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Assignment No. 2 Group #4 Rachelle Manalo Ronna Anne Oyco Jonas Jane Aguila Albert Jim Gonzales Joeward Manaig
Problem 4 ,[object Object]
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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4a group4 edited

  • 1. Assignment No. 2 Group #4 Rachelle Manalo Ronna Anne Oyco Jonas Jane Aguila Albert Jim Gonzales Joeward Manaig
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  • 3. 11 cm 1 m 1.2 m 0.75 m DIAGRAM
  • 4. 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
  • 5. 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
  • 6. 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
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  • 9. 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
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  • 13. 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
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