Assignment No. 2

Group #4Rachelle Manalo

Ronna Anne OycoJonas Jane Aguila

Albert Jim GonzalesJoeward Manaig

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.

11 cm

1 m

1.2 m

0.75 m

DIAGRAM

q

q = ΔT

Rtotal

q =200 - 32

11/100

(1.4 • 0.75 • 1.2)

1855.6363 J/s x

= 1855.6363 J/s

3600 s1 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 hrx 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 hrx 24 hr

=213769309.1 J

qtotal = 2 (160326981.8) + 133605818.2 + 2(213769309.1)

qtotal = 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

SOLUTION

q = kAm (T1 - T2)

ΔxΔx = 0.6m

Am = a2 b2 – a1 b1

ln a2 b2

a1 b1

= (0.13)(0.18) – (0.08)(0. 10)

ln (0.13)(0.18)

(0.08)(0. 10)

= 0.01435 m2

SOLUTION

Get k:Using A = kdT Get Area under the curve using trapezoidal rule: A = 1798.7836 m2

1798.7836 m2 = k (1673 – 473)

k = 1.49899

SOLUTION

q = kAm (T1 - T2)

Δx

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

0.6

q = 39.44 W

PROBLEM NO. 4.3-1Insulation 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 m2 q= 586 W Δx1= 0.0191 m

Δx2= ?

Δx3= 0.0508 m

k1= 0.151 W/mK

k2= 0.0433 W/mK

k3= 0.762 W/mK

T1=-17.8°C

T2=29.4°C

0.0191 m 0.0508 mΔx2

29.4°C-17.8°C

pin

ewoo

d

Cor

k b

oard

con

cret

e

PROBLEM NO. 1 - Solution

R1 = Δx1 = 0.0191 = 0.003243 K/W k1 A 0.151(39)R2 = Δx2 = Δx2 k2 A 0.0433(39)R1 = Δx1 = 0.0508 = 0.001709 K/W k1 A 0.762(39)q = ΔT = 302.4 – 255.2 RT R1 + R2 + R3

586 W= 302.4 – 255.2 0.003243 + Δx2 + 0.001709 0.0433(39) Δx2 = 0.128 m

PROBLEM 4.3-5

Heat Loss with Trial-and-Error SolutionThe 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 T1 = 588.7 K, and the outside surface temperature of the insulation is T3 = 311 K. Calculate the heat loss for 1.5 m of duct and the interface temperature T2 between the ceramic and the insulation. [Hint: The correct value of km for the insulation is that evaluated at the mean temperature of say, 448 K. Then calculate the heat loss and T2. Using this new T2, calculate a new mean temperature and proceed as before.]

GIVEN

D1 = 0.1143 m thick2 = 0.0064 m thick3 = 0.102 m K2 = 1.52 W/mK K3 = 0.046 W/mK T1 = 588.7 K T2 = ? T3 = 311 K L = 1.5

SOLUTION

Rc = =

First, we assume :Tm = 448 K = 175 °CT2 = 585 KKm = 0.0733 W/mKDi2 = 0.1143 mDo2 = 0.1143 + 2(0.0064) = 0.1271 m

Rc = 7.41 x 10-3

Δx 6.4/1000

KmAm (1.52)(1.5)(Π) 0.1271-0.1143

ln (0.1271/0.1143)

SOLUTIONDi3 = 0.1271 mDo3 = 0.1271 + 2(0.102)

Do3 = 0.3311 m

Rr = 1.3859

Δx 102/1000

KrAr (0.0733)(1.5)(Π) 0.3311-0.1271

ln (0.3311/0.1271)

Rr = =

SOLUTION

q = (588.7 – 311)

(0.00741 +1.3859)

q = 199.3096 W

199.31 W = (588.7 –T2)

0.0074

T2 = 587.2251 °C ~ 588.7 °C

=> q = 199.31 W

q = -ΔT

RT

SOLUTION

Using T2 = 587.23 K

Tm = 449.12 K = 176.12°C

Km = 0.0735 W/mK

Δx 102/1000

KrAr (0.0735)(1.5)(Π) 0.3311-0.1271

ln (0.3311/0.1271)

R = =

R = 1.3822

SOLUTION

q = (588.7 – 311)

(0.0074 + 1.3822)

q = 199.8417 W

199.8417 = (588.7 – T2)

0.0074

T2 = 587.2212 K

ANSWER:

T2 = 587.2212 K q = 199.8417 W