diff options
Diffstat (limited to '2510/CH22')
| -rwxr-xr-x | 2510/CH22/EX22.10/Ex22_10.sce | 25 | ||||
| -rwxr-xr-x | 2510/CH22/EX22.12/Ex22_12.sce | 22 | ||||
| -rwxr-xr-x | 2510/CH22/EX22.13/Ex22_13.sce | 16 | ||||
| -rwxr-xr-x | 2510/CH22/EX22.23/Ex22_23.sce | 23 | ||||
| -rwxr-xr-x | 2510/CH22/EX22.24/Ex22_24.sce | 46 | ||||
| -rwxr-xr-x | 2510/CH22/EX22.6/Ex22_6.sce | 20 | ||||
| -rwxr-xr-x | 2510/CH22/EX22.7/Ex22_7.sce | 23 | ||||
| -rwxr-xr-x | 2510/CH22/EX22.8/Ex22_8.sce | 17 | ||||
| -rwxr-xr-x | 2510/CH22/EX22.9/Ex22_9.sce | 71 |
9 files changed, 263 insertions, 0 deletions
diff --git a/2510/CH22/EX22.10/Ex22_10.sce b/2510/CH22/EX22.10/Ex22_10.sce new file mode 100755 index 000000000..2cea58d8a --- /dev/null +++ b/2510/CH22/EX22.10/Ex22_10.sce @@ -0,0 +1,25 @@ +//Variable declaration: +//From example 22.9: +t1 = 23.5 //Initial temperature of oil ( C) +t2 = 27 //Final temperature of oil ( C) +T1 = 93 //Water heating temperature of water ( C) +T2 = 88.16 //Minimum temperature of heating water ( C) +U = 34.6 //Overall heat transfer coefficient (W/m^2. C) +Q = 7227.2 //Duty of exchanger (W) +D = 6*0.0254 //Inside diameter of %pipe (m) +l = 6.68 //Previous heat transfer length (m) + +//Calculation: +DT1 = T1-t1 //Inlet temperature difference ( C) +DT2 = T2-t2 //Outlet temperature difference ( C) +DTlm = (DT1-DT2)/log(DT1/DT2) //Log mean temperature difference ( C) +A = Q/(U*DTlm) //Required heat transfer area (m^2) +L = A/(%pi*D) //Required heat transfer length (m) + +//Result: +printf("The length of the parallel %%pipe heat exchanger is : %.2f ",L) +if L>l then + printf("The tube length would increase slightly.") +elseif L<l then + printf("The tube length would decrease slightly.") +end diff --git a/2510/CH22/EX22.12/Ex22_12.sce b/2510/CH22/EX22.12/Ex22_12.sce new file mode 100755 index 000000000..30538292d --- /dev/null +++ b/2510/CH22/EX22.12/Ex22_12.sce @@ -0,0 +1,22 @@ +//Variable declaration: +T = 80 //Pipe surface temperature ( F) +t1 = 10 //Inlet temperature of brine solution ( F) +m = 1200 //mass flowrate of solution (kg/s) +c = 0.99 //Heat capacity of brine solution (Btu/lb. F) +A = 2.5 //Heat transfer area (ft^2) +U1 = 150 //Overall heat transfer coefficient at temperature approach (Btu/h.ft^2. F) +U2 = 140 //Overall heat transfer coefficient at inlet brine temperature (Btu/h.ft^2. F) + +//Calculation: +DT1 = T-t1 //Temperature approach at the pipe entrance ( F) + +function [ans] = equation(DT2) + Q1 = m*c*(DT1-DT2) //Energy balance to the brine solution across the full length of the pipe (Btu/h) + DTlm = (DT1-DT2)*log(DT2/DT1) //Log mean temperature difference ( F) + Q2 = A*(U2*DT1-U1*DT2)/log((U2*DT1)/(U1*DT2)) //Heat transfer rate (Btu/h) + ans = Q2-Q1 +endfunction +t2 = T-fsolve(1,equation) //The temperature of the brine solution ( F) + +//Results: +printf("The temperature of brine solution is: %.0f C",(t2-32)/1.8) diff --git a/2510/CH22/EX22.13/Ex22_13.sce b/2510/CH22/EX22.13/Ex22_13.sce new file mode 100755 index 000000000..dec3f79be --- /dev/null +++ b/2510/CH22/EX22.13/Ex22_13.sce @@ -0,0 +1,16 @@ +//Variable declaration: +m = 1200 //mass flowrate of solution (kg/s) +c = 0.99 //Heat capacity of brine solution (Btu/lb. F) +DT1 = 70 //Temperature approach at the pipe entrance ( F) +DT2 = 51.6 //Temperature difference at the pipe exit ( F) + +//Calculation: +Q = m*c*(DT1-DT2) //Heat transfer rate (Btu/h) +DTlm = (DT1-DT2)/log(DT1/DT2) //Log mean temperature difference ( F) +Q1 = round(Q*10**-1)/10**-1 + +//Result: +printf("1. The rate of heat transfer is : %f Btu/h.",Q1) +printf("Or, the rate of heat transfer is : %.0f W.",Q/3.412) +printf("2. The log mean temperature difference is : %.1f F.",DTlm) +printf("Or, the log mean temperature difference is : %.1f C.",DTlm/1.8) diff --git a/2510/CH22/EX22.23/Ex22_23.sce b/2510/CH22/EX22.23/Ex22_23.sce new file mode 100755 index 000000000..a9416287d --- /dev/null +++ b/2510/CH22/EX22.23/Ex22_23.sce @@ -0,0 +1,23 @@ +//Variable declaration: +Too = 100 //Steam temperature ( C) +Ti = 18 //Initial temperature of liquid TCA ( C) +Tf = 74 //Final temperature of liquid TCA ( C) +t = 180 //Heating time (s) +p = 87.4 //Density of TCA (lb/ft^3) +V = 18 //Kinematic viscosity of TCA (m^2/s) +cp = 0.23 //Heat capacity of TCA (Btu/lb. F) +U = 200 //Overall heat transfer coefficient (Btu/h.ft^2. F) + +//Calculation: +ui = Too-Ti //Initial excess temperature ( C) +uf = Too-Tf //Final excess temperature ( C) +R = log(ui/uf) //Ratio t/r +r = t/R //Thermal time constant (s) +A = p*V*cp/(3600*U*r) //Required heating area (ft^3) +Ti_F = Ti*9/5+32 //Initial temperature in fahrenheit scale ( F) +Tf_F = Tf*9/5+32 //Final temperature in fahrenheit scale ( F) +Q = p*V*cp*(Tf_F-Ti_F) //Total amount of heat added (Btu) + +//Result: +printf("1. The required surface area of the heating coil is : %e ft^3",A) +printf("2. The total heat added to the liquid TCA is : %.0f Btu",Q) diff --git a/2510/CH22/EX22.24/Ex22_24.sce b/2510/CH22/EX22.24/Ex22_24.sce new file mode 100755 index 000000000..365c0072e --- /dev/null +++ b/2510/CH22/EX22.24/Ex22_24.sce @@ -0,0 +1,46 @@ +//Variable declaration: +m1 = 62000 //Mass flowrate of alcohol (lb/h) +h1 = 365 //Enthalpy of vapour (Btu/lb) +cp = 1 //Heat capacity of water (Btu/lb. F) +T1 = 85 //Entering temperature of water ( F) +T2 = 120 //Exit temperature of water ( F) +a1 = 2.11 //Flow area for the shell side (ft^2) +N = 700 //Total number of tubes +a2 = 0.546 //Flow area per tube (in^2/tube) +n = 4 //Number of tube passes +p = 62.5 //Density of water (lb/ft^3) +L = 16 //Length of condenser (ft) +hio = 862.4 //Cooling water inside film coefficient (Btu/h.ft^2. F) +g = 9.8 //Gravitational accleration (m^2/s) +Rf = 0.003 //Fouling factor (Btu/h.ft^2. F) + +//Calculation: +Q1 = m1*h1 //Heat loss from alcohol (Btu/h) +Q2 = Q1 //Heat gained by water (Btu/h) +DT = T2-T1 //Temperature difference ( F) +m2 = Q2/(cp*DT) //Water mass flow rate (lb/h) +LMTD = ((T2-32)-(T1-32))/log((T2-32)/(T1-32)) //Log mean temperature difference ( F) +at = (N*a2)/(144*n) //Total flow area for tube side (ft^2) +G1 = m1/a1 //Mass velocity of flow in shell side (lb/h.ft^2) +G2 = m2/at //Mass velocity of flow in tube side (lb/h.ft^2) +V = G2/(3600*p) //Velocity of water (ft/s) +G3 = m1/(L*N)**(2/3) //Loading G (lb/h.ft) +//For alcohol: +kf = 0.105 //Thermal conductivity (Btu/h.ft. F) +muf = 0.55*2.42 //Dynamic viscosity (lb/ft.h) +sf = 0.79 // +pf = sf*p //Density (lb/ft^3) +h = 151*(((kf**3)*(pf**2)*g*muf)/((muf**2)*n*G3))**(1/3) //Heat transfer coefficient for the shell side (Btu/h.ft^2. F) +ho = h //Outside heat transfer coefficient of the tube bundle (Btu/h.ft^2. F) +Uc = (hio*ho)/(hio+ho) //Overall heat transfer coefficient for a