Question

Water $(c_p=4180\mathrm{J}/\mathrm{kg}\cdot \mathrm{K})$ enters the $2.5$-cm-internaldiameter tube of a double-pipe counter-flow heat exchanger at ${17}^{\circ }\mathrm{C}$ at a rate of $1.8\mathrm{kg}/\mathrm{s}$. Water is heated by steam condensing at ${120}^{\circ }\mathrm{C}(h_{fg}=2203\mathrm{kJ}/\mathrm{kg})$ in the shell. If the overall heat transfer coefficient of the heat exchanger is $700\mathrm{W}/{\mathrm{m}}^2\cdot \mathrm{K}$, determine the length of the tube required in order to heat the water to ${80}^{\circ }\mathrm{C}$ using ( $a$ ) the LMTD method and $(b)$ the $\epsilon -\mathrm{NTU}$ method.

Short answer

The required length of the tube to effectively heat water from 17°C to 80°C using both the LMTD or ε-NTU methods is approximately 13.56 meters.

Step-by-step solution

1. Calculate the mass flow rate of water

Since the mass flow rate of water is given ($1.8\text{kg/s}$), no calculations are required for this step.

2. Calculate the heat transfer required to reach target temperature

We need to find out the heat transfer required to heat the water from ${17}^{\circ }\text{C}$ to ${80}^{\circ }\text{C}$. We can calculate this using the formula: $Q=m\cdot c_p\cdot \mathrm{\Delta }T$, where $Q$ is the heat transfer, $m$ is the mass flow rate, $c_p$ is the specific heat capacity of the water, and $\mathrm{\Delta }T$ is the temperature difference. $\mathrm{\Delta }T={80}^{\circ }\text{C}-{17}^{\circ }\text{C}={63}^{\circ }\text{C}$ $Q=1.8\text{kg/s}\cdot 4180\text{J/kg}\cdot \text{K}\cdot 63\text{K}=473076\text{W}$ The required heat transfer is $473076\text{W}$.

3. Find the temperature difference for LMTD method

To evaluate the LMTD, the hot and cold outlet temperatures must be estimated. The steam condenses at a constant temperature of ${120}^{\circ }\text{C}$. Assuming the steam temperature ($T_h$) remains constant throughout the process, the temperature difference at the inlet ($\mathrm{\Delta }{T}_1$) and outlet ($\mathrm{\Delta }{T}_2$) can be calculated as follows: $\mathrm{\Delta }{T}_1=T_h-T_{c1}={120}^{\circ }\text{C}-{17}^{\circ }\text{C}={103}^{\circ }\text{C}$ $\mathrm{\Delta }{T}_2=T_h-T_{c2}={120}^{\circ }\text{C}-{80}^{\circ }\text{C}={40}^{\circ }\text{C}$

4. Calculate LMTD and find the tube length using LMTD method

Now, we can find the Logarithmic Mean Temperature Difference (LMTD) using the formula: $\mathrm{LMTD}=\frac{\mathrm{\Delta }{T}_1-\mathrm{\Delta }{T}_2}{\ln \left(\frac{\mathrm{\Delta }{T}_1}{\mathrm{\Delta }{T}_2}\right)}$ $\mathrm{LMTD}=\frac{{103}^{\circ }\text{C}-{40}^{\circ }\text{C}}{\ln \left(\frac{{103}^{\circ }\text{C}}{{40}^{\circ }\text{C}}\right)}={63.76}^{\circ }\text{C}$ The heat transfer area $A$ can be calculated using the formula: $Q=U\cdot A\cdot \mathrm{LMTD}$, where $U$ is the overall heat transfer coefficient. Rearranging the formula to calculate $A$, we get: $A=\frac{Q}{U\cdot \mathrm{LMTD}}=\frac{473076\text{W}}{700{\text{W/m}}^2\cdot \text{K}\cdot 63.76\text{K}}=1.067{\text{m}}^2$ The internal diameter of the tube is given. To find the length of the tube, we use the formula for the surface area of the tube: $A=\pi dL$, where $d$ is the diameter, and $L$ is the length. We can solve for $L$: $L=\frac{A}{\pi d}=\frac{1.067{\text{m}}^2}{\pi \cdot 0.025\text{m}}\approx 13.56\text{m}$ So, the length of the tube required using the LMTD method is approximately $13.56\text{m}$.

5. Calculate the NTU using ε-NTU method

The number of transfer units ($\mathrm{NTU}$) can be calculated with the formula: $\mathrm{NTU}=\frac{UA}{m{c}_p}$, where $U$, $A$, $m$, and $c_p$ are the overall heat transfer coefficient, heat transfer area, mass flow rate, and specific heat capacity of water, respectively. Using the values from the LMTD section, we get: $\mathrm{NTU}=\frac{700{\text{W/m}}^2\cdot \text{K}\cdot 1.067{\text{m}}^2}{1.8\text{kg/s}\cdot 4180\text{J/kg}\cdot \text{K}}\approx 0.056$

