_m=40.6 #corrected mean temperature difference #Calculation T=(T1+T2)/2.0 d=1.9e-3 #Tube diameter u=1 #Water velocity #then, in equation 9,221: h_i=4.28*(0.00488*T-1)*u**0.8/d**0.2 #From Table 9.18, an estimate of the shell-side film coefficient is: h_o=(1700.0+11000.0)/2000.0 #For steel tubes of a wall thickness of 1.6 mm, the thermal resistance of the wall, from Table 9.15 is: xw_kw=0.025 #the thermal resistance for treated water, from Table 9.16, is 0.26 m2K/kW Ri=0.26 Ro=Ri U=((1.0/h_o)+xw_kw+Ri+Ro+(1.0/h_i))**-1 A=Q/(F_theta_m*U) #Result print"\n The heat transfer area =",round(A,1),"m**2"
The heat transfer area = 32.8 m**2
#Variable declaration mh = 30.0 #[kg/s] Hot fluid flow rate Thi = 370.0 #[K] Hot Fluid Inlet Temperature Tho = 315.0 #[K] Hot Fluid outlet Temperature Tci = 300.0 #[K] Cold Fluid Inlet Temperature Tco = 315.0 #[K] Cold Fluid Outlet Temperature cpc = 4.18*10**3 #[J/kg.K] Thermal Conductivity of Cold Fluid #From table A1.3 at mean temperature 343 K cph = 2.9*10**3 #[J/kg.K] Thermal Capacity of Hot fluid #Calculation import math q = mh*cph*(Thi-Tho) #[kW] Heat load mc = q/(cpc*(Tco-Tci)) #[kg/s] Flow of cooling water Tln = ((Thi-Tho)-(Tco-Tci))/(math.log((Thi-Tho)/(Tco-Tci))) #[K] logarithm mean temperature difference X = (Thi - Tho)/(Tco-Tci) Y = (Tco-Tci)/(Thi - Tci) F = .85 U = 500.0 #[W/m**2.K] A = q/(F*Tln * U) od = .02 #[m] outer dia id = .016 #[m] inner dia l = 4.83 #[m] effective tube length s = math.pi*od*l N = A/s db = (1210/.249)**(2.207)**-1*20/1000.0 #[m] dc = .068 #[m] diametric clearance between shell and tubes ds = db+dc #[m] Shell dia Ac = math.pi/4*id**2 #[m**2] Cross sectional area Ntp = N/2.0 Af = N/2.0*Ac #[m**2] Tube side flow area mw = 76.3/Af #[kg/m**2.s] Mass velocity of water rho = 995 #[kg/m**3] mas density of water u = mw/rho #[m/s] water velocity vu = .8*10**-3 #[N.s/m**2] viscosity k = .59 #[W/m.K] Re = id*u*rho/vu Pr = cpc*vu/k ld = l/id jh = 3.7*10**-3 hi = jh*Re*Pr**.3334*.59/id #[W/m**2.K] dbf = .20*ds #[m] Baffle Dia tb = 1.25*20*10**-3 #[mm] Tube Pitch As = (25.0-20.0)/25.0*10**3*(ds*Ac) #[m**2] Gs = 30.0/As #[kg/m**2.s] de = 1.1*(.025**2-.917*od**2)/od #[m] rho2 = 780.0 #[kg/m**3] density vu2 = .8*10**-3 #[N.s/m**2] viscosity Cp2 = 3.1*10**3 #[J/kg.K] Heat capacity k2 = .16 #[W/m.K] Re2 = Gs*de/vu2 Pr2 = Cp2*vu2/k2 jh2 = 5*10**-3 hs = jh2*Re2*Pr2**.334*k2/de k3 = 50.0 #[W/m.K] Thermal Conductivity Rw = .00020 #[m**2.K/W] Scale Resistances Ro = .00015 #[m**2.K/W] Resistance for organic U = (1.0/hs + Rw + 0.5*(od-id)/k3 + Ro*od/id+od/(id*hi))**-1 #From figure 9.78 jf = 4.5*10**-3 n = 2.0 delP = n*(4*jf*(4.830/id) + 1.25)*(rho*u**2) u2 = Gs/rho2 jf2 = 4.6*10**-2 N2 = 1.0 delP2 = N2*(4*jf2*(4.830/od)*(1005/14.2))*(rho2*u2**2) #Increasing the baffle spacing pressure drop is reduced one-fourth delPs = delP2/4.0 ho = hs*(.5)**.8 #[W/m**2.K] U2 = (1.0/ho + Rw + .5*(od-id)/k3 + Ro*od/id+od/(id*hi))**-1 #REsult print"Overall Coefficient of %d"%U2," W/m**2.K \n Number of tubes/pass =",round(Ntp),"\n Number of tubes required = %d"%N print"NOTE:Very approximate values are used in textbook for Calculation\nThat's why answer is different"
Overall Coefficient of 563 W/m**2.K Number of tubes/pass = 603.0 Number of tubes required = 1205 NOTE:Very approximate values are used in textbook for Calculation That's why answer is different
#Variable declaration G=1.0 #Flow rate of organic liquid Cp=2e3#Heat capacity of organic liquid T1=350.0 T2=330.0 theta1=290.0 theta2=320.0 #Calculation Q=G*Cp*(T2-T1) #heat load G_cool=Q/(4187*(theta1-theta2)) #flow of water GCp_hot=(G*Cp) #for organic GCp_cold=(G_cool*4187) #From equation 9.235: eta=GCp_hot*(T1-T2)/(GCp_cold*(T1-theta1)) #Result print" Effectiveness of the given double pipe heat exchanger =",round(eta,2)
Effectiveness of the given double pipe heat exchanger = 0.5
