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
xcelluloid
Heat transfer coefficient, hs
From Appendix 5:
kcardboard
kcelluloid
=
=
=
3.2 x 10-3 m
0.1 x 10-3 m
11 Jm-2s-1 oC-1
=
=
0.07 Jm-1s-1 oC-1
0.21 Jm-1s-1 oC-1
1/U
1/U
=
=
=
1/hs + x1/k1 + x2/k2
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
3. 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
4. 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
-1
-1
q
= 5 x 10 Js
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
5. 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 Plate
Aluminium
Soup layer
Bulk soup
T1
T2
T3
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-2 oC-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.
6. Blanching peas in water
(a) r = 0.0024m
k = 0.48 Jm-1s-1 oC-1
= 0.48 Wm-1oC-1
-2 -1 o -1
h = 400 Jm s C
= 400 Wm-2oC-1
Bi
= hr/k = (400 x 0.0024)/0.48
=2
1/Bi
= 0.5
o
Temperature of water T0 = 85 C,
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
= F0 ρcr2/k
= (0.46 x 990x 3510x 0.00242)/0.48
= 19.2s
[kgm-3Jkg-1oC-1 m2/Jm-1s-1 oC-1 ]
2
(b)
F0 = kt/ρcr
where 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
= (T - T0 )/(Ti – T0 )
= T 0 + (Ti – T0 )0.45
= 85 + (18 - 85) 0.45
= 55 oC
7. Metals in pans for heating of food products
Assuming a steady state:
1/U
1/U
From Appendix 5
kstainless steel
kcopper
kmild steel
Ustainless steel
Ucopper
Umild steel
= 1/hS + x/k + 1/hF
= 1/10,000 + 1.6 x x10-3 /k + 1/700
= 21 Jm-1s-1oC-1
= 388 Jm-1s-1oC-1
= 45 Jm-1s-1 oC-1
= 623 Jm-2s-1oC-1 or Wm-2 oC-1
= 652 Jm-2s-1oC-1 or Wm-2 oC-1
= 638 Jm-2s-1oC-1 or Wm-2 oC-1
Mild steel is 2% worse than copper
Stainless steel is 4.5% worse than copper
8. Heating of cylinder of aluminium
For a 7.5cm diameter cylinder of aluminium:
From Appendix 5
kaluminum = 220 Jm-1s-1oC-1, caluminum = 0.87 kJkg-1oC-1, ρ = 2640kgm-3
r = 3.75x 10-2m
t = 85s
Ti = 5oC
T = 47.5.oC
T0 = 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(-hs At) / (cV)
Therefore
Now
And
(47.5 – 100)/(5 -100)
52.5/ 95 = 0.552
ln 0.552 = - 0.593
hs
=
exp((-hs x 4 x 85) / (870 x 220 x 0.075))
=
=
(0.593 x 870 x 220 x 0.075)/( 4 x 85)
25 Jm-2s-1 oC-1 or W m-2 oC-1
9. 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
Min s
0
0
10 600
20 1200
30 1800
40 2400
50 3000
60 3600
70 4200
80 4800
Fo
Slab
kt/ρcL2
0
0.037
0.074
0.111
0.148
0.186
0.223
0.260
0.297
Fo
Cylinder
kt/ρcr2
1
0.064
0.127
0.199
0.254
0.310
0.382
0.445
0.509
F(x)
Slab
From Fig.5.3
1
1
1
0.97
0.90
0.80
0.78
0.70
0.62
F(r) cylinder
Cylinder
From Fig. 5.3
1
0.95
0.80
0.55
0.40
0.27
0.19
0.14
0.07
F(x,r)
T
CanFo (T–To)+T0
F(x)xF(r)
1
20
0.95
25
0.80
40
0.53
67
0.36
85
0.22
99
0.15
106
0.10
111
0.04
116
By interpolation or plotting, the temperature at the centre of the can is at 115oC at 79 minutes.
10. Steam boiler insulation
Steam temperature from Appendix 8 at 150 kPa(abs.)
= 112oC
Air Temperature
= 18oC
0.25
Using Equation 5.17
hc
= 1.3(ΔT/L)
hc
= 1.3{( 112 – 18)/1.3}}0.25
= 3.8
≈ 4 Jm-2s-1 oC-1 or W m-2 oC-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
Check ΔT between outside of vessel and air:
q/A = 62.7
Tw
check
hc
= x/k( ΔT) = x/k( 112 - Tw )
= 112 – 62.7 x 0.04/0.05
= 61oC
= 1.3(ΔT/L)0.25
= 3 Jm-2s-1 oC-1 or W m-2 oC-1
which is close enough
11. Chilling of water by ammonia coils
V = 3m3 h-1 = 8.3 x 10-4 m3 s-1
v = 0.8ms-1
-2
Outer D = 2.13cm = 2.13 x 10 m Inner D = 1.71 cm = 1.71 x 10-2 m
ΔT = 8oC
Tw1 = 4oC
Tw2 = -11 oC
o
From Appendix 6, for water at 0 C
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-2 oC-1
≈ 3600
h value for ammonia side:
hh = 0.72[ (k3 ρ2g/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-2 oC-1
ksteel
1/U
U
A
= 45 Jm-1s-1 oC-1 or W m-1 oC-1
= 1/ hammonia + x/ksteel + 1/hwater
= 1/6000 + (0.0213 – 0.0171)/45 + 1/3600
= 0.000538
= 1860 Jm-2s-1 oC-1 or W m-2 oC-1
= 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