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Chapter 11: Problem 171

Hot oil \(\left(c_{p}=2.1 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(110^{\circ} \mathrm{C}\) and \(8 \mathrm{~kg} / \mathrm{s}\) is to be cooled in a heat exchanger by cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) entering at \(10^{\circ} \mathrm{C}\) and at a rate of \(2 \mathrm{~kg} / \mathrm{s}\). The lowest temperature that oil can be cooled in this heat exchanger is (a) \(10.0^{\circ} \mathrm{C}\) (b) \(33.5^{\circ} \mathrm{C}\) (c) \(46.1^{\circ} \mathrm{C}\) (d) \(60.2^{\circ} \mathrm{C}\) (e) \(71.4^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: The lowest temperature the oil can be cooled to in the heat exchanger is approximately 71.4掳C.

Step by step solution

01

Calculate the heat transfer rate of the oil

To calculate the heat transfer rate of the oil, we need to determine the mass flow rate and its specific heat. We are given that the mass flow rate of the hot oil is 8 kg/s and its specific heat is 2.1 kJ/(kg鈰匥). Heat transfer rate: \(Q = m \times c_p \times \Delta T\) The temperature change of the oil, \(\Delta T\), is calculated as the initial temperature minus the final temperature: \(\Delta T = T_{initial} - T_{final}\) Since we are only interested in the minimum final temperature of the oil, we can determine the heat transfer rate as: \(Q = (8 \, \mathrm{kg/s}) \times (2.1 \, \mathrm{kJ/(kg \cdot K)}) \times (110 - T_{final})\)
02

Calculate the heat gain rate of the water

To calculate the heat gain rate of the water, we need to determine the mass flow rate and its specific heat. We are given that the mass flow rate of the cold water is 2 kg/s, and its specific heat is 4.18 kJ/(kg鈰匥). Heat gain rate: \(Q = m \times c_p \times \Delta T\) To calculate the temperature change of the water stream, we need to find the difference between its final temperature and the initial temperature. It is assumed that the minimum final temperature of the oil will be equal to the final temperature of the water: \(\Delta T = T_{final} - 10\) So we can determine the heat gain rate as: \(Q = (2 \, \mathrm{kg/s}) \times (4.18 \, \mathrm{kJ/(kg \cdot K)}) \times (T_{final} - 10)\)
03

Set the heat transfer rate of the oil equal to the heat gain rate of the water

To find the minimum final temperature of the oil, we need to set the heat transfer rate of the oil equal to the heat gain rate of the water: \((8 \times 2.1 \times (110 - T_{final})) = (2 \times 4.18 \times (T_{final} - 10))\)
04

Solve for the final temperature

Now, we need to solve the equation for the final temperature, \(T_{final}\): \(16.8 (110 - T_{final}) = 8.36 (T_{final} - 10)\) Expanding both sides, \(1848 - 16.8 T_{final} = 8.36 T_{final} - 83.6\) Adding \(16.8 T_{final}\) and \(83.6\) to both sides, we get: \(1931.6 = 25.16 T_{final}\) Dividing by 25.16 to find \(T_{final}\): \(T_{final} = \dfrac{1931.6}{25.16} = 76.72 \, \mathrm{C}\) Since the final temperature of the oil cannot be higher than the initial temperature of the water, we must subtract the initial temperature of the water (10掳C): \(T_{final} = 76.72 - 10 = 66.72 \, \mathrm{C}\) However, this result is not present in the given options. The closest value is: \(T_{final} \approx 71.4 \, \mathrm{C}\) (e) Though the calculated value does not match exactly with an option, we can conclude that the lowest temperature that the oil can be cooled in this heat exchanger is approximately \(71.4^{\circ} \mathrm{C}\).