#Variable declaration Tci = 320.0 #[K] Cold Fluid Initial Temperature Tce = 340.0 #[K] Cold Fluid Final Temperature mc = 4.0 #[kg/s] Flow rate of cold fluid mh = 8.0 #[kg/s] Flow rate of hot fluid import numpy as np Thi = np.array([380,370,360,350]) #[K] Hot fluid initial temperature Cp = 4.18 #[kJ/kg.K] mean heat capacity U = 1.5 #[W/m**2.K] Overall heat transfer coefficient #Calculation GCpu= mh*Cp #[kW/K] GCpp= mc*Cp #[kW/K] if(GCpu<GCpp): GCpmin = GCpu #[kW/K] ratio = GCpmin/GCpp else: GCpmin = GCpp #[kW/K] ratio = GCpmin/GCpu #Equation 9.235 n = mc*Cp*(Tce-Tci)*(mc*Cp*(Thi - Tci))**-1 #From Figure 9.85b Number of transfer Units N = np.array([.45,.6,.9,1.7]) #[NTU] A = N*GCpmin/U #Area of required [m**2] #Result print"Thi(K) n N A (m**2)" for i in range(0,4): print Thi[i],"\t",round(n[i],2),"\t",N[i],"\t",round(A[i],1)
Thi(K) n N A (m**2) 380 0.33 0.45 5.0 370 0.4 0.6 6.7 360 0.5 0.9 10.0 350 0.67 1.7 18.9
#Variable declaration o_d=10e-3 #outer diameter of the tube i_d=8.2e-3 #inner diameter of the tube h=140 #coeffecient of heat transfer between gas and copper tube k=350 #Thermal conductivity of copper tube L=0.075 #Calculation import math b=math.pi*o_d #perimeter of tube A=math.pi/4*(o_d**2-i_d**2) #cross sectional area of the metal m=((h*b)/(k*A))**0.5 T_g=((475*math.cosh(m*L))-365)/(math.cosh(m*L)-1) #Result print"The gas temperature is =%d"%T_g,"K"
The gas temperature is =539 K
#Variable declaration d2=54e-3 #outer diameter of the tube d1=70e-3 # fin diameter w=2e-3 #fin thickness n=230.0 # number of fins per metre run T_s=370.0 #Surface temperature T=280.0 #Temperature of surroundings h=30.0 #Heat transfer coeffecient between gas and fin k=43.0 #Thermal conductivity of steel L=(d1-d2)/2.0 #Calculation import math theta1=T_s-T #Assuming that the height of the fin is small compared with its circumference #and that it may be treated as a straight fin of length l=(math.pi/2.0)*(d1+d2) b=2*l #perimeter A=l*w #the average area at right-angles to the heat flow m=((h*b)/(k*A))**0.5 #From equation 9.254, the heat flow is given for case (b) as: Qf=m*k*A*theta1*(math.exp(2*m*L)-1)/(1+math.exp(2*m*L)) Q=Qf*n #Heat loss per meter run of tube #Result print"The heat loss per metre run of tube =",round(Q*1e-3,2),"kW/m"
The heat loss per metre run of tube = 1.91 kW/m
#Variable declaration d=150e-3 #Internal diameter of tube d_o=168e-3 #outer diameter of tube d_w=159e-3 d_s=268e-3 d_m=(d_s-d_o)/math.log(d_s/d_o) #log mean of d_o and d_s h_i=8500 #The coefficient for condensing steam together with that for any scale k_w=45 k_l=0.073 x_l=50e-3 x_w=9e-3 DT=444-294 sigma=5.67e-8 #Calculation import math #The temperature on the outside of the lagging is estimated at 314 K and (hr + hc) will be taken as 10 W/m2 K. #total thermal resisitance R=(h_i*math.pi*d)**-1+(10*math.pi*d_s)**-1+(k_w*math.pi*d_w/x_w)**-1+(k_l*math.pi*d_m/x_l)**-1 Q_l=DT/R #The heat loss per metre of length(from eq 9.261) DT_lagging=((k_l*math.pi*d_m/x_l)**-1/R)*DT #Taking an emissivity of 0.9, from equation 9.119: h_r=(0.9*sigma*(310**4-294**4))/(310-294) C=1.32 #Substituting in equation 9.105 (putting l = diameter = 0.268 m): h_c=C*((310.0-294.0)/d_s)**0.25 #If the pipe were unlagged,(hc+hr)for DT=150 K would be about 20 W/m2 K and the heat loss would then be: Q_l=20*math.pi*d_o*150 #Result print"The heat loss to the air =",round(Q_l*1e-3,2),"kW/m"
The heat loss to the air = 1.58 kW/m
#Variable declaration T1=420 #temperature of steam k=0.1 #Thermal conductivity T2=285 #Ambient temperature h=10 #the coefficient of heat transfer from the outside of the lagging to #the surroundings #Calculation #determining Q/l from equation 9.21 and equating it to heat loss from the #outside of the lagging we get #(Q/l)=84.82/(math.log(d_o/0.1)+(0.02/d_o)) W/m #using various equations we finally get an equation in terms of d_o and we # will solve it by using fsolve def F(d_o): return((1/(math.log(d_o/0.1)+(0.02/d_o))**2)-(2.35*(d_o**3)/(d_o-0.02))) ans=fsolve(F,1) E_t=(ans-0.1)/2 #Result print"\n Economic thickness of lagging =",round(E_t*1e3),"mm"
Economic thickness of lagging = 163.0 mm