UNIT UNIT OPERATIONS OPERATIONS Group 2 4 ChE BGroup 2 4 ChE B

Set 7Set 7

Bacal, NathanielBacal, Nathaniel

Cueto, AnaCueto, Ana

De Vera, BernadetteDe Vera, Bernadette

Mercado, RommelMercado, Rommel

Nieva, JaredNieva, Jared

# 7 # 7 A plane wall is composed of an 20 cm layer of A plane wall is composed of an 20 cm layer of refractory brick (k = 1.3W/mK) and a 5 cm layer of refractory brick (k = 1.3W/mK) and a 5 cm layer of insulating material with k for the insulating material insulating material with k for the insulating material varying linearly as k = 0.034 + 0.00018 t where t is the varying linearly as k = 0.034 + 0.00018 t where t is the temperature in °C. The inside surface temperature of temperature in °C. The inside surface temperature of the brick is 1100 °C and the outside surface the brick is 1100 °C and the outside surface temperature of the insulating material is 38 °C. temperature of the insulating material is 38 °C. Calculate the temperature at the boundary of the brick Calculate the temperature at the boundary of the brick and insulation.and insulation.

20 cm 5 cm

38 °C1100 °CT’ = ?

Refractory brick ; k = 1.3 W/m-k

Insulating brick ; k = 0.034 + 0.00018t

Basis: 1 m2 of cross-sectional area

Given: T1 = 1100 °C ; X = 20 / 100 m

T2T1

T2 = 38 °C ; X = 5 /100 m

T’ = ?

Required: T’ = ?

q = Σ ΔT = 1100 + 38

RT R1 + R2

Solution:

R = ΔX

KmAm

R1 = 20 / 100 = 0.1538 K / W

(1.3) x (1m)2

R2 = 5 / 100 = ?

(0.034 + 0.00018t) Eqn. 1

Since: q = q1 = q2

T’ - 38 = 1100 + 38

R2 R1 + R2

Eqn. 2

Assume : T’ = 600 °C ; T ave = (600 + 38) / 2 = 319 °C

T ave = t

Using Eqn. 1: R2 = 5 / 100 = 0.5469 K / W (0.034 + [0.00018][319])

Using Eqn. 2 : T’ = 926.21 °C

% difference = (600 – 926.21) / 600 x 100 = 54.37%

Assume : T’ = 926.21 °C ; T ave = (926.21 + 38) / 2 = 482.105 °C

Using Eqn. 1: R2 = 5 / 100 = 0.5469 K / W (0.034 + [0.00018][482.105])

Using Eqn. 2 : T’ = 867.75 °C

% difference = (926.21 – 867.75) / 926.21 x 100 = 6.31%

% difference is less than 10% so T’ ≈ 867.75 °C

15. (US) A large sheet of glass 50 cm 15. (US) A large sheet of glass 50 cm thick is initially at 150°C throughout. thick is initially at 150°C throughout. It is plunged into a stream of running It is plunged into a stream of running water having a temperature of 15°C. water having a temperature of 15°C. How long will it take to cool the glass How long will it take to cool the glass to an average temperature of 38°C? to an average temperature of 38°C? For glass: For glass: κκ = 0.70 W/mK; = 0.70 W/mK; ρρ= 2480 = 2480 kg/m3, Cp = 0.84 kJ/kgKkg/m3, Cp = 0.84 kJ/kgK

Given:Given:

TTo = o = 150°C150°C ; T ; T1 1 = = 15°C15°C ; ;

15 cm

T= 38°C when x=15cm

κ = 0.70 W/mK;

ρ= 2480 kg/m3;

Cp = 0.84 kJ/kgK

Required: time to cool the glass to an average temperature of 38°C?

SolutionSolution

( )( ) s

mx

KkgJxmkg

KmW

Cp

k 27

331036.3

/1084.0/2480

/70.0 −=⋅

⋅=⋅

=ρ

α 17.015015

3815 =−−=

−−

=oi

i

TT

TTY

( )2

27

2

100

50

)/1036.3(63.0

=

=

−

m

tsmx

a

tX

α

Using the Average Temperature Table (Fig. 5.3-13 on page 377)

At Y = 0.17, X = 0.63

T= 468750 s or 130.21 hr

Geankoplis 4.1-2

Determination of Thermal Conductivity. In determining the thermal conductivity of an insulating material, the temperatures were measured on both sides of a flat slab of 25 mm of the material and were 318.4 and 303.2 K. the heat flux was measured as 35.1 W/m2. Calculate the thermal conductivity in BTU/h-ft-ºF and in W/m-K

q = 35.1 W/m2

T1 = 318.4 K

T2 = 303.2 K

25 mm

T1 = 318.4 K

T2 = 303.2 K

25 mm

q = -k dTdx

In W/m-K:

35.1 W/m2 = - k (303.2–318.4) K 0.025 m

0.0577 W/m-K = k

q = 35.1 W/m2

In BTU/hr-ft-ºF:

0.0577 W 1 = 0.0333 BTU m-K 1.73 hr-ft- ºF

Geankoplis 5.3-7

Cooling a Steel Rod. A long steel rod 0.305 m in diameter is initially at a temperature of 588 K. It is immersed in an oil bath maintained at 311 K. The surface convective coefficient is 125 W/m2-K. Calculate the temperature at the center of the rod after 1 hour. The average physical properties of the steel are k=38 W/m-K and α=0.0381 m2-h

0.305 mX

X1 = 0.1525

0.305 mx

x1 = 0.1525 m Given:D = 0.305 mx1= D/2 = 0.1525 mx = 0To = 588 KT1 = 311 Kh = 125 W/m2 –Kt = 1 hourk = 38 W/m-Kα = 0.0381 m2-h

m = k = 38 = 1.99 ~ 2 h x1 (125)(0.1525)

n = x = 0 = 0 x1 0.1525

X = αt = (0.0381)(1) = 1.64 x1 0.1525 Using Fig. 5.3-7 of Geankoplis (Gurney - Lurie Chart for cylinders)

Y = 0.29 = T1 – T = 311 – T T1 - To 311 – 588

T = 391.33 K , where T is the temperature at the center of the cylinder.