4a group4 edited

Assignment No. 2 Group #4 Rachelle Manalo Ronna Anne Oyco Jonas Jane Aguila Albert Jim Gonzales Joeward Manaig

11 cm 1 m 1.2 m 0.75 m DIAGRAM

q q = Δ T R total q = 200 - 32 11/100 (1.4 • 0.75 • 1.2) 1855.6363 J/s x = 1855.6363 J/s 3600 s 1 hr x 24 hr = 160326981.8 J

q q = 200 - 38 11/100 (1.4 • 0.75 • 1) 1546.3636 J/s x = 1546.3636 J/s 3600 s 1 hr x 24 hr = 133605818.2 J

q q = 200 - 38 11/100 (1.4 • 1.2 • 1) 2474.1818 J/s x = 2474.1818 J/s 3600 s 1 hr x 24 hr =213769309.1 J q total = 2 (160326981.8) + 133605818.2 + 2(213769309.1) q total = 881798399.8 J

PROBLEM 12 ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],8 cm 10 cm 18 cm 13 cm 60 cm 8 cm 10 cm

SOLUTION ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

SOLUTION Get k: Using A = kdT Get Area under the curve using trapezoidal rule: A = 1798.7836 m 2 1798.7836 m 2 = k (1673 – 473) k = 1.49899

SOLUTION ,[object Object],[object Object],[object Object],[object Object],[object Object]

PROBLEM NO. 4.3-1 ,[object Object],[object Object]

PROBLEM Given ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],0.0191 m 0.0508 m Δx 2 29.4°C -17.8°C pinewood Cork board concrete

PROBLEM NO. 1 - Solution R 1 = Δx 1 = 0.0191 = 0.003243 K/W k 1 A 0.151(39) R 2 = Δx 2 = Δx 2 k 2 A 0.0433(39) R 1 = Δx 1 = 0.0508 = 0.001709 K/W k 1 A 0.762(39) q = ΔT = 302.4 – 255.2 R T R 1 + R 2 + R 3 586 W= 302.4 – 255.2 0.003243 + Δx 2 + 0.001709 0.0433(39) Δx 2 = 0.128 m

PROBLEM 4.3-5 ,[object Object],[object Object]

GIVEN ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

SOLUTION ,[object Object],First, we assume : T m = 448 K = 175 ° C T 2 = 585 K K m = 0.0733 W/mK D i2 = 0.1143 m D o2 = 0.1143 + 2(0.0064) = 0.1271 m R c = 7.41 x 10 -3 Δ x 6.4/1000 K m A m (1.52)(1.5)( Π ) 0.1271-0.1143 ln (0.1271/0.1143)

SOLUTION ,[object Object],[object Object],[object Object],[object Object],Δ x 102/1000 K r A r (0.0733)(1.5)( Π ) 0.3311-0.1271 ln (0.3311/0.1271) R r = =

SOLUTION ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],q = - Δ T R T

SOLUTION ,[object Object],[object Object],[object Object],Δ x 102/1000 K r A r (0.0735)(1.5)( Π ) 0.3311-0.1271 ln (0.3311/0.1271) R = = R = 1.3822

SOLUTION ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],ANSWER: T 2 = 587.2212 K q = 199.8417 W

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