Chapter 5 Unit Operations Problems

Chapter 5 Unit Operations Problems

1. Heat loss from polystyrene wall

q=(k/x) AΔT

k = 0.036 Jm-1s-1oC-1 from Appendix 5

A= 1m2

q= 8 Js-1

ΔT= 20 – (-18) = 38oC

Therefore x= (kA ΔT)/ q

= (0.036 x 1 x 38)/8 [Jm-1s-1oC-1 x m2 x oC/ Js-1]

= 0.171 m

= 17cm

2.Heat transfer in package

xcardboard=3.2 x 10-3 m xcelluloid = 0.1 x 10-3 m

Heat transfer coefficient, hs=11 Jm-2s-1 oC-1

From Appendix 5:

kcardboard= 0.07 Jm-1s-1 oC-1

kcelluloid=0.21 Jm-1s-1 oC-1

1/U =1/hs+ x1/k1 + x2/k2

1/U=1/11+ (3.2 x 10-3)/0.07 + (0.1 x 10 -3)/0.21

=0.1371 J-1m2soC

U =7.29 Jm-2s-1oC-1 or Wm-2oC-1

1. Oven rate of heat transfer

The rate of heat transfer is calculated from outside surface heat convection.

q/A=hsΔT

hs= 15 Jm-2s-1 oC-1

Maximum internal oven wall temperature=300oC

Maximum external oven wall temperature=50oC

Air temperature =25oC

q/A=15 (50 –25)

= 375 J s-1 m-2

This rate flow is equal to the overall rate through the wall.

Uoverall=(q/A)/ ΔToverall

=375/(300 –25)

= 1.364 J m-2s-1oC-1

kinsulation=0.18 Jm-1s-1oC-1

ksteel=45 Jm-1s-1oC-1

hs= 15 Jm-2s-1 oC-1

1/U = 1/15 + Thickness steel/45 + x/0.18 + Thickness steel/45 + 1/15

The effect of the steel can be ignored because it provides a very low barrier to the heat transfer.

1/1.364=1/15 + x/0.18 +1/15

0.733=0.133 + x/0.18

where x is the thickness of the insulation

x=0.6 x 0.18

=0.108 m

=10.8cm

1. Thermal conductivity of uncooked pastry

Area, A= 10cm2= 10x 10-4 m2= 1 x 10-3 m2

Thickness x=1.3cm= 1.3 x 10-2m

q=5 x 10-1 Js-1

ΔT= 17oC

k=qx/AΔT

=(5 x 10-1 x 1.3 x 10-2)/ (10-3 x 17)

=0.38 Jm-1s-1 oC-1

1. Thick soup boiled in a pan

Heat transfer from plate to pan h1= 600 Jm-2s-1oC-1

Heat transfer from soup layer to soup h1= 1400 Jm-2s-1oC-1

Thickness of aluminium x1= 2mm = 0.002m

Thickness of soup layer x2= 2mm = 0.002m

Temperature of hot plate= 500oC

Hot PlateAluminiumSoup layerBulk soup

T1T2T3 T4 T5

ΔT1 ΔT2 ΔT3 ΔT4

For the combined conduction to the liquid soup:

1/U=1/h1+ x1/k1 + x2/k2 + 1/h2

We need to estimate thermal conductivity of the stagnant thick soup layer; can take this as approximating to water i.e. k≈ 0.5 Jm-1s-1 oC-1 (from Appendix 7) and for aluminium k= 220 Jm-1s-1 oC-1 (Appendix 5)

so

1/U = 1/h1+ x1/k1 + x2/k2 + 1/h2

= 1/600 + 0.002/220 + 0.002/0.5 + 1/1400

= 1.67 x 10-3 + 0.0091 x 10-3 + 4 x10-3 +0.71 x10-3

= 6.389 x 10-3

U= 1.57 x 102 Jm-2s-1oC-1 or Wm-2oC-1

Now the temperature drops are proportional to the resistances so if ΔT1 drop across aluminium/plate interface, ΔT2 drop across the aluminium, ΔT3 drop across stagnant soup layer, ΔT4 is drop across the liquid film drop, and ΔT the total temperature drop = 500-100 = 400oC.

