Question 8.3.6: Energy Balance on a Waste-Heat Boiler Eliminating discharges......
Energy Balance on a Waste-Heat Boiler
Eliminating discharges of hot streams to the environment has two beneficial effects: the temperature of the receiving entity (e.g., a lake, a river, or the atmosphere) is not raised, which can avoid violating an EPA regulation, and the energy contained in the discharged stream is not wasted by being dissipated in the environment. For example, a gas stream at 500°C containing 8.0 mol% CO and 92.0% CO_{2} that was originally going to be sent up a stack is instead sent to a heat exchanger and flows across tubes through which water is flowing. The water enters at 25°C and is fed at a ratio of 0.200 mol water/mol hot gas, is heated to its boiling point, and forms saturated steam at 5.0 bar. The steam may be used for heating or power generation in the plant or as the feed to another process unit. The heat exchanger can be assumed to operate adiabatically—that is, all heat transferred from the hot gas goes to heat the water, as opposed to some of it leaking through the heat exchanger walls to the environment. The flowchart for an assumed basis of 1.00 mol feed gas is shown below. Calculate the temperature of the gas leaving the heat exchanger (a) using data from Tables B.1 and B.2 but not using APEx; (b) using APEx.
Since no material balances are required in this problem, we may proceed directly to the energy balance, which for this adiabatic unit reduces to
\Delta H = \sum\limits_{out}{\dot{n}_{i}\hat{H}_{i}} – \sum\limits_{in}{\dot{n}_{i}\hat{H}_{i}} = 0
Note that we do not write \Delta \dot{H} and \dot{n}_{i}since a quantity (1 mol feed gas) and not a flow rate has been assumed as a basis of calculation.
(Exercise: What assumptions have been made in writing the energy balance?)
The solution strategy will be to calculate \hat{H}_{1}(T) and \hat{H}_{2}(T) by integrating the heat capacity formulas of Table B.2 from the reference temperature (500°C) to the unknown T at the gas outlet, look up \hat{H}_{3} and \hat{H}_{4} in the steam tables, substitute for \hat{H}_{1} through \hat{H}_{4} in the energy balance, and solve the resulting equation for T using a spreadsheet.
Note the following points about the enthalpy table:
• We chose the reference states for CO and CO_{2} as the gas inlet temperature and 1 atm. We assume ideal-gas behavior so that deviations of the pressure from 1 atm have no effect on enthalpies, and accordingly set the inlet enthalpies of the gas species equal to zero.
• We will find the enthalpies of the feedwater and product steam in the steam tables. Knowing this, we chose the reference state for the steam tables (liquid water at the triple point) as our reference for water, and knowing that the enthalpies in the steam tables are in kJ/kg, we list the quantity of water in kg (m = 0.200 mol H_{2}O × 0.0180 kg/mol = 0.00360 kg).
• We will integrate the heat capacity formulas of Table B.2 for CO and CO_{2} even though enthalpies for those species are listed in Table B.8, since we do not know the temperature at which to look them up.
(a) The specific enthalpies are
\hat{H}_{3} = \hat{H}[H_{2}O(l, 25°C, 5 bar)] ≈ 105 kJ/kg (Table B.5: neglect effect of pressure on \hat{H})
\hat{H}_{4} = \hat{H}[H_{2}O(v, 5 bar, sat’d)] = 2747.5 kJ/kg (Table B.6)
Integrating the expressions for \hat{H}_{1} and \hat{H}_{2} and substituting the resulting expressions and the values of \hat{H}_{3} and \hat{H}_{4} into the energy balance (ΔH = 0) yields the following equation:
1.672 × 10^{-12} T^{4} – 0.8759 × 10^{-8} T^{3} + 1.959 × 10^{-5} T^{2} + 0.03554T – 12.16 = 0
The problem is to find the value of T(°C) that satisfies this equation. The equation can be solved by trial-and-error using any mathematical software program, or using Excel’s Solver or Goal Seek. The solution is \boxed{ T = 299°C} . The heat transferred from the specified quantity of gas as it cools from 500°C to 299°C goes to convert the specified amount of feed water into steam.
(b) Following is a portion of an Excel spreadsheet with the APEx add-on activated. The n values in Columns C and E and the 0s (zeroes) in Column D are entered as numbers, and the T value in Cell C8 is an initial guess at the temperature of the outlet gas. The other values in Columns C through H result from entering the formulas shown on the next page. We strongly recommend that you attempt to replicate the spreadsheet for yourself.
Cell Entries
[D5]= SteamSatT(25, “T”, “H”, “L”) [Enthalpy of saturated H_{2}O(l) at T= 25°C]; “T” is the first argument, “H” is the quantity to be provided by the spreadsheet (i.e., enthalpy), “L” is the phase of the water.
[F3]=Enthalpy(“CO”, 500, C$8) [Integral of (C_{p})CO from 500°C to T in Cell C8]
[F4]=Enthalpy(“CO2”, 500, C$8) [As above for CO_{2}]
[F5]=SteamSatP(5,“P”, “H”, “V”) [Enthalpy of saturated H_{2}O(v) at P=5 bar]
[G3]= C3*D3 [H3]= E3*F3
[G4]= C4*D4 [H4]= E4*F4
[G5]= C5*D5 [H5]= E5*F5
[G6]= SUM(G3.G5) [H6]= SUM(H3:H5)
[C9]= H6–G6 [ΔH, which must equal zero when the correct temperature is found]
Once all of the formulas have been entered, open Solver; set the target cell to C9 (or $C$9), check “Value of ” and enter 0, insert C8 as the cell to be changed, and then click on “Solve.” Cell C8 should then contain the solution \boxed{ T = 299°C} .
References: CO(g, 500°C, 1 atm), CO_{2}(g, 500°C, 1 atm), H_{2}O(l, triple point)
| Substance | n_{in} | \hat{H}_{in} | n_{out} | \hat{H}_{out} |
| CO | 0.080 mol | 0 kJ/mol | 0.080 mol | \hat{H}_{1} (kJ/mol) |
| CO_{2} | 0.920 mol | 0 kJ/mol | 0.920 mol | \hat{H}_{2} (kJ/mol) |
| H_{2}O | 0.00360 kg | \hat{H}_{3} (kJ/kg) | 0.00360 kg | \hat{H}_{4} (kJ/kg) |
| A | B | C | D | E | F | G | H | |
| 1 | Solution to Example 8.3-6b | |||||||
| 2 | Substance | n(in) | H(in) | n(out) | H(out) | nH(in) | nH(out) | |
| 3 | CO | 0.08 | 0 | 0.08 | -12.1852 | 0 | -0.97482 | |
| 4 | CO2 | 0.92 | 0 | 0.92 | -18.4467 | 0 | -16.971 | |
| 5 | H2O | 0.0036 | 104.8 | 0.0036 | 2547.3 | 0.37728 | 9.891 | |
| 6 | Sum-> | 0.37728 | -8.05482 | |||||
| 7 | ||||||||
| 8 | T(gas out) | 100 | ||||||
| 9 | Delta H | -8.4321 | ||||||