In the Psy. Ch., plot point o, r, h r = 51.0 kj/kg, h o = 85 kj/kg

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1 & Marking Scheme (Page 1 of 8 ) 1. (a)(i) In the Psy. Ch., plot point o, r, h r = 51.0 kj/kg, h o = 85 kj/kg SHR = rs/rt = 63/70 = 0.9 Draw a line with SHR= 0.9 from r and intercept at s or cc in such a way that s is 95%RH. From the chart, t s = 14 C, h s = h cc = 38 kj/kg and v s = m 3 /kg (ii) (iii) Volume flow rate, V s = Q s *v s /( C pa (t r t s )) = 63*0.826/(1.02(25-14)) = m 3 /s Outdoor ventilation air required = 30 x 50 x = 0.75 m 3 /s t m = (V r * t r + V o * t o )/ V m, V r + V o =V m,v s =V m V r = V s - V o = = m 3 /s t m = (3.888* *32)/ = C h m = (3.888* *85)/ = kj/kg [Use of graphical methos is also accepted] (iv) (b)(i) (ii) Cooling Coil Load, Q cc = V s (h m h cc ) /v s = 4.638( )/0.826 = kw When the outdoor ventilation air is increased by 50%, no change in t o and h o. While, t m and h m will become higher and the cooling coil load Q cc will become higher. When the outdoor ventilation air condition is changed to 34 C and 80% relative humidity, t o and h o.will become higher. Also, t m and h m will become higher and the cooling coil load Q cc will become higher.

2 & Marking Scheme (Page 2 of 8 )

3 & Marking Scheme (Page 3 of 8) Question No. 2 (a) (b) Solar radiation and conduction through window, and conduction through external walls and roof. People, equipment, lighting and infiltration. [0.5 mark for each item] (i) From table, Zone types B, C and C for Solar, People/Equipment and Lighting respectively (ii) t O = 35 and t r = 23 t m = t o - (Daily range/2) = 35 (7/2) = 31.5 Correct Factor = (25.5-t r ) + (t m -29.4) = ( ) + ( ) = 4.6 (iii) (1) q = A x SCL x SC SCL = 523 W/m 2 = 40 x 523 x 0.65 = W (2) q = A x CLTDc x U CLTD = 9 = 80 x (9+4.6) x 3.5 = 3808 W (3) q = W F ul F al (CLF el ) CLF el = 0.92 = (500x20) x 1.0 x 1.2 x 0.92 = W (c) Heavier mass implies greater thermal storage leading to reduction in peak value and its occurrence shifting to some later time.

4 & Marking Scheme (Page 4 of 8 ) Question No. 3 (a) Sizing of Air Ducts Duct Section Flow Rate (L/s) Friction Loss (Pa/m) Duct Dia. (mm) Air Velocity (m/s) Length (m) OA AB CD DE EF EI FH FG GJ (b) Velocity Pressures and Dynamic Losses Junction Air Velocity (m/s) Velocity Pressure at Junctions Pv,c = 0.5 ρv2 Dynamic Loss Coefficient Dynamic Loss (Pa) A D E (C c,s ) F (C c,s ) G ΔP atraight duct from O to J = ( ) x 0.9 = 27.9 Pa ΔPO-A-B-C-D-E-F-G-J = ΔPO + ΔP (O to J) +ΔPA +ΔPD + ΔPE(Cc,s) + ΔPF(Cc,s) + ΔPG + ΔPJ = = Pa (c) Install volume control dampers at the non-critical paths O to I and O to H so that all the three paths have the same pressure loss..

5 & Marking Scheme (Page 5 of 8 ) Question No. 4(a) 4(b) (i)linear dimension, D served by an ceiling diffuser < < 8.7 m Along 38m length : 38/8.7 = nos Along 30m length : 30/8.7 = nos. So, no. of diffuser = 20 (5x4 arrangement) (ii) D1=38/5 = 7.6 D2 = 30/4 = 7.5 D max / D min = 7.6 / 7.5 = 1.01 < 1.3 (acceptable) Characteristic Length L for 38m length = 7.6 / 2 = 3.8m Characteristic Length L for 30m length = 7.5 / 2 = 3.75m For ceiling diffusers, T 0.25 / L = 0.8 for ADPI > 80% T 0.25 for 38m length = = 3.04 T 0.25 for 30m length = = 3 Assume equal load at each outlet, 3.59 / 20 = m 3 /s Processing 0.18 m 3 /s and T 0.25 (3.04; 3)-- 250x250 neck size is suggested. (NC=24; velocity= criteria check OK) Use ADPI as a tool Average velocity = 23.9 o C Point Temp Velocity θ OK OK OK OK OK OK OK Number of points where -1.7 o C θ 1.1 o C and velocity 0.35m/s = 14 ADPI = 7 /10 = 70% As ADPI < 80%, i.e. ineffective space air diffusion system

6 & Marking Scheme (Page 6 of 8 ) Question No. 5 (a) 5 (b) Superheat : to prevent liquid refrigerant from entering the compressor ; and to increase refrigeration effect; Subcooling : to ensure complete condensation; to increase refrigeration effect (i) Condenser Pressure : 1.2MPa Evaporator Pressure: 0.41MPa (ii) h 1 = 410 kj/kg h 3 = h 4, h 4 = 258 kj/kg R.E. = h 1 h 4 = = 152 kj/kg (iii) η = isentropic work h h = 1 actual work actual work 2 S 1 = S 2 (ideal compression), h 2 = 428 kj/kg ( ) Actual work = = 20 kj/kg 0.9 (iv) Refrigeration Capacity = RE m r = = 608 kw (v) The lower temperature of cooling water results in a lower condenser pressure. For a given evaporator pressure, the power input to the compressor will be reduced leading to an increase in COP. There will be an overall energy saving and improved in system performance.

7 & Marking Scheme (Page 7 of 8 ) Question No. 5 (b)

8 & Marking Scheme (Page 8 of 8 ) Question No. 6 (a) 6 (b) Primary air fan coil system is preferred. It is because PAU can be used to carry out dehumidification of humid outdoor air and to reduce noise from outside by installing silencer. There will be less condensate at FCU/indoor, less water leakage problem and lower indoor noise level. Lw = log P + 10 log Ps Lw = log (30) + 10 log (200) = = db For backward curve fan, correction factor at 2kHz = -16 SWL = = db 6 (c) (i) SPLt = SPL d + SPLr = 10 Log(10^(55/10) +10^(58/10)) = db (ii) For a general office, NC 35 to 45 is required and hence SPL NC = 34 to 44 db at 2 khz. As the resultant SPL = db >34 to 44 db, the ventilation fan is not suitable for use in this general office. 6 (d) By Allen formula, SWL = SWL B D R + 10log(S w /A) S w = ( )*2 *8 = 16 m 2 A = (0.6) *(0.4) = 0.24 m 2 SWL = log(16/0.24) = db B