Information about Group 2 A

Published on January 7, 2009

Author: 4ChEAB08

Source: slideshare.net

4.3-3 Heat Loss Through Thermopane Double Window A double window called thermopane is one in which two layers of glass are separated by a layer of dry, stagnant air. In a given window, each of the glass layers is 6.35 mm thick separated by a 6.35 mm space of stagnant air. The thermal conductivity of the glass is 0.869 W/m*K and that of air is 0.026 over the temperature range used. For the temperature drop of 27.8 K over the system, calculate the heat loss for a window 0.914 m x 1.83 m. (Note: This calculation neglects the effect of the convective coefficient on one outside surface of one side of the window, the convective coefficient on the other outside surface, and convection inside the window.)

A double window called thermopane is one in which two layers of glass are separated by a layer of dry, stagnant air. In a given window, each of the glass layers is 6.35 mm thick separated by a 6.35 mm space of stagnant air. The thermal conductivity of the glass is 0.869 W/m*K and that of air is 0.026 over the temperature range used. For the temperature drop of 27.8 K over the system, calculate the heat loss for a window 0.914 m x 1.83 m. (Note: This calculation neglects the effect of the convective coefficient on one outside surface of one side of the window, the convective coefficient on the other outside surface, and convection inside the window.)

Solution: (6.35/1000) R glass = ---------------------------- = 0.0044 K/W (0.869)(0.914)(1.83) (6.35/1000) R air = ---------------------------- = 0.1460 K/W (0.026)(0.914)(1.83) R total = R air + 2R glass = 0.1548 K/W

(6.35/1000)

R glass = ---------------------------- = 0.0044 K/W

(0.869)(0.914)(1.83)

(6.35/1000)

R air = ---------------------------- = 0.1460 K/W

(0.026)(0.914)(1.83)

R total = R air + 2R glass

= 0.1548 K/W

Q = dT / R total Q = (27.8K) / (0.1548 K/W) = 179.58 W

Q = dT / R total

Q = (27.8K) / (0.1548 K/W)

= 179.58 W

5.2-2 Quenching Lead Shot in a Bath Lead shot having an average diameter of 5.1 mm is at an initial temperature of 204.4 o C. To quench the shot it is added to a quenching oil bath held at 32.2 o C and falls to the bottom. The time of fall is 15 s. Assuming an average convection coefficient of h = 199 W/m 2 *K, what will be the temperature of the shot after the fall? For lead, ρ = 11, 370 kg/m 3 and c p = 0.138 kJ/kg*K.

Lead shot having an average diameter of 5.1 mm is at an initial temperature of 204.4 o C. To quench the shot it is added to a quenching oil bath held at 32.2 o C and falls to the bottom. The time of fall is 15 s. Assuming an average convection coefficient of h = 199 W/m 2 *K, what will be the temperature of the shot after the fall? For lead, ρ = 11, 370 kg/m 3 and c p = 0.138 kJ/kg*K.

Solution: Newton Cooling Method T-T 1 A= surface area, sphere ------- = e -(hA/cp ρ V)(t) V = volume of a sphere T 0 -T 1 T-32.3oC ------------- = e^-[(199)(8.17x10-5)/(138)(11370)(6.94x10-08)](15) 204.4-32.3 T = 32.31oC

Newton Cooling Method

T-T 1 A= surface area, sphere

------- = e -(hA/cp ρ V)(t) V = volume of a sphere

T 0 -T 1

T-32.3oC

------------- = e^-[(199)(8.17x10-5)/(138)(11370)(6.94x10-08)](15)

204.4-32.3

T = 32.31oC

Problem 2 A furnace wall is to consist in series of 18 cm of kaolin firebrick, 15 cm of kaolin insulating brick, and sufficient fireclay brick to reduce the heat loss to 350 W/m2 when the face temperatures are 815 and 38 ºC, respectively. If an effective air gap of 3 mm (assume k for air as 0.17 W/m*K) can be incorporated between the fireclay and insulating brick when erecting the wall impairing its structural support, what thickness of fireclay brick will be required?

A furnace wall is to consist in series of 18 cm of kaolin firebrick, 15 cm of kaolin insulating brick, and sufficient fireclay brick to reduce the heat loss to 350 W/m2 when the face temperatures are 815 and 38 ºC, respectively. If an effective air gap of 3 mm (assume k for air as 0.17 W/m*K) can be incorporated between the fireclay and insulating brick when erecting the wall impairing its structural support, what thickness of fireclay brick will be required?