new (clean) heat exchanger (Btu/h.ft^2. F) +A = N*L*0.2618 //Area for heat transfer (ft^2) +Ud = Q1/(A*DT) //Design (D) overall heat transfer coefficient (Btu/h.ft^2. F) +Rd = (Uc-Ud)/(Uc*Ud) //Dirt (d) factor (Btu/h.ft^2. F) + +//Result: +printf("The dirt (d) factor is : %.4f Btu/h.ft^2. F .",Rd) +if (Rd>Rd) then + printf("Therefore, the exchanger as specified is unsuitable for these process conditions since the fouling factor is above the recommended value. Cleaning is recommended.") +else + printf("Therefore, the exchanger as specified is suitable for these process conditions since the fouling factor is below the recommended value. Cleaning is not recommended.") +end diff --git a/2510/CH22/EX22.6/Ex22_6.sce b/2510/CH22/EX22.6/Ex22_6.sce new file mode 100755 index 000000000..892d35303 --- /dev/null +++ b/2510/CH22/EX22.6/Ex22_6.sce @@ -0,0 +1,20 @@ + //Variable declaration: +//From steam tables: +h1 = 1572 //Enthalpy for super heated steam at (P = 40 atm, T = 1000 F) (Btu/lb) +h2 = 1316 //Enthalpy for super heated steam at (P = 20 atm, T = 600 F) (Btu/lb) +h3 = 1151 //Enthalpy for saturated steam (Btu/lb) +h4 = 28.1 //Enthalpy for saturated water (Btu/lb) +m1 = 1000 //Mass flowrate of steam (lb/h) +syms m //Mass flow rate of steam (lb/h) + +//Calculation: +Dh1 = m1*(h3-h4) //The change in enthalpy for the vaporization of the water stream (Btu/h) +Dh2 = m*(h1-h2) //The change in enthalpy for the cooling of the water stream (Btu/h) +x = eval(solve(Dh1-Dh2,m)) //Mass flowrate of steam (lb/h) +m2 = x; //Mass flowrate of steam (lb/h) + +//Result: +disp("The mass flowrate of the utility steam required is : ") +disp(m2) +disp(" lb/h.") + diff --git a/2510/CH22/EX22.7/Ex22_7.sce b/2510/CH22/EX22.7/Ex22_7.sce new file mode 100755 index 000000000..ca5c03fc0 --- /dev/null +++ b/2510/CH22/EX22.7/Ex22_7.sce @@ -0,0 +1,23 @@ +//Variable declaration: +//From table 22.1: +QH1 = 12*10**6 //Heat duty for process unit 1 (Btu/h) +QH2 = 6*10**6 //Heat duty for process unit 2 (Btu/h) +QH3 = 23.5*10**6 //Heat duty for process unit 3 (Btu/h) +QH4 = 17*10**6 //Heat duty for process unit 4 (Btu/h) +QH5 = 31*10**6 //Heat duty for process unit 5 (Btu/h) +T1 = 90 //Supply water temperature ( F) +T2 = 115 //Return water temperature ( F) +cP = 1 //Cooling water heat capacity (Btu/(lb. F)) +p = 62*0.1337 //Density of water (lb/gal) +BDR = 5/100 //Blow-down rate + +//Calculation: +QHL = (QH1+QH2+QH3+QH4+QH5)/60 //Heat load (Btu/min) +DT = T2-T1 //Change in temperature ( F) +qCW = round(QHL*10**-5)/10**-5/(DT*cP*p) //Required cooling water flowrate (gpm) +qBD = BDR*qCW //Blow-down flow (gpm) +qCW = round(qCW*10**-1)/10**-1 + +//Result: +printf("The total flowrate of cooling water required for the services is : %f gpm.",qCW) +printf("The required blow-down flow is : %.0f gpm.",qBD) diff --git a/2510/CH22/EX22.8/Ex22_8.sce b/2510/CH22/EX22.8/Ex22_8.sce new file mode 100755 index 000000000..59da2015f --- /dev/null +++ b/2510/CH22/EX22.8/Ex22_8.sce @@ -0,0 +1,17 @@ +//Variable declaration: +Q1 = 10*10**6 //Unit heat duty for process unit 1 (Btu/h) +Q2 = 8*10**6 //Unit heat duty for process unit 2 (Btu/h) +Q3 = 12*10**6 //Unit heat duty for process unit 3 (Btu/h) +Q4 = 20*10**6 //Unit heat duty for process unit 4 (Btu/h) +hv = 751 //Enthalpy of vaporization for pressure 500 psig (Btu/lb) + +//Calculation: +mB1 = Q1/hv //Mass