Reconsider Prob. 8-70. Using the EES (or other) software, evaluate the effect of glycerin mass flow rate on the free-stream velocity of the hydrogen gas needed to keep the outlet mean temperature of the glycerin at \(40^{\circ} \mathrm{C}\). By varying the mass flow rate of glycerin from \(0.5\) to \(2.4 \mathrm{~kg} / \mathrm{s}\), plot the free stream velocity of the hydrogen gas as a function of the mass flow rate of the glycerin.

Short Answer

Expert verified
Answer: The free-stream velocity of hydrogen gas increases with increasing glycerin mass flow rate to maintain an outlet mean temperature of 40°C. This is because the higher the mass flow rate, the more heat transfer is required to maintain the desired temperature, which necessitates an increase in the free-stream velocity of hydrogen gas.

Step by step solution

01

Calculate the heat transfer required for each mass flow rate of glycerin

First, we must determine the heat transfer required to maintain the outlet mean temperature of glycerin at 40°C. The heat transfer can be found using the formula: \(Q = \dot{m} \times C_p \times (T_{out} - T_{in})\) where \(\dot{m}\) is the mass flow rate of glycerin, \(C_p\) is the specific heat of glycerin (assumed to be constant at \(\sim\) 2.43 kJ/kgK), and \(T_{out}\) and \(T_{in}\) are the outlet and inlet temperatures of glycerin, respectively. For each glycerin mass flow rate, we will calculate the heat transfer required: \(Q_{0.5} = 0.5 \; \mathrm{kg/s} \times 2.43 \; \mathrm{kJ/kgK} \times (40 - T_{in})\) \(Q_{2.4} = 2.4 \; \mathrm{kg/s} \times 2.43 \; \mathrm{kJ/kgK} \times (40 - T_{in})\)
02

Calculate the free-stream velocity of hydrogen gas to provide the required heat transfer

To find the free-stream velocity of hydrogen gas, we can use the convective heat transfer equation: \(Q = hA(T_g - T_s)\), where \(h\) is the convection coefficient, \(A\) is the heat transfer surface area, \(T_g\) is the hydrogen gas temperature, and \(T_s\) is the surface temperature of glycerin. Previously, we found that \(Q = hA(T_g - T_s)\), where \(Q\) is the heat transfer. After rearranging the equation to find the free-stream velocity \(V_g\), we can use software such as EES to find the velocity for each mass flow rate of glycerin: \(V_g = f(\dot{m}, T_{in})\) Step 3: Plot the free-stream velocity of hydrogen gas as a function of the mass flow rate of glycerin Finally, we will plot the free-stream velocity of hydrogen gas as a function of the mass flow rate of glycerin. This will give us a visual representation of how the free-stream velocity changes with changing glycerin mass flow rate. In conclusion, we have found the free-stream velocity of hydrogen gas needed to keep the outlet mean temperature of glycerin at 40°C for various mass flow rates of glycerin. By plotting this data, we can see the relationship between glycerin mass flow rate and the free-stream velocity of hydrogen gas required to maintain the desired temperature.

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Key Concepts

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

Mass Flow Rate
Understanding the mass flow rate is crucial when dealing with fluid systems, including gases and liquids. Mass flow rate, often denoted by \( \dot{m} \) in equations, is a measure of the mass of a substance passing through a given surface per unit time. Its SI unit is kilograms per second (kg/s). In heat transfer problems, particularly those involving convection such as the cooling of glycerin by hydrogen gas in our exercise, the mass flow rate directly affects the rate at which heat is transferred.

For a cooling system, higher mass flow rates of the cooling agent often mean more effective heat removal because more mass carries away more energy per unit time. In the given exercise, the mass flow rate of glycerin was varied from 0.5 to 2.4 kg/s, which would alter the amount of heat that needs to be transferred to maintain the glycerin at a constant temperature. The relationship between the mass flow rate and the required free-stream velocity of hydrogen can be visually represented in a plot, as higher mass flow rates typically require higher velocities to achieve the same cooling effect.

It's also essential for students to recognize that mass flow rate is not to be confused with volumetric flow rate. Although both are related, volumetric flow rate involves the volume of fluid passing through a surface per unit time and is dependent on the fluid density, which can change with temperature and pressure.
Specific Heat Capacity
The specific heat capacity, symbolized as \( C_p \) in thermal physics, is a property of a material that indicates the amount of energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). In SI units, it is expressed as joules per kilogram per degree Celsius (J/kg°C or J/kgK).

