Cooling a Hot and Dry Gas Stream A 10,000 ACFM dry (0% water vapor) discharge gas stream containing small particles exits a metallurgical kiln at T1 = 600. °C and p1 = 101. kPa. To capture the particles it is necessary to reduce the gas temperature to T3 = 200. °C and p3 = 101. kPa before passing

Question

Cooling a Hot and Dry Gas Stream
A 10,000 ACFM dry (0% water vapor) discharge gas stream containing small particles exits a metallurgical kiln at [latex]T_1[/latex] = 600. °C and [latex]P_1[/latex] = 101. kPa. To capture the particles it is necessary to reduce the gas temperature to [latex]T_3[/latex] = 200. °C and [latex]P_3[/latex] = 101. kPa before passing the gas into a baghouse. To keep the bags from blinding (being totally blocked by particles), it has been found necessary to maintain the humidity ratio, [latex]ω_3[/latex] = (mass of water vapor) / (mass of dry air), below 0.65. The gas stream enters the cooler at state (1), is cooled by material entering at state (2), and leaves the cooler to enter the baghouse at state (3). Three cooling methods are to be considered:

- Method 1: cool with dried air (0% water vapor) at [latex]T_2[/latex] = 25. °C
- Method 2: cool with a water spray at [latex]T_2[/latex] = 25. °C
- Method 3: cool with ambient air at [latex]T_2[/latex] = 25. °C in which the relative humidity, Φ, is 50.% and [latex]ω_2[/latex] = 0.010
Humidity ratio (ω) is related to the partial pressure of water vapor [latex](P_{water vapor})[/latex] and that of dry air [latex](P_{dry air})[/latex]. Using relationships developed for gas mixtures in Chapter 1,

[latex]\omega =\frac{m_{water vapor}}{m_{dry air}}=\frac{M_{water vapor}n_{water vapor}}{M_{dry air}n_{dry air}}=\frac{m_{water}}{M_{dry air}}\frac{n_{water vapor}}{n}\frac{n}{n_{dry air}}=\frac{M_{water }}{M_{dry air}}\frac{P_{water vapor}}{P}\frac{P}{P_{dry air}}[/latex]

where n is the total number of mols in the gas mixture and P is its total pressure. Since P is the sum of partial pressures [latex]P_{dry air}[/latex] and [latex]P_{water vapor}[/latex], the above reduces to

[latex]\omega =\frac{M_{water}}{M_{dry air}} \frac{P_{water vapor}}{P-P_{water vapor}}=\frac{18.02}{28.97} \frac{P_{water vapor}}{P-P_{water vapor}} =0.662\frac{P_{water vapor}}{P-P_{water vapor}}[/latex]

Thus a maximum value of ω = 0.65 corresponds to a water vapor partial pressure of [latex]P_{water}[/latex] = 51.6 kPa. For purposes of calculation one can round this to a maximum value of [latex]P_{water}[/latex] ≈ 50 kPa.
To do: Analyze the three cooling methods.

Accepted Answer

The data are summarized below:

[latex]ρ_1 = P_1(/RT_1) = 0.4031 kg/m^3[/latex] [latex]Q_1 = Q_{dry air,1} = 10,000 ACFM (283.39 actual m^3 /min)[/latex]
[latex]ρ_3 = P_3/RT_3 = 0.7440 kg/m^3[/latex] [latex]\dot{m} _1 = \dot{m} _{dry air,1} = Q_1ρ_1 = 114.23 kg/min[/latex]

water vapor specific enthalpy: [latex]\hat{h} _{w,3}[/latex] (200. °C, 50. kPa) = 2877.7 kJ/kg
Method 1 (dilution with dried air) - Cool with dry air [latex]T_2[/latex] = 25. °C. Defining the cooler as an adiabatic open system (a control volume) and applying conservation of mass,

[latex]\dot{m} _3=\dot{m} _1+\dot{m_2} =\dot{m}_{dry air .1} +\dot{m} _{dry air .2}[/latex]

and conservation of energy,

[latex]0=\sum{(\dot{m}\hat{h} )} _{out}-\sum{(\dot{m}\hat{h} )} _{in}=\dot{m}_3 \hat{h} _3-\dot{m} _{dry air .1}\hat{h} _{dry air .1}-\dot{m} _{dry air .2}\hat{h} _{dry air .2}[/latex]

