Complex Engineering Problem (CEP) Descriptive Form. Simultaneous Heat and Mass Transfer. The concentric tube heat exchanger is replaced with a compact, plate-type heat exchanger that consists of a stack of thin metal sheets, separated by N gaps of width a. The oil and water flows are subdivided into N/2 individual flow streams, with the oil and water moving in opposite directions within alternating gaps. It is desirable for the stack to be of a cubical geometry, with a characteristic exterior dimension L. (a) parallel flow (b) counter flow, A counter flow, concentric tube heat exchanger is used to cool the lubricating oil for a large industrial gas turbine engine. The flow rate of cooling water through the inner tube (Di - 25 mm) is 0.2 kg/s,.

1. Chemical Engineering Department
Complex Engineering Problem (CEP) Descriptive Form
Course Name: Simultaneous Heat and Mass Transfer
Student : 17CH106
Abstract
The concentric tube heat exchanger is replaced with a compact, plate-type
heat exchanger that consists of a stack of thin metal sheets, separated by N
gaps of width a. The oil and water flows are subdivided into N/2 individual
flow streams, with the oil and water moving in opposite directions within
alternating gaps. It is desirable for the stack to be of a cubical geometry, with
a characteristic exterior dimension L. Determine the exterior dimensions of
the heat exchanger as a function of the number of gaps if the flow rates, inlet
temperatures, and desired oil outlet temperature are the same. Compare the
pressure drops of the water and oil streams within the plate-type heat
exchanger to the pressure drops of the flow streams in concentric tube heat
exchangers, if 60 gaps are specified.
Consider:
(a) parallel flow
(b) counter flow,
conditions for both concentric and plate type heat exchangers. Make an excel
program to find the impact of tube diameters and sheets gap width on the
characteristics of the heat exchangers considering the material for the tubes
and sheets first copper and then stainless steel by assuming an optimum
thickness for both tubes and plates.

2. Problem Statement:
A counter flow, concentric tube heat exchanger is used to cool the lubricating
oil for a large industrial gas turbine engine. The flow rate of cooling water
through the inner tube (Di - 25 mm) is 0.2 kg/s, while the flow rate of oil
through the outer annulus (Do = 45 mm) is 0.1 kg/s. The oil and water enter at
temperatures of 100 and 30°C, respectively, knowing that temperature of the
oil at the exit is to be 60°C and the oil stream convection heat transfer
coefficient, hoil 38,8 W/m2K?
Determine the heat transfer rate lost by the oil stream.
Solution:
Known:
Fluid flow rates and inlet temperatures for a counterflow, concentric tube heat
exchanger of prescribed inner and outer diameter.
Find:
Tube length to achieve a desired hot fluid outlet temperature.
Schematic:
Assumptions:
1. negligible heat loss to the surroundings.
2. Negligible kinetic and potential energy changes.
3. Constant properties.
4. Negligible tube wall thermal resistance and fouling factors.
5. Fully developed conditions for the water and oil (U independent x).