Therefore ΔT1/ΔT= 1.7/6.4 = 0.27, ΔT2/ΔT = 0.009/6.4  0, ΔT3/ΔT = 4/6.4 = 0.63

and ΔT4/ΔT= 0.71/6.4 = 0.11

Therefore ΔT1 =0.27 x 400 = 108, ΔT2 = 0, ΔT3= 0.63x 400 = 252, ΔT4= 0.11 x 400=44 oC

T1 = 500oC,T2 = 392oC, T3 = 392oC, T4 = 144oC, T5 = 100oC

So temperature of the surface of the soup layer attached to the pan = 392oC and the soup there will burn.

1. Blanching peas in water

(a) r = 0.0024mk = 0.48 Jm-1s-1 oC-1 = 0.48 Wm-1oC-1

h = 400 Jm-2s-1 oC-1= 400 Wm-2oC-1

Bi= hr/k= (400 x 0.0024)/0.48

= 2

1/Bi= 0.5

Temperature of water T0 = 85oC,

final pea temperature T = 70oC, initial pea temperature T = 18oC

(T – T0) /(Ti – T0)= (70-85)/(18-85)

= 0.224

From Figure 5.3 for sphere with unaccomplished temperature change of 0.224, and 1/Bi 0.5,

F0 = 0.46

F0= kt/ρcr2

t= F0ρcr2/k

= (0.46 x 990x 3510x 0.00242)/0.48

= 19.2s [kgm-3Jkg-1oC-1 m2/Jm-1s-1 oC-1 ]

(b) F0 = kt/ρcr2where r = 0.00315m

F0= (0.48 x 19.2)/(990 x 3510 x 0.003152)

= 0.267

Bi =hr/k= (400 x 0.00315)/0.48

= 2.625

1/Bi= 0.38

From Figure 5.3, unaccomplished temperature change ≈ 0.45

0.45= (T - T0)/(Ti – T0)

T= T 0+ (Ti – T0)0.45

= 85 + (18 - 85) 0.45

= 55 oC

1. Metals in pans for heating of food products

Assuming a steady state:

1/U= 1/hS+ x/k+ 1/hF

1/U= 1/10,000 + 1.6 x x10-3 /k + 1/700

From Appendix 5

kstainless steel = 21 Jm-1s-1oC-1

kcopper =388 Jm-1s-1oC-1

kmild steel = 45Jm-1s-1 oC-1

Ustainless steel = 623 Jm-2s-1oC-1 or Wm-2oC-1

Ucopper= 652 Jm-2s-1oC-1 or Wm-2oC-1

Umild steel= 638 Jm-2s-1oC-1 or Wm-2oC-1

Mild steel is 2% worse than copper

Stainless steel is 4.5% worse than copper

1. Heating of cylinder of aluminium

For a 7.5cm diameter cylinder of aluminium:

From Appendix 5 kaluminum=220Jm-1s-1oC-1, caluminum= 0.87 kJkg-1oC-1, ρ = 2640kgm-3

r = 3.75x 10-2mt = 85s

Ti = 5oC T = 47.5.oCT0= 100 oC

Assuming the likely heat transfer coefficient to be around 20 Jm-2s-1 oC-1

Bi=hD/2k

= (20 x 0.075)/(2x220)

=3.4 x 10-3

2

and therefore Equation 5.6 can be applied, and assuming the cylinder is long enough to neglect the ends:

A/V(DL) / (D 2L/4)

=4/D

(T - T0 )/( Ti - T0)=exp(-hsAt) / (cV)

Therefore (47.5 – 100)/(5 -100)= exp((-hsx 4 x 85) / (870 x 220 x 0.075))

Now 52.5/ 95 = 0.552

And ln 0.552 = - 0.593

hs= (0.593 x 870 x 220 x 0.075)/( 4 x 85)

