/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 The oil in an engine is cooled b... [FREE SOLUTION] | 91Ó°ÊÓ

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The oil in an engine is cooled by air in a cross-flow heat exchanger where both fluids are unmixed. Atmospheric air enters at \(30^{\circ} \mathrm{C}\) and \(0.53 \mathrm{~kg} / \mathrm{s}\). Oil at \(0.026 \mathrm{~kg} / \mathrm{s}\) enters at \(75^{\circ} \mathrm{C}\) and flows through a tube of 10-mm diameter. Assuming fully developed flow and constant wall heat flux, estimate the oil- side heat transfer coefficient. If the overall convection coefficient is \(53 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the total heat transfer area is \(1 \mathrm{~m}^{2}\), determine the effectiveness. What is the exit temperature of the oil?

Short Answer

Expert verified
The oil-side heat transfer coefficient is \(53 W/m^2 \cdot K \), and the heat exchanger effectiveness is \(\frac{0.026 \times c_{p,oil} \times (75 - T_{outlet})}{2.34}\). The exit temperature of the oil is \(30^{\circ} \mathrm{C}\).

Step by step solution

01

Find the flow area and heat flux for oil

First, we will need to calculate the flow area (A) and heat flux (q) for the oil using the given flow rate of the oil (\(0.026 kg/s\)) and tube diameter (10mm): Flow Area (A): \(A = \pi(D^2)/4\) Heat Flux (q): \(q = \dot{m}_{oil} \cdot c_{p,oil} \cdot (T_{inlet} - T_{outlet})\)
02

Find the Convection Heat Transfer Coefficient for Oil

The oil-side heat transfer coefficient (h_oil) can be estimated using the formula for the overall convection heat transfer coefficient (h_o): \( h_{o} = \dfrac{1}{(\dfrac{1}{h_{oil}})}\) Now, we will rearrange the formula to find h_oil: \(h_{oil} = \dfrac{1}{(\dfrac{1}{h_{o}})}\)
03

Calculate the effectiveness of the Heat Exchanger

We have the overall heat transfer coefficient and the heat transfer area. Now, we can calculate the effectiveness of the heat exchanger (ε) using the following formula: \(\epsilon = \dfrac{q}{q_{max}}\) The maximum heat transfer (q_max) can be calculated using the inlet temperatures of both fluids and their respective heat capacities (c_p): \(q_{max} = min(\dot{m}_{air} \cdot c_{p,air}, \dot{m}_{oil} \cdot c_{p,oil}) \cdot (T_{inlet,hot} - T_{inlet,cold})\)
04

Determine the Exit Temperature of the Oil

The exit temperature of the oil (T_outlet) can be calculated using the heat transfer rate (q) and the inlet temperature of the oil (T_inlet): \(T_{outlet} = T_{inlet} - \dfrac{q}{\dot{m}_{oil} \cdot c_{p,oil}}\) Solution:
05

Step 1

Calculate the flow area (A): A = \(\dfrac{\pi(0.01)^2}{4}\) = \(7.854 \times 10 ^{-5} m^2 \) Calculate the heat flux (q) assuming fully developed flow and constant wall heat flux: q = \(0.026 kg/s \cdot c_{p,oil} \cdot (75^{\circ} \mathrm{C} - T_{outlet})\)
06

Step 2

Find the oil-side heat transfer coefficient (h_oil) using the given overall convection heat transfer coefficient (53 W/m²·K): h_oil = \(\dfrac{1}{(1/53)}\) = \(53 W/m² \cdot K \)
07

Step 3

Calculate the effectiveness of the heat exchanger (ε): \(q_{max} = min(0.53 \times c_{p,air}, 0.026 \times c_{p,oil}) \times (75-30)\) Assuming heat capacities of air and oil as 1 kJ/kg·K and 2 kJ/kg·K respectively, we rewrite the above equation: \(q_{max} = min(0.53,0.052) \times (75-30)\) \(q_{max} = 0.052 \times 45\) \(q_{max} = 2.34 kW\) Now, find the effectiveness (ε): \(\epsilon = \dfrac{q}{q_{max}} = \dfrac{0.026 \times c_{p,oil} \times (75 - T_{outlet})}{2.34}\)
08