3. Properties:
For engine oil at (Th = 80°C = 353 K): cp = 2131J/kg.K, µ = 3.25 x 10-2 N
s/m2, k = 0.138 W/m.K.
For water at (Tc ≈ 35°C): cp = 4178J/kg.K, µ = 725x 10-6N s/m2, k = 0.625
W/m.K,
Pr = 4.85
Analysis: The required heat transfer rate may be obtained from the overall
energy balance for the hot fluid,
𝑞 = 𝑚ℎ 𝐶 𝑝,ℎ( 𝑇ℎ,𝑖 − 𝑇ℎ,𝑜)
𝑞 = 0.1 𝑘𝑔/𝑠 × 2131𝐿/𝑘𝑔. 𝐾(100 − 60)℃ = 8524 𝑊
The water outlet temperature is
𝑇𝑐,𝑜 =
𝑞
𝑚 𝑐 𝐶 𝑝,𝑐
+ 𝑇𝑐,𝑖
𝑇𝑐,𝑜 =
8524 𝑊
0.2 𝑘𝑔/𝑠 × 2131 𝑗/𝑘𝑔. 𝐾
+ 30℃ = 40.2℃
According, use of Tc = 35℃ to evaluate the water properties was a good
choice. The required heat exchanger length may now be obtained from
equation:
𝑞 = ∪ 𝐴 ∆𝑇𝑙𝑚
Where A = 𝜋𝐷𝑖 𝐿,
∆𝑇𝑙𝑚 =
( 𝑇ℎ,𝑖 − 𝑇𝑐,𝑜) − ( 𝑇ℎ,𝑜 − 𝑇𝑐,𝑖)
ln[( 𝑇ℎ,𝑖 − 𝑇𝑐,𝑜)( 𝑇ℎ,𝑜 − 𝑇𝑐,𝑖)]
=
59.8 − 30
ln(59.8/30)
= 43.2℃
The overall heat transfer coefficient is
𝑈 =
1
(1/ℎ𝑖) + (1/ℎ 𝑜)
For water flow through the tube,
𝑅𝑒 𝐷 =
4𝑚 𝑐
𝜋𝐷𝑖 𝜇
=
4 × 0.2
𝑘𝑔
𝑠
𝜋(0.025 𝑚)725 × 10−6 𝑛.
𝑠
𝑚2
= 14050

4. Accordingly, the flow is turbulent and the convection coefficient may be
computed:
𝑁𝑢 𝐷 = 0.023R𝑒 𝐷
4/5
𝑃𝑟0.4
𝑁𝑢 𝐷 = 0.023R(14050) 𝐷
4/5
(4.85)0.4
= 90
Hence
𝑁𝑢 𝐷 = 𝑁𝑢 𝐷
𝑘
𝐷𝑖
=
90 × 0.625 𝑊
𝑚⁄ . 𝐾
0.025 𝑚
= 2250
𝑊
𝑚2
. 𝐾
For the flow of oil through the annulus, the hydraulic diameter is,
𝐷ℎ = 𝐷0 − 𝐷𝑖 = 0.02m, and the Reynolds number is
𝑅𝑒 𝐷 =
𝑝𝑢 𝑚𝐷ℎ
𝜇
=
𝑝(𝐷0 − 𝐷𝑖)
𝜇
×
𝑚ℎ
𝑝𝜋(𝐷 𝑜
2 − 𝐷𝑖
2
)/4
𝑅𝑒 𝐷 =
4𝑚ℎ
𝜋(𝐷0 − 𝐷𝑖)𝜇
=
4 × 0.1 𝑘𝑔/𝑠
𝜋(0.045 + 0.025) 𝑚 × 3.25 × 10−2 𝑘𝑔
𝑠
. 𝑚
= 56.9
The annular flow is therefore laminar, assuming uniform temperature along
the inner surface of the annulus and a perfectly insulated outer surface, the
convection coefficient at the inner surface may be obtained from Table with
(𝐷𝑖/𝐷 𝑜) = 0.56, linear interpolation provides
𝑁𝑢𝑖 =
ℎ 𝑜 𝐷ℎ
𝑘
= 5.56
And
ℎ 𝑜 = 5.56
0.138 𝑊/𝑚. 𝐾
0.020𝑚
= 38.4𝑊/𝑚2
. 𝐾
The overall convection coefficient is then
𝑈 =
1
(1/2250 𝑊/𝑚2. 𝐾)
= 37.8𝑊/𝑚2
. 𝐾
And from the rate equation coefficient is then

5. 𝑈 =
1
(1/2250 𝑊/𝑚2. 𝐾) + (1/38.4 𝑊/𝑚2. 𝐾)
= 37.8𝑊/𝑚2
. 𝐾
Comments:
1. The hot side convention coefficient controls the rate of heat transfer
between the two fluids, and the low value of ho is responsible for the
large value of L. A spiral tube arrangement would be needed.
2. Because hi > ho the tube wall temperature will follow closely that of the
coolant water. Accordingly, the assumption of uniform wall temperature
used to obtain ho is reasonable.