4B group 4

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Published on January 8, 2009

Author: 4ChEAB08

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4B group 4

4ChE B Group 4 Almazar, Karina Ascalon, Kristine Malana, Rebecca Rozario, Vera Cresta Tancio, Romar

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. 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.

Solution Given: K = 1.4 W/m.K T1 =200 0 C T2 =38 0 C 38 0 C Req’d: q tot in 24 hrs. 1.2m 1 m 0.11 m 3 1 2 0.75m 200 0 C

Given:

K = 1.4 W/m.K

T1 =200 0 C T2 =38 0 C

38 0 C Req’d: q tot in 24 hrs.

q 1 = (200-38) 0 C = 1855.64 W 0.11m (1.4 W/m.K)(1.2) (0.75)m 2 q 1 =1855.64 J/s (3600s/1hr) (24hr) =160.33 MJ Solution 1.2m 1 0.11 m 0.75m q

q 1 = (200-38) 0 C = 1855.64 W

0.11m

(1.4 W/m.K)(1.2) (0.75)m 2

q 1 =1855.64 J/s (3600s/1hr) (24hr) =160.33 MJ

q 2 = (200-38) 0 C 0.11m (1.4 W/m.K) (1) (0.75)m 2 = 1546.36 W q 2 = 1546.36 J/s (3600s/1hr) (24hr) =133.61 MJ Solution q 1 m 0.75m 0.11m 2

q 2 = (200-38) 0 C

0.11m

(1.4 W/m.K) (1) (0.75)m 2

= 1546.36 W

q 2 = 1546.36 J/s (3600s/1hr) (24hr) =133.61 MJ

Solution 3 1 m 1.2m 0.11m q q 3 = 200-38 = 2474.18 W 0.11 (1)(1.2)(1.4) q 3 = 2474.18 J/s (3600s/1 hr) (24hr) =213.77 MJ q tot = (2 x q 1 ) + (2x q 2 ) + q 3 = (133.61)(2) + (160.33)(2) +(213.77) = 801.65 MJ

12. Given the following diagram: 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 Problem

12. Given the following diagram:

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

Plotting the given points: Solution

Plotting the given points:

Solution To solve for the area under the curve, use trapezoidal rule: But: , solving for km:

To solve for the area under the curve, use trapezoidal rule:

But: , solving for km:

Substituting the values and solving for q: Solution

Substituting the values and solving for q:

Problem 4.3-5. Heat Loss with Trial-and-Error Solution. The exhaust duct from a heater has an inside diameter of 114.3 mm with ceramic walls 6.4 mm thick. The average k = 1.52 W/mK. Outside this wall, an insulation of rock wool 102 mm thick is installed. The thermal conductivity of the rock wool is k = 0.046 + 1.56x10-4 T O C (W/mK). The inside surface temperature of the ceramic is T 1 = 588.7 K, and the outside surface temperature of the insulation T 3 = 311 K. Calculate the heat loss for 1.5 m duct and the interface temperature T 2 between the ceramic and the insulation. [Hint: the correct value of km for the insulation is the elevated at the mean temperature of ( T 2 + T 3 )/2. Hence, for the first trial assume a 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.]

4.3-5. Heat Loss with Trial-and-Error Solution. The exhaust duct from a heater has an inside diameter of 114.3 mm with ceramic walls 6.4 mm thick. The average k = 1.52 W/mK. Outside this wall, an insulation of rock wool 102 mm thick is installed. The thermal conductivity of the rock wool is k = 0.046 + 1.56x10-4 T O C (W/mK). The inside surface temperature of the ceramic is T 1 = 588.7 K, and the outside surface temperature of the insulation T 3 = 311 K. Calculate the heat loss for 1.5 m duct and the interface temperature T 2 between the ceramic and the insulation. [Hint: the correct value of km for the insulation is the elevated at the mean temperature of ( T 2 + T 3 )/2. Hence, for the first trial assume a 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.]

