## Chapter 13

## Q. 13.1

A parallel-flow air washer is designed as shown in Fig. 13-1. Find the basic dimensions for the air washer. The design conditions are as follows:

t_{l1} = 90 F | Water temperature at the inlet |

t_{l2} = 75 F | Water temperature at the outlet |

t_{a1} = 60 F | Air dry bulb temperature at the inlet |

t_{wb1} = 42 F | Air wet bulb temperature at the inlet |

G_{a} = 1250 lbm/(hr-ft²) | Air mass flow rate per unit area |

G_{l}/G_{a} = 0.75 | Spray ratio |

h_{a} a_{h} = 50 Btu/(hr-F-ft³) | Air heat-transfer coefficient per unit volume |

h_{l} a_{h} = 666 Btu/(hr-F-ft³) | Liquid heat-transfer coefficient per unit volume |

Q = 7000 cfm | Air volume flow rate |

## Step-by-Step

## Verified Solution

The mass flow rate of the dry air is given by

\dot{m}_{a} = \frac{\dot{Q}}{v_{1}} = \frac{7000}{13.1} = 534 lb/min

Then the spray chamber must have a cross-sectional area of

A_c = \frac{\dot{m}_{a} }{G_{a}} = \frac{534(60)}{1250} = 25.6 ft^{2}

The Colburn analogy of Eq. 13-18 with Le = 1 will be used to obtain the masstransfer coefficient (assuming a_{m} = a_{h}):

\frac{h}{h_{d} c_{pa}} = Le^{2/3} (13-18)

h_{d} a_{m} = \frac{h_{a} a_{h}}{c_{pa}} = \frac{50}{0.24} = 208 lbm/(hr-ft³)

In parallel flow the air entering at 60 F is in contact with the water at 90 F, and the air leaving at state 2 is in contact with the water at 75 F. This helps to understand the construction of Fig. 13-3, which shows the graphical solution for the interface states and the process path for the air passing through the air washer. The solution is carried out as follows:

1. Locate state 1 as shown at the intersection of t_{a1} and i_{1}. Point A is a construction point defined by the entering water temperature t_{l1} and i_{1}. Note that the temperature scale is used for both the air and water.

2. The energy balance line is constructed from point A to point B and is defined by Eq. 13-23:

G_{a} d_{i} = ±G_{l} c_{l} dt_{l} + c_{l} t_{l} dG_{l} (13-23)

\frac{d_{i}}{d_{t_{l}}} = -\frac{G_{l} c_{l}}{G_{a}} = -\frac{G_{l}}{G_{a}} = -0.75

since c_{l} = 1 Btu/(lbmw-F). Point B is determined by the temperature of leaving water, t_{l2} = 75 F. The negative slope is a consequence of parallel flow. The line AB has no physical significance, except that it depends on the energy balance between the water and air.

3. The line A1_{i} is called a tie line and is defined by Eq. 13-25:

\frac{i – i_{i}}{t_{l} – t_{i} } = -\frac{h_{l} a_{h}}{h_{d} a_{m}} = -\frac{h_{l}}{h_{d}} (13-25)

\frac{i – i_{i}}{t_{l} – t_{i} } = \frac{-h_{l} a_{h}}{h_{d} a_{m}} = \frac{-666}{208} = -3.2

The intersection of the tie line having slope –3.2 with the saturation curve defines the interface state 1_{i}. The combination of the line AB and line A1_{i} represents graphical solutions of Eqs. 13-23 and 13-25.

4. The initial slope of the air process path is then given by a line from state 1 to state 1_{i}. The length of the line 1a depends on the required accuracy of the solution and the rate at which the curvature of the path is changing.

5. The procedure is repeated by constructing the line aM and the tie line Ma_{i}, which has the same slope as A1_{i}. The path segment ab is on a line from a to a_{i}. Continue in the same manner until the final state of the air at point 2 is reached. State point 2 is on a horizontal line passing through point B. The final state of the air is defined by _{ta2} = 68 F, i_{2} = 28.0 Btu/lbma, and t_{wb2} = 61 F.

6. To complete the solution, it is necessary to determine the length of the air washer. Equation 13-22b gives

G_{a} d_{i} = h_{d} a_{m} (i_{i} – i)dL (13-22b)

dL = \frac{G_{a}}{h_{d} a_{m}} \frac{d_{i}}{(i_{i} – i)} (13-22c)

or

L = \frac{G_{a}}{h_{d} a_{m}} \int_{1}^{2}{\frac{d_{i}}{(i_{i} – i)}} (13-22d)

Equation 13-22d can be evaluated graphically or numerically. A plot of 1/(i_{i} – i) versus i is shown in Fig. 13-4, where i_{i} – i is found from Fig. 13-3. The area under the curve represents the value of the integral. Using Simpson’s rule with four equal increments yields

y = \int_{1}^{2}{\frac{d_{i}}{i_{i} – i} } \approx \frac{Δi}{3} (y_{1} + 4y_{2} + 2y_{3} + 4y_{4} + y_{5})

with

Δi = \frac{i_{2} – i_{1}}{4} = \frac{28.0 – 16.3}{4} = 2.93 Btu/lbm

y = \frac{2.93}{3} [0.036 + 4(0.044) + 2(0.057) + 4(0.076) + 0.12] = 0.733

The design length is then

L = \frac{1250}{208} (0.733) = 4.4 ft

The graphical procedure can be programmed for a digital computer. Use of a computer also permits refinements and variation of design parameters.