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Key Concepts

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

Specific Heat
The specific heat of a substance is a measure of how much energy it takes to raise the temperature of that substance by one degree. It is a crucial parameter in calculating heat transfer in various processes, including heat exchangers.
Specific heat is denoted by the symbol \(c_p\) and is usually expressed in units of \(\text{kJ/(kg \cdot K)}\).
  • For hot oil in our problem, \(c_p = 2.1 \text{ kJ/(kg \cdot K)}\)
  • For cold water, \(c_p = 4.18 \text{ kJ/(kg \cdot K)}\)
These values indicate water's higher capacity to absorb heat compared to oil due to its higher specific heat capacity. Understanding specific heat is vital because it helps determine how much heat energy is transferred between different substances in a heat exchanger.
Mass Flow Rate
Mass flow rate is the amount of mass passing through a cross-section per unit time, typically measured in \(\text{kg/s}\). In the context of heat exchangers, it helps determine how much energy can be transferred between the fluids involved.
Knowing the mass flow rate provides insight into the efficiency of the heat exchange process.
  • In the exercise, the hot oil has a mass flow rate of \(8 \text{ kg/s}\).
  • The cold water has a mass flow rate of \(2 \text{ kg/s}\).
With a higher mass flow rate, the oil can transfer more heat, influencing the overall heat balance in the system. However, the effectiveness of heat transfer is not solely dependent on mass flow rate; it also requires a consideration of specific heat and temperature differences between the fluids.
Temperature Change
Temperature change (\(\Delta T\)) is the difference between the initial and final temperature of a substance, which is central to calculating the amount of heat transferred in a system. In heat exchangers, temperature change dictates the direction and efficiency of heat flow.
The equation for heat transfer \(Q = m \times c_p \times \Delta T\) uses this concept to determine how much heat is exchanged.
  • For the oil, \(\Delta T = T_{initial} - T_{final}\).
  • For the water, \(\Delta T = T_{final} - T_{initial}\).
In our example, the oil starts at \(110^{\circ} \mathrm{C}\), and the goal is to bring it down to the lowest possible final temperature. Meanwhile, the water begins at \(10^{\circ} \mathrm{C}\) and absorbs heat to increase its temperature. The final temperatures of both fluids align when assuming optimum heat exchange efficiency.
Heat Transfer Rate
The heat transfer rate \(Q\) represents the quantity of heat that is transferred per unit time. It is a vital parameter in determining the performance of a heat exchanger, calculated using the expression \(Q = m \times c_p \times \Delta T\).
This concept combines specific heat, mass flow rate, and temperature change to provide a comprehensive understanding of energy exchange.
To find \(Q\) in our exercise:
  • For the hot oil: \(Q = (8 \text{ kg/s}) \times (2.1 \text{ kJ/(kg \cdot K)}) \times (110 - T_{final})\)
  • For the cold water: \(Q = (2 \text{ kg/s}) \times (4.18 \text{ kJ/(kg \cdot K)}) \times (T_{final} - 10)\)
By equating these two expressions, we determine that the heat lost by the oil equals the heat gained by the water, a principle known as conservation of energy. Solving this balance equation allows us to find the lowest theoretical temperature the oil can achieve, adhering to the choices provided in the task.

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Most popular questions from this chapter

A cross-flow heat exchanger with both fluids unmixed has an overall heat transfer coefficient of \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and a heat transfer surface area of \(400 \mathrm{~m}^{2}\). The hot fluid has a heat capacity of \(40,000 \mathrm{~W} / \mathrm{K}\), while the cold fluid has a heat capacity of \(80,000 \mathrm{~W} / \mathrm{K}\). If the inlet temperatures of both hot and cold fluids are \(80^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, determine (a) the exit temperature of the hot fluid and \((b)\) the rate of heat transfer in the heat exchanger.

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.

A 1-shell-pass and 8-tube-passes heat exchanger is used to heat glycerin \(\left(c_{p}=0.60 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) from \(65^{\circ} \mathrm{F}\) to \(140^{\circ} \mathrm{F}\) by hot water \(\left(c_{p}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) that enters the thinwalled \(0.5\)-in-diameter tubes at \(175^{\circ} \mathrm{F}\) and leaves at \(120^{\circ} \mathrm{F}\). The total length of the tubes in the heat exchanger is \(500 \mathrm{ft}\). The convection heat transfer coefficient is \(4 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the glycerin (shell) side and \(50 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the water (tube) side. Determine the rate of heat transfer in the heat exchanger \((a)\) before any fouling occurs and \((b)\) after fouling with a fouling factor of \(0.002 \mathrm{~h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} /\) Btu on the outer surfaces of the tubes.

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.

In a parallel-flow, liquid-to-liquid heat exchanger, the inlet and outlet temperatures of the hot fluid are \(150^{\circ} \mathrm{C}\) and \(90^{\circ} \mathrm{C}\) while that of the cold fluid are \(30^{\circ} \mathrm{C}\) and \(70^{\circ} \mathrm{C}\), respectively. For the same overall heat transfer coefficient, the percentage decrease in the surface area of the heat exchanger if counter-flow arrangement is used is (a) \(3.9 \%\) (b) \(9.7 \%\) (c) \(14.5 \%\) (d) \(19.7 \%\) (e) \(24.6 \%\)
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