Solution 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 q = (588.7 – 311) / RT Assume : T m = 448 K = 175 O 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 2 = D i3 = 0.1271 m D o3 = 0.1271 + 2(0.102) = 0.3311 m

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 q = (588.7 – 311) / RT

Assume : T m = 448 K = 175 O 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 2 =

D i3 = 0.1271 m D o3 = 0.1271 + 2(0.102) = 0.3311 m



Solution R 3 = q = (588.7 – 311)/(0.0074 +1.3859) = 199.31 W 199.31 W = (588.7 –T 2 )/0.0074 T 2 = 587.23 K * 5% difference is valid Therefore: T 2 = 585 K , q = 199.31 W

R 3 =

q = (588.7 – 311)/(0.0074 +1.3859) = 199.31 W

199.31 W = (588.7 –T 2 )/0.0074

T 2 = 587.23 K * 5% difference is valid

Therefore:

T 2 = 585 K , q = 199.31 W

Solution Use: T2 = 587.23 K Tm = 449.12 K = 176.12 OC Km = 0.0735 W/mK R3 = q = (588.7 – 311)/(0.0074 + 1.3822) = 199.84 W 199.84 = (588.7 – T 2 )/0.0074 T 2 = 587.22 K Therefore: T 2 = 587.23 K q = 199.84 W

Use: T2 = 587.23 K

Tm = 449.12 K = 176.12 OC

Km = 0.0735 W/mK

R3 =

q = (588.7 – 311)/(0.0074 + 1.3822) = 199.84 W

199.84 = (588.7 – T 2 )/0.0074

T 2 = 587.22 K

Therefore:

T 2 = 587.23 K

q = 199.84 W

4.3-1. Insulation Needed for Food Cold Storage Room. A food cold storage room is to be constructed of an internal layer of 19.1 mm of pine wood, a middle layer of cork board, and an outer layer of 50.8 mm of concrete. The inside wall surface temperature is -17.8 O C and the outside surface temperature is 29.4 O 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 39m2 (neglecting corner and end effects). What thickness of cork board is needed to keep the heat loss to 586 W? Problem

4.3-1. Insulation Needed for Food Cold Storage Room. A food cold storage room is to be constructed of an internal layer of 19.1 mm of pine wood, a middle layer of cork board, and an outer layer of 50.8 mm of concrete. The inside wall surface temperature is -17.8 O C and the outside surface temperature is 29.4 O 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 39m2 (neglecting corner and end effects). What thickness of cork board is needed to keep the heat loss to 586 W?

Given: A = 39 m 2 q = 586W ΔX 1 = 0.0191 m K 1 = 0.151 W/m.k ΔX 2 = ? K 2 = 0.0433 W/m.k ΔX 3 = 0.0508 m K 3 = 0.762 W/m.k T 1 = -17.8 0 C T 3 = 29.4 0 C R 1 = 0.0191 = 0.003243K/W 0.151 (39) R 2 = ΔX 2 0.0433 (39) R 3 = 0.0508 = 0.001709 K/W 0.762(39) Solution

Given: A = 39 m 2 q = 586W

ΔX 1 = 0.0191 m K 1 = 0.151 W/m.k

ΔX 2 = ? K 2 = 0.0433 W/m.k

ΔX 3 = 0.0508 m K 3 = 0.762 W/m.k

T 1 = -17.8 0 C T 3 = 29.4 0 C

R 1 = 0.0191 = 0.003243K/W

0.151 (39)

R 2 = ΔX 2

0.0433 (39)

R 3 = 0.0508 = 0.001709 K/W

0.762(39)

q = ΔT = 302.4-255.2 R t R 1 + R 2 + R 3 586W = 302.4 – 255.2 0.003243 + ΔX 2 + 0.001709 0.0433(39) ΔX 2 = 0.1277m Solution

q = ΔT = 302.4-255.2

R t R 1 + R 2 + R 3

586W = 302.4 – 255.2

0.003243 + ΔX 2 + 0.001709

0.0433(39)

ΔX 2 = 0.1277m

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