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) \(0.318 \mathrm{~kg} / \mathrm{s}\)

Short Answer

Expert verified
Answer: The rate at which steam condenses in the heat exchanger is \(0.290 \: \mathrm{kg/s}\).

Step by step solution

01

Calculate the Enthalpy Difference for the Water Stream

First, we need to find the enthalpy difference (∆H) for the water entering and leaving the condenser. To do this, we can use the formula: \(\Delta H = m_w \cdot c_p \cdot (T_{out} - T_{in})\) Where \(m_w\) is the water mass flow rate, \(c_p\) is the specific heat capacity of the water, and \(T_{out}\) and \(T_{in}\) are the outlet and inlet water temperatures, respectively. Using the given values, \(\Delta H = 20 \: kg/s \cdot 4.18 \: kJ/kg \cdot K \cdot (25^{\circ}C - 16^{\circ}C)\) \(\Delta H = 1502.8 \: \mathrm{kJ/s}\)
02

Apply the Conservation of Energy Equation

Now we will apply the conservation of energy equation to the problem. The sum of the heat inputs must equal the sum of the heat outputs. In this case, the heat input is from the condensing steam, while the heat outputs are due to the cooling water and the heat loss to surrounding air. The equation for this is: \(q_{steam} = \Delta H + q_{loss}\) Where \(q_{steam}\) is the heat transfer from the condensing steam, \(\Delta H\) is the enthalpy difference calculated in step 1, and \(q_{loss}\) is the heat loss to the surroundings. Using the values calculated above as well as the given 8 kW heat loss, \(q_{steam} = 1502.8 \: \mathrm{kJ/s} + 8 \: \mathrm{kW}\) \(q_{steam} = 1510.8 \: \mathrm{kJ/s}\)
03

Calculate the Condensation Rate of Steam

The final step is to calculate the condensation rate (\(m_s\)) of the steam, which could be found using the formula: \(m_s = \frac{q_{steam}}{c_p \cdot (T_s - T_{in})}\) Using the values calculated above, \(m_s = \frac{1510.8 \: \mathrm{kJ/s}}{4.18 \: \mathrm{kJ/kg} \cdot K \cdot (45^{\circ}C-16^{\circ}C)}\) \(m_s = 0.290 \: \mathrm{kg/s}\) Therefore, the correct answer is (b) \(0.290 \: \mathrm{kg/s}\).

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Key Concepts

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

Enthalpy Difference Calculation
Understanding the concept of enthalpy difference is crucial when studying heat exchangers and other thermodynamic systems. Enthalpy, often denoted as 'H', is a measure of total energy in a thermodynamic system. It includes both internal energy and the energy required to make space for it in the environment (pressure-volume work).

When cold water flows through a heat exchanger to absorb heat from steam, its temperature increases. This process changes the water's enthalpy. The enthalpy difference calculation is a way to quantify the energy exchanged via heat between two substances at different temperatures.

In the given exercise, we established the enthalpy difference by multiplying the mass flow rate of water (\(m_w\text{ in kg/s}\)), the specific heat capacity of water (\(c_p\text{ in kJ/kg·K}\)), and the temperature change of water (\(T_{out} - T_{in}\text{ in K}\)). Specific heat capacity is an intensive property that represents the amount of heat required to raise the temperature of a unit mass of a substance by one degree.

Conservation of Energy Equation
The conservation of energy principle, also known as the first law of thermodynamics, asserts that energy cannot be created or destroyed in an isolated system. In the context of the exercise, we're dealing with a heat exchanger where energy is transferred, not created or lost. The energy that goes into the system has to equal the energy that leaves it. This is crucial for calculating the efficiency of the heat exchanger.

The equation applied in the exercise puts this principle into practice. It sets the heat absorbed by the cold water (\(ΔH\text{ in kJ/s}\)) plus the heat loss to the surroundings (\(q_{loss}\text{ in kW}\)) equal to the heat released by the steam (\(q_{steam}\text{ in kJ/s}\)). Heat loss can occur due to lack of insulation or other inefficiencies.

An important takeaway is that in real-world scenarios, not all the heat from the steam is transferred to the cold water. Some energy is always lost to the surroundings, and an effective heat exchanger design aims to minimize such losses to increase efficiency.

