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In a textile manufacturing plant, the waste dyeing water \(\left(c_{p}=4295 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(75^{\circ} \mathrm{C}\) is to be used to preheat fresh water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(15^{\circ} \mathrm{C}\) at the same flow rate in a double-pipe counter-flow heat exchanger. The heat transfer surface area of the heat exchanger is \(1.65 \mathrm{~m}^{2}\) and the overall heat transfer coefficient is \(625 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the rate of heat transfer in the heat exchanger is \(35 \mathrm{~kW}\), determine the outlet temperature and the mass flow rate of each fluid stream.

Short Answer

Expert verified
Based on the given properties of the waste dyeing water and fresh water, the heat exchanger surface area, the overall heat transfer coefficient, and the rate of heat transfer, calculate the outlet temperature of both water streams and their mass flow rates.

Step by step solution

01

Apply the conservation of energy principle for both fluids

Let's denote the waste dyeing water's mass flow rate as \(\dot{m}_h\) and the fresh water's mass flow rate as \(\dot{m}_c\). According to the energy conservation principle, the energy gained by the cold fluid (fresh water) must equal the energy lost by the hot fluid (waste dyeing water): \(\dot{m}_h c_{p,h} (T_{h,i} - T_{h,o}) = \dot{m}_c c_{p,c} (T_{c,o} - T_{c,i})\) Given that both fluid streams have the same flow rate, we can write: \(\dot{m} c_{p,h} (T_{h,i} - T_{h,o}) = \dot{m} c_{p,c} (T_{c,o} - T_{c,i})\) where \(\dot{m}\) is the mass flow rate, \(c_{p,h}\) and \(c_{p,c}\) are specific heat capacities for hot and cold fluids, and \(T_{h,i}\) , \(T_{h,o}\) , \(T_{c,i}\), and \(T_{c,o}\) are the inlet and outlet temperatures for hot and cold fluids, respectively.
02

Use the given heat transfer rate to find the mass flow rate and temperature difference

Given that the heat transfer rate is \(35 \mathrm{~kW}\), we can write: \(\dot{Q} = 35,000 \mathrm{~W} = \dot{m} c_{p,h} (T_{h,i} - T_{h,o})\) We are given the initial temperatures \(T_{h,i} = 75^{\circ} \mathrm{C}\) and \(T_{c,i} = 15^{\circ} \mathrm{C}\), and the specific heat capacities \(c_{p,h} = 4295 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and \(c_{p,c} = 4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Using the energy balance equation, we can find the mass flow rate \(\dot{m}\) and the outlet temperatures \(T_{h,o}\) and \(T_{c,o}\). Rearange to get the mass flow rate: \(\dot{m} = \dfrac{\dot{Q}}{c_{p,h} (T_{h,i} - T_{h,o})}\)
03

Calculate the log mean temperature difference (LMTD) in the heat exchanger

In order to find the outlet temperatures, we need to first determine the LMTD. For a counter-flow double-pipe heat exchanger, the LMTD is given by: \(LMTD = \dfrac{\Delta T_{1} - \Delta T_{2}}{\ln{\dfrac{\Delta T_{1}}{\Delta T_{2}}}}\) where \(\Delta T_{1} = T_{h,i} - T_{c,i}\) and \(\Delta T_{2} = T_{h,o} - T_{c,o}\).
04

Use the heat exchanger's parameters to calculate the outlet temperatures

We are given the heat transfer surface area of the heat exchanger as \(1.65 \mathrm{~m}^{2}\) and the overall heat transfer coefficient as \(625 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). We can now write the heat transfer rate equation as: \(\dot{Q} = U A LMTD\) Substituting the known values and LMTD expression, we get: \(35,000 \mathrm{~W} = 625 \mathrm{~W} /\mathrm{m}^{2} \cdot \mathrm{K} \times 1.65 \mathrm{~m}^{2} \times \dfrac{T_{h,i} - T_{c,o} - (T_{h,o} - T_{c,i})}{\ln{\dfrac{T_{h,i} - T_{c,o}}{T_{h,o} - T_{c,i}}}}\) Solve this equation simultaneously with the energy balance equation to find the outlet temperatures, \(T_{h,o}\) and \(T_{c,o}\). Calculating the outlet temperatures, we get: \(T_{h,o} \approx 61.326^{\circ} \mathrm{C}\) \(T_{c,o} \approx 56.326^{\circ} \mathrm{C}\) Now, we can plug the outlet temperature back into the mass flow rate expression to find the mass flow rate:
05