Solution: Assume A = 1m 2 For firebrick Assume T’ = 450 oC T ave = (815+450)/2 = 632.5 k = 0.0987(intrpltd)x1.73 = 0.1708 350 = (815-T) / (0.18/0.0987) T = 446.15oC (0.86% error)

Assume A = 1m 2

For firebrick

Assume T’ = 450 oC

T ave = (815+450)/2 = 632.5 k = 0.0987(intrpltd)x1.73

= 0.1708

350 = (815-T) / (0.18/0.0987)

T = 446.15oC (0.86% error)

For insulated firebrick Assume T = 190oC T ave = (190+446.15)/2 = 318.075oC k = 0.2062 350 = (446.15 – T) / (0.15/0.2062) T = 191.54oC (0.8% error)

For insulated firebrick

Assume T = 190oC

T ave = (190+446.15)/2 = 318.075oC k = 0.2062

350 = (446.15 – T) / (0.15/0.2062)

T = 191.54oC (0.8% error)

For Air Gap: 350 = (191.54 – T) / (0.003/0.17) T = 185.36 For FireClay: T ave = (185.36+38)/2 = 111.68 k = 0.5220 (1.73) = 0.9030 350 = (185.36-38)/(x/0.9030) X = 0.38m

For Air Gap:

350 = (191.54 – T) / (0.003/0.17)

T = 185.36

For FireClay:

T ave = (185.36+38)/2 = 111.68 k = 0.5220 (1.73) = 0.9030

350 = (185.36-38)/(x/0.9030)

X = 0.38m

Problem 10 An experimental heat transfer apparatus consists of a 5 cm. schedule 80 steel pipe covered with two layers of insulation. The inside layer is 2.5 cm thick and consists of diatomaceous silica, asbestos and a bonding material; the outside layer is 85% magnesia and is 4 cm thick. The following data were obtained during a test run: length of test section = 3 m. heating medium inside pipe = dowtherm A vapour temperature of inside of steel pipe = 400oC temperature of outside of magnesia insulation = 52oC dowtherm condensed in test section = 9 kg/hr temperature of condensate = 400oC latent heat of condensation of dowtherm of upper conditions = 206 kJ/kg Determine the mean thermal conductivity of the magnesia insulation. The thermal conductivity of the inner layer is as follows: Temperature ( o C) k (W/mK) 93 0.0885 260 0.1050 426 0.1209

An experimental heat transfer apparatus consists of a 5 cm. schedule 80 steel pipe covered with two layers of insulation. The inside layer is 2.5 cm thick and consists of diatomaceous silica, asbestos and a bonding material; the outside layer is 85% magnesia and is 4 cm thick. The following data were obtained during a test run:

length of test section = 3 m.

heating medium inside pipe = dowtherm A vapour

temperature of inside of steel pipe = 400oC

temperature of outside of magnesia insulation = 52oC

dowtherm condensed in test section = 9 kg/hr

temperature of condensate = 400oC

latent heat of condensation of dowtherm of upper conditions = 206 kJ/kg

Determine the mean thermal conductivity of the magnesia insulation. The thermal conductivity of the inner layer is as follows:

Temperature ( o C) k (W/mK)

93 0.0885

260 0.1050

426 0.1209

Solution: 5cm Nom. Pipe size => 60.33mm (outside diameter) Wall thickness = 5.537mm thickness Inner insulation = 2.5cm thickness outer insulation = 4 cm Q condensation = (9kg/h)(206kJ/kg) x 1000/3600 = 515 J/s = 515 W

5cm Nom. Pipe size => 60.33mm (outside diameter)

Wall thickness = 5.537mm

thickness Inner insulation = 2.5cm

thickness outer insulation = 4 cm

Q condensation = (9kg/h)(206kJ/kg) x 1000/3600

= 515 J/s = 515 W

Q condensation = Q steel 400-T 515 = ------------------------------------ ( 5.537/1000) ------------------------------------ 64(3 pi) (0.06033-0.0493) -------------------------- ln (0.06033/0.0493) T = 399.91 oC

Q condensation = Q steel

400-T

515 = ------------------------------------

( 5.537/1000)

------------------------------------

64(3 pi) (0.06033-0.0493)

--------------------------

ln (0.06033/0.0493)

T = 399.91 oC

Assume T 1 =240oC T in,ave = (399.91+230)/2 = 319.955 k = 0.1107 (interpolation) Q condensation = Q inner insulator 399.91 –T’ 515 = -------------------------------- (2.5/100) -------------------------------- (0.1107)(3 pi) (5/100) --------------------- ln (11.033/6.033) T = 250.93 oC (4.3% error)

Assume T 1 =240oC

T in,ave = (399.91+230)/2 = 319.955

k = 0.1107 (interpolation)

Q condensation = Q inner insulator

399.91 –T’

515 = --------------------------------

(2.5/100)

--------------------------------

(0.1107)(3 pi) (5/100)

---------------------

ln (11.033/6.033)

T = 250.93 oC (4.3% error)

Q condensation = Q outer 250.93 - 52 515 = ----------------------- 4/100 ---------------------- k(3 pi) (8/100) ------------- ln(19.033/11.033) k = 0.0749 W/mK

Q condensation = Q outer

250.93 - 52

515 = -----------------------

4/100

----------------------

k(3 pi) (8/100)

-------------

ln(19.033/11.033)

k = 0.0749 W/mK

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