PROBLEM 3: CALCULATION OF AREA OF THE HEAT EXHCHANGER
Hot exhaust gases are used in a finned-tube crossflow heat exchanger to heat 2.5 kg/s of water from 35 to
85 ◦ C. The gases [cp =1.09 kJ/kg · ◦ C] enter at 200 and leave at 93 ◦ C. The overall heat-transfer
coefficient is 180 W/m2 · ◦ C. Calculate the area of the heat exchanger using (a) the LMTD approach and
(b) the effectiveness-NTU method.
Schematic Diagram
Figure 1. Schematic Diagram of a Crossflow both fluids unmixed
Let:
𝑇1 = 𝐼𝑛𝑙𝑒𝑡 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝐻𝑜𝑡 𝐹𝑙𝑢𝑖𝑑
𝑇2 = 𝑂𝑢𝑡𝑙𝑒𝑡 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝐻𝑜𝑡 𝐹𝑙𝑢𝑖𝑑
𝑡1 = 𝐼𝑛𝑙𝑒𝑡 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝐶𝑜𝑙𝑑 𝐹𝑙𝑢𝑖𝑑
𝑡2 = 𝑂𝑢𝑡𝑙𝑒𝑡 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝐶𝑜𝑙𝑑 𝐹𝑙𝑢𝑖𝑑
𝐹 = 𝐶𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝐹𝑎𝑐𝑡𝑜𝑟
𝑈 = 𝑂𝑣𝑒𝑟𝑎𝑙𝑙 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡
𝐶𝑝𝑤 = 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐻𝑒𝑎𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝐶ℎ𝑖𝑙𝑙𝑒𝑑 𝑤𝑎𝑡𝑒𝑟
𝐶𝑝𝑔 = 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐻𝑒𝑎𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝐻𝑜𝑡 𝑔𝑎𝑠
𝑚𝑤 = 𝑀𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝐶ℎ𝑖𝑙𝑙𝑒𝑑 𝑊𝑎𝑡𝑒𝑟
Assumption:




The overall heat transfer coefficient remains the same
Steady State Steady Flow
Fluid and material properties are constant
No Losses
Given:
𝑇1 = 200 𝐶
𝑇2 = 93 𝐶
𝑡1 = 35 𝐶
𝑡2 = 85 𝐶
𝑈 = 180 𝑊/𝑚2 𝐾
𝑚𝑤 = 2.5 𝑘𝑔/𝑠
𝐶𝑝𝑤 = 4180 𝐽/𝑘𝑔 𝐾
Required:
Calculate the Area of Heat Exchanger using
a. LMTD Method
Calculate the log mean temperature difference
𝛥𝑇𝐿𝑀𝑇𝐷 =
𝛥𝑇1 − 𝛥𝑇2
𝛥𝑇1
ln(𝛥𝑇2)
when,
𝛥𝑇1 = 𝑇1 − 𝑡2
𝛥𝑇2 = 𝑇2 − 𝑡1
Substitute the given values