[latex]=(\dot{m}_{rdy air .1}+\dot{m}_{dry air .2} )\hat{h} _{dry air .3}-\dot{m} _{dry air 1}\hat{h} _{dry air .1}-\dot{m}_{dry air .2} \hat{h} _{dry air .2}[/latex]

[latex]=\dot{m} _{dry air .1}(\hat{h}_{dry air .3}-\hat{h}_{dry air .1} )+\dot{m} _{dry air .2}(\hat{h}_{dry air.3}-\hat{h}_{dry air .2} )[/latex]

Since air behaves as an ideal gas, each specific enthalpy difference can be replaced by the specific heat at constant pressure [latex](C_P)[/latex] times the temperature difference,

[latex]0=\dot{m} _{dry air .1}c_p(T_3-T_1)+\dot{m} _{dry air .2}c_p(T_3-T_2)[/latex]

[latex]c_P[/latex] can be removed from the above equation since it appears in both terms on the right. Solving for the mass flow rate at state 2,

[latex]\dot{m} _{dry air .2}=- \frac{\dot{m}_{dry air .1}(T_3-T_1) }{(T_3-T_2)}[/latex]

Plugging in the values,

[latex]\dot{m} _{dry air .2}=-\frac{\dot{m}_{dry air .1}(T_3-T_1) }{(T_3-T_2)} =-114.23\frac{Kg}{min} \frac{(200.-600.)K}{(200.-25.0)K}\simeq 261.\frac{Kg}{min}[/latex]

The volumetric flow rate exiting the cooler can now be calculated.

[latex]Q_3=\frac{\dot{m_3} }{\rho _3} =\frac{\dot{m}_{dry air .1}+\dot{m}_{dry air .2} }{\rho _3} =\frac{114.2\frac{Kg}{min} +261.\frac{Kg}{min} }{0.744\frac{Kg}{m^3} } =504.\frac{m^3}{min} [/latex]

This flow rate is about 78 percent larger than the input flow rate, [latex]Q_1[/latex].

[latex]Q_3[/latex] (Method 1) = 504. m³/min

Method 2 (evaporative cooling) - Cool with a water spray at [latex]T_2[/latex] = 25. °C in which [latex]\hat{h} _{w ,2}[/latex] = [latex]\hat{h} _f[/latex] (25. °C) = 104.9 kJ/kg. Apply conservation of mass and energy, as previously. Although the air can be treated as an ideal gas, the water vapor cannot. Conservation of mass becomes

[latex]\dot{m} _3=\dot{m} _1+\dot{m_2} =\dot{m}_{dry air .1} +\dot{m} _{dry air .2}[/latex]

and conservation of energy becomes

[latex]0=\sum{(\dot{m}\hat{h} )} _{out}-\sum{(\dot{m}\hat{h} )} _in=\dot{m}_3 \hat{h} _3-\dot{m} _{dry air .1}\hat{h} _{dry air .1}-\dot{m} _{dry air .2}\hat{h} _{dry air .2}[/latex]

[latex]=(\dot{m} _{dry air.1}\hat{h}_{dry air.3}+\dot{m}_{water .2}\hat{h}_{water.3} )-\dot{m}_{dry air.1} \hat{h} _{dry air.1}-\dot{m} _{water.2}\hat{h} _{water.2}[/latex]

Solving for the mass flow rate of cooling water,

[latex]\dot{m} _{water.2}=\frac{\dot{m}_{dry air .1}(\hat{h}_{dry air.3}-\hat{h}_{dry air.1} ) }{(\hat{h}_{water.2}-\hat{h}_{wter.3}) } =\frac{\dot{m}_{dry air.1}c_p(_3-T_1) }{(\hat{h} _{water.2}-\hat{h}_{water.3} )} [/latex]

[latex]=\frac{114.2\frac{Kg}{min}1.004\frac{KJ}{Kg.K}(200.-600.)K }{(104.9-2877.7)\frac{KJ}{Kg} } =16.54\frac{Kg}{min} [/latex]

The volumetric flow rate exiting the cooler is

[latex]Q_3=\frac{\dot{m}_3 }{\rho _3} =\frac{\dot{m}_{dry air .1} +\dot{m}_{water .2} }{\rho _3} =\frac{114.2\frac{Kg}{min}+16.54\frac{Kg}{min} }{0.744\frac{Kg}{m^3} } =176.\frac{m^3}{min} [/latex]

This flow rate is about 38 percent smaller than the input flow rate, [latex]Q_1[/latex].

