Heat and mass transfer calculation for carbon capture pilot plant - Assignment Example

Heat transfer aspects Experiment Temperature reduction for the Hot Fluid ∆t Hot = T1 – T2 = 57.5 – 53.2 = 4.3KTemperature increase for the Cold Fluid ∆t Cold = T4 – T3 = 25.4 – 13.2 = 12.2KTable below shows the summary of the observations and calculations Obs. No.Vhot(g/s)Vcold(g/s)T1°CT2°CT3°CT4°C∆t Hot (T1-T2) K∆t Cold (T4-T3) K1501557.553.213.225.44.312.22502057.252.713.323.34.5103502756.952.413.120.84.57.74503156.751.413.321.55.38.2Experiment 2Mean Temperature = (T1 + T2)/2 = (57.5 + 53.2)/2 = 55.

350CCp of Water at Temperature T equals6x10-9 T4 – 1x10-6 T3 + 7.0487x10-5T2 – 2.4403x10-3 T +4.2113Cp at 55.350C = 4.1789J/gThe power emission rate from the hot water stream Qhot is = Vhot x Cphot(T1 – T2) J/s (Incropera and DeWitt 236) = 50 x 4.1789 (57.5 – 53.2) = 898.5J/sCold waterMean Temperature = (T3 + T4)/2 = (13.2+ 25.4)/2 = 19.30CAt 19.30C the specific heat of water calculated using the above equation is Cp at 19.30C = 4.1841J/gThe power absorption rate from the cold water stream Qcold is = Vcold x Cpcold(T4 – T3) J/s = 15 x 4.1841 (25.4 – 13.2) = 765.

7J/sTherefore the overall exchange efficiency ƞ = 85.2%The calculations for all observations are presented in Table-2 (a) and (b)Ob. No.Vhot (g/s)Vcold (g/s)∆t Hot (T1-T2) K∆t Cold (T4-T3) KAverage Temp Hot StreamAverage Temp Cold StreamCp of the hot streamCp of the cold stream150154.312.255.3519.34.17894.1841250204.51054.9518.34.17884.1848350274.57.754.6516.954.17884.1858450315.38.254.0517.44.17864.1855Calculation of Cp for both streams using the prescribed formulaObs. No.Heat loss rate for hot streamHeat gain rate for cold streamThermal Efficiency1898.5765.785.2%2940.2837.089.0%3940.2870.292.6%41107.31063.996.1%Calculation of Thermal EfficienciesWe observe two important things from our findings.

The first is that holding the mass flow rate and temperature of the hot stream steady and increasing the flow of the cold stream led to a steady improvement of the thermal efficiency of the apparatus. This supports the theory that increased flow creates more turbulence and improvement in the heat transfer coefficient. The second observation is that the improvement of efficiency might also have resulted in a gradual reduction of heat losses from the system as the experiments were performed one after the other and despite the fact that sufficient time was allowed for the system to become stable.

Experiment 3A sample of the calculations is as follows:Temperature reduction for the Hot Fluid ∆t Hot = T1 – T2 = 56.4 – 48.3 = 8.1KTemperature increase for the Cold Fluid ∆t Cold = T4 – T3 = 42.9 – 21.3 = 21.6Kthe summary of the observations and calculationsObs. No.Vhot(g/s)Vcold(g/s)T1°CT2°CT3°CT4°C∆t Hot (T1-T2) (K)∆t Cold (T4-T3) (K)1401456.448.321.342.98.121.62342055.543.421.340.712.119.43282656.440.421.137.416.016.34223256.835.821.234.921.013.7The calculations show that with increase in the flow rate of both streams the temperature drop of the hot stream increased steadily.

On the other hand, the temperature rise of the cold stream decreased substantially. The clear jacket allowed the observation that the jacket was continuously full at all times and because neither stream had any color, it was not possible to notice whether the turbulence of the two streams changed at all. In future experiments it might be a good idea to introduce a small stream of dye into both streams to observe turbulence. However, this would lead to a change in the mass flow rate and more importantly in the specific heat capacity, which would have to be accounted for in subsequent calculations.

Experiment 4Mean Temperature = (T1 + T2)/2 = (56.4 + 48.3)/2 = 52.350CCp of Water at Temperature T equals6x10-9 T4 – 1x10-6 T3 + 7.0487x10-5T2 – 2.4403x10-3 T +4.2113Cp at 52.350C = 4.1783J/gThe power emission rate from the hot water stream Qhot is = Vhot x Cphot(T1 – T2) J/s = 40 x 4.1783 (56.4 – 48.3) = 1353.8J/sCold StreamMean Temperature = (T3 + T4)/2 = (21.3+ 42.9)/2 = 21.60CAt 21.60C the specific heat of water calculated using the above equation is Cp at 21.60C = 4.

