Information about Group 5-4ChEA

Published on January 5, 2009

Author: 4ChEAB08

Source: slideshare.net

Problem Set #2 5. The following data was obtained in a test on a flat-walled furnace, the linings of which consist of a 11.5 cm. non-corrosive brick of unknown conductivity and the outer wall of 20 cm clay brick, also of unknown conductivity. The temperature of the inner wall (flame side) was found to be 735°C and that of the outer wall 185°C. This furnace was lagged with 5 cm of magnesia (k=0.07 W/mK) thermocouples inserted at various points and the ff. data taken: Temp of inner wall (flame side)735°C; Temp at the junction of brick layers700°C Temp at the junction of ordinary brick and magnesia475°C Temp of the outer surface of magnesia88°C Calculate the percentage of heat loss that is saved by the lagging. Solution: Without Lagging 735 °C 185 °C Basis: A=1m2 q = Σ ∆T/Rt= 735-185 q=q1=q2 R1 + R2 k1 = 0.69 (W/mK) R1 = (11.5/100) = 0.167 (0.69)(1) Find k2: k(W/mK) T (°C) 1.00 200 1.47 600 1 2 11.5 20

Problem Set #2

5. The following data was obtained in a test on a flat-walled furnace, the linings of which consist of a 11.5 cm. non-corrosive brick of unknown conductivity and the outer wall of 20 cm clay brick, also of unknown conductivity. The temperature of the inner wall (flame side) was found to be 735°C and that of the outer wall 185°C. This furnace was lagged with 5 cm of magnesia (k=0.07 W/mK) thermocouples inserted at various points and the ff. data taken:

Temp of inner wall (flame side)735°C; Temp at the junction of brick layers700°C

Temp at the junction of ordinary brick and magnesia475°C Temp of the outer surface of magnesia88°C Calculate the percentage of heat loss that is saved by the lagging.

Solution: Without Lagging

735 °C 185 °C

Basis: A=1m2

q = Σ ∆T/Rt= 735-185 q=q1=q2

R1 + R2

k1 = 0.69 (W/mK)

R1 = (11.5/100) = 0.167

(0.69)(1)

Find k2:

k(W/mK) T (°C)

1.00 200

1.47 600

Iteration 1: Iteration 2: Let T = 215 °C Let T = 484.73 °C Tave= 200 °C Tave= 334.87 °C k2 = 1 (W/mK) k T R2 = (20/100) = 0.2 1.00 200 (1)(1) x 334.87 1.47 600 q = 735-185 __ = 735 – T 334.87-200 = x-1 0.167 + 0.2 0.167 600-200 1.47-1 T = 484.73 °C x = 1.16 R2 = 0.172 q = 735-185 __ = 735 – T 0.167 + 0.172 0.167 T = 464.06 °C Iteration 3: Let T = 464.06 °C Tave= 324.53 °C k T 1.00 200 x 324.53 1.47 600

Iteration 1: Iteration 2:

Let T = 215 °C Let T = 484.73 °C

Tave= 200 °C Tave= 334.87 °C

k2 = 1 (W/mK) k T

R2 = (20/100) = 0.2 1.00 200

(1)(1) x 334.87

1.47 600

q = 735-185 __ = 735 – T 334.87-200 = x-1

0.167 + 0.2 0.167 600-200 1.47-1

T = 484.73 °C x = 1.16

R2 = 0.172

q = 735-185 __ = 735 – T

0.167 + 0.172 0.167

T = 464.06 °C

Iteration 3:

Let T = 464.06 °C

Tave= 324.53 °C

k T

1.00 200

x 324.53

1.47 600

324.53-200 = x-1 600-200 1.47-1 x = 1.15 R2 = 0.174 q = 735-185__ = 735 – T 0.167 + 0.174 0.167 T = 465.65 °C q = 735 – 465.65 = 1612.87 W 0.167 With Lagging: q = q1 = q2 = q3 q = Σ∆T/Rt using q = q3 735- 88 = 475-88 R3 = ( 5/100) = 0.714 Rt R3 (0.07) (1) q = 475-88 = 542 W 0.714 % heat loss saved = (542/1612.87) x (100) = 33.61 % 1 2 3 700 0 C 475 0 C 735 0 C 88 0 C 11.5 20cm 5cm Magnesia k = 0.07 (W/mK)

324.53-200 = x-1

600-200 1.47-1

x = 1.15

R2 = 0.174

q = 735-185__ = 735 – T

0.167 + 0.174 0.167

T = 465.65 °C

q = 735 – 465.65 = 1612.87 W

0.167

With Lagging:

q = q1 = q2 = q3 q = Σ∆T/Rt

using q = q3

735- 88 = 475-88 R3 = ( 5/100) = 0.714

Rt R3 (0.07) (1)

q = 475-88 = 542 W

0.714

% heat loss saved = (542/1612.87) x (100)

= 33.61 %

.13 ( US) What length of time is required to heat a 2.5 cm diameter, 5 cm long copper cylinder ( k= 380 W/mk) from 16deg C to 93 deg C in a furnace whose temperature is 315 deg C? The value of h at the cylinder surface is 204 KJ/hm 2 K and its thermal diffusivity is 0.186 m 2 /s. Given: D= 2.5cm X 1 = .0125 m L=5cm ; k= 380 W/mk To= 16 deg C T 1 = 315 deg C T=93deg C h= 204 KJ/hm 2 K or 56.67 W/m 2 K; α = 0.186 m 2 /s Required t in sec. Solution: Solution: Y = T1- T = 315-93 = 0.74 m = k__ = (380) _________ = 536.43 T1 – To 315-16 Һ x 1 ( 56.67 ) (.0125) Using Heisler chart: X = the obtained slope does not fit any of the available given chart t = X x 1 2 = ( 2.8) (.0125) 2 α (0.186)

.13 ( US) What length of time is required to heat a 2.5 cm diameter, 5 cm long copper cylinder ( k= 380 W/mk) from 16deg C to 93 deg C in a furnace whose temperature is 315 deg C? The value of h at the cylinder surface is 204 KJ/hm 2 K and its thermal diffusivity is 0.186 m 2 /s.

