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Chapter 11: Problem 27

In a dairy operation, milk at a flow rate of \(250 \mathrm{~L} / \mathrm{h}\) and a cow-body temperature of \(38.6^{\circ} \mathrm{C}\) must be chilled to a safe-to-store temperature of \(13^{\circ} \mathrm{C}\) or less. Ground water at \(10^{\circ} \mathrm{C}\) is available at a flow rate of \(0.72 \mathrm{~m}^{3} / \mathrm{h}\). The density and specific heat of milk are \(1030 \mathrm{~kg} / \mathrm{m}^{3}\) and \(3860 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. (a) Determine the UA product of a counterflow heat exchanger required for the chilling process. Determine the length of the exchanger if the inner pipe has a 50 -mm diameter and the overall heat transfer coefficient is \(U=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) Determine the outlet temperature of the water. (c) Using the value of \(U A\) found in part (a), determine the milk outlet temperature if the water flow rate is doubled. What is the outlet temperature if the flow rate is halved?

Short Answer

Expert verified
In a counterflow heat exchanger for chilling milk, with given conditions, we determined that: 1. The UA product required for the chilling process is 110037.3 W/K, and the length of the exchanger with a 50-mm inner pipe diameter is 7.013 meters. 2. The outlet temperature of the ground water is 26.5掳C. 3. If the water flow rate is doubled, the milk outlet temperature is 13.97掳C, and if the water flow rate is halved, the milk outlet temperature is 16.96掳C.

Step by step solution

01

Part A - Determine the UA product and Length of the Exchanger

First, we need to calculate the mass flow rate and heat capacity rate of milk using its properties and flow rate: 1. Convert the milk flow rate from L/h to m鲁/h: \( 250 L/h = \frac{250}{1000} m^3/h = 0.25 m^3/h \) 2. Calculate the mass flow rate of milk: \( \dot{m}_{milk} = \rho_{milk} \times \dot{V}_{milk} = 1030 kg/m^3 \times 0.25 m^3/h = 257.5 kg/h \) 3. Calculate the heat capacity rate of milk: \( C_{milk} = \dot{m}_{milk} \times c_{p,milk} = 257.5 kg/h \times 3860 J/kgK = 994025 J/hK \) Now let's calculate the mass flow rate and heat capacity rate of the ground water: 1. Calculate the mass flow rate of water: \( \dot{m}_{water} = \rho_{water} \times \dot{V}_{water} = 1000 kg/m^3 \times 0.72 m^3/h = 720 kg/h (assume water density as 1000 kg/m^3) \) 2. Calculate the heat capacity rate of water: \( C_{water} = \dot{m}_{water} \times c_{p,water} = 720 kg/h \times 4180 J/kgK = 3011040 J/hK (assume specific heat capacity of water as 4180 J/kgK) \) Using the heat transfer equation: \( q = U A \Delta T_{lm} \) We have all variables except UA and \(\Delta T_{lm}\). So, we can use the heat capacity rate to find the required heat transfer, then use the LMTD method to determine the temperature difference, and finally calculate UA. Calculate the required heat transfer: \( q = \min(C_{milk},C_{water})\times(\Delta T_1-\Delta T_2) \). Since \(C_{milk}<C_{water}\), the heat transfer is determined by the capacity of milk. Let \(\Delta T_1 = 38.6^{\circ}C - 10^{\circ}C = 28.6^{\circ}C\) and \(\Delta T_2 = 38.6^{\circ} C - 13^{\circ} C = 25.6^{\circ}C\). \(q= 994025 J/hK \times (28.6 K - 25.6 K)=2982075 J/h \) . Determine the LMTD: \(\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{ln(\frac{\Delta T_1}{\Delta T_2})} = \frac{28.6 - 25.6}{ln(\frac{28.6}{25.6})}= 27.1^{\circ}C\). Calculate UA: \(UA = \frac{q}{\Delta T_{lm}} = \frac{2982075 J/h} {27.1 K} = 110037.3 W/K\). Finally, let's find the length of the exchanger: 1. Calculate the inner pipe area per unit length: \( A = \pi D = \pi \times 0.05 m = 0.1571 m^2/m \) 2. Calculate the exchanger length: \(L = \frac{UA}{U \times A} = \frac{110037.3 W/K}{1000 W/m^2K \times 0.1571 m^2/m} = 7.013 m\) So, the UA product of the counterflow heat exchanger required for the chilling process is 110037.3 W/K, and the length of the exchanger if the inner pipe has a 50-mm diameter is 7.013 meters.
02

