Group 7 4ChE A
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  • 1. Group 1 4ChEB Gamboa, Ma.Raisa Faye N. Aňonuevo, Noelle P. Balane, Jun Elwene Raz, Alyssa Leigh Mesina, Renz Ronel
  • 2. 1. A Steam pipe 5 cm outside diameter has an outside surface temperature of 175 ºC. The pipe is covered with a coating material 5 cm, thick. The thermal conductivity of the coating varies with the temperature such that, k= 0.89 + 0.0015T where T is in ºC and k in W/mK. The outside surface of the coating is at 38 ºC. Calculate the heat loss per meter of pipe length. Required: q/meter of pipe k coat = 0.89 + 0.0015T Given: pipe coating pipe coating 5cm T 2 = 175 ºC T 3 =38 ºC 5cm coat = 0.0254m
  • 3. q=q 1 =q 2 R T = R 1 +R 2 K m = 1.04975 W/m-K Assumption: L pipe = 1m q 2 =T 2 -T 3 R 2 = 5/100 K m2 A m 2 R 2 D lm2 = (5/100) + 2(5/100) – (5/100) ln(0.15/(5/100)) = 0.1820m = π (1)(0.1820) = 7.97m 2 = 5/100 1.04975 (7.97) R 2 = R 2 5.98E-3 q 2 =175-38 5.98E-3 q=q 2 = 22909.7W/m
  • 4. 4.3-4 Heat Loss from Steam Pipeline . A steel pipeline, 2-in. schedule 40 pipe, contains saturated steam at 121.1 ⁰ C. The line is covered with 25.4mm of insulation. Assuming that the inside surface temperature of the metal wall is at 121.1 ⁰ C and the outer surface of the insulation is 26.7 ⁰ C, calculate the heat loss for 30.5m of pipe. Also, calculate the kg of steam condensed per hour in the pipe due to heal loss. The average k for steel from Appendix A.3 is 45 W/m-K and the k for the insulation is 0.182. Given: T 1 = 121.1⁰C. T 3 = 26.7 ⁰ C 30.5m D o steel = 2.375”(0.0254) = 0.0603 m D i steel = 0.0603 -2(3.91E-3 ) = 0.0525m steel = 0.154”(0.0254) = 3.91E-3 m k steel =45 W/m-K k insulation = 0.182 M/m-K ins = 0.0254m
  • 5. = 1.6120E-5 = 0.0175 = 3.91E-3 m 45 A m steel = 0.0603-0.0525 ln(0.0603/0.0525) = 0.0563m 3.91E-3 m = 45(5.39) = π (30.5)(0.0563) = 5.39m 2 = 0.0254 m 0.182A m ins = 0.0254 m 0.182(7.97) D lm2 = 0.0603 + 2(0.0254) – 0.0603 ln(0.111124/0.0525) = 0.0832m = π (30.5)(0.0832) = 7.97m 2 = 1.6120E-5 + 0.0175 = 0.01752
  • 6. calculate the heat loss for 30.5m of pipe calculate the kg of steam condensed per hour in the pipe due to heat loss 2260 J/g water 5388.13 J/s = 2.3841g/s q = 121.1-26.7 0.01752 q= 5388.13 W 8.58 kg/h
  • 7. 5.2-1 Temperature Response in Cooling a Wire . A small copper wire with a diameter of 0.792mm and initially at 366.5 K is suddenly immersed in a liquid held constant at 311K. The convection coefficient h=85.2W/m 2 -K. The physical properties can be assumed constant and are k=374 W/m-K, c p = 0.389kJ/kg-K, ρ =8890 kg/m 3 . a.) Determine the time in seconds for the average temperature of the wire to drop to T= 338.8K (one-half the initial temperature difference). <0.1; use lumped capacity method = 389J/kg-K = 7.92E-4 /4 = 1.98E-4 Given: D = 0.792mm =7.92E-4 m T 0 =366.5 K T 1 =311K h =85.2W/m 2 -K k =374 W/m-K c p = 0.389kJ/kg-K ρ =8890 kg/m 3 For long a long cylinder , x 1 = V = π D 2 L/4 = D A π DL 4 N Bi = hx 1 /k = 85.2(7.92E-4 /4) 374 = 4.51E-5
  • 8. b.) Do the same but for h= 11.36 W/m 2 -K hA C p ρ V = 85.2 (389)(8890)( 1.98E-4 ) = 0.1244s -1 338.8-311 366.5-311 = e -0.1244t t=5.57s T 0 -T 1 T 0 -T 1 = e -(hA/C p ρ V)t hA C p ρ V = 11.36 (389)(8890)( 1.98E-4 ) = 0.0166s -1 338.8-311 366.5-311 = e -0.0166t t=41.65s
  • 9. c.) For part (b), calculate the total amount of heat removed for a wire 1.0m long. Q = C p ρ V (T 0 -T 1 ) 1- e -(hA/C p ρ V)t Q = (389 x 8890)( π D 2 L ) (366.5-311) (1- e -0.0166(41.65) ) 4 Q = (389 x 8890) π (7.92E-4) 2 (1) (366.5-311) (1- e -0.0166(41.65) ) 4 Q = 47.19 J
  • 10. 9. A small electric furnace is 15 by 15 by 30cm. inside dimensions and has fire-brick walls (k=1.12 W/mK) 2m thick. The front of the furnace is a movable wall which permits entry into the furnace. In this section is by a 5 by 5 by 0.6 cm quartz observation windows (k=0.07 W/mK). The inner surface temperature for all sides is 1100oC, and the outer surface temperature is 121oC. Assuming all joints perfectly made and neglecting the influence of the corners on the temperature distribution, what is the heat loss from this furnace? Inside dimensions: 15 x 15 x 30 cm K brick = 1.12 W/mK For quartz: Dimension: 5 x 5 x 30 cm K quartz = 0.07 W/ mK Inner temp = 1100 °C For all sides Outer temp = 121 °C For all sides
  • 11.
    • Assumption:
    • Uniform inside temperature distribution
    • use inside dimensions to compute for the total inside area
    • 2[(0.15)(0.15)]+4[(0.15)(.3)] = 0.225 sq. m.
    • inside area of brick = total area – area of quartz
    • area of quartz = (0.05)(0.05) = 2.5 x 10^-3
    • inside area of brick = 0.225- 2. x 10^-3 = 0.2225
    • R brick = 2/ (1.12)(0.2225)
    • R brick = 8.026
    • R quartz = .6/ (100) (.07) (2.5 x 10^-3)
    • R quartz = 34.286
    R total = 34.286(8.026)/[34.286 + 8.026] = 6.504 q = (1100-121)/ 6.504 q = 150.523W