Condensation Rate of Steam
The condensation rate of steam is a term that describes how quickly steam turns into liquid water when it loses energy. This is an essential variable in many industrial applications, including power plants and HVAC systems.

In the exercise, we calculated the condensation rate by utilizing the conservation of energy equation and the concept of enthalpy. By dividing the heat transfer from the condensing steam (\(q_{steam}\text{ in kJ/s}\)) by the product of the specific heat capacity of steam at constant pressure (\(c_p\text{ in kJ/kg·K}\)) and the temperature difference between the steam and entering water (\(T_s - T_{in}\text{ in K}\)), we determine the mass of steam that can condense per second.

The calculated condensation rate connects the thermal energy transfer with the phase change of the steam. Phase change, in this case, the process of condensation, is heavily influenced by both temperature and pressure. The specific enthalpy of condensation is a property that is unique to the substance undergoing the phase change and the conditions under which it condenses.

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Most popular questions from this chapter

A shell-and-tube heat exchanger with 2-shell passes and 12 -tube passes is used to heat water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) with ethylene glycol \(\left(c_{p}=2680 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\). Water enters the tubes at \(22^{\circ} \mathrm{C}\) at a rate of \(0.8 \mathrm{~kg} / \mathrm{s}\) and leaves at \(70^{\circ} \mathrm{C}\). Ethylene \(\mathrm{glycol}\) enters the shell at \(110^{\circ} \mathrm{C}\) and leaves at \(60^{\circ} \mathrm{C}\). If the overall heat transfer coefficient based on the tube side is \(280 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the heat transfer surface area on the tube side.

A company owns a refrigeration system whose refrigeration capacity is 200 tons ( 1 ton of refrigeration = \(211 \mathrm{~kJ} / \mathrm{min}\) ), and you are to design a forced-air cooling system for fruits whose diameters do not exceed \(7 \mathrm{~cm}\) under the following conditions: The fruits are to be cooled from \(28^{\circ} \mathrm{C}\) to an average temperature of \(8^{\circ} \mathrm{C}\). The air temperature is to remain above \(-2^{\circ} \mathrm{C}\) and below \(10^{\circ} \mathrm{C}\) at all times, and the velocity of air approaching the fruits must remain under \(2 \mathrm{~m} / \mathrm{s}\). The cooling section can be as wide as \(3.5 \mathrm{~m}\) and as high as \(2 \mathrm{~m}\). Assuming reasonable values for the average fruit density, specific heat, and porosity (the fraction of air volume in a box), recommend reasonable values for the quantities related to the thermal aspects of the forced-air cooling, including (a) how long the fruits need to remain in the cooling section, \((b)\) the length of the cooling section, \((c)\) the air velocity approaching the cooling section, \((d)\) the product cooling capacity of the system, in \(\mathrm{kg}\) fruit/h, \((e)\) the volume flow rate of air, and \((f)\) the type of heat exchanger for the evaporator and the surface area on the air side.

Consider the flow of saturated steam at \(270.1 \mathrm{kPa}\) that flows through the shell side of a shell-and-tube heat exchanger while the water flows through 4 tubes of diameter \(1.25 \mathrm{~cm}\) at a rate of \(0.25 \mathrm{~kg} / \mathrm{s}\) through each tube. The water enters the tubes of heat exchanger at \(20^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). Due to the heat exchange with the cold fluid, steam is condensed on the tubes external surface. The convection heat transfer coefficient on the steam side is \(1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the fouling resistance for the steam and water may be taken as \(0.00015\) and \(0.0001 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), respectively. Using the NTU method, determine \((a)\) effectiveness of the heat exchanger, \((b)\) length of the tube, and \((c)\) rate of steam condensation.

Hot water coming from the engine is to be cooled by ambient air in a car radiator. The aluminum tubes in which the water flows have a diameter of \(4 \mathrm{~cm}\) and negligible thickness. Fins are attached on the outer surface of the tubes in order to increase the heat transfer surface area on the air side. The heat transfer coefficients on the inner and outer surfaces are 2000 and \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. If the effective surface area on the finned side is 10 times the inner surface area, the overall heat transfer coefficient of this heat exchanger based on the inner surface area is (a) \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(857 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(1075 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(2150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Explain how you can evaluate the outlet temperatures of the cold and hot fluids in a heat exchanger after its effectiveness is determined.

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