Winter plant with heating coil and steam humidifier: 22-Mec-B2 Environmental Control in Buildings, December 2014, Q2

Question 2: Winter plant with heating coil and steam humidifier (20 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A once-through-plus-recirculation winter air-handling unit serving a single space whose load, state and supply air rate are all specified.

Given data — Problem 2
Quantity	Symbol	Value
Space heating load	$q_t$	30 kW
Sensible heat ratio	SHR	0.60
Space state	$t_r$ / RH	$21^\circ\text{C}$ DB / 50 % RH
Outdoor state	$t_o$ / RH	$10^\circ\text{C}$ DB / 10 % RH
Outdoor-air fraction (dry-air mass)	$f_o$	0.40
Supply air rate	$\dot m$	7,000 kg/hr
Humidifier steam	$t_g$ / $h_g$	$120^\circ\text{C}$ saturated, 2,706 kJ/kg

Find. A plant schematic and the corresponding chart cycle with every state point identified, then the heating-coil duty and the humidifier moisture rate.

Problem 2 (a) and (c) — plant schematic. Outdoor air and return air mix ahead of the filter (state 1); the heating coil raises the dry-bulb temperature at constant humidity ratio (state 2); the fan adds a small amount of heat (state 3); the steam humidifier adds moisture at essentially constant dry-bulb temperature (state 4), and the air is delivered to the space as state 5.

Problem 2 (b) and (c) — the same cycle on the SI psychrometric chart. The mixing line runs from O to R through state 1; the coil process 1→2 is horizontal (sensible heating); the humidification 2→5 is very nearly vertical because saturated steam at $120^\circ\text{C}$ enters at close to the air's own dry-bulb enthalpy per unit moisture; and the room process 5→R has the slope set by SHR = 0.60.

Approach. Convert the load and the supply rate into the required supply state, mix outdoor and return air to fix the coil entering state, obtain the moisture addition from the humidity-ratio difference across the humidifier, and back out the coil leaving enthalpy from an energy balance on the humidifier — the coil duty is then the enthalpy rise from the mixed state to that leaving state.

Reduce the given data to rates. The dry-air mass flow is $\dot m = 7{,}000/3{,}600 = 1.944\ \text{kg}/\text{s}$, and the load splits as $q_s = 0.60(30) = 18.0\ \text{kW}$ sensible and $q_l = 0.40(30) = 12.0\ \text{kW}$ latent. Because this is a heating load, the supply air must be both warmer and wetter than the room.
Fix the room and outdoor states. At $21^\circ\text{C}$ and 50 % RH, $W_r = 0.00773\ \text{kg}/\text{kg}$; at $10^\circ\text{C}$ and 10 % RH the outdoor air is extremely dry, $W_o = 0.00076\ \text{kg}/\text{kg}$. That dryness is what makes the humidifier necessary.
Supply state (point 5). The supply air gives up its sensible heat and its moisture to the space, so $$t_5 = t_r + \frac{q_s}{\dot m\,c_p} = 21 + \frac{18.0}{1.944(1.0204)} = 30.1^\circ\text{C}, \qquad W_5 = W_r + \frac{q_l}{\dot m\,h_{fg}} = 0.00773 + \frac{12.0}{1.944(2501)} = 0.01020\ \text{kg}/\text{kg}$$ giving $h_5 = 56.33\ \text{kJ}/\text{kg}$.
(c) Mixed air (point 1). Mixing on a dry-air mass basis, $$t_1 = 0.40(10)+0.60(21) = 16.6^\circ\text{C}, \qquad W_1 = 0.40(0.00076)+0.60(0.00773) = 0.00494\ \text{kg}/\text{kg}$$ so $h_1 = 29.21\ \text{kJ}/\text{kg}$ and the mixed air stands at about 42 % RH.
(f) Moisture addition by the humidifier. Only the humidifier changes the humidity ratio, and it must carry the air from $W_1$ to $W_5$: $$\dot m_w = \dot m\,(W_5-W_1) = 1.944\,(0.01020-0.00494) = 0.01022\ \text{kg}/\text{s} = \boxed{36.8\ \text{kg}/\text{hr}}$$
State leaving the heating coil (point 2). The humidifier adds steam carrying $h_g = 2{,}706\ \text{kJ}/\text{kg}$, so an energy balance across it gives $h_2 = h_5 - (W_5-W_1)h_g = 56.33 - 0.00526(2706) = 42.10\ \text{kJ}/\text{kg}$ at the unchanged humidity ratio $W_1$, which is $t_2 = 29.3^\circ\text{C}$. The humidifier therefore raises the dry-bulb temperature by less than one degree — steam humidification is very nearly a vertical line on the chart.
(e) Heating-coil duty. The coil carries the air from state 1 to state 2 at constant humidity ratio, $$q_{\mathrm{coil}} = \dot m\,(h_2-h_1) = 1.944\,(42.10-29.21) = \boxed{25.1\ \text{kW}}$$
Check the whole plant against the room load. The supply air delivers $\dot m (h_5 - h_r) = 1.944(56.33-40.76) = 30.3\ \text{kW}$ to the space against the stated 30 kW — agreement to about 1 %, which is the expected residual from reading chart properties. The coil duty plus the steam enthalpy, $25.1 + 27.7 = 52.8\ \text{kW}$, exceeds the room load because the plant is also warming and wetting 40 % outdoor air from $10^\circ\text{C}$ and 10 % RH.