Part B - Determine the Outlet Temperature of the Water

To find the outlet temperature of the water, we can use the following energy balance equation for water: \( \dot{m}_{water}c_{p,water}(T_{out,water} - T_{in,water}) = \dot{m}_{milk}c_{p,milk}(T_{in,milk} - T_{out,milk})\), where \(T_{in,water} = 10^{\circ}C\), \(T_{in,milk} = 38.6^{\circ}C\), and \(T_{out,milk} = 13^{\circ}C\). Now solve for \(T_{out,water}\): \(T_{out,water} = T_{in,water} + \frac{\dot{m}_{milk}c_{p,milk}(T_{in,milk} - T_{out,milk})}{\dot{m}_{water}c_{p,water}} \) \(T_{out,water} = 10 + \frac{257.5 \times 3860 (38.6 - 13)}{720 \times 4180} = 10 + 16.5 = 26.5^{\circ}C \) The outlet temperature of the water is 26.5掳C.
03

Part C - Determine Milk Outlet Temperature for Different Water Flow Rates

To find the milk outlet temperature for different water flow rates, we can use the effectiveness-NTU method of heat exchanger analysis. The effectiveness (\(\epsilon\)) is defined as the actual heat transfer divided by the maximum possible heat transfer. First, we need to calculate the effectiveness of the original system when water flow rate is not changed: \(\epsilon = \frac{q}{\min(C_{milk}, C_{water})(T_{h, in} - T_{c, in})} = \frac{2982075}{994025 \times 28.6} = 0.979\) Now, let's examine the two scenarios: 1. If the water flow rate is doubled: \(\dot{m'}_{water} = 2 \times \dot{m}_{water} = 1440 kg/h\) Calculate the new heat capacity rate of water: \( C'_{water} = \dot{m'}_{water} \times c_{p, water} = 1440 kg/h \times 4180 J/kgK = 6022080 J/hK \) Since \(C_{milk}C''_{water}\), the heat transfer is determined by the capacity of water. Using the same effectiveness, we can calculate the new outlet temperature of milk: \(T''_{out,milk} = T_{in,milk} - \frac{\epsilon \times C''_{water}}{C_{milk}}(T_{in,milk}-T_{in,water}) = 38.6 - \frac{0.979 \times 1505520}{994025}(38.6 - 10) = 16.96^{\circ}C\) In conclusion, when the water flow rate is doubled, the milk outlet temperature is 13.97掳C, and when the water flow rate is halved, the milk outlet temperature is 16.96掳C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Counterflow Heat Exchanger
Understanding a counterflow heat exchanger is critical when we talk about heat transfer systems. It's a type of heat exchanger where fluids move parallel to each other but in opposite directions. In our dairy operation example, the milk (hot fluid) and the groundwater (cold fluid) are flowing in opposing directions.

This configuration is highly efficient because it maintains a greater temperature difference between the fluids over the length of the heat exchanger, which drives the heat transfer. It's like a race where the hot and cold streams are running towards each other, and as they pass by, heat is exchanged. In this setup, the milk cools down significantly because it continually encounters colder water, as opposed to a co-current or parallel flow system where the temperatures of the two streams equalize quicker.