Given:

D= 2.5cm X 1 = .0125 m

L=5cm ; k= 380 W/mk

To= 16 deg C T 1 = 315 deg C T=93deg C

h= 204 KJ/hm 2 K or 56.67 W/m 2 K; α = 0.186 m 2 /s

Required t in sec.

Solution:

Solution:

Y = T1- T = 315-93 = 0.74 m = k__ = (380) _________ = 536.43

T1 – To 315-16 Һ x 1 ( 56.67 ) (.0125)

Using Heisler chart:

X = the obtained slope does not fit any of the available given chart

t = X x 1 2 = ( 2.8) (.0125) 2

α (0.186)

Geankoplis 4.2-3. Removal of Heat from a Bath. Repeat Problem 4.2-2 but for a cooling coil made of 308 stainless steel having an average thermal conductivity of 15.23 W/mK. Problem 4.2-2. A cooling coil of 1.0 ft of 304 stainless-steel tubing having an inside diameter of 0.25 in and an outside diameter of 0.40 in is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40 °F and is 80 °F on the outside. Calculate the heat removal in btu/s and watts. Given: do = 0.40 in or 0.03 ft di = 0.25 in or 0.02 ft L = 1.0 ft km = 15.23 W/mK or 8.8 btu/h ft °F q = ∆T R = ∆X R kmAm Am = ((π)(1)(0.03-0.02))/(ln (0.03/0.02)) ∆ X = [(.4/12)- (.25/12)] / 2 = 6.25x10-3 q = (80 – 40) __ (6.25x10-3)_______ (8.8) ((π)(1)(0.03-0.02)) ln (0.03/0.02) 1 btu = 1054.368 J q = 4705.67 btu x 1h = 1.3071 btu/s or 1378.16 W h 3600s

Geankoplis

4.2-3. Removal of Heat from a Bath. Repeat Problem 4.2-2 but for a cooling coil made of 308 stainless steel having an average thermal conductivity of 15.23 W/mK.

Problem 4.2-2. A cooling coil of 1.0 ft of 304 stainless-steel tubing having an inside diameter of 0.25 in and an outside diameter of 0.40 in is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40 °F and is 80 °F on the outside. Calculate the heat removal in btu/s and watts.

Given:

do = 0.40 in or 0.03 ft

di = 0.25 in or 0.02 ft

L = 1.0 ft

km = 15.23 W/mK or 8.8 btu/h ft °F

q = ∆T R = ∆X

R kmAm

Am = ((π)(1)(0.03-0.02))/(ln (0.03/0.02))

∆ X = [(.4/12)- (.25/12)] / 2 = 6.25x10-3

q = (80 – 40) __

(6.25x10-3)_______

(8.8) ((π)(1)(0.03-0.02))

ln (0.03/0.02)

1 btu = 1054.368 J

q = 4705.67 btu x 1h = 1.3071 btu/s or 1378.16 W

h 3600s

5.3-5.Cooking a Slab of Meat . A slab of meat 25.44 mm thick originally at a uniform temperature of 10 °C is to be cooked from both sides until the center reaches 121 °C in an oven at 177 °C. The convection coefficient can be assumed constant at 25.6 W/m2K. Neglect any latent heat changes and calculate the time required. The thermal conductivity is 0.69 W/mK and thermal diffusivity 5.85 x 10-4m2/h. x1 = 25.4/2 = 12.7 mm Given: To = 10 °C T1 = 177 °C T = 121 °C x = 0 Һ = 25.6 W/m2K k = 0.69 W/mK α = 5.85 x 10 -4 m2/h Required: t in s Solution: Y = T1- T = 177-121 = 0.34 m = k__ = (0.69) _________ = 2.12 T1 – To 177-10 Һ x1 (25.6) (12.7/1000) Using Heisler chart: X = 2.8 t = X x 1 2 = ( 2.8) (12.7/1000) 2 α (5.85x10-4/3600) t = 2779.15 s

5.3-5.Cooking a Slab of Meat . A slab of meat 25.44 mm thick originally at a uniform temperature of 10 °C is to be cooked from both sides until the center reaches 121 °C in an oven at 177 °C. The convection coefficient can be assumed constant at 25.6 W/m2K. Neglect any latent heat changes and calculate the time required. The thermal conductivity is 0.69 W/mK and thermal diffusivity 5.85 x 10-4m2/h.

x1 = 25.4/2 = 12.7 mm

Given:

To = 10 °C T1 = 177 °C

T = 121 °C x = 0

Һ = 25.6 W/m2K

k = 0.69 W/mK

α = 5.85 x 10 -4 m2/h

Required: t in s

Solution:

Y = T1- T = 177-121 = 0.34 m = k__ = (0.69) _________ = 2.12

T1 – To 177-10 Һ x1 (25.6) (12.7/1000)

Using Heisler chart:

X = 2.8

t = X x 1 2 = ( 2.8) (12.7/1000) 2

α (5.85x10-4/3600)

t = 2779.15 s

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