Note on the lettering. The examination paper labels this problem's parts a, b, c, e, f — there is no part (d) in the printed paper. The five printed parts are answered as printed.

First, the coil duty (25.1 kW) is smaller than the space load (30 kW) even though the plant also conditions outdoor air, because the humidifier's steam carries in roughly 28 kW of enthalpy of its own; a system using an adiabatic washer instead would need a much larger coil, since the evaporating water would take its latent heat out of the air stream. Second, the supply temperature of 30.1°C is high for a ceiling-diffuser system and would stratify in a tall space, so on a real project this would prompt either a higher supply air rate or perimeter heating — the arithmetic is only the beginning of the design decision.

Problem 2 — answers
Quantity	Symbol	Result
Dry-air mass flow	$\dot m$	1.944 kg/s (7,000 kg/hr)
Sensible / latent split of the load	$q_s$ / $q_l$	18.0 kW / 12.0 kW
(c) State 1, mixed air	$t_1$ / $W_1$	$16.6^\circ\text{C}$ / 0.00494 kg/kg
(c) State 2, leaving the heating coil	$t_2$ / $W_2$	$29.3^\circ\text{C}$ / 0.00494 kg/kg
(c) State 5, supply air	$t_5$ / $W_5$	$30.1^\circ\text{C}$ / 0.01020 kg/kg
(c) Room state R	$t_r$ / $W_r$	$21.0^\circ\text{C}$ / 0.00773 kg/kg
(e) Heating-coil duty	$q_{\mathrm{coil}}$	25.1 kW
(f) Moisture addition	$\dot m_w$	36.8 kg/hr (0.0102 kg/s)

Topic. Sequential psychrometric processes in a winter air-handling unit: mixing, sensible heating and steam humidification.

Key relations. Mixing: $t_1 = f_o t_o + (1-f_o)t_r$ and likewise for $W$. Sensible heating: $q = \dot m\,c_p\,\Delta t$ at constant $W$, with $c_p = 1.006 + 1.86W$. Steam humidification: $\dot m_w = \dot m\,\Delta W$ and $h_{out} = h_{in} + \Delta W\,h_g$. Room: $q_s = \dot m c_p (t_5-t_r)$, $q_l = \dot m h_{fg}(W_5-W_r)$.

Why this works. Each device changes the air state in one characteristic direction, and the chart makes those directions visible: mixing moves along the straight line joining the two entering states, in the ratio of the mass flows; a heating coil moves horizontally to the right; a steam humidifier moves almost vertically upward, because the enthalpy the steam brings in per kilogram of moisture is close to the enthalpy the air already assigns to its moisture. Because the humidity ratio is only changed by one device, the moisture balance can be closed before the energy balance, and the coil duty then falls out of the enthalpy difference without any iteration.

Common pitfall. Taking the coil duty as $\dot m(h_5 - h_1)$, i.e. crediting the coil with the humidifier's steam enthalpy as well. That gives about 52.8 kW instead of 25.1 kW — a coil more than twice its correct size. The other common error is mixing on a volume basis when the question states mass fractions.

Source. W. P. Jones, Air Conditioning Engineering, Ch. 3 (the psychrometry of plant processes) and Ch. 6 (humidifiers); ASHRAE Handbook — Fundamentals Ch. 1.

Question 2 of 8: Winter plant with heating coil and steam humidifier

Question 2: Winter plant with heating coil and steam humidifier (20 points)