1783J/gThe power absorption rate from the cold water stream Qcold is = Vcold x Cpcold(T4 – T3) J/s = 14 x 4.1783 (42.9 – 21.3) = 1263.7J/sTherefore the overall thermal efficiency ƞ = 93.3%The calculations for all observations are presented in Table-4 (a) and (b)Obs. No.Vhot (g/s)Vcold (g/s)∆t Hot (T1-T2) (K)∆t Cold (T4-T3) (K)Average Temp Hot StreamAverage Temp Cold StreamCp of the hot streamCp of the cold stream140148.121.652.3532.14.17834.17892342012.119.449.45314.17794.

17913282616.016.348.429.254.17794.17964223221.013.746.328.054.17774.1800Calculation of Cp for both streams using the prescribed formulaObs. No.Heat loss rate for hot streamHeat gain rate for cold streamThermal Efficiency11353.81263.793.3%21718.81621.594.3%31871.71771.394.6%41930.11832.594.9%Important findings of this experiment are that the thermal efficiency of the plate and frame heat exchanger is better than that of the concentric tube heat exchanger and secondly, with the increase in flow rate the efficiency improved only marginally.

This is perhaps because with the multiple pass arrangement where the two streams changed direction frequently an element of turbulence was already present that did not change much with the change in flow rates. However, the observations taken are too few to arrive at this conclusion with confidence. Experiment 5Mean Temperature = (T1 + T2)/2 = (57.9 + 53.6)/2 = 55.750CCp of Water at Temperature T equals6x10-9 T4 – 1x10-6 T3 + 7.0487x10-5T2 – 2.4403x10-3 T +4.2113Cp at 55.750C = 4.1790J/gThe power emission rate from the hot water stream Qhot is = Vhot x Cphot(T1 – T2) J/s = 50 x 4.1790 (57.9 – 53.6) = 898.

5J/sCold StreamMean Temperature = (T3 + T4)/2 = (14.2+ 25.5)/2 = 19.850CAt 19.850C the specific heat of water calculated using the above equation is Cp at 19.850C = 4.1837J/gThe power absorption rate from the cold water stream Qcold is = Vcold x Cpcold(T4 – T3) J/s = 16 x 4.1837 (25.5 – 14.2) = 756.4J/sTherefore the overall thermal efficiency ƞ (Dincer and Konoglu 136) = 84.2%The calculations for all observations are presented in Tables below Calculation of Cp Obs. No.Vhot (g/s)Vcold (g/s)∆t Hot (T1-T2) (K)∆t Cold (T4-T3) (K)Average Temp Hot StreamAverage Temp Cold StreamCp of the hot streamCp of the cold stream150164.311.355.7519.854.17904.1837250204.610.155.519.154.17904.1842350264.98.355.3518.054.17894.1850450345.26.85516.904.17884.

1859Calculation of exchange EfficienciesObs. No.Heat loss rate for hot streamHeat gain rate for cold streamThermal Efficiency1898.5756.484.2%2961.2845.287.9%31023.8903.188.2%41086.5967.889.1%Calculation of exchange EfficienciesImportant findings of this experiment are that the thermal efficiency of the plate and frame heat exchanger is comparable to that of the concentric tube heat exchanger but lower than that of the plate and frame heat exchanger. Secondly, with the increase in flow rate of the cold stream the efficiency improved marginally.

In this case again, the conclusion can be that the design allows for introducing greater turbulence in the shell-side stream through the baffles provided.Mass transfer aspects Mass transfer is calculated as shown below; Mass transfer = Overall mass transfer coefficient x Actual driving force mol/m3 Overall mass transfer coefficient= (Kreith, Manglik and Bohn 175)KG Mass transfer coefficient in the gas phase m/sKL Mass transfer coefficient in the liquid phase m/sKov Overall mass transfer coefficient m/sm Solubility of carbon dioxide at equilibriumE = = 1.

011Ha = = 0.382Ha -Hatta modulesD CO2 ,am - CO2 diffusivity in the MEA solution m2/sk2 Forward second order reaction rate constant m3/mol.sCMEA - MEA concentration mol/m3E Enhancement factorE = = 57.25 Overall mass transfer coefficient= = 0.716Mass transfer = 0.716 x 0.278 = 0.1991Works Cited Incropera, Frank and David, DeWitt. Fundamentals of Heat and Mass Transfer. New York: John Wiley & Sons, 2002. PrintDincer, Ibrahim & Mehmet Konoglu. Refrigeration system and application. Chichester : John Wiley & sons ltd, 2011.

PrintKreith, Frank, Raj, Manglik & Mark, Bohn. Principles of heat transfer. Stanford: Cengage Learning, Stanford UK, 2011. Print