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Information about 4a Group4

Published on January 8, 2009

Author: 4ChEAB08

Source: slideshare.net

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Problem 4 The inside dimensions of a furnace 1m x 1.2m x 0.75m. The walls are 11 cm thick and made from a refractory material having an average thermal conductivity of 1.4 W/mK. Assuming that the inner and outer surface temperatures are 200°C and 38°C respectively, compute the heat loss by conduction in 24 hr. Neglect heat loss on the floor of the furnace.

The inside dimensions of a furnace 1m x 1.2m x 0.75m. The walls are 11 cm thick and made from a refractory material having an average thermal conductivity of 1.4 W/mK. Assuming that the inner and outer surface temperatures are 200°C and 38°C respectively, compute the heat loss by conduction in 24 hr. Neglect heat loss on the floor of the furnace.

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 Where the T1 = 1500K and the upper face is at 400 K, find the heat loss if T varies with K as follows: T in K K in (W/m-K) 473 1.003168 873 1.47016 1273 1.64312 1673 1.76419 8 cm 10 cm 18 cm 13 cm 60 cm 8 cm 10 cm

Where the T1 = 1500K and the upper face is at 400 K, find the heat loss if T varies with K as follows:

T in K K in (W/m-K)

473 1.003168

873 1.47016

1273 1.64312

1673 1.76419

SOLUTION q = kA m (T 1 - T 2 ) Δx Δx = 0.6m A m = a 2 b 2 – a 1 b 1 ln a 2 b 2 a 1 b 1 = (0.13)(0.18) – (0.08)(0. 10) ln (0.13)(0.18) (0.08)(0. 10) = 0.01435 m 2

q = kA m (T 1 - T 2 )

Δx

Δx = 0.6m

A m = a 2 b 2 – a 1 b 1

ln a 2 b 2

a 1 b 1

= (0.13)(0.18) – (0.08)(0. 10)

ln (0.13)(0.18)

(0.08)(0. 10)

= 0.01435 m 2

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 q = kA m (T 1 - T 2 ) Δx = (1.49899)(0.01435)(1500-400) 0.6 q = 39.44 W

q = kA m (T 1 - T 2 )

Δx

= (1.49899)(0.01435)(1500-400)

0.6

q = 39.44 W

PROBLEM NO. 4.3-1 Insulation Needed for Food Cold Storage Room A food cold storage room is to be constructed of an inner layer of 19.1mm of pine wood, a middle layer of cork board, and an outer layer of 50.8mm of concrete. The inside wall surface temperature is -17.8 C and the outside surface temperature is 29.4 C at the outer concrete surface. The mean conductivities are for pine, 0.151;cork, 0.0433; and concrete, 0.762 W/mK. The total inside surface area of the room to use in the calculation is approximately 39 m2 (neglecting corner and end effects). What thickness of cork board is needed to keep the heat loss to 586 W?

Insulation Needed for Food Cold Storage Room

A food cold storage room is to be constructed of an inner layer of 19.1mm of pine wood, a middle layer of cork board, and an outer layer of 50.8mm of concrete. The inside wall surface temperature is -17.8 C and the outside surface temperature is 29.4 C at the outer concrete surface. The mean conductivities are for pine, 0.151;cork, 0.0433; and concrete, 0.762 W/mK. The total inside surface area of the room to use in the calculation is approximately 39 m2 (neglecting corner and end effects). What thickness of cork board is needed to keep the heat loss to 586 W?

PROBLEM Given A= 39 m 2 q= 586 W Δx 1 = 0.0191 m Δx 2 = ? Δx 3 = 0.0508 m k 1 = 0.151 W/mK k 2 = 0.0433 W/mK k 3 = 0.762 W/mK T 1 =-17.8°C T 2 =29.4°C 0.0191 m 0.0508 m Δx 2 29.4°C -17.8°C pinewood Cork board concrete

A= 39 m 2

q= 586 W

Δx 1 = 0.0191 m

Δx 2 = ?

