## Chapter 13

## Q. 13.2

A counterflow spray dehumidifier is to be designed as shown schematically in Fig. 13-5. These are the design conditions:

t_{l2} = 7 C | Water temperature at the inlet |

t_{l1} = 15 C | Water temperature at the outlet |

t_{a1} = 28 C | Air dry bulb temperature at the inlet |

t_{wb1} = 22 C | Air wet bulb temperature at the inlet |

G_{a} = 1.36 kg/(s-m²) | Air mass flow rate per unit area |

G_{l} c_{l}/G_{a} = 3.25 | Spray ratio |

h_{a} a_{h} = 1210 W/(C-m³) | Air heat-transfer coefficient per unit volume |

h_{l} a_{h} = 14,700 W/(C-m³) | Liquid heat-transfer coefficient per unit volume |

Q = 2.83 m³/s | Air volume flow rate |

Find the cross-sectional area and final state of the air.

## 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{2.83}{0.875} = 3.24 kg/s

The chamber cross-sectional area is then

A_c = \frac{\dot{m}_{a} }{G_{a}} = \frac{3.24}{1.36} = 2.38 m^{2}

Using the Colburn analogy and assuming that a_{m} = a_{h},

h_{d} a_{m} = \frac{h_{a} a_{m}}{c_{pa}} = \frac{1210}{1000} = 1.21 kg/(s-m³)

The graphical solution for the interface states and the process path is shown in Fig.13-6. In the case of counterflow, the air entering at 28 C is in contact with the water at 15 C, and the air leaving is in contact with the water at 7 C. The procedure is quite similar to that given in Example 13-1. Note, however, that the energy balance line AB has a positive slope because of the counterflow arrangement. Notice also that because of counterflow, point A corresponds to t_{l1} and point B corresponds to t_{l2}. The final state of the air is defined by t_{a2} = 14 C and t_{wb2} = 13.5 C.

The height of the spray chamber is determined by using Eq. 13-22b with the procedure given in Example 13-1.

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