𝛥𝑇1 = 200 𝐶 − 85 𝐶 = 115 𝐶
𝛥𝑇2 = 93 𝐶 − 35 𝐶 = 58 𝐶
Solve for 𝛥𝑇𝐿𝑀𝑇𝐷
𝛥𝑇𝐿𝑀𝑇𝐷 =
115 𝐶 − 58 𝐶
= 83.27 𝐶
115 𝐶
ln(
)
58 𝐶
To identify the Correction Factor F, calculate the P and R value
𝑃=
𝑡2 − 𝑡1
85 𝐶 − 35 𝐶
=
= 0.30
𝑇1 − 𝑡1 200 𝐶 − 35 𝐶
𝑅=
𝑇1 − 𝑇2 200 𝐶 − 93 𝐶
=
= 2.14
𝑡2 − 𝑡1
85 𝐶 − 35 𝐶
Figure 2. Plot of Correction Factor for Heat Exchanger
Through estimation based on the graph, we able to get the correction factor, 𝐹 = 0.92
Since,
𝑄ℎ𝑜𝑡 𝑔𝑎𝑠 = 𝑄𝑐𝑜𝑙𝑑 𝑤𝑎𝑡𝑒𝑟
𝑄 = 𝑚𝑤𝐶𝑝𝑤(𝑡2 − 𝑡1)
𝑄 = 𝑈𝐴𝐹𝛥𝑇𝐿𝑀𝑇𝐷
Then,
𝑚𝑤𝐶𝑝𝑤(𝑡2 − 𝑡1) = 𝑈𝐴𝐹𝛥𝑇𝐿𝑀𝑇𝐷
Substitute the values to the equation
2.5
𝑘𝑔
𝐽
𝑊
× 4180
− 𝐾 × (85 𝐶 − 35 𝐶) = 180 2 − 𝐾 × 𝐴 × 0.9191 × 83.27𝐶
𝑠
𝑘𝑔
𝑚
Solve for Surface Area
𝑨 = 𝟑𝟕. 𝟗𝟑 𝒎𝟐
b. Effectiveness-NTU Method
Since, we need to solve the problem of a counterflow exhanger using Effectiveness-NTU Method, use
equation 10-15
𝑄ℎ𝑜𝑡 𝑔𝑎𝑠 = 𝑄𝑐𝑜𝑙𝑑 𝑤𝑎𝑡𝑒𝑟
𝐶𝑚𝑖𝑛 = 𝑚𝑔 × 𝐶𝑝𝑔 (𝑔𝑎𝑠)
𝐶𝑚𝑎𝑥 = 𝑚𝑤 × 𝐶𝑝𝑤 (𝑤𝑎𝑡𝑒𝑟)
𝐶𝑚𝑖𝑛
Determine the value of C, when 𝐶 = 𝐶𝑚𝑎𝑥
For 𝐶𝑚𝑖𝑛
𝑚𝑔 × 𝐶𝑝, 𝑔 × (𝑇1 − 𝑇2) = 𝑚𝑤 × 𝐶𝑝𝑤 × (𝑡2 − 𝑡1)
𝑚𝑔 × 𝐶𝑝𝑔 × (200 𝐶 − 93 𝐶) = 2.5 𝑘𝑔/𝑠 × 4180𝐽/𝑘𝑔 − 𝐾 × (85 𝐶 − 35 𝐶)
𝐶𝑚𝑖𝑛 = 𝑚𝑔 𝐶𝑝𝑔 = 4883.18 𝑊/ °𝐶
For 𝐶𝑚𝑎𝑥
𝐶𝑚𝑎𝑥 = 𝑚𝑤 × 𝐶𝑝𝑤 = 2.5𝑘𝑔/𝑠 × 4180𝐽/𝑘𝑔 − 𝐾
𝐶𝑚𝑎𝑥 = 𝑚𝑔 𝐶𝑝𝑔 = 104500 𝑊/ °𝐶
Therefore,
𝐶=
𝐶𝑚𝑖𝑛 4883.18 𝑊/ °𝐶
=
= 0.4673
𝐶𝑚𝑎𝑥 104500 𝑊/ °𝐶
From Equation 10-17, we can able to solve for 𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒𝑛𝑒𝑠𝑠, 𝜀
𝜀=
𝑇1 − 𝑇2 200 𝐶 − 93 𝐶
=
= 0.648
𝑇1 − 𝑡2 200 𝐶 − 35 𝐶
Using the Effectiveness-NTU Chart where C=0.4673 and 𝜀 = 0.93 𝑜𝑟 93%
Figure3.0 Plot for NTU
Through estimation based on the chart above, we able to get 𝑁𝑇𝑈 = 1.4 and finally, solve for the surface
area of heat exchanger when
𝑁𝑇𝑈 =
𝑈𝐴
𝐶𝑚𝑖𝑛
𝑊
𝑁𝑇𝑈 × 𝐶𝑚𝑖𝑛 1.4 × 4883.18 °𝐶
𝐴=
=
𝑊
𝑈
180 2 − 𝐾
𝑚
𝑨 = 𝟑𝟕. 𝟗𝟖 𝒎𝟐