group 2 problem set 7
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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.