For the dairy operation, we maximize thermal efficiency by using this counterflow method. It allows us to reach the desired chilling temperature for the milk with a relatively short length of the exchanger, which we calculated to be about 7 meters. This is a fantastic application of physics and engineering to meet the needs of safe milk storage.
Log Mean Temperature Difference (LMTD)
Log Mean Temperature Difference, or LMTD, is a superb way to average the temperature difference between the hot and cold streams within a heat exchanger. It's not just an ordinary average; it's a logarithmic one, which smooths out the immediate ups and downs over the temperature gradient. To compute the LMTD, we focus on the difference in temperatures between the two fluids at each end of the heat exchanger.

The formula for LMTD is:
\[ \Delta T_{\text{lm}} = \frac{\Delta T_1 - \Delta T_2}{\ln(\frac{\Delta T_1}{\Delta T_2})} \]
Where \( \Delta T_1 \) and \( \Delta T_2 \) are the temperature differences at each end. In the case of the dairy operation, we used these values to solve for the LMTD and obtained 27.1掳C. It's the LMTD that then allows us to determine how effectively our exchanger is doing its job by combining it with the overall heat transfer coefficient, U, to solve for the UA product. This step is essential because it ties the actual physical length required for the exchanger to the thermal performance needed for safe milk storage.
Effectiveness-NTU Method
When it comes to fine-tuning the performance of our heat exchanger, we employ the Effectiveness-NTU method. NTU stands for Number of Transfer Units, which reflects the heat exchanger鈥檚 size relative to the capacity rate of the fluids. Effectiveness, on the other hand, is the measure of how well the heat exchanger performs relative to its maximum potential. In essence, it tells us how close we are to the ideal scenario.

The formula that articulates this relationship is:
\[ \epsilon = \frac{q}{\min(C_{\text{milk}}, C_{\text{water}}) (T_{\text{h,in}} - T_{\text{c,in}})} \]
Where \( \epsilon \) is the effectiveness, \( q \) is the actual heat transfer, \( C \) is the heat capacity rate of each fluid, and \( T_{\text{h,in}} \) and \( T_{\text{c,in}} \) are the inlet temperatures of the hot and cold fluids.

Using this method, we can predict the impact of changing water flow rates on the cooling process, as was done for the dairy operation exercise. By understanding the effectiveness and how changing flow rates adjust heat capacity, it's like having a thermostat at our disposal. We controlled the milk's finished temperature by the flow rate and confirmed the effectiveness of our heat exchanger, ensuring that the milk reached safe storage temperature under different conditions.

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Most popular questions from this chapter

Saturated process steam at 1 atm is condensed in a shell-and-tube heat exchanger (one shell, two tube passes). Cooling water enters the tubes at \(15^{\circ} \mathrm{C}\) with an average velocity of \(3.5 \mathrm{~m} / \mathrm{s}\). The tubes are thin walled and made of copper with a diameter of \(14 \mathrm{~mm}\) and length of \(0.5 \mathrm{~m}\). The convective heat transfer coefficient for condensation on the outer surface of the tubes is \(21,800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Find the number of tubes/pass required to condense \(2.3 \mathrm{~kg} / \mathrm{s}\) of steam. (b) Find the outlet water temperature. (c) Find the maximum possible condensation rate that could be achieved with this heat exchanger using the same water flow rate and inlet temperature. (d) Using the heat transfer surface area found in part (a), plot the water outlet temperature and steam condensation rate for water mean velocities in the range from 1 to \(5 \mathrm{~m} / \mathrm{s}\). Assume that the shell-side convection coefficient remains unchanged.

