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mars1129 [50]
3 years ago
13

A coal-fired power plant is burning bituminous coal that has an energy content of 12,000 Btu/lb. The power plant is burning the

coal at a rate of 110 lb/s. The efficiency of the powerplant is 33%, which means that 67% of the total energy from the coal is lost as waste heat.
a. A new heat exchanger is installed that is capable of using 0.12% of the waste heat to heat an air stream for use in another process in the plant. The air enters the heat exchanger at a volumetric flow rate of 2,000 acfm at 298 K and 1 atm. Assuming the heat exchanger is 90% efficient (i.e., 90% of the heat supplied to the heat exchanger is transferred to the air stream), at what temperature will the air leave the heat exchanger, assuming an exit pressure of 1 atm? Use an average specific heat of 0.244 Btu/lb-°R.
b. What will be the volumetric flow rate of the gas leaving the heat exchanger (in units of acfm)?
c. If the air exiting the heat exchanger has a moisture content of 12% (v/v), what is the moisture-corrected dry flow rate (in units of dry standard cubic meters per minute) at STP (1 atm, 298 K)?
Engineering
1 answer:
dybincka [34]3 years ago
3 0

Answer:

a) T_{out} = 2190.455 ^{\textdegree}R b) \dot V_{air,out} = 8352.941 acfm  c) \dot V_{air,out,corr} = 7350.588 afcm

Explanation:

a) The heat lost by the power plant is:

\dot Q_{loss} = (0.67) \cdot (110 \frac{lbm}{s} ) \cdot (12000 \frac{BTU}{lbm} )\\\dot Q_{loss} = 884400 BTU

The waste heat used by heat exchanger is:

\dot Q_{used} = (0.0012) \cdot \dot Q_{loss}\\\dot Q_{used} = 1061.28 BTU

Assuming that air behaves as an ideal gas, density is given by following expression:

\rho_{air} = \frac{P \cdot M_{air}}{R_{u} \cdot T}

\rho_{air} = \frac{(101.325 kPa) \cdot (28 \frac{kg}{kmol})}{(8.314 \frac{kPa\cdot m^{3}}{kmol \cdot K} )(298 K)}\\\rho_{air} = 1.145 \frac{kg}{m^{3}}

The density unit is converted to pounds per cubic feet:

\rho_{air} = 1.145 \frac{kg}{m^{3}} \cdot (\frac{1 lb}{0.453 kg}) \cdot (\frac{0.304 m}{1 ft} )^{3} \\\rho_{air} = 0.071 \frac{lb}{ft^{3}}

The heat received by air flow through heat exchanger is:

\dot Q_{air} = (0.90) \cdot \dot Q_{used}\\\dot Q_{air} = 955.152 BTU

Outlet temperature can isolated from the following formula:

\dot Q_{air} = \rho_{air} \cdot \dot V_{air} \cdot c_{p,air} \cdot (T_{out} - T_{in})

T_{out} =T_{in} + \frac{\dot Q_{air}}{\rho_{air} \cdot \dot V_{air} \cdot c_{p,air}}

T_{out} = 536.4 ^{\textdegree}R + \frac{955.152 BTU}{(0.071\frac{lb}{ft^{3}})\cdot (33.333 \frac{ft}{s} )\cdot(0.244 \frac{BTU}{lb ^{\textdegree}R})}\\T_{out} = 2190.455 ^{\textdegree}R

b) Due to the compressibility of air, density has to be calculated by using the approach from section a).

\rho_{air,out} = \frac{(101.325 kPa) \cdot (28 \frac{kg}{kmol})}{(8.314 \frac{kPa\cdot m^{3}}{kmol \cdot K} )(1217 K)}\\\rho_{air,out} = 0.280 \frac{kg}{m^{3}}

\rho_{air,out} = 0.017 \frac{lb}{ft^{3}}

Volumetric flow can be found by Principle of Mass Conservation:

\dot V_{air, out} = \frac{\rho_{air}}{\rho_{air,out}}\cdot \dot V_{air}

\dot V_{air,out} = 8352.941 acfm

c) The moisture component indicates that 88 percent of volume is occupied by dry-air. The moisture-corrected dry flow rate is:

\dot V_{air,out,corr} = 0.88 \cdot (8352.941 afcm)\\\dot V_{air,out,corr} = 7350.588 afcm

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Answer:

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Explanation:

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The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.

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Now remember that mass flow is given by the product of density, cross-sectional area and velocity

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3 years ago
Determine the general lighting load for a two-story office building that measures 125 feet by 150 feet.
Irina-Kira [14]

The general lighting load for a two-story office building that measures 125 feet by 150 feet is 112, 500 sq ft.

<h3>What is lighting load?</h3>

Lighting loads are the energy used to power electric lights and they make up nearly a third of US commercial building energy use.

Lighting load = n(LW)

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For one story building, = 3

For two story building, n = 6

Lighting load = 6 x 125 x 150 = 112, 500 sq ft.

Learn more about lighting load here: brainly.com/question/14070748

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Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 440°C.
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Explanation:

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For a bronze alloy, the stress at which plastic deformation begins is 275 MPa, and the modulus of elasticity is 115 GPa. (a) Wha
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Answer:

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Explanation:

Rearrange the formula for normal stress for F:

\sigma=\frac{F}{A}

F=\sigma*A

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325 mm^{2} = 0.000325 m^{2}

Substituting in given values:

F = (275*10^{6})*(0.000325)=89375 N

3 0
2 years ago
Why is it important to know the accuracy and precision of a measuring device? Do you think that the dial caliper manufacturer’s
Leto [7]

Answer:

Accuracy and precision allow us to know how much we can rely on a measuring device readings. ±.001 as a "accuracy" claim is vague because there is no unit next to the figure and the claim fits better to the definition of precision.

Explanation:

Accuracy and Precision: the golden couple.

Accuracy and precision are key elements to define if a measuring device is reliable or not for a specific task. Accuracy determines how close are the readings from the ideal/calculated values. On the other hand, precision refers to repeatability, that is to say how constant the readings of a device are when measuring the same element at different times. One of those two key concepts may not fulfill the criteria for measuring tool to be used on certain engineering projects where lack of accuracy (disntant values from real ones) or precision (not constant readings) may lead to malfunctons and severe delays on the project development.

±.001 what unit?

The manufacturer says that is an accuracy indicator, nevertheless there is now unit stated so this is not useful to see how accurate the device is. Additionally, That notation is more used to refer to device tolerances, that is to say the range of possible values the instrument may show when reading and element. It means it tells us more about the device precision during measurments than actual accuracy. I would recommend the following to the dial calipers manufacturers to better explain its measurement specifications:

  1. Use  ±.001 as  a reference for precision. It is important to add the respective unit for that figure.
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