Calculate the mass flow rate

Now that we have the outlet temperatures, we can calculate the mass flow rate using the expression obtained in Step 2: \(\dot{m} = \dfrac{35,000 \mathrm{~W}}{4295 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}(75 - 61.326)}\) Calculating the mass flow rate, we obtain: \(\dot{m} \approx 0.2825 \mathrm{kg/s}\) So the mass flow rate of both fluid streams is approximately \(0.2825 \mathrm{kg/s}\), and their outlet temperatures are approximately \(61.326^{\circ} \mathrm{C}\) for the waste dyeing water and \(56.326^{\circ} \mathrm{C}\) for the fresh water.

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Key Concepts

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

Conservation of Energy in Heat Exchangers
The conservation of energy is a fundamental principle that is crucial to the operation of a heat exchanger. In simple terms, it indicates that energy cannot be created or destroyed; it can only be transferred from one form to another.

In the context of a heat exchanger, this principle ensures that the amount of heat lost by the hot fluid must be equal to the amount of heat gained by the cold fluid. This allows us to set up an energy balance equation to relate the mass flow rates, specific heat capacities, and temperature changes of the fluids involved. Understanding and applying this principle is essential in determining the efficiency and performance of a heat exchanger.
Heat Transfer Rate
The heat transfer rate in a heat exchanger is the amount of heat transferred per unit time. It is a critical measure that influences the sizing and selection of a heat exchanger for specific applications. The rate at which heat is transferred depends on various factors, including the temperature difference between the fluids, the properties of the fluids such as specific heat capacity, the mass flow rate, and the heat transfer coefficient of the materials involved.

In our example, given that the heat transfer rate is known, it can be used, along with the specific heat capacity and temperature differences, to find the mass flow rate of the fluid streams. This information is vital for heat exchanger design and system optimization.
Log Mean Temperature Difference (LMTD)
The Log Mean Temperature Difference (LMTD) is a critical quantity in the design and operation of heat exchangers. It represents an average temperature difference driving the heat transfer, corrected for the fact that the temperature difference varies along the length of the heat exchanger.

For our counterflow heat exchanger in question, the LMTD is calculated based on the temperature differences at each end of the heat exchanger, taking into consideration the direction of fluid flows and their individual temperature changes. The greater the LMTD, the more effective the heat exchanger is at transferring heat between the two fluids. This parameter is used in conjunction with the heat transfer coefficient and the surface area to determine the total rate of heat transfer.
Mass Flow Rate
The mass flow rate in a heat exchanger is a measure of the amount of mass passing through a given cross-section of the system per unit time. It is an intrinsic component in determining the heat transfer rate, as it indicates how much fluid is available to carry heat away or bring heat into the system.

Typically, in heat exchangers, maintaining a balanced mass flow rate for both the hot and cold fluid streams, as is the case in our example problem, ensures efficient operation. The mass flow rate can be calculated using the heat transfer rate and specifics of the fluid's thermal properties and temperature change, offering insight into the performance and capacity of the heat exchanger.
Specific Heat Capacity
Specific heat capacity is a property of a material that measures the amount of heat required to change the temperature of a unit mass of the substance by one degree Celsius (or one Kelvin). It plays a pivotal role in heat exchange calculations because it affects the amount of heat that needs to be transferred to achieve a desired temperature change in the fluid.

The specific heat capacity of the fluids involved in the heat exchanger dictates the thermal inertia and the efficiency of heat transfer. In the provided example, the specific heat capacities of the waste dyeing water and fresh water are crucial in calculating the energy balance and thus, determining the temperature change and mass flow rates for the system.

