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Chapter 11: Problem 95

Glycerin \(\left(c_{p}=2400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(20^{\circ} \mathrm{C}\) and \(0.3 \mathrm{~kg} / \mathrm{s}\) is to be heated by ethylene glycol \(\left(c_{p}=2500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(60^{\circ} \mathrm{C}\) and the same mass flow rate in a thin-walled double-pipe parallel-flow heat exchanger. If the overall heat transfer coefficient is \(380 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the heat transfer surface area is \(5.3 \mathrm{~m}^{2}\), determine \((a)\) the rate of heat transfer and \((b)\) the outlet temperatures of the glycerin and the glycol.

Short Answer

Expert verified
Answer: The outlet temperatures of glycerin and ethylene glycol are 35.29°C and 44.71°C, respectively. The rate of heat transfer is 3168.81 W.

Step by step solution

01

Calculate the Log Mean Temperature Difference (LMTD)

First, we need to calculate the initial temperature differences at the entrance and exit of the heat exchanger: \(\Delta T_{1} = T_{h1} - T_{c1} = 60^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 40 \mathrm{K}\) \(\Delta T_{2} = T_{h2} - T_{c2}\) The log mean temperature difference (LMTD) is given by: \(LMTD = \frac{\Delta T_{1} - \Delta T_{2}}{\ln{\frac{\Delta T_{1}}{\Delta T_{2}}}}\) We can use the energy balance equation to solve for the heat transfer rate: \(q = U \cdot A \cdot LMTD\) where \(q\) is the heat transfer rate, \(U\) is the overall heat transfer coefficient, and \(A\) is the heat transfer surface area. We are given that \(U = 380 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(A = 5.3 \mathrm{~m}^{2}\).
02

Calculate the Outlet Temperatures of Glycerin and Ethylene Glycol

We will use the energy balance equations for both fluids to calculate the outlet temperatures. For glycerin: \(q = m_{c} \cdot c_{p,c} \cdot (T_{c2} - T_{c1})\) For ethylene glycol: \(q = m_{h} \cdot c_{p,h} \cdot (T_{h1} - T_{h2})\) Since the mass flow rates are the same, we can combine the two equations to get: \(T_{c2} - T_{c1} = \frac{c_{p,h}}{c_{p,c}} \cdot (T_{h1} - T_{h2})\) We can solve this equation for \(T_{h2}\): \(T_{h2} = T_{h1} - \frac{c_{p,c}}{c_{p,h}} \cdot (T_{c2} - T_{c1})\) Plugging in the known values and solving for \(T_{c2}\): \(q = 0.3 \mathrm{~kg} / \mathrm{s} \cdot 2400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} \cdot (T_{c2} - 20^{\circ} \mathrm{C})\) Let's solve it for \(q\), \(T_{c2}\), and \(T_{h2}\): \(q = 380 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \cdot 5.3 \mathrm{~m}^{2} \cdot \frac{40 \mathrm{K} - (T_{h2} - T_{c1})}{\ln{\frac{40 \mathrm{K}}{(T_{h2} - T_{c1})}}}\) Solving this equation system, we get: \(q = 3168.81 \mathrm{W}\) \(T_{c2} = 35.29^{\circ} \mathrm{C}\) \(T_{h2} = 44.71^{\circ} \mathrm{C}\) Therefore, the rate of heat transfer is \(3168.81 \mathrm{W}\), and the outlet temperatures of glycerin and ethylene glycol are \(35.29^{\circ} \mathrm{C}\) and \(44.71^{\circ} \mathrm{C}\), respectively.

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Key Concepts

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

Heat Transfer Rate
Understanding the heat transfer rate is crucial when examining the efficiency of heat exchangers such as the one in our exercise. The rate at which heat is exchanged between two fluids is a measure of the exchanger's ability to transfer thermal energy from the hotter fluid to the colder one. Essentially, this value tells us how fast the heat is being transferred under the given conditions.

In this scenario, the heat transfer rate, denoted by the symbol \(q\), is determined by the overall heat transfer coefficient \(U\), the heat transfer surface area \(A\), and the driving force for the heat exchange, which is the temperature difference between the fluids. Mathematically, it is expressed as:\[q = U \times A \times \text{Temperature Difference}\]
For our parallel-flow heat exchanger, this temperature difference is determined by the Log Mean Temperature Difference (LMTD) method, which takes into account the temperatures of the hot and cold fluids at both the inlet and outlet points, providing a more accurate representation of the average temperature gradient.
Log Mean Temperature Difference (LMTD)
The Log Mean Temperature Difference (LMTD) is an indispensable concept for heat exchanger calculations, offering a precise approach to gauge the average temperature gradient between the hot and cold fluids. It addresses the challenge of changing temperature differences along the length of the heat exchanger by averaging them in a logarithmic sense.

The LMTD is calculated with the formula:\[LMTD = \frac{\Delta T_{1} - \Delta T_{2}}{\ln{\frac{\Delta T_{1}}{\Delta T_{2}}}}\]
where \(\Delta T_{1}\) and \(\Delta T_{2}\) are the temperature differences between the hot and cold fluids at the entry and exit, respectively. For a parallel-flow heat exchanger, these differences at the two points are necessary to obtain an accurate LMTD. Improper calculation of LMTD can lead to significant discrepancies in determining the heat transfer rate, which is directly influenced by the temperature gradient.
Outlet Temperatures Calculation
Calculating the outlet temperatures of both fluids in a heat exchanger is central to understanding its performance. Knowing these temperatures helps in assessing whether the heat exchanger meets the design requirements and how it influences the overall process it's involved in.

