problems-1-2009

problems-1-2009 - Problem 1-4 Pressure drop = 4/12 (ft) x...

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AE 3111 Basic Principles of HVAC Review Problems Set # 1 Problems 2-11 and 2-18 are drawing exercises meant to make you research different systems (discussed within the chapter), and mix different components appropriately to answer the question. Problem 1-1 (a) 98 Btu/hr.ft.F x 1.7307 = 170 W/m.K (b) 0.24 Btu/lbm.F x 4186.8 = 1.0 kJ/kg.K (c) 0.04 lbm/ft.hr/(3600 s/hr) x 1.488 = 16.5x10 -6 N.s/m 2 (d) 1050 Btu/lbm x (1/9.48x10 -4 ) J/Btu x (2.20462 lbm/kg) = 2.44 MJ/kg (e) 12,000 Btu/lbm x 1/3.412 = 3.52 kW (f) 14.7 lbf/in 2 x 6894.76 = 101 kPa Problem 1-2 (a) 120 kPa x (lbf/ in 2 )/ 6.89476 kPa = 17.4 lbf/ in 2 (b) 100 W/m.k x 0.5778 = 57.8 Btu/hr.ft.F (c) 0.8 W/m 2 .K x 0.1761 = 0.14 Btu/hr.ft 2 .F (d) 10 -6 N.s/m 2 x 1/1.488 = 6.7 x 10 -7 lbm/ft.s (e) 1200 kW x 3412 = 4.1 X 10 -6 Btu/hr (f) 1000 kJ/kg x (1 Btu/1.055 kJ) x (1 kg/2.2046 lbm) = 430 Btu/lbm

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Unformatted text preview: Problem 1-4 Pressure drop = 4/12 (ft) x 0.3048 (m/ft) x 9.807 (m/s 2 ) x 1000 (kg/m 3 ) = 996 Pa a 1.0 kPa. Problem 1-12 air water q q i = (heat loss from water = heat gain by air) Note: Use the Ideal Gas relationship to get the specific volume v, which is 1/ . ( 29 ( 29 2 1 2 1 1 p p p water air air Q P c T T m c T Q c T T R T - =-= &amp; 11,200 x 1 x 10 = 5000x60x14.7x144x0.24 (T 2- 50)/((53.35 x (50+460)) 112,000 = 5601.5 (T 2 50) T 2 = 70 F. Problem 1-17 ( 29 ( 29 1 3 1 2 3 2 p p m c T T m c T T i-+-= 0 1 1 2 2 3 1 2 m T m T T m m i i + = + 1 1 1 m Q = = 1000 x (14.7 x 144)/(53.35 x (460 +50)) = 73.5 lbm/min 2 2 2 m Q = = 6000 x (14.7 x 144)/((53.35 x (460+80)) = 46.7 lbm/min T 3 = ((73.5 x 80) + (46.7 x 50)) / (73.5 + 46.7) = 68.3 F....
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This note was uploaded on 08/28/2009 for the course MECH&AE 465 taught by Professor Moham during the Summer '09 term at UCLA.

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problems-1-2009 - Problem 1-4 Pressure drop = 4/12 (ft) x...

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