Question 2.2: An air-conditioning unit has a fan that delivers a mass of a...
An air-conditioning unit has a fan that delivers a mass of atmospheric air. The mass flow rate of the dry air in this is m_{a}, which is 0.5 kg/s. The specific humidity is 0.01 kg/kg air, and it can be concluded that the mass flow rate of the water vapour, m_{s2}, is 0.005 kg/s. The ingoing air temperature is 30°C.
The unit causes a drop in specific humidity to 0.006 kg/kg. What is the rate of heat taken out of the air in the cooler section of the air conditioning unit, \dot{q}?
What is the pressure of the dry air, p_{a}, and of the water vapour, p_s, at exit?
(1) Calculate p_{s2} and p_{s3} using specific humidity–vapour pressure formula.
(2) Calculate the mass flow rate of dry air and of water vapour at entry to, and at the exit from, the cooler.
(3) What is the dew point temperature corresponding to p_{g} = p_{s3}?
(4) Find c_{pair} from the tables at the appropriate mean temperature.
(5) Find, from the tables, the specific enthalpy of liquid water at the dew point temperature, and of the vapour at points 2 and 3.
(6) Use the SFEE across the cooler to find \dot{Q} the rate of heat delivery to the air stream.
(1) using \omega = \frac{0.622p_{s}}{p – p_{s}} = \frac{0.622p_{s}}{1.01325 – p_{s}} 0.01 \Rightarrow p_{s2} \Rightarrow 0.01 \times 1.01325 – 0.01p_{s2} = 0.622p_{s2}
\Rightarrow 0.01013 = 0.632p_{s2} \Rightarrow p_{s2} = 0.016 bar
p_{s3} must be 100% Φ, ∴ p_{g} = p_{SAT} = p_{s3}
\omega = \frac{0.622p_{s3}}{p – p_{s3}} \Rightarrow 0.006 = \frac{0.622p_{s3}}{1.01325 – p_{s3}}\Rightarrow 0.00608 – 0.006p_{s3} = 0.622p_{s3}\Rightarrow 0.00608 = 0.628p_{S3} \Rightarrow p_{S3} = 0.0097 bar
(2) Atmospheric air mass flow rate at entry is given as 0.5 kg/s. We know
\omega \Rightarrow \frac{\dot{m}_{W} }{\dot{m}_{DRYAIR}} = 0.01 and \dot{m}_{DRYAIR} + \dot{m}_{W} =0.5 kg/s \therefore \dot{m}_{W} = 0.005 kg/sand
\dot{m}_{DRYAIR} = 0.495 kg/sAt exit, we must have \dot{m}_{DRYAIR} = 0.495 kg/s still – it can only go out one way.We know that ω = 0.006, and therefore we say
\omega = \frac{\dot{m}_{W}}{0.495} = 0.006 \Rightarrow \dot{m}_{W} =0.003 kg/s(3) p_{s3}, t_{SAT}\sim 6.5°C from tables of saturated steam and water.
(4) The mean temperature of 6.5°C and 30°C is 18.25°C or 291 K. Using the data for dry air at low pressure,
| \frac{T}{[K]} | c_{p} [KJ/kgK] | c_{v} [KJ/kgK] | \gamma [-] | \mu \times 10^{-5} [kg/m.s.] | K \times 10^{-5} [kW/m.K.] | Pr [-] |
| 175 | 1.0023 | 0.7152 | 1.401 | 1.182 | 1.593 | 0.74 |
| 200 | 1.0025 | 0.7154 | 1.401 | 1.329 | 1.809 | 0.73 |
| 225 | 1.0027 | 0.7156 | 1.401 | 1.467 | 2.020 | 0.72 |
| 250 | 1.0031 | 0.716 | 1.401 | 1.599 | 2.227 | 0.72 |
| 275 | 1.0038 | 0.7167 | 1.401 | 1.725 | 2.428 | 0.71 |
| 300 | 1.0049 | 0.7178 | 1.400 | 1.846 | 2.624 | 0.70 |
| 325 | 1.0063 | 0.7192 | 1.400 | 1.962 | 2.816 | 0.70 |
Table 2.1 Dry air at low pressure (Rogers and Mayhew, 1995)
c_{p,air)291K} could be interpolated for accuracy, but inspection shows it is 1.0045 kJ/kgK.