A recuperator is a heat exchanger that heats the air used in a combustion process by extracting energy from the products of combustion (the flue gas). Consider using a single-pass, cross-flow heat exchanger as a recuperator. Eighty \((80)\) silicon carbide ceramic tubes \((k=20\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) of inner and outer diameters equal to 55 and \(80 \mathrm{~mm}\), respectively, and of length \(L=1.4 \mathrm{~m}\) are arranged as an aligned tube bank of longitudinal and transverse pitches \(S_{L}=100 \mathrm{~mm}\) and \(S_{T}=120 \mathrm{~mm}\), respectively. Cold air is in cross flow over the tube bank with upstream conditions of \(V=1 \mathrm{~m} / \mathrm{s}\) and \(T_{c i}=300 \mathrm{~K}\), while hot flue gases of inlet temperature \(T_{\mathrm{h}, \mathrm{I}}=1400 \mathrm{~K}\) pass through the tubes. The tube outer surface is clean, while the inner surface is characterized by a fouling factor of \(R_{f}^{N \prime}=2 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The air and flue gas flow rates are \(\dot{m}_{c}=1.0 \mathrm{~kg} / \mathrm{s}\) and \(m_{\mathrm{h}}=1.05 \mathrm{~kg} / \mathrm{s}\), respectively. As first approximations, (1) evaluate all required air properties at \(1 \mathrm{~atm}\) and \(300 \mathrm{~K},(2)\) assume the flue gas to have the properties of air at \(1 \mathrm{~atm}\) and \(1400 \mathrm{~K}\), and (3) assume the tube wall temperature to be at \(800 \mathrm{~K}\) for the purpose of treating the effect of variable properties on convection heat transfer. (a) If there is a \(1 \%\) fuel savings associated with each \(10^{\circ} \mathrm{C}\) increase in the temperature of the combustion air \(\left(T_{c o}\right)\) above \(300 \mathrm{~K}\), what is the percentage fuel savings for the prescribed conditions? (b) The performance of the recuperator is strongly influenced by the product of the overall heat transfer coefficient and the total surface area, UA. Compute and plot \(T_{c, \infty}\) and the percentage fuel savings as a function of UA for \(300 \leq U A \leq 600 \mathrm{~W} / \mathrm{K}\). Without changing the flow rates, what measures may be taken to increase \(U A^{*}\) ?

A counterflow, concentric tube heat exchanger is designed to heat water from 20 to \(80^{\circ} \mathrm{C}\) using hot oil, which is supplied to the annulus at \(160^{\circ} \mathrm{C}\) and discharged at \(140^{\circ} \mathrm{C}\). The thin-walled inner tube has a diameter of \(D_{i}=20 \mathrm{~mm}\), and the overall heat transfer coefficient is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The design condition calls for a total heat transfer rate of \(3000 \mathrm{~W}\). (a) What is the length of the heat exchanger? (b) After 3 years of operation, performance is degraded by fouling on the water side of the exchanger, and the water outlet temperature is only \(65^{\circ} \mathrm{C}\) for the same fluid flow rates and inlet temperatures. What are the corresponding values of the heat transfer rate, the outlet temperature of the oil, the overall heat transfer coefficient, and the water- side fouling factor, \(R_{f,}^{n}\) ?

Consider a concentric tube heat exchanger characterized by a uniform overall heat transfer coefficient and operating under the following conditions: \begin{tabular}{lccrc} \hline & \(\dot{m}\) \((\mathbf{k g} / \mathbf{s})\) & \(c_{p}\) \((\mathbf{J} / \mathbf{k g} \cdot \mathbf{K})\) & \(T_{i}\) \((\boldsymbol{C})\) & \(T_{o}\) \((\mathbf{C})\) \\ \hline Cold fluid & \(0.125\) & 4200 & 40 & 95 \\ Hot fluid & \(0.125\) & 2100 & 210 & \(-\) \\ \hline \end{tabular} What is the maximum possible heat transfer rate? What is the heat exchanger effectiveness? Should the heat exchanger be operated in parallel flow or in counterflow? What is the ratio of the required areas for these two flow conditions?

Consider Problem 11.36. (a) For \(\dot{m}_{c \mathrm{CA}}=\dot{m}_{\mathrm{h}, \mathrm{B}}=10 \mathrm{~kg} / \mathrm{s}\), determine the outlet air and ammonia temperatures, as well as the heat transfer rate. (b) Plot the outlet air and outlet ammonia temperatures versus the water flow rate over the range \(5 \mathrm{~kg} / \mathrm{s} \leq \dot{m}_{c, \mathrm{~A}}=m_{h, \mathrm{~B}} \leq 50 \mathrm{~kg} / \mathrm{s}\).
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