=25 Jm-2s-1 oC-1 or W m-2oC-1

1. Pumpkin puree

Assume h is very high so that Bi →∞,1/Bi → 0

From Appendix 8, Steam temperature = 121oC

Can initial temperature is 20oC

Setting up a table

Fo FoF(x)F(r) cylinderF(x,r) T Slab Cylinder Slab Cylinder CanFo (T–To)+T0

Min s kt/ρcL2 kt/ρcr2From Fig.5.3From Fig. 5.3F(x)xF(r)

0 00111120

10 6000.0370.06410.950.9525

20 12000.0740.12710.800.8040

30 18000.1110.1990.970.550.5367

40 24000.1480.2540.900.400.3685

50 30000.1860.3100.800.270.22 99

60 36000.2230.3820.780.190.15 106

70 42000.2600.4450.700.140.10 111

80 4800 0.2970.5090.620.070.04 116

By interpolation or plotting, the temperature at the centre of the can is at 115oC at 79 minutes.

1. Steam boiler insulation

Steam temperature from Appendix 8 at 150 kPa(abs.)= 112oC

Air Temperature= 18oC

Using Equation 5.17hc= 1.3(ΔT/L) 0.25

hc= 1.3{( 112 – 18)/1.3}}0.25

= 3.8

≈ 4 Jm-2s-1 oC-1 or W m-2oC-1

Ignoring heat losses from top and bottom.

Rate of heat loss from non-insulated boiler is:

q/A= h ΔT

= 4(112-18)

= 376 Jm-2s-1 or Wm-2

If boiler is insulated:

1/U= x/k + 1/h

= 5 x 10-2 /0.04 + 1/4

= 1.5

U= 0.67

q/A for insulated tank= 0.67(112-18)

= 62.7 Jm-2s-1 or Wm-2

Therefore 83% savings in energy

CheckΔT between outside of vessel and air:

q/A = 62.7= x/k(ΔT) = x/k(112 - Tw )

Tw = 112 – 62.7 x 0.04/0.05

= 61oC

checkhc= 1.3(ΔT/L)0.25

= 3 Jm-2s-1 oC-1 or W m-2oC-1

which is close enough

1. Chilling of water by ammonia coils

V = 3m3 h-1 = 8.3 x 10-4m3 s-1 v = 0.8ms-1

Outer D = 2.13cm = 2.13 x 10-2m Inner D = 1.71 cm = 1.71 x 10-2m

ΔT = 8oC Tw1 = 4oC Tw2= -11 oC

From Appendix 6, for water at 0oC

µ = 1.87 x 10-3 Nsm-2

c = 4230 J kg-1 oC-1

ρ= 1000kgm-3

k = 0.57 J m-1 s-1 oC-1

q = VρcΔT

= 8.3 x 10-4 x 1000 x 4.23 x 103 x 15

= 5.287 x104 Js-1 [m3 s-1kg m-3 J kg-1 oC-1 oC]

= 52.87 kW

Refrigerant must be approx. – 4oC

Re = Dvρ/µ= (0.0213 x 0.8 x1000)/1.87 x 10-3

= 9112.3

≈ 9112

Pr = cpµ/k= 4230 x 1.87 x 10-3 /0.57

=13.9

Nu = 0.26(Re)0.6 (Pr)0.3

=136

Nu = hD/k

h =Nu k/D

= (136 x 0.57)/0.0213

= 3639 Jm-2s-1 oC-1 or W m-2oC-1

≈ 3600

h value for ammonia side:

hh = 0.72[ (k3ρ2g/µ ) x (λ/ρΔT)]0.25

But from Example 5.13, for condensing ammonia assume h = 6000 Jm-2s-1 oC-1 or W m-2oC-1

ksteel = 45 Jm-1s-1 oC-1 or W m-1oC-1

1/U= 1/ hammonia + x/ksteel + 1/hwater

= 1/6000 + (0.0213 – 0.0171)/45 + 1/3600

= 0.000538

U= 1860 Jm-2s-1 oC-1 or W m-2oC-1

A = q/UΔT

= 52.87 x 103/(1860 x 8)

= 3.55 m2

But A = DL

And so L =A/D

= 3.55/ (3.14 x 0.0213)

= 53.1m