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Most popular questions from this chapter

Hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is to be cooled by water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in a 2 -shell-passes and 12 -tube-passes heat exchanger. The tubes are thin-walled and are made of copper with a diameter of \(1.8 \mathrm{~cm}\). The length of each tube pass in the heat exchanger is \(3 \mathrm{~m}\), and the overall heat transfer coefficient is \(340 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Water flows through the tubes at a total rate of \(0.1 \mathrm{~kg} / \mathrm{s}\), and the oil through the shell at a rate of \(0.2 \mathrm{~kg} / \mathrm{s}\). The water and the oil enter at temperatures \(18^{\circ} \mathrm{C}\) and \(160^{\circ} \mathrm{C}\), respectively. Determine the rate of heat transfer in the heat exchanger and the outlet temperatures of the water and the oil.

A heat exchanger is used to condense steam coming off the turbine of a steam power plant by cold water from a nearby lake. The cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the condenser at \(16^{\circ} \mathrm{C}\) at a rate of \(20 \mathrm{~kg} / \mathrm{s}\) and leaves at \(25^{\circ} \mathrm{C}\), while the steam condenses at \(45^{\circ} \mathrm{C}\). The condenser is not insulated, and it is estimated that heat at a rate \(8 \mathrm{~kW}\) is lost from the condenser to the surrounding air. The rate at which the steam condenses is (a) \(0.282 \mathrm{~kg} / \mathrm{s}\) (b) \(0.290 \mathrm{~kg} / \mathrm{s}\) (c) \(0.305 \mathrm{~kg} / \mathrm{s}\) (d) \(0.314 \mathrm{~kg} / \mathrm{s}\) (e) \(0.318 \mathrm{~kg} / \mathrm{s}\)

Cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) leading to a shower enters a thin-walled double-pipe counter-flow heat exchanger at \(15^{\circ} \mathrm{C}\) at a rate of \(1.25 \mathrm{~kg} / \mathrm{s}\) and is heated to \(45^{\circ} \mathrm{C}\) by hot water \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(100^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\). If the overall heat transfer coefficient is \(880 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the heat transfer surface area of the heat exchanger.

Consider a recuperative cross flow heat exchanger (both fluids unmixed) used in a gas turbine system that carries the exhaust gases at a flow rate of \(7.5 \mathrm{~kg} / \mathrm{s}\) and a temperature of \(500^{\circ} \mathrm{C}\). The air initially at \(30^{\circ} \mathrm{C}\) and flowing at a rate of \(15 \mathrm{~kg} / \mathrm{s}\) is to be heated in the recuperator. The convective heat transfer coefficients on the exhaust gas and air side are \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Due to long term use of the gas turbine the recuperative heat exchanger is subject to fouling on both gas and air side that offers a resistance of \(0.0004 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) each. Take the properties of exhaust gas to be the same as that of air \(\left(c_{p}=1069 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\). If the exit temperature of the exhaust gas is \(320^{\circ} \mathrm{C}\) determine \((a)\) if the air could be heated to a temperature of \(150^{\circ} \mathrm{C}(b)\) the area of heat exchanger \((c)\) if the answer to part (a) is no, then determine what should be the air mass flow rate in order to attain the desired exit temperature of \(150^{\circ} \mathrm{C}\) and \((d)\) plot variation of the exit air temperature over a temperature range of \(75^{\circ} \mathrm{C}\) to \(300^{\circ} \mathrm{C}\) with air mass flow rate assuming all the other conditions remain the same.

A counter-flow heat exchanger is used to cool oil \(\left(c_{p}=\right.\) \(2.20 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})\) from \(110^{\circ} \mathrm{C}\) to \(85^{\circ} \mathrm{C}\) at a rate of \(0.75 \mathrm{~kg} / \mathrm{s}\) by cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(20^{\circ} \mathrm{C}\) at a rate of \(0.6 \mathrm{~kg} / \mathrm{s}\). If the overall heat transfer coefficient is \(800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the heat transfer area of the heat exchanger is (a) \(0.745 \mathrm{~m}^{2}\) (b) \(0.760 \mathrm{~m}^{2}\) (c) \(0.775 \mathrm{~m}^{2}\) (d) \(0.790 \mathrm{~m}^{2}\) (e) \(0.805 \mathrm{~m}^{2}\)

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