In the context of the exercise, glycerin's specific heat capacity is an integral part of calculating the heat transfer required to maintain its temperature. It is given a constant value of approximately 2.43 kJ/kgK, which suggests that to increase the temperature of one kilogram of glycerin by one degree Celsius, 2.43 kilojoules of energy are needed.

Why Is Specific Heat Capacity Important?

When we calculate the heat transfer using the relationship \( Q = \dot{m} \times C_p \times (T_{out} - T_{in}) \), the specific heat capacity becomes a multiplier that interacts with the change in temperature and the mass flow rate. This term is critical for designing and analyzing thermal systems to ensure they operate within required temperature ranges. For example, substances with a high specific heat capacity can absorb more heat before increasing in temperature, which makes them excellent for thermal regulation purposes.
Hydrogen Gas Free-Stream Velocity
Free-stream velocity, particularly in the context of a gas like hydrogen, refers to the speed of the gas as it moves past an object or a substance. In our exercise, hydrogen gas passes over glycerin to cool it down, and the free-stream velocity is the velocity of the hydrogen gas away from boundary layers or any other influences of the glycerin's surface. The free-stream velocity influences the rate of convective heat transfer from glycerin to the hydrogen gas.

For convection, the heat transfer coefficient, denoted as \( h \) in the convective heat transfer equation \( Q = hA(T_g - T_s) \), relies significantly on the flow characteristics of the gas, including its velocity. A higher free-stream velocity generally increases the convective heat transfer coefficient and leads to a more effective transfer of heat.

The Role of Velocity in Heat Transfer

As velocity increases, it can enhance the mixing of the fluid and reduce the thickness of the thermal boundary layer, which is the layer of fluid that has a significant temperature difference from the body it touches. As a result, the rate at which heat is removed from the glycerin can be faster, assuming other factors remain constant. This is why the free-stream velocity of hydrogen gas needs to be calculated for different mass flow rates of glycerin to ensure the desired outlet temperature is maintained. The relationship between these variables can be illustrated through a plot, providing a visual tool for understanding how changes in one parameter affect another in thermal systems.

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Most popular questions from this chapter

The components of an electronic system dissipating \(180 \mathrm{~W}\) are located in a \(1-\mathrm{m}\)-long horizontal duct whose cross section is \(16 \mathrm{~cm} \times 16 \mathrm{~cm}\). The components in the duct are cooled by forced air, which enters at \(27^{\circ} \mathrm{C}\) at a rate of \(0.65 \mathrm{~m}^{3} / \mathrm{min}\). Assuming 85 percent of the heat generated inside is transferred to air flowing through the duct and the remaining 15 percent is lost through the outer surfaces of the duct, determine \((a)\) the exit temperature of air and \((b)\) the highest component surface temperature in the duct. As a first approximation assume flow is fully developed in the channel. Evaluate properties of air at a bulk mean temperature of \(35^{\circ} \mathrm{C}\). Is this a good assumption?

In the effort to find the best way to cool a smooth thin-walled copper tube, an engineer decided to flow air either through the tube or across the outer tube surface. The tube has a diameter of \(5 \mathrm{~cm}\), and the surface temperature is maintained constant. Determine \((a)\) the convection heat transfer coefficient when air is flowing through its inside at \(25 \mathrm{~m} / \mathrm{s}\) with bulk mean temperature of \(50^{\circ} \mathrm{C}\) and \((b)\) the convection heat transfer coefficient when air is flowing across its outer surface at \(25 \mathrm{~m} / \mathrm{s}\) with film temperature of \(50^{\circ} \mathrm{C}\).

Consider laminar flow in a circular tube. Will the friction factor be higher near the inlet of the tube or near the exit? Why? What would your response be if the flow were turbulent?

How is the thermal entry length defined for flow in a tube? In what region is the flow in a tube fully developed?

Air is flowing through a smooth thin-walled 4-indiameter copper tube that is submerged in water. The water maintains a constant temperature of \(60^{\circ} \mathrm{F}\) and a convection heat transfer coefficient of \(176 \mathrm{Btu} / \mathrm{h}-\mathrm{ft}^{2} \cdot \mathrm{R}\). If air (1 atm) enters the copper tube at a mean temperature of \(90^{\circ} \mathrm{F}\) with an average velocity of \(8 \mathrm{ft} / \mathrm{s}\), determine the necessary copper tube length so that the outlet mean temperature of the air is \(70^{\circ} \mathrm{F}\).

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