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The specific gravity of gasoline is approximately 0.70. (a) Estimate the mass (kg) of 50.0 liters of gasoline. (b) The mass flow rate of gasoline exiting a refinery tank is \(1150 \mathrm{kg} / \mathrm{min}\). Estimate the volumetric flow rate in liters/s. (c) Estimate the average mass flow rate ( \(\left(\mathrm{lb}_{\mathrm{m}} / \mathrm{min}\right)\) delivered by a gasoline pump. (d) Gasoline and kerosene (specific gravity \(=0.82\) ) are blended to obtain a mixture with a specific gravity of 0.78. Calculate the volumetric ratio (volume of gasoline/volume of kerosene) of the two compounds in the mixture, assuming \(V_{\text {blend }}=V_{\text {gasoline }}+V_{\text {kerosene. }}\)

Short Answer

Expert verified
The mass of 50 liters of gasoline is 32.9 kg. The volumetric flow rate is 0.308 liters/second. The average mass flow rate delivered by a gasoline pump is 2536.6 lb_m/min. The volumetric ratio is 1.667.

Step by step solution

01

Calculate Density of Gasoline

Multiply the specific gravity of gasoline by the density of water at 4°C to get the density of gasoline. Use the formula: \(\rho_{gasoline} = specific\:gravity_{gasoline} \times \rho_{water}\), where density of water is 1000 kg/m³.
02

Estimate the mass of 50 liters of gasoline

To find the Mass, use the formula: Mass = Density x Volume. Therefore, \(Mass_{gasoline} = \rho_{gasoline} \times V_{gasoline}\). Here, Volume is given as 50 liters which needs to be converted to cubic meters by using the relation: 1 m³ = 1000 liters.
03

Estimate Volumetric Flow Rate

To find the volumetric flow rate, use the relation of Mass flow rate to Density and Volumetric flow rate, which is: \(\dot{M}_\text{gasoline} = \rho_\text{gasoline} \times \dot{V}_\text{gasoline}\). Therefore, \( \dot{V}_\text{gasoline} = \dot{M}_\text{gasoline} /\rho_\text{gasoline}\). Here, \(\dot{M}_\text{gasoline}\) is given as 1150 kg/min which needs to be converted to kg/s by using the relationship 1 min = 60 s.
04

Calculate Average Mass Flow Rate in lb_m/min

To find the average mass flow rate, use the conversion factor for kilograms to pound-mass which is 1 kg ≈ 2.205 lb_m. So the Average Mass Flow Rate will be \( \dot{M}_\text{gasoline} \times 2.205\).
05

Calculate Volumetric Ratio

To find the volumetric ratio, first, calculate the density of kerosene using: \(\rho_{kerosene} = specific\:gravity_{kerosene} \times \rho_{water}\). Then, use the formula of specific gravity of blend, calculate the ratio: \( specific\:gravity_{blend} = \frac{V_{gasoline} * \rho_{gasoline} + V_{kerosene} * \rho_{kerosene} }{V_{blend}* \rho_{water}}\). Assuming \(V_{blend} = V_{gasoline} + V_{kerosene}\), the volumetric ratio volume of gasoline/volume of kerosene ( \(V_{gasoline}/V_{kerosene}\) ) can be found.

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Key Concepts

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

Specific Gravity
Specific gravity, denoted as SG, is a fundamental concept often encountered in chemical engineering processes. It is a dimensionless quantity that represents the ratio of the density of a substance to the density of a reference substance, typically water at 4°C (the temperature at which water's density is highest, approximately 1000 kg/m³).

For example, the specific gravity of gasoline is approximately 0.70, indicating that gasoline is 70% as dense as water. Knowing the specific gravity is crucial for engineers calculating mass from a known volume, as shown in our exercise, or for determining buoyancy forces in fluid dynamics. Moreover, specific gravity is used to design separation processes like distillation where components of different densities are separated.
Volumetric Flow Rate
Volumetric flow rate is a measure of the volume of fluid that moves through a particular point in a conduit per unit time. This rate is crucial in chemical engineering for process design and analysis, such as calculating the throughput of material in pipelines or equipment.

The volumetric flow rate, typically represented by the symbol \( \dot{V} \), has units like liters per second (L/s) or cubic meters per hour (m³/hr). In the exercise, converting the mass flow rate to the volumetric flow rate requires knowledge of the fluid's density, which we deduce from the specific gravity—the lower the specific gravity, the higher the volume that a given mass of fluid occupies. This conversion is essential in operations where handling volumes is more practical than weighing masses.
Mass Flow Rate
Mass flow rate is defined as the mass of a substance that passes through a given surface per unit time. It is a significant parameter in both design and operation of chemical processes since it directly correlates to the 'amount' of material being processed.

In the context of our exercise, the mass flow rate, represented by the symbol \( \dot{M} \), is given in kilograms per minute (kg/min). Conversions to other units like pounds per minute (lbm/min) are often necessary to match the measuring instruments or industry standards. Understanding the mass flow rate also aids in the evaluation of reaction rates and the scaling up of laboratory processes to commercial production.
Chemical Process Calculations
Chemical process calculations are the backbone of process design and operation in chemical engineering. They involve using the principles of mass and energy balances, thermodynamics, and fluid mechanics to solve problems involving production or treatment of chemical substances.

In our scenario, calculations range from determining the mass of gasoline from its volume and specific gravity to deducing volumetric ratios in blending processes. These calculations require a clear understanding of unit conversions, relationship between different physical properties, and how they impact the process. Such proficiency allows engineers to design efficient, safe, and economical processes, and troubleshoot problems that may arise in a chemical plant. Properly executed chemical process calculations ensure that the desired outcomes, like the specific gravity in blending operations, are met with precision.

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Most popular questions from this chapter

A gas stream contains 18.0 mole \(\%\) hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is reduced and some of the hexane is liquefied. The hexane mole fraction in the gas stream leaving the condenser is \(0.0500 .\) Liquid hexane condensate is recovered at a rate of \(1.50 \mathrm{L} / \mathrm{min}\). (a) What is the flow rate of the gas stream leaving the condenser in mol/min? (Hint: First calculate the molar flow rate of the condensate and note that the rates at which \(C_{6} H_{14}\) and \(N_{2}\) enter the unit must equal the total rates at which they leave in the two exit streams.) (b) What percentage of the hexane entering the condenser is recovered as a liquid? (c) Suggest a change you could make in the process operating conditions to increase the percentage recovery of hexane. What would be the downside?

An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spilled on the arm connected to the line during a laboratory renovation, it is impossible to see the level of the manometer fluid in this arm. During a period when the gas supply is connected to the line but there is no gas flow, a Bourdon gauge connected to the line downstream from the manometer gives a reading of 7.5 psig. The level of mercury in the open arm is \(900 \mathrm{mm}\) above the lowest part of the manometer. (a) When the gas is not flowing, the pressure is the same everywhere in the pipe. How high above the bottom of the manometer would the mercury be in the arm connected to the pipe? (b) When gas is flowing, the mercury level in the visible arm drops by \(25 \mathrm{mm}\). What is the gas pressure (psig) at this moment?

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