Question 1 of 8: Summer air-conditioning plant for a department store

Question 1: Summer air-conditioning plant for a department store (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 single-zone mixed-air plant serves a store held at 75 °F dry bulb and 50 % relative humidity against the design cooling load below, with ventilation air fixed at 30 % of the supply air by mass.

Find. The system diagram and the cycle on the chart, the dry- and wet-bulb temperature of every significant state, the supply air rate, the coil capacity in kilowatts, the apparatus dew point and the coil bypass factor.

Approach. Fix the two known states from the ASHRAE moist-air relations, mix them in the stated mass ratio to get the coil entering state, close the one remaining degree of freedom with an assumed coil bypass factor, use the Carrier effective-sensible-heat-factor line to locate the apparatus dew point, and then read the air quantity, the supply state and the coil duty straight off the balances.

1. Fix the room and outdoor states. The humidity ratio follows from the saturation pressure and the relative humidity, and the enthalpy from the standard inch-pound moist-air relation:$$W=0.622\,\frac{\phi\,p_{ws}}{p-\phi\,p_{ws}},\qquad h=0.240\,t+W\,(1061+0.444\,t)$$At 75 °F and 50 %, $W_{R}=0.009236\ \text{lb/lb}$ (64.6 grains per pound) and $h_{R}=28.11\ \text{Btu/lb}$; at 90 °F and 60 %, $W_{O}=0.018268\ \text{lb/lb}$ and $h_{O}=41.71\ \text{Btu/lb}$. The corresponding thermodynamic wet bulbs are 62.56 °F and 78.35 °F.
2. Mix 30 % outdoor air with 70 % return air. Adiabatic mixing conserves moisture and enthalpy exactly, so both are mass-weighted and the dry bulb is recovered from them — never weighted directly, because the $W\,t$ cross term in the enthalpy relation would then leave $h(t_{M},W_{M})$ disagreeing with the mixed enthalpy:$$W_{M}=x\,W_{O}+(1-x)\,W_{R}=0.011945\ \text{lb/lb},\qquad h_{M}=x\,h_{O}+(1-x)\,h_{R}=32.19\ \text{Btu/lb}$$$$t_{M}=\frac{h_{M}-1061\,W_{M}}{0.240+0.444\,W_{M}}=79.55\ ^\circ\text{F}$$State M therefore sits at 79.55 °F dry bulb, 67.92 °F wet bulb — the air the coil actually sees.
3. Recognise the missing degree of freedom, and close it. The room sensible heat factor is $\mathrm{RSHF}=450{,}000/570{,}000=0.7895$, which fixes the direction of the supply-to-room line but not how far down it the supply state lies. The paper gives no supply temperature, so one assumption is unavoidable; cover-page instruction 1 invites it. Assuming a coil bypass factor of 0.10, typical of a five- or six-row chilled-water coil at 500 fpm face velocity, is the better choice than assuming a supply temperature difference, because the bypass factor then reappears as an independent output in step 6 and verifies the whole construction.
4. Form the effective heat factor and locate the apparatus dew point. The bypassed fraction of the outdoor air reaches the room unprocessed, so it is charged to the room rather than to the coil:$$\mathrm{ERSH}=\mathrm{RSH}+\mathrm{BF}\cdot\mathrm{OASH},\qquad \mathrm{ERLH}=\mathrm{RLH}+\mathrm{BF}\cdot\mathrm{OALH}$$with $\mathrm{OASH}=1.10\,Q_{O}(t_{O}-t_{R})=88,903$ Btu/h and $\mathrm{OALH}=4840\,Q_{O}(W_{O}-W_{R})=235,541$ Btu/h on the 5,388 cfm of outdoor air found below. That gives $\mathrm{ERSH}=458,890$ Btu/h, $\mathrm{ERLH}=143,554$ Btu/h and $\mathrm{ESHF}=0.7617$. Dropping a line of that slope from the room state until it meets the saturation curve gives$$\boxed{\ \mathrm{ADP}=49.19\ ^\circ\text{F}\ }$$The line must be drawn from R, not from M: the room-ratio line to saturation would give the room apparatus dew point, several degrees adrift of the coil's.
5. Size the air supply rate. Only the $(1-\mathrm{BF})$ share of the air is brought all the way to the apparatus dew point, so the effective room sensible heat is carried by$$Q=\frac{\mathrm{ERSH}}{1.10\,(1-\mathrm{BF})\,(t_{R}-\mathrm{ADP})}=\frac{458,890}{1.10\times0.90\times(75-49.19)}$$$$\boxed{\ Q=17,960\ \text{cfm}\ }$$That is 8.48 m³/s, or about 82,463 lb of dry air per hour, of which 5,388 cfm is outdoor air — the 30 % the ventilation requirement calls for.
6. Recover the supply state and check the bypass factor. The room balances give the supply condition directly:$$t_{S}=t_{R}-\frac{\mathrm{RSH}}{1.10\,Q}=52.22\ ^\circ\text{F},\qquad W_{S}=W_{R}-\frac{\mathrm{RLH}}{4840\,Q}=0.007855\ \text{lb/lb}$$so the supply air leaves the coil at 52.22 °F dry bulb and 51.40 °F wet bulb, a supply differential of about 23 F°. The bypass factor now follows from the finished state points, independently of the value assumed in step 3:$$\mathrm{BF}=\frac{t_{S}-\mathrm{ADP}}{t_{M}-\mathrm{ADP}}=\frac{52.22-49.19}{79.55-49.19}$$$$\boxed{\ \mathrm{BF}=0.100\ }$$It reproduces the assumed 0.10 to within a quarter of one per cent, which is what turns the assumption into a verified result.
7. Compute the coil capacity. The coil takes the whole supply stream from M down to S, so its grand total heat is$$\mathrm{GTH}=4.5\,Q\,(h_{M}-h_{S})=4.5\times17,960\times(32.19-21.05)=900,213\ \text{Btu/h}$$$$\boxed{\ \dot{Q}_{\text{coil}}=263.8\ \text{kW}\ (75.0\ \text{tons})\ }$$As an independent check the same duty must equal the room load plus the whole outdoor-air load: $570{,}000+4.5\times5,388\times(41.71-28.11)=899,883$ Btu/h, agreeing to better than 0.1 %. Note how much of the plant the ventilation air costs: the room asks for 570,000 Btu/h and the coil has to be a third larger again.

Part (c) asks for each significant point to be identified with its dry- and wet-bulb temperature, which the table below collects.