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Chapter 5: Problem 14

A stream of air enters a \(7.50-\mathrm{cm}\) ID pipe with a velocity of \(60.0 \mathrm{m} / \mathrm{s}\) at \(27^{\circ} \mathrm{C}\) and 1.80 bar (gauge). At a point downstream, the air flows through a \(5.00 \mathrm{cm}\) ID pipe at \(60^{\circ} \mathrm{C}\) and 1.53 bar (gauge). What is the average velocity of the gas at this point.

Short Answer

Expert verified
After calculating the densities and areas using the given conditions and applying the mass flow rate conservation principle, the average velocity at the second point downstream can be found. The specific number will depend on the calculated densities and areas.

Step by step solution

01

Calculate the density at the first point

First, the ideal gas equation is used to calculate the air density (蟻1) at the initial temperature and pressure. The ideal gas equation is \( 蟻 = \frac{P}{RT} \), where P is the absolute pressure (absolute pressure = gauge pressure + atmospheric pressure), R is the specific gas constant for dry air (287 J/kgK) and T is the absolute temperature in Kelvin (T(K) = T(C) + 273).
02

Calculate the density at the second point

Next, we calculate the air density (蟻2) downstream at the second point under the conditions of 60 degrees Celsius and 1.53 bar (gauge pressure) using the same ideal gas equation.
03

Calculate the velocity at the second point

Since the mass flow rate is conserved, we equate the mass flow rate at point 1 and point 2 (i.e. \( 蟻1*A1*V1 = 蟻2*A2*V2 \)). Our aim is to find \( V2 \), which can be found by rearranging the equation to give \( V2 = \frac{蟻1*A1*V1}{蟻2*A2} \). Here, \( A = \frac{\pi *D^2}{4} \) is the cross-sectional area of the pipe at that point with D being the inner diameter. Therefore, we calculate \( A1 \) and \( A2 \), from their respective diameters, and finally solve for \( V2 \)

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Key Concepts

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

Understanding the Ideal Gas Law
When we talk about the ideal gas law, we're referring to a fundamental equation used to approximate the behavior of most gases under many conditions. It's given by the formula: \( PV = nRT \). Here's what each symbol represents:
  • \( P \) is the absolute pressure of the gas
  • \( V \) is the volume it occupies
  • \( n \) is the amount of substance in moles
  • \( R \) is the ideal gas constant
  • \( T \) is the temperature in Kelvin
For our purpose in flow velocity calculation, we modify the equation in terms of density (\( \rho \)), which gives us \( \rho = \frac{P}{RT} \) by incorporating the mass of the gas (m) and dividing by the volume (V), so the equation nR becomes m (mass) and thus \( \rho = \frac{Pm}{RTV} \) simplifies to \( \rho = \frac{P}{RT} \) when considering one unit of volume. This step is crucial as it allows us to relate pressure and temperature of the gas to its density, setting the stage for further calculations in determining flow velocity.
Mass Flow Rate Consistency in Pipe Flow
The concept of mass flow rate is central when dealing with fluid flow through pipes. It's a measure of the quantity of mass passing through a given surface per time unit and is constant for a closed system. The mass flow rate can be expressed as \( \dot{m} = \rho AV \), where:
  • \( \dot{m} \) is the mass flow rate,
  • \( \rho \) is the fluid density,
  • \( A \) is the cross-sectional area, and
  • \( V \) is the flow velocity.
From the conservation of mass, we know that the mass flow rate at the inlet \( \dot{m}_1 \) must equal the mass flow rate at the outlet \( \dot{m}_2 \) for an incompressible flow. That is, \( \rho_1 A_1 V_1 = \rho_2 A_2 V_2 \). In our exercise, we're using this principle to find the unknown velocity at the second point in the pipe system, leveraging the fact that despite changes in temperature and pressure, the quantity of mass flowing through must remain constant.
Density Calculation for Gases
Density calculation for gases is often done using the ideal gas law since it's easily influenced by changes in pressure and temperature. As applied in our exercise, the density of a gas is calculated with the formula \( \rho = \frac{P}{RT} \) as mentioned earlier. The first step involves determining the absolute pressure, which requires adding atmospheric pressure to the gauge pressure, as gauge pressure is measured relative to atmospheric pressure. Next, we consider the specific gas constant for air, and the temperature is adjusted to Kelvin by adding 273 to the Celsius value.

With these values, we can calculate the density at both initial and downstream conditions, taking note of the fact that air density decreases with increases in temperature or decreases in pressure. Therefore, any change to either of these conditions will lead to a change in air's density. This variable is vital for not only establishing the mass flow rate but also in various industrial applications where gas volume and pressure readings are crucial for safe and efficient operations.

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Most popular questions from this chapter

A process stream flowing at \(35 \mathrm{kmol} / \mathrm{h}\) contains 15 mole \(\%\) hydrogen and the remainder 1 -butene. The stream pressure is 10.0 atm absolute, the temperature is \(50^{\circ} \mathrm{C}\), and the velocity is \(150 \mathrm{m} / \mathrm{min}\). Determine the diameter (in \(\mathrm{cm}\) ) of the pipe transporting this stream, using Kay's rule in your calculations.

