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
In the construction of a large reactor pressure vessel, a new steel alloy with a plane strain fracture toughness of 55 MPa-m1/2
sp2606 [1]

Answer:

l=24mm

Explanation:

From the question we are told that:

Plane strain fracture toughness of T=55 MPa-m1/2

Y value Y=1.0

Stress level of\sigma =200 MPa

Generally the equation for length of a surface crack is mathematically given by

l=\frac{1}{\pi}(\frac{T}{Y*\sigma})^2

l=\frac{1}{3.142}(\frac{55}{1*200})^2

l=0.024m

Therefore

in mm

l=24mm

6 0
3 years ago
A complete mix of an activated sludge system without primary clarification is used for treatment of municipal wastewater with a
Hitman42 [59]

Answer:

sorry di ko alam

Explanation:

4 0
2 years ago
I need your help to answer in the picture plss help me!​
Aloiza [94]

this looks like its the different phases of a single cylinder 4 stroke engine what are you doing in the picture or assignment though matching the numbers to the descriptions on the side?

3 0
3 years ago
Calculate the molar heat capacity of a monatomic non-metallic solid at 500K which is characterized by an Einstein temperature of
aleksandr82 [10.1K]

Answer:

Explanation:

Given

Temperature of solid T=500\ K

Einstein Temperature T_E=300\ K

Heat Capacity in the Einstein model is given by

C_v=3R\left [ \frac{T_E}{T}\right ]^2\frac{e^{\frac{T_E}{T}}}{\left ( e^{\frac{T_E}{T}}-1\right )^2}

e^{\frac{3}{5}}=1.822

Substitute the values

C_v=3R\times (\frac{300}{500})^2\times (\frac{1.822}{(1.822-1)^2})

C_v=3R\times \frac{9}{25}\times \frac{1.822}{(0.822)^2}

C_v=0.97\times (3R)            

6 0
3 years ago
While discussing what affects the amount of pressure exerted by the brakes: Technician A says that the shorter the line, the mor
harina [27]

Answer:

Only Technician B is right.

Explanation:

The cylindrical braking system for a car works through the mode of pressure transmission, that is, the pressure applied to the brake pedals, is transmitted to the brake pad through the cylindrical piston.

Pressure applied on the pedal, P(pedal) = P(pad)

And the Pressure is the applied force/area for either pad or pedal. That is, P(pad) = Force(pad)/A(pad) & P(pedal) = F(pedal)/A(pedal)

If the area of piston increases, A(pad) increases and the P(pad) drops, Meaning, the pressure transmitted to the pad reduces. And for most cars, there's a pressure limit for the braking system to work.

If the A(pad) increases, P(pad) decreases and the braking force applied has to increase, to counter balance the dropping pressure and raise it.

This whole setup does not depend on the length of the braking lines; it only depends on the applied force and cross sectional Area (size) of the piston.

5 0
3 years ago
Other questions:
  • A water pump delivers 6 hp of shaft power when operating. The pressure differential between the outlet and the inlet of the pump
    9·1 answer
  • Using Von Karman momentum integral equation, find the boundary layer thickness, the displacement thickness, the momentum thickne
    14·1 answer
  • What are the causes of kickback on a table-saw?
    13·1 answer
  • What is centrifugal force with respect to unbalance? What is the formula used to determine centrifugal force?
    12·1 answer
  • A piston/cylinder contains 1.5 kg of water at 200 kPa, 150°C. It is now heated by a process in which pressure is linearly relate
    14·1 answer
  • Module 42 Review and Assessment
    7·1 answer
  • Water discharging into a 10-m-wide rectangular horizontal channel from a sluice gate is observed to have undergone a hydraulic j
    12·1 answer
  • What is the angular velocity (in rad/s) of a body rotating at N r.p.m.?
    13·1 answer
  • Implement a program that manages shapes. Implement a class named Shape with a method area() which returns the double value 0.0.
    8·1 answer
  • What is a splitter gearbox​
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!