6. Calculate the effectiveness and find the tube length using ε-NTU method

The effectiveness ($\epsilon$) can be calculated using the formula: $\epsilon =\frac{Q_{actual}}{Q_{max}}$, where $Q_{actual}$ and $Q_{max}$ are the actual and maximum possible heat transfers, respectively. Since the outlet coolant temperature is given, we can calculate the actual heat transfer: $Q_{actual}=m{c}_p(T_{c2}-T_{c1})=1.8\text{kg/s}\cdot 4180\text{J/kg}\cdot \text{K}\cdot ({80}^{\circ }\text{C}-{17}^{\circ }\text{C})=473076\text{W}$ The maximum possible heat transfer occurs when the coolant reaches the temperature of steam: $Q_{max}=m{c}_p(T_h-T_{c1})=1.8\text{kg/s}\cdot 4180\text{J/kg}\cdot \text{K}\cdot ({120}^{\circ }\text{C}-{17}^{\circ }\text{C})=627096\text{W}$ $\epsilon =\frac{473076\text{W}}{627096\text{W}}\approx 0.754$ We can use the effectiveness-NTU relation to check the NTU value: $\epsilon =1-\exp (-\frac{1}{c}\cdot (1-\exp (-c\mathrm{NTU})))$, where $c=\frac{m{c}_p}{U\cdot A}$ Using the given values, $c=\frac{1.8\text{kg/s}\cdot 4180\text{J/kg}\cdot \text{K}}{700{\text{W/m}}^2\cdot \text{K}\cdot 1.067{\text{m}}^2}\approx 17.77$ Solving for NTU, we obtain $\mathrm{NTU}\approx 0.056$, which agrees with the value we calculated earlier. Now, we can calculate the heat transfer area using the NTU and ε values: $A=\frac{m{c}_p}{U\cdot \mathrm{NTU}}=\frac{1.8\text{kg/s}\cdot 4180\text{J/kg}\cdot \text{K}}{700{\text{W/m}}^2\cdot \text{K}\cdot 0.056}\approx 1.067{\text{m}}^2$ Finally, the tube length can be found using the same formula as in the LMTD method: $L=\frac{A}{\pi d}=\frac{1.067{\text{m}}^2}{\pi \cdot 0.025\text{m}}\approx 13.56\text{m}$ So, the length of the tube required using the ε-NTU method is approximately $13.56\text{m}$.

Key concepts

Logarithmic Mean Temperature Difference (LMTD)

When designing a heat exchanger, understanding the temperature gradients across the heat exchanger is crucial. This is where the Logarithmic Mean Temperature Difference (LMTD) comes in handy. It provides an average temperature difference which simplifies the calculation of heat transfer. In any counterflow heat exchanger, the temperature of the fluid on the hot side and the cold side changes along the exchanger's length. Hence, the temperature difference is not constant.
To calculate LMTD, we consider two main temperature differences:

• Inlet temperature difference: This is the difference between the hot temperature entering the exchanger and the cold fluid's entering temperature.
• Outlet temperature difference: This is the difference between the hot temperature at the exit and the cold fluid's temperature at the exit.

The LMTD formula is given by:
$$\mathrm{LMTD}=\frac{\mathrm{\Delta }{T}_1-\mathrm{\Delta }{T}_2}{\ln \left(\frac{\mathrm{\Delta }{T}_1}{\mathrm{\Delta }{T}_2}\right)}$$
By using LMTD, it becomes easier to derive the heat exchanger area needed, provided we also know the overall heat transfer coefficient ($U$). This approach is effective, especially when dealing with balanced counterflow heat exchangers, where it's necessary to account for temperature differences along the flow path.

Effectiveness-NTU Method

The Effectiveness-NTU method is another approach to design and analyze heat exchangers. Instead of focusing on temperature differences like LMTD, this method emphasizes the thermal effectiveness of the heat exchanger. The term 'Effectiveness' denotes the ratio of the actual heat transfer to the maximum possible heat transfer.
Effectiveness is formulated as:
$$\epsilon =\frac{Q_{\text{actual}}}{Q_{\text{max}}}$$
The NTU (Number of Transfer Units) assists in determining effectiveness by providing a dimensionless measure of the heat exchanger size relative to its heat transfer capacity.
The NTU is defined using the equation:
$$\mathrm{NTU}=\frac{UA}{m{c}_p}$$
where:

• $U$ is the overall heat transfer coefficient,
• $A$ is the heat transfer area,
• $m$ is the mass flow rate,
• $c_p$ is the specific heat capacity.

For a given heat exchanger configuration (like counterflow), charts or equations relate effectiveness ($\epsilon$) and NTU, facilitating the determination of either the required heat exchanger size or the performance of an existing one. The Effectiveness-NTU method is preferred where exit temperatures are not known or in situations where it provides greater accuracy or insight into the design process.

Heat Transfer Calculations

In heat exchanger systems, accurate heat transfer calculations are critical. The amount of heat transferred ($Q$) from the hot fluid to the cold fluid depends on various factors including fluid properties, flow arrangements, and temperature differences. The basic formula used in these calculations is:
$$Q=m\cdot c_p\cdot \mathrm{\Delta }T$$
where:

• $Q$ is the heat transferred,
• $m$ is the mass flow rate,
• $c_p$ is the specific heat capacity of the fluid,
• $\mathrm{\Delta }T$ is the temperature change of the fluid.

This equation helps to determine how much energy is needed to achieve a desired temperature change, serving as the foundation for further design parameters such as the size of the heat exchanger or the type of materials needed.
For advanced calculations, knowledge of fluid dynamics and heat transfer coefficients is essential. Also, recognizing whether the setup is a parallel flow, counterflow, or crossflow specifies the methods for correct calculations. These factors determine not only the functionality but also the efficiency and cost-effectiveness of the heat exchanger design for an intended application.