Step 4

Determine the exit temperature of the oil (T_outlet): Solving the equation for T_outlet: \(T_{outlet} = 75^{\circ} \mathrm{C} - \dfrac{2.34 \times 1000}{0.026 \times 2000}\) N.B.: We multiplied by 1000 to convert kW to W. \(T_{outlet} = 75^{\circ} \mathrm{C} - 45^{\circ} \mathrm{C}\) \(T_{outlet} = 30^{\circ} \mathrm{C}\) The exit temperature of the oil is \(30^{\circ} \mathrm{C}\).

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Key Concepts

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

Convective Heat Transfer Coefficient
The convective heat transfer coefficient is a critical factor in heat exchanger design and analysis. It represents the heat transfer rate per unit area per unit temperature difference between the solid surface and the fluid nearby. A high coefficient indicates an efficient transfer of heat. For fluids flowing through a tube, as in the problem statement, the coefficient depends on factors like fluid properties, flow velocity and regime (laminar or turbulent), and the geometry of the heat exchange surfaces.

Improving the convective heat transfer coefficient can be achieved by increasing the flow rate, using a fluid with better thermal conductivity, or creating a turbulent flow which usually enhances the heat transfer rate compared to a laminar one.
Exit Temperature Calculation
The exit temperature of a fluid in a heat exchanger is a crucial result for engineers to determine. It tells us how effective the heat exchanger is at cooling or heating a particular fluid. In our exercise, we obtained the exit temperature of the oil by accounting for the heat transferred out of it and its flow rate. This relationship is an application of energy conservation, where the heat lost by the oil equals the change in its thermal energy.

To calculate the exit temperature, we need to know the mass flow rate of the fluid, its specific heat capacity, the inlet temperature, and the heat transfer rate. From a practical perspective, ensuring the accuracy of these parameters is vital for an accurate exit temperature prediction.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient is a measure of a heat exchanger's ability to transfer heat between two fluids that are separated by a solid barrier. This value encapsulates all modes of heat transfer involved: conduction through the heat exchanger material and convection to and from the fluid. It's a compound coefficient that incorporates the resistance to heat flow on all sides of the heat exchanger material. The larger this coefficient, the more effective the exchanger.

The overall heat transfer coefficient is influenced by the materials used, the thickness of the walls of the exchanger, and the nature of the fluid flow. Ensuring the best choice of materials and optimizing the flow dynamics can greatly enhance the overall heat transfer coefficient, thereby maximizing the heat exchanger's effectiveness.
Flow Area and Heat Flux
Flow area refers to the cross-sectional area in a tube or duct available for fluid flow. It's directly related to the fluid velocity—another crucial factor for convective heat transfer. For a given volumetric flow rate, a smaller area leads to higher fluid velocity, which can enhance the convection heat transfer coefficient.

Heat flux is the rate of heat energy transferred per unit area. It's defined as the amount of heat transferred per unit area per unit time. In the context of our exercise, we calculated the heat flux by considering the constant wall heat flux assumption, which means that the temperature gradient in the direction of heat flow remains constant across the surface of the heat exchanger. Managing heat flux is important in avoiding local overheating or under heating within the system.

Optimizing Flow Area for Heat Transfer Efficiency

Narrowing or widening the flow area allows control over the velocity and turbulence of the fluid, which directly influences the heat transfer rate. In tube heat exchangers, the diameter of the tube is adjusted as per the requirement to maintain the desired heat transfer rate. A balance must be struck between pressure drop, which increases with decreasing flow area, and heat transfer efficiency to achieve an optimal design.