flowrate of 500 psig steam through unit 1 (lb/h) +mB2 = Q2/hv //Mass flowrate of 500 psig steam through unit 2 (lb/h) +mB3 = Q3/hv //Mass flowrate of 500 psig steam through unit 3 (lb/h) +mB4 = Q4/hv //Mass flowrate of 500 psig steam through unit 4 (lb/h) +mBT = mB1+mB2+mB3+mB4 //Total steam required (lb/h) +mBT = round(mBT*10**-1)/10**-1 + +//Result: +printf("The total steam required is : %f lb/h.",mBT) diff --git a/2510/CH22/EX22.9/Ex22_9.sce b/2510/CH22/EX22.9/Ex22_9.sce new file mode 100755 index 000000000..13a64b8e8 --- /dev/null +++ b/2510/CH22/EX22.9/Ex22_9.sce @@ -0,0 +1,71 @@ +//Variable declaration: +po = 53*16.0185 //Density of oil (kg/m^3) +co = 0.46*4186.7 //Heat capacity of oil (J/kg. C) +pi = %pi +muo = 150/1000 //Dynamic viscosity of oil (kg/m.s) +ko = 0.11*1.7303 //Thermal conductivity of oil (W/m. C) +qo = 28830*4.381*10**-8 //Volumetric flowrate of oil (m^3/s) +pw = 964 //Density of water (kg/m^3) +cw = 4204 //Heat capacity of water (J/kg. C) +muw = 0.7/3600*1.4881 //Dynamic viscosity of water (kg/m.s) +kw = 0.678 //Thermal conductivity of water (W/m. C) +qw = 8406*4.381*10**-8 //Volumetric flowrate of water (m^3/s) +t1 = 23.5 //Initial temperature of oil ( C) +t2 = 27 //Final temperature of oil ( C) +T1 = 93 //Water heating temperature of water ( C) +syms T2 //Minimum temperature of heating water ( C) +syms A //Heat transfer area (m^2) +Uc = 35.4 //Clean heat transfer coefficient (W/m^2.K) +Rf = 0.0007 //Thermal resistance (m^2.K/W) +D = 6*0.0254 //Inside diameter of pipe (m) + +//Calculation: +vo = muo/po //Kinematic viscosity of oil (m^2/s) +mo = po*qo //Mass flowrate of oil (kg/s) +vw = muw/pw //Kinematic viscosity of (m^2/s) +mw = pw*qw //Masss flow rate of water (kg/s) +Q1 = mo*co*(t2-t1) //Duty of exchanger of oil (W) +T2m = t1 //Lowest possible temperature of the water ( C) (part 1) +Qmw = mw*cw*(T1-T2m) //Maximum duty of exchanger of water (W) (part 2) +Q2 = mw*cw*(T1-T2) //Duty of exchanger of water in terms of T2 (W) +x = eval(solve(Q1-Q2,T2)) //Solving value for T2 ( C) +T3 = x; //Minimum temperature of heating water ( C) +DT1 = T3-t1 //Inlet temperature difference ( C) +DT2 = T1-t2 //Outlet temperature difference ( C) +DTlm = (DT1-DT2)/log(DT1/DT2) //Log mean temperature difference ( C) +Ud1 = 1/Uc+Rf //Dirty heat transfer coefficient (W/m^2.K) (part 3) +Ud2 = 34.6 //Dirty heat transfer coefficient (W/m^2. C) +Q3 = Ud2*A*DTlm //Duty of exchanger (W) (part 4) +y = eval(solve(Q1-Q3,A)) //Heat transfer area (m^2) +A1 = y //Required heat transfer area (m^2) +L = A1/(pi*D) //Required heat transfer length (m) +Qmo = mo*co*(T1-t1) //Maximum duty of exchanger of oil (W) (part 5) +Qm = Qmw //Maximum duty of exchanger (W) +E = Q1/Qm*100 //Effectiveness (%) +NTU = Ud2*A1/(mw*cw) //Number of transfer units + +//Result: +disp("1. The lowest possible temperature of the water is :") +disp(T2m) +disp(" C .") + +disp("2. The log mean temperature difference is : ") +disp (DTlm) +disp(" C .") + +disp("3. The overall heat transfer coefficient for the new clean exchanger is : ") +disp (Ud2) +disp ("W/m^2. C .") + +disp("4. The length of the double pipe heat exchanger is : ") +disp(L) +disp (" m .") + +disp("5. The effectiveness of the exchanger is : ") +disp(E) +disp("%") + +disp("The NTU of the exchanger is : ") +disp(NTU) + +// Answers are correct. Please calculate manually. |