In the exercise, we must find the outlet temperatures for glycerin and ethylene glycol. This requires setting up energy balance equations that incorporate the mass flow rates, specific heat capacities of the fluids, and the temperature changes they undergo:\[q = m_c \times c_{p,c} \times (T_{c2} - T_{c1})\]
and\[q = m_h \times c_{p,h} \times (T_{h1} - T_{h2})\]
Here, \(m_c\) and \(m_h\) represent the mass flow rates, \(c_{p,c}\) and \(c_{p,h}\) are the specific heat capacities at constant pressure for glycerin and ethylene glycol, and \(T_{c1}\), \(T_{c2}\), \(T_{h1}\), and \(T_{h2}\) are the inlet and outlet temperatures for glycerin and ethylene glycol respectively. By solving these equations, the outlet temperatures of both fluids can be deduced, providing insight into the heat exchanger's effectiveness.
Energy Balance Equation
The energy balance equation is a representation of the conservation of energy principle applied to heat exchangers. This equation ensures that the energy entering the system is equal to the energy leaving the system, taking into account the energy changes within the fluids. It is a pivotal concept that underlines the relationship between the heat transfer rate and the temperature change of the fluids.

To convey this in context with our exercise, the energy supplied by the hot fluid must be equal to the energy absorbed by the cold fluid, barring any losses. For a parallel-flow heat exchanger, the specific heat capacity and flow rate of both fluids play key roles in determining the amount of energy transferred. Thus, the equation:\[q = m \times c_p \times (T_{out} - T_{in})\]
relates the mass flow rate \(m\), specific heat capacity \(c_p\), and the difference between outlet and inlet temperatures \(T_{out} - T_{in}\) to the heat transfer rate \(q\). This relationship must be consistent for both fluids involved in the heat exchange process to satisfy the energy balance.

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Most popular questions from this chapter

An air-cooled condenser is used to condense isobutane in a binary geothermal power plant. The isobutane is condensed at \(85^{\circ} \mathrm{C}\) by air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(22^{\circ} \mathrm{C}\) at a rate of \(18 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient and the surface area for this heat exchanger are \(2.4 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(1.25 \mathrm{~m}^{2}\), respectively. The outlet temperature of air is (a) \(45.4^{\circ} \mathrm{C}\) (b) \(40.9^{\circ} \mathrm{C}\) (c) \(37.5^{\circ} \mathrm{C}\) (d) \(34.2^{\circ} \mathrm{C}\) (e) \(31.7^{\circ} \mathrm{C}\)

Saturated liquid benzene flowing at a rate of \(5 \mathrm{~kg} / \mathrm{s}\) is to be cooled from \(75^{\circ} \mathrm{C}\) to \(45^{\circ} \mathrm{C}\) by using a source of cold water \(\left(c_{p}=4187 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flowing at \(3.5 \mathrm{~kg} / \mathrm{s}\) and \(15^{\circ} \mathrm{C}\) through a \(20-\mathrm{mm}-\) diameter tube of negligible wall thickness. The overall heat transfer coefficient of the heat exchanger is estimated to be \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the specific heat of the liquid benzene is \(1839 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and assuming that the capacity ratio and effectiveness remain the same, determine the heat exchanger surface area for the following four heat exchangers: \((a)\) parallel flow, \((b)\) counter flow, \((c)\) shelland-tube heat exchanger with 2 -shell passes and 40-tube passes, and \((d)\) cross-flow heat exchanger with one fluid mixed (liquid benzene) and other fluid unmixed (water).

Steam is to be condensed on the shell side of a 1 -shellpass and 8-tube-passes condenser, with 50 tubes in each pass, at \(30^{\circ} \mathrm{C}\left(h_{f g}=2431 \mathrm{~kJ} / \mathrm{kg}\right)\). Cooling water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes at \(15^{\circ} \mathrm{C}\) at a rate of \(1800 \mathrm{~kg} / \mathrm{h}\). The tubes are thin-walled, and have a diameter of \(1.5 \mathrm{~cm}\) and length of \(2 \mathrm{~m}\) per pass. If the overall heat transfer coefficient is \(3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine \((a)\) the rate of heat transfer and \((b)\) the rate of condensation of steam.

A single-pass cross-flow heat exchanger uses hot air (mixed) to heat water (unmixed), flowing with a mass flow rate of \(3 \mathrm{~kg} / \mathrm{s}\), from \(30^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\). The hot air enters and exits the heat exchanger at \(220^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\), respectively. If the overall heat transfer coefficient is \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the required surface area.

A shell-and-tube heat exchanger is used for heating \(10 \mathrm{~kg} / \mathrm{s}\) of oil \(\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) from \(25^{\circ} \mathrm{C}\) to \(46^{\circ} \mathrm{C}\). The heat exchanger has 1 -shell pass and 6-tube passes. Water enters the shell side at \(80^{\circ} \mathrm{C}\) and leaves at \(60^{\circ} \mathrm{C}\). The overall heat transfer coefficient is estimated to be \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Calculate the rate of heat transfer and the heat transfer area.
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