(5) The saturated water and steam table is required again.
| \frac{T}{C} | \frac{P_{s}}{[bar]} | \frac{V_{g}}{[m^{3}/kg]} | \frac{h_{f}}{[kJ/kg]} | \frac{h_{fg}}{[kJ/kg]} | \frac{h_{g}}{[kJ/kg]} |
| 0.01 | 0.006112 | 206.1 | 0 | 2500.8 | 2500.8 |
| 1 | 0.006566 | 192.6 | 4.2 | 2498.3 | 2502.5 |
| 2 | 0.007054 | 179.9 | 8.4 | 2495.9 | 2504.3 |
| 3 | 0.007575 | 168.2 | 12.6 | 2493.6 | 2506.2 |
| 4 | 0.008129 | 157.3 | 16.8 | 2491.3 | 2508.1 |
| 5 | 0.008719 | 147.1 | 21.0 | 2488.9 | 2509.9 |
| 6 | 0.009346 | 137.8 | 25.2 | 2486.6 | 2511.8 |
| 7 | 0.01001 | 129.1 | 29.4 | 2484.3 | 2513.7 |
| 8 | 0.01072 | 121.0 | 33.6 | 2481.9 | 2515.5 |
| 9 | 0.01147 | 113.4 | 37.8 | 2479.6 | 2517.4 |
Table 2.2 Saturated water and steam (Rogers and Mayhew, 1995)
From this it can be seen that h_{f} of liquid water at 6.5°C is about 27 kJ/kg, hg at 6.5°C is 2512 kJ/kg and at 30°C, h_{g} is 2555 kJ/kg.
| 29 | 0.04004 | 34.77 | 121.5 | 2432.4 | 2553.9 |
| 30 | 0.04242 | 32.93 | 125.7 | 2430.0 | 2555.7 |
| 32 | 0.04754 | 29.57 | 134.0 | 2425.3 | 2559.3 |
Table 2.3
(6) Now use the SFEE, which is the statement of the first law (conservation of energy) for moving fluids, and ignore the terms from kinetic and potential energy as being negligible, Q + W = ΔH. In the air-conditioning unit between 2 and 3, there is no fan or other working device, so W = 0. We need the change in enthalpy of the fluids in section 2 to 3 in order to work out the heat transfer to cause it.
Referring to the schematic, the enthalpy flows coming in at 2 and out at 3 and through the condensate collection chute, are all indicated in terms of mass flow rate and specific enthalpy. The flow of enthalpy can then be compared in the SFEE:
\dot{Q} = \dot{m}_{DRYAIR}h_{DRYAIR,2} – \dot{m}_{DRYAIR}h_{DRYAIR,3} + \dot{m}_{W,2}h_{g,2} – \dot{m}_{W,3}h_{g,3} – \dot{m}_{COND}h_{f,COND}We can use ΔH = mCpΔT for the enthalpy change of dry air, which is
0.495 × 1.004 × (30 – 6.5) = 11.68 kW
just to cool the dry air. The change in enthalpy of the water vapour carried in the air is mainly due to the loss of mass as vapour that condenses. Therefore, with the data for h_{g} and mass flow rate of vapour going in at 2 and out at 3, we have
0.006 × 2555 – 0.003 × 2485 = 5.32 kW
The condensate is assumed to leave at 6.5°C, and it has a mass rate of 0.003 kg/s, therefore enthalpy flow rate is
0.003 × 27 = 1.27 kW
Putting all this in the SFEE, we have
\dot{Q} = 11.68 + 5.32 – 1.27 = 15.73 kW