The concentration of oxygen in a 5000 -liter tank containing air at 1 atm is to be reduced by pressure purging prior to charging a fuel into the tank. The tank is charged with nitrogen up to a high pressure and then vented back down to atmospheric pressure. The process is repeated as many times as required to bring the oxygen concentration below 10 ppm (i.c., to bring the mole fraction of \(\mathrm{O}_{2}\) below \(10.0 \times 10^{-6}\) ). Assume that the temperature is \(25^{\circ} \mathrm{C}\) at the beginning and end of each charging cycle. When doing \(P V T\) calculations in Parts (b) and (c), use the generalized compressibility chart if possible for the fully charged tank and assume that the tank contains pure nitrogen. (a) Speculate on why the tank is being purged. (b) Estimate the gauge pressure (atm) to which the tank must be charged if the purge is to be done in one charge-vent cycle. Then estimate the mass of nitrogen (kg) used in the process. (For this part, if you can't find the tank condition on the compressibility chart, assume ideal-gas behavior and state whether the resulting estimate of the pressure is too high or too low.) (c) Suppose nitrogen at 700 kPa gauge is used for the charging. Calculate the number of charge-vent cycles required and the total mass of nitrogen used. (d) Use your results to explain why multiple cycles at a lower gas pressure are preferable to a single cycle. What is a probable disadvantage of multiple cycles?

As everyone who has used a fireplace knows, when a fire burns in a furnace, a draft, or slight vacuum, is induced that causes the hot combustion gases and entrained particulate matter to flow up and out of the stack. The reason is that the hot gas in the stack is less dense than air at ambient temperature, leading to a lower hydrostatic head inside the stack than at the furnace inlet. The theoretical draft \(D\left(\mathrm{N} / \mathrm{m}^{2}\right)\) is the difference in these hydrostatic heads; the actual draft takes into account pressure losses undergone by the gases flowing in the stack. Let \(T_{\mathrm{s}}(\mathrm{K})\) be the average temperature in a stack of height \(L(\mathrm{m})\) and \(T_{\mathrm{a}}\) the ambient temperature, and let \(M_{\mathrm{s}}\) and \(M_{\mathrm{a}}\) be the average molecular weights of the gases inside and outside the stack. Assume that the pressures inside and outside the stack are both equal to atmospheric pressure, \(P_{\mathrm{a}}\left(\mathrm{N} / \mathrm{m}^{2}\right)\) (In fact, the pressure inside the stack is normally a little lower.) (a) Use the ideal-gas equation of state to prove that the theoretical draft is given by the expression $$D\left(\mathrm{N} / \mathrm{m}^{2}\right)=\frac{P_{\mathrm{a}} L g}{R}\left(\frac{M_{\mathrm{a}}}{T_{\mathrm{u}}}-\frac{M_{\mathrm{s}}}{T_{\mathrm{s}}}\right)$$ (b) Suppose the gas in a 53 -m stack has an average temperature of \(655 \mathrm{K}\) and contains 18 mole\% \(\mathrm{CO}_{2}\) \(2 \% \mathrm{O}_{2},\) and \(80 \% \mathrm{N}_{2}\) on a day when barometric pressure is \(755 \mathrm{mm}\) Hg and the outside temperature is \(294 \mathrm{K}\). Calculate the theoretical draft \(\left(\mathrm{cm} \mathrm{H}_{2} \mathrm{O}\right)\) induced in the furnace.

An oxygen tank with a volume of \(2.5 \mathrm{ft}^{3}\) is kept in a room at \(50^{\circ} \mathrm{F}\). An engineer has used the idealgas equation of state to determine that if the tank is first evacuated and then charged with \(35.3 \mathrm{lb}_{\mathrm{m}}\) of pure oxygen, its rated maximum allowable working pressure (MAWP) will be attained. Operation at pressures above this value is considered unsafe. (a) What is the maximum allowable working pressure (psig) of the tank? (b) You suspect that at the conditions of the fully charged tank, ideal-gas behavior may not be a good assumption. Use the SRK equation of state to obtain a better estimate of the maximum mass of oxygen that may be charged into the tank. Did the ideal-gas assumption lead to a conservative estimate (on the safe side) or a nonconservative estimate of the amount of oxygen that could be charged? (c) Suppose the tank is charged and ruptures before the amount of oxygen calculated in Part (b) enters it. (It should have been able to withstand pressures up to four times the MAWP.) Think of at least five possible explanations for the failure of the tank below its rated pressure limit.

The product gas from a coal gasification plant consists of 60.0 mole \(\%\) CO and the balance \(\mathrm{H}_{2}\); it leaves the plant at \(150^{\circ} \mathrm{C}\) and 135 bar absolute. The gas expands through a turbine, and the outlet gas from the turbine is fed to a boiler furnace at \(100^{\circ} \mathrm{C}\) and 1 atm at a rate of \(425 \mathrm{m}^{3} / \mathrm{min}\). Estimate the inlet flow rate to the turbine in \(\mathrm{ft}^{3} / \mathrm{min},\) using Kay's rule. What percentage error would result from the use of the ideal-gas equation of state at the turbine inlet?
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