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Most popular questions from this chapter

Water at a rate of \(45,500 \mathrm{~kg} / \mathrm{h}\) is heated from 80 to \(150^{\circ} \mathrm{C}\) in a heat exchanger having two shell passes and eight tube passes with a total surface area of \(925 \mathrm{~m}^{2}\). Hot exhaust gases having approximately the same thermophysical properties as air enter at \(350^{\circ} \mathrm{C}\) and exit at \(175^{\circ} \mathrm{C}\). Determine the overall heat transfer coefficient.

A boiler used to generate saturated steam is in the form of an unfinned, cross-flow heat exchanger, with water flowing through the tubes and a high- temperature gas in cross flow over the tubes. The gas, which has a specific heat of \(1120 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and a mass flow rate of \(10 \mathrm{~kg} / \mathrm{s}\), enters the heat exchanger at \(1400 \mathrm{~K}\). The water, which has a flow rate of \(3 \mathrm{~kg} / \mathrm{s}\), enters as saturated liquid at \(450 \mathrm{~K}\) and leaves as saturated vapor at the same temperature. If the overall heat transfer coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and there are 500 tubes, each of \(0.025-\mathrm{m}\) diameter, what is the required tube length?

In open heart surgery under hypothermic conditions, the patient's blood is cooled before the surgery and rewarmed afterward. It is proposed that a concentric tube, counterflow heat exchanger of length \(0.5 \mathrm{~m}\) be used for this purpose, with the thin-walled inner tube having a diameter of \(55 \mathrm{~mm}\). The specific heat of the blood is \(3500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (a) If water at \(T_{h j}=60^{\circ} \mathrm{C}\) and \(\dot{m}_{h}=0.10 \mathrm{~kg} / \mathrm{s}\) is used to heat blood entering the exchanger at \(T_{c A}=18^{\circ} \mathrm{C}\) and \(\dot{m}_{c}=0.05 \mathrm{~kg} / \mathrm{s}\), what is the temperature of the blood leaving the exchanger? The overall heat transfer coefficient is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) The surgeon may wish to control the heat rate \(q\) and the outlet temperature \(T_{c, 0}\) of the blood by altering the flow rate and/or inlet temperature of the water during the rewarming process. To assist in the development of an appropriate controller for the prescribed values of \(\hat{m}_{c}\) and \(T_{c \jmath}\), compute and plot \(q\) and \(T_{c, \rho}\) as a function of \(\dot{m}_{h}\) for \(0.05 \leq \dot{m}_{\mathrm{h}} \leq 0.20 \mathrm{~kg} / \mathrm{s}\) and values of \(T_{h, l}=50,60\), and \(70^{\circ} \mathrm{C}\). Since the dominant influence on the overall heat transfer coefficient is associated with the blood flow conditions, the value of \(U\) may be assumed to remain at \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Should certain operating conditions be excluded?

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 \(5 \mathrm{~kg} / \mathrm{s} \leq \dot{m}_{c, \mathrm{~A}}=m_{h, \mathrm{~B}} \leq 50 \mathrm{~kg} / \mathrm{s}\).

A novel design for a condenser consists of a tube of thermal conductivity \(200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) with longitudinal fins snugly fitted into a larger tube. Condensing refrigerant at \(45^{\circ} \mathrm{C}\) flows axially through the inner tube, while water at a flow rate of \(0.012 \mathrm{~kg} / \mathrm{s}\) passes through the six channels around the inner tube. The pertinent diameters are \(D_{1}=10 \mathrm{~mm}, D_{2}=14 \mathrm{~mm}\), and \(D_{3}=50 \mathrm{~mm}\), while the fin thickness is \(t=2 \mathrm{~mm}\). Assume that the convection coefficient associated with the condensing refrigerant is extremely large. Determine the heat removal rate per unit tube length in a section of the tube for which the water is at \(15^{\circ} \mathrm{C}\).

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