[latex]Q_3[/latex] (Method 2) = 176. m³/min

Method 3 (dilution with moist ambient air) - Cool with ambient air [latex]T_2[/latex] = 25. °C containing water vapor at 50% relative humidity, [latex]ω_2[/latex] = 0.01, and [latex]\hat{h} _{w ,2}[/latex] ≅ [latex]\hat{h}_g[/latex] (25. °C) = 2547.2 kJ/kg. Applying conservation of mass,

[latex]\dot{m} _3=\dot{m} _{dry air.3}+\dot{m} _{water .3}=\dot{m_1} +\dot{m_2} =\dot{m}_{dry air.1}+\dot{m}_{dry air.2} +\dot{m} _{water.2}[/latex]

Using the definition of humidity ratio (ω),

[latex]\dot{m} _{water.2}=\omega _2\dot{m} _{dry air .2}[/latex]

and the above becomes

[latex]\dot{m} _{dry air .3}+\dot{m} _{water.3}=\dot{m_{dry air .1}} +\dot{m} _{dry air .2}(1+\omega _2)[/latex]

Applying conservation of energy,

[latex]0=\sum({\dot{m}\hat{h} } )_{out}-\sum{({\dot{m}\hat{h} })} _{in}[/latex]

[latex]=\dot{m} _{dry air .3}\hat{h} _{dry air .3}+\dot{m} _{water.3}\hat{h} _{water.3}-\dot{m} _{dry air .1}\hat{h} _{dry air .1}-\dot{m} _{dry air .2}\hat{h} _{dry air .2}-\dot{m} _{water .2}\hat{h} _{water .2}[/latex]

[latex]=\dot{m} _{dry air .1}(\hat{h}_{dry air .3}- \hat{h}_{dry air .1} )+\dot{m} _{dry air .2}(\hat{h}_{dry air .3} - \hat{h}_{dry air . 2} )+\dot{m} _{water .2}(\hat{h}_{water.3}-\hat{h}_{water .2} )[/latex]

[latex]=\dot{m} _{dry air .1}(\hat{h}_{dry air .3}- \hat{h}_{dry air .1} )+\dot{m} _{dry air .2}(\hat{h}_{dry air .3} - \hat{h}_{dry air . 2} )+\omega _2\dot{m} _{dry air .2}(\hat{h}_{water.3}-\hat{h}_{water .2} )[/latex]

Solving for the mass flow rate of cooling water,

[latex]\dot{m} _{dry air .2}=-\frac{\dot{m}_{dry air .1}(\hat{h}_{dry air .3} -\hat{h}_{dry air .1} ) }{(\hat{h}_{dry air .3} -\hat{h}_{dry air .2} )+\omega _2(\hat{h}_{water .3} -\hat{h}_{water .2} )} =\frac{\dot{m}_{dry air .1}c_P(T_3-T_1) }{c_p(T_3-T_2)+\omega _2(\hat{h}_{water .3} -\hat{h}_{water.2} )} [/latex]

[latex]=-\frac{114.2\frac{Kg}{min}1.004\frac{KJ}{Kg.K}(200.-600.)K }{1.004\frac{KJ}{Kg.K}(200.-25.)K+0.010(2877.7-2547.2)\frac{KJ}{Kg} }=256.2\frac{Kg}{min} [/latex]

The volumetric flow rate exiting the cooler is

[latex]Q_3=\frac{\dot{m_3} }{\rho _3} =\frac{\dot{m}_{dry air .1}+\dot{m} _{dry air .2} }{\rho _3} =\frac{114.2\frac{Kg}{min}+256.2\frac{Kg}{min} }{0.744\frac{Kg}{m^3} } =498.\frac{m^3}{min} [/latex]

This flow rate is about 76 percent larger than the input flow rate, [latex]Q_1[/latex].

[latex]Q_3[/latex] (Method 3) = 498. m³/min

Discussion: It is clear that a water spray reduces both the temperature and the volumetric flow rate of gas entering the baghouse. If cooling is attempted by using either bone dry air at 25. °C or ambient air at 25. °C, 50% relative humidity, the required volumetric flow rate of cooling gas entering the baghouse must be considerably larger than that of the original hot gas stream.

Question 9.6: Cooling a Hot and Dry Gas Stream A 10,000 ACFM dry (0% water...

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