Chemical Engineering Material Balance Problem?
Please help me I am so confused! The indicator dilution method is a technique used to determine flow rates of fluids in channels for which devices like rotameters and orifice meters cannot be used. A stream of an easily measured substance (the tracer) is injected into the channel at a known rate and the tracer concentration is measured at a point far enough downstream of the injection point for the tracer to be completely mixed with the flowing fluid. The larger the flow rate of the fluid, the lower the tracer concentration at the measurement point. A gas stream that contains 1.50 mole% CO2 flows through a pipeline. 20.0 kg/min of CO2 is injected into the line. A sample of the gas is drawn from a point in the line 10 meters downstream of the injection point and found to contain 2.3 mole% CO2. Estimate the gas flow rate (kmol/min) upstream of the injection point.
Pls help material balance question in chemical engineering. Ammonia is removed from a stream of air by absorbt?
I'll only solve the second part of the problem and in order to do so, let's take a look at the flux of gas passing each second: Volume of gas = 200 m^3 Initial concentration of NH3 = 5.0% (v/v) = (5 m^3 NH3)/(100 m^3 air) Final concentration of NH3 = 0.05% (v/v) = (0.05 m^3 NH3)/(100 m^3 air) Calculate the amount of ammonia absorbed by the column: (200 m^3 air) x [(5 m^3 NH3)/(100 m^3 air)] x [(0.73 kg NH3) / (1 m^3 NH3)] = 7.3 kg NH3 (200 m^3 air) x [(0.05 m^3 NH3)/(100 m^3 air)] x [(0.73 kg NH3) / (1 m^3 NH3)] = 0.073 kg NH3 Thus the difference (7.2 kg NH3) is how much NH3 is absorbed per second
Chemical Engineering Material Balance Problem?
Brackish water is being desalinated for irrigation purposes by evaporation. The feedwater (the brackish water) has 600 ppm salt and the irrigation water (the outlet) is allowed to have 50 ppm salt. To accomplish this, a bypass line is set upstream to the evaporator. Per a basis of 1 kg brackish feedwater: 1. Draw a flow chart for the process. On the flow chart, label all variables and unknowns. 2. Calculate Degrees of Freedom around evaporator, mixing points, and entire process. 3. List material balances and the order for which you solve for the fraction of water fed that can bypass the evaporator. DO NO CALCULATIONS.
Chemical engineering material balance?
First forget the 100 mol butane. The combustion equation is: 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O First we calculate the theoretical air supply n_th . For the combustion of 1 mol butane 6.5 mol O2 are needed and 4 mol CO2 and 5 mol H2O are formed. Keep in mind that 1 mol of air consist of 0.21 mol O2 and 0.79 mol N2 or n(N2) = 0.79 n(O2)/0.21 = 3.76 * 6.5 mol= 24.44 mol and n_th = 4.76 n(O2) = 30.94 mol. 1) n_th, 100 % conversion Number of mol of exhaust gas n_ex = n(But)+n(O2)+n(N2)+n(CO2)+n(H2O) = 0 + 0 + 30.94 mol + 4 mol + 5 mol = 33.4 mol Exhaust gas composition ( = n(X)*100 % /n_ex ; X stands for a component): O2: 0 % ; N2: 73.2 % ; CO2: 11.9 %; H2O: 14.9 % 2) n(air) = 1.2 n_th ; 100% conversion n(air) = 37.12 mol containing 7.79 mol O2 and 29.32 mol N2. After combustion 7.79 mol - 6.5 mol O2= 1.29 mol O2 remain in the exhaust gas so n_ex = n(O2) + n(N2) +n(CO2) +n(H2O) // n(But)= 0 because of 100 % conversion n_ex = 1.29 + 29.32 +4 + 5 mol = 39.61 mol Composition of exhaust gas: O2: 3.3 % ; N2: 74 % ; CO2: 10.1 % ; H2O: 12.6 % 3) n(air) = 1.2 n_th ; 90 % conversion n(air) = 37.12 mol containing 29.32 mol N2 and 7.79 mol O2 For the combustion of 0.9 mol Butane 5.85 mol O2 are needed and 3.6 mol CO2 and 4.5 mol H2O are produced. In the exhaust gas there is 0.1 mol butane and 1.94 mol O2. The number of mol of the gas is n_ex= n(But)+n(O2) + n(CO2)+n(H2O) n_ex= 0.1 + 29.32 + 1.94 +3.6 +4.5= 39.46 mol Composition: But: 0.25 % ;O2: 4.91 %; N2: 74.3 % ; CO2: 9.12 % ; H2O: 11.4 % 4) Mass m= Q/ΔH = 20.32 g butane ( molar mass M = 58 g/mol). n(but) =m/M = 0.35 mol From the combustion equation n(CO2) = 4 n(but) = 1.4 mol CO2 Molar mass for CO2= 48 g / mol -> m(CO2) = 67.2 g
Material balance question?
Basis: 100 moles of stack gas. Moles of C (from butane) = 6.14 + 2.27 = 8.41 moles Moles of O2 (actual) = 73.12 (21/79) = 19.44 moles 1C + 1O2 -------> 1CO2 theo. O2 (for C from butane) = 8.41 moles H2 + (1/2)O2 --------> H2O Since no H2 is present in the stack gas, then ALL hydrogen has been burned. total H2 = moles H20 = 12.62 moles theo. O2 (for H2) = 12.62 moles (1/2) = 6.31 moles theo. O2 (total) = 8.41 + 6.31 moles = 14.72 moles Excess O2 = [(actual O2 - theo O2)/theo O2] * 100% = [(19.44 - 14.72)/14.72] * 100% Excess O2 = 32%
Material balance engineering problem?
Material balancing in Engineering plays vital role for strengthening the property of the product.As my own work experience in civil Engineering field,I would like to prefer the several checks on material ration balancing. Because of proper material balancing,the overall quality and strength will increase abruptly. For example,when i am going to make a M20 grade of concrete,i would like to balance the proportion of cement,sand and aggregate at 1:2:4 ratio.If there is improper balancing in ratio of any one of its constituents then you will get inferior quality of concrete which ultimately reduces the life of structures.