1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
Select the correct answer.
Ulleksa [173]
The answer is A. Immediately inform her colleague
4 0
3 years ago
Given below are the measured streamflows in cfs from a storm of 6-hour duration on a stream having a drainage area of 185 mi^2.
sertanlavr [38]

Answer:

33.56 ft^3/sec.in

Explanation:

Duration = 6 hours

drainage area = 185 mi^2

constant baseflow = 550 cfs

<u>Derive the unit hydrograph using the inverse procedure </u>

first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below

Vdrh = sum of drh *  duration

        = 29700 * 6 hours ( 216000 secs )

        = 641,520,000 ft^3.

next step : Calculate the volume of runoff in equivalent depth

Vdrh / Area = 641,520,000  / 185 mi^2

                    = 1.49 in

Finally derive the unit hydrograph

Unit of hydrograph = drh /  volume of runoff in equivalent depth

                                = 50 ft^3 / 1.49 in  =  33.56 ft^3/sec.in

5 0
3 years ago
Two particles have a mass of 7.8 kg and 11.4 kg , respectively. A. If they are 800 mm apart, determine the force of gravity acti
aleksley [76]

Answer:

A) About 9.273 \times 10^{-9} newtons

B) 76.518 newtons

C) 111.834 newtons

Explanation:

A) F_g=\dfrac{GM_1M_2}{r^2} , where G is the universal gravitational constant, M 1 and 2 are the masses of both objects in kilograms, and r is the radius in meters. Plugging in the given numbers, you get:

F_g=\dfrac{(6.67408 \times 10^{-11})(7.8)(11.4)}{(0.8)^2}\approx 9.273 \times 10^{-9}

B) You can find the weight of each object on Earth because you know the approximate acceleration due to gravity is 9.81m/s^2. Multiplying this by the mass of each object, you get a weight for the first particle of 76.518 newtons.

C) You can do a similar thing to the previous particle and find that its weight is 11.4*9.81=111.834 newtons.

Hope this helps!

3 0
3 years ago
A highway reconstruction project is being undertaken to reduce crash rates. The reconstruction involves a major realignment of t
CaHeK987 [17]

Answer:

The provided length of the vertical curve is satisfactory for the reconstruction design speed of 60 mi/h

Explanation:

The explanation is shown on the first uploaded image

8 0
3 years ago
Line.
Veronika [31]

Air supplied to a pneumatic system is supplied through the C. Actuator

Explanation

Pneumatic systems are like hydraulic systems, it is just that these systems uses compressed air rather than hydraulic fluid.  Pneumatic systems are used widely across the industries. these pneumatic systems needs a constant supply of compressed air to operate. This is provided by an air compressor. The compressor sucks in air at a very high rate from the environment and stores it in a pressurized tank. the Air is supplied thereafter with the help of a actuator valve that is a more sophisticated form of a valve.

From the above statement it is clear that Air supplied to a pneumatic system is supplied through the  Actuator

7 0
3 years ago
Other questions:
  • 5. Assume that you and your best friend ench have $1000 to invest. You invest your money
    8·1 answer
  • function summedValue = SummationWithLoop(userNum) % Summation of all values from 1 to userNum summedValue = 0; i = 1; % Write a
    11·1 answer
  • Researchers compared protein intake among three groups of postmenopausal women: (1) women eating a standard American diet (STD),
    14·1 answer
  • 12. The small space above the piston in which fuel is burned is called the
    10·1 answer
  • How do I calculate the gear ratio​
    6·1 answer
  • Air is to be heated steadily by an 8-kW electric resistance heater as it flows through an insulated duct. If the air enters at 5
    10·1 answer
  • A nozzle receives an ideal gas flow with a velocity of 25 m/s, and the exit at 100 kPa, 300 K velocity is 250 m/s. Determine the
    14·1 answer
  • A column carries 5400 pounds of load and is supported on a spread footing. The footing rests on coarse sand. Design the smallest
    10·1 answer
  • .If aligned and continuous carbon fibers with a diameter of 6.90 micron are embedded within an epoxy, such that the bond strengt
    11·1 answer
  • What test should be performed on abrasive wheels
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!