Δx 3 = 0.0508 m

k 1 = 0.151 W/mK

k 2 = 0.0433 W/mK

k 3 = 0.762 W/mK

T 1 =-17.8°C

T 2 =29.4°C

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 Heat Loss with Trial-and-Error Solution The exhaust duct from the heater has an inside diameter of 114.3 mm with ceramic walls 6.4 mm thick. The average k = 1.52 W/m-K. Outside this wall, an insulation of rock wool 102 mm thick is installed. The thermal conductivity of the rock wool is k = 0.046 +1.56 x 10 -4 T ○ C (W/m-K). The inside surface temperature of the ceramic is T 1 = 588.7 K, and the outside surface temperature of the insulation is T 3 = 311 K. Calculate the heat loss for 1.5 m of duct and the interface temperature T 2 between the ceramic and the insulation. [Hint: The correct value of k m for the insulation is that evaluated at the mean temperature of say, 448 K. Then calculate the heat loss and T 2. Using this new T 2, calculate a new mean temperature and proceed as before.]

Heat Loss with Trial-and-Error Solution

The exhaust duct from the heater has an inside diameter of 114.3 mm with ceramic walls 6.4 mm thick. The average k = 1.52 W/m-K. Outside this wall, an insulation of rock wool 102 mm thick is installed. The thermal conductivity of the rock wool is k = 0.046 +1.56 x 10 -4 T ○ C (W/m-K). The inside surface temperature of the ceramic is T 1 = 588.7 K, and the outside surface temperature of the insulation is T 3 = 311 K. Calculate the heat loss for 1.5 m of duct and the interface temperature T 2 between the ceramic and the insulation. [Hint: The correct value of k m for the insulation is that evaluated at the mean temperature of say, 448 K. Then calculate the heat loss and T 2. Using this new T 2, calculate a new mean temperature and proceed as before.]

GIVEN D 1 = 0.1143 m thick 2 = 0.0064 m thick 3 = 0.102 m K 2 = 1.52 W/mK K 3 = 0.046 W/mK T 1 = 588.7 K T 2 = ? T 3 = 311 K L = 1.5

D 1 = 0.1143 m

thick 2 = 0.0064 m

thick 3 = 0.102 m

K 2 = 1.52 W/mK

K 3 = 0.046 W/mK

T 1 = 588.7 K

T 2 = ?

T 3 = 311 K

L = 1.5

SOLUTION R c = = 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)

R c = =

SOLUTION D i3 = 0.1271 m D o3 = 0.1271 + 2(0.102) D o3 = 0.3311 m R r = 1.3859 Δ x 102/1000 K r A r (0.0733)(1.5)( Π ) 0.3311-0.1271 ln (0.3311/0.1271) R r = =

D i3 = 0.1271 m

D o3 = 0.1271 + 2(0.102)

D o3 = 0.3311 m

R r = 1.3859

SOLUTION q = (588.7 – 311) (0.00741 +1.3859) q = 199.3096 W 199.31 W = (588.7 –T 2 ) 0.0074 T 2 = 587.2251 ° C ~ 588.7 ° C => q = 199.31 W q = - Δ T R T

q = (588.7 – 311)

(0.00741 +1.3859)

q = 199.3096 W

199.31 W = (588.7 –T 2 )

0.0074

T 2 = 587.2251 ° C ~ 588.7 ° C

=> q = 199.31 W

SOLUTION Using T 2 = 587.23 K T m = 449.12 K = 176.12 ° C K m = 0.0735 W/mK 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

Using T 2 = 587.23 K

T m = 449.12 K = 176.12 ° C

K m = 0.0735 W/mK

SOLUTION q = (588.7 – 311) (0.0074 + 1.3822) q = 199.8417 W 199.8417 = (588.7 – T 2 ) 0.0074 T 2 = 587.2212 K ANSWER: T 2 = 587.2212 K q = 199.8417 W

q = (588.7 – 311)

(0.0074 + 1.3822)

q = 199.8417 W

199.8417 = (588.7 – T 2 )

0.0074

T 2 = 587.2212 K

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