Group # 34ChEA

Members:Balancio, Manuel Johnson

Brillantes, RodulfoSibug, Anna Kristina

Yatco, Shirley Mae

Problem 3:

Compute the heat loss per square meter of surface for a furnace wall 23 cm. thick. The inner and outer surface temperature are 315°C and 38°C respectively. The variation of the thermal conductivity in W/mK, with temperature in °C is given by the following relation: k= 0.006T-1.4x10-6 T2.

Solution:

23 cm

q

T1 = 315 oC

T2 = 38 oC

k= 0.006T-1.4x10-6 T2.

Required: q / m2

Solution:

315

Km (dT) =[(0.006T^2)/2 – (1.4x10-6T^3)/3]

38

= 278.7825

Q/A = Km (dT) / dX

= (278.7825 W/m) / (0.23m)

= 1212.097 W/m2

Problem 11:

An insulated steam pipe having an outside diameter of 0.0245 m is to be covered w/ two layers of insulation each having a thickness of 0.0245 m. The average thermal conductivity of one material is approximately four times that of the other. Assuming that the inner and outer surface temperature of the composite insulation are fixed, how much will the heat transfer be reduced when the better insulating material is next to the pipe then when it is the outer layer?

Solution:

Ro = 0.01225m

T1 = 0.0245 m

Where: Do = 0.0245 m

Thickness of insulation (T1) = T2 = 0.0245 m

T2 = 0.0245 m

Solution:

Aouter insulator = 2 pi L (rlm)

r2 = 0.06125m r1=0.03675m

Aoi = 2 pi L (0.048m)

Ainside insulator = 2 pi L (rlm)

r1 = 0.03675m r0=0.01225m

Aii = 2 pi L (0.0223m)

dX dx

Resistance = +

k1 (Aii) k2 (Aoi)

0.0245m 0.0245m

Resistance = +

k1(2 pi L)[0.0223m] k2(2 pi L)[0.048m]

If k1 = 4k and k2 = k [better insulation is outside]

then resistance = 0.7850 / (2 k pi L)

if k2 = 4k and k1 = k [better insulation is near pipe]

then resistance = 1.2263 / (2 k pi L)

dT dT 2 dT k pi L

Q = ----------------- = ----------------- = -------------------

resistance x / 2 k pi L x

where x = coefficient of the resistance

let 2 dt k pi L = A

Q = A/x

case 1: better insulator outside Q1 = 1.2739 A

case 2: better insulator near pipe Q2 = 0.8155 A

Q1 – Q2

% reduced Q = -------------- 100% Q1

%reduced Q = 35.98% less when the better insulator is near the pipe

4.3-2 Insulation of a Furnace

A wall of a furnace 0.244 m thick is constructed of material having a thermal conductivity 0f 1.30 W/m*K. The wall will be insulated on the outside with material having an average k of 0.346 W/m*K, so the heat loss from the furnace will be equal to or less than 1830 W/m2. The inner surface temperature is 1588 K and the outer 299 K. Calculate the thickness of insulation required.

Ans: 0.179 m

Solution:

0.244 m

K inner = 1.30 W/m*K

T1 = 1588 K

T2 = 299 K

K ave outer = 0.346 W/m*K

q <= 1830 W/m2

Required: thickness of insulation required

Given:

Solution:

1588K – 299K

Q/A = ----------------------- = 1830 W/m2 (1)

total resistance

dx1 dx2 0.244m dx2

Total resistance = ------+ -------- = ----------- + ------------ (2)

k1 k2 1.30W/mK 0.346W/mK

Substitute (2) in (1)

dx2 = 0.178m

5.3-3 Cooling a Slab of Aluminum

A large piece of aluminum that can be considered a semi-infinite solid initially has a uniform temperature of 505.4 K. The surface is suddenly exposed to an environment at 338.8 K with a surface convection coefficient of 455 W/m2*K. Calculate the time in hours for the temperature to reach 388.8 K at a depth of 25.4 mm. The average physical properties are ά = 0.340 m2/h and k = 208 W/m*K.

Solution:

q

T1 = 505.4 K

T2 = 338.8 K

k = 208 W/m*K

25.4 mm

ά = 0.340 m2/h

Given:

Required: time in h when T=388.8K

h = 455 W/m2*K

Solution:

X =άt/X12

Y= T2-T/ T2-T1

= [(338.8-388.8)/(338.8-505.4)] = 0.30M= k/hX1

= 208 / (455 x 0.0127) = 36

X= άt/X12

52 = 0.340 t / 0.0127 2

t=0.025 hrs.