Question

Partially saturated air at \(1\ \text{bar}\) and \(50^\circ\text{C}\) is contacted with water in an adiabatic saturator. The air is cooled and humidified to saturation, and exits at \(25^\circ\text{C}\) with an absolute humidity of \(0.02\ \text{kg water per kg dry air}\). Use latent heat of vaporization of water as \(2450\ \text{kJ}\,\text{kg}^{-1}\), and average specific heats \(c_{p,a}=1.01\ \text{kJ}\,\text{kg}^{-1}\text{K}^{-1}\) (dry air) and \(c_{p,w}=4.18\ \text{kJ}\,\text{kg}^{-1}\text{K}^{-1}\) (liquid water). If the absolute humidity of air entering the adiabatic saturator is \(\mathcal{H}\times 10^{-3}\ \text{kg water per kg dry air}\), the value of \(\mathcal{H}\) is ____________ (rounded off to two decimal places).

Hint:

For an adiabatic saturator with water at exit temperature, \(c_{p,w}\) does not appear and the key relation is \(c_{p,a}(T_1-T_2)=(w_2-w_1)h_{fg}\).
Keep units in kJ/kg consistently when using \(h_{fg}\) and \(c_{p,a}\).

Answer 1

Correct Answer: 9.1

Step 1: Perform an energy balance for the adiabatic saturator on a basis of \(1\) kg dry air. With water introduced at the exit temperature \(T_2\), the adiabatic-saturation relation reduces to \[ c_{p,a}(T_1-T_2) = (w_2-w_1)\,h_{fg}(T_2), \] where \(w_1\) and \(w_2\) are inlet and saturated exit humidities (kg/kg dry air). Step 2: Substitute data: \(T_1=50^\circ\text{C}\), \(T_2=25^\circ\text{C}\), \(w_2=0.02\), \(c_{p,a}=1.01\ \text{kJ kg}^{-1}\text{K}^{-1}\), \(h_{fg}=2450\ \text{kJ kg}^{-1}\). \[ w_1 = w_2 - \frac{c_{p,a}(T_1-T_2)}{h_{fg}} = 0.02 - \frac{1.01\times 25}{2450} = 0.02 - 0.01031 = 0.00969\ \text{kg/kg dry air}. \] Step 3: Express as \(\mathcal{H}\times 10^{-3}\): \(w_1=\mathcal{H}\times 10^{-3}\Rightarrow \mathcal{H}=9.69\). \[ \boxed{\mathcal{H}=9.69}. \]