ENV 3001 Introduction to Environmental Engineering - PowerPoint PPT Presentation

1 / 29
About This Presentation
Title:

ENV 3001 Introduction to Environmental Engineering

Description:

Iso-Octane balance: 0.92 G1 0.63 G2 = 0.83 (10,000) G2= 900/0.29 = 3,103 gal of straight run. G1 = 10,000 3,103 = 6,897 gal of Iso-Octane. HW 3.15 ... – PowerPoint PPT presentation

Number of Views:393
Avg rating:5.0/5.0
Slides: 30
Provided by: nch2
Category:

less

Transcript and Presenter's Notes

Title: ENV 3001 Introduction to Environmental Engineering


1
ENV 3001 Introduction to Environmental Engineering
Lecture III (b) Review Session Instructor Dr.
Ni-Bin Chang Spring, 2006
2
HW 3.1
Five million kilograms per day of coal are
burned in an electric power plant. The coal
has an ash content of 12 by mass. Forty percent
of the ash falls out the bottom of the furnace.
The rest of the ash is carried out of the
furnace with the hot gases into an electrostatic
precipitator (ESP). The ESP is 99.5 efficient
in removing the ash that comes into it. Draw a
diagram representing this process and calculate
the mass emissions rate of ash into the
atmosphere from this plant.
3
HW 3.1
4
HW 3.1
Solution
5
HW 3.3
clarifier
thickener
The above is a clarifier-thickener system to
separate solids and liquids. Ignore the effect of
solids content on density of the streams (i.e.,
assume density of solids density water).
6
HW 3.3
  • Complete the following material balance table.
  • Stream Flow rate, L/s Solids, mg/L
  • A 100 3000
  • B 95 15
  • C 6000
  • D 50
  • E
  • A

7
HW 3.3
  • First Mass Balance

8
HW 3.3
  • Second Mass Balance

9
HW 3.3
  • Third Mass Balance

10
HW 3.4
  • A 600 MW coal burning power plant is burning
    Illinois bituminous coal with 8 ash content.
  • The plant is 39 efficient.
  • 35 of the ash drops out in the furnace
  • The electrostatic precipitator is 99.0
    efficient.
  • A simplified sketch appears below

11
HW 3.4
  • a. Draw an energy balance diagram for the plant
    and calculate the rate of heat emitted to the
    environment, in J/s.
  • b. Calculate the rate of coal input to the
    furnace, in kg/day.
  • c. Calculate the rate of ash emissions to the
    atmosphere, in kg/s.

12
HW 3.4
13
HW 3.4
  • Energy Balance
  • Waste heat1579-600979MW

Heat in
14
HW 3.4
  • Coal input
  • Ash to atmosphere

15
HW 3.7
  • An electrostatic precipitator removes 1000 kg of
    particulates per day.
  • The stack gas from a process contains 4.0 g
    particulates per m3 of gas.
  • The gas flow rate is 5 m3/s.
  • What is the emission rate of particulates? Give
    answer in kg/day.

16
HW 3.7
INOUT or ABC CA-B
17
HW 3.9
  • A mining processes a silty water into a clean
    brook at the rate of 5 m3/min
  • The silt content is 200 mg/L.
  • The brook flows at 50 m3/min in rainy season and
    5 m3/min in dry season.
  • What is the concentration in the brook during the
    rainy and dry seasons?

18
HW 3.9
In rainy season 50m3/min5 m3/min55 m3/min 50
(0)5(200)55(Ccs) Ccs18.2 mg/L In dry season
5m3/min5 m3/min10 m3/min 5 (0)5(200)10(Ccs) Cc
s100 mg/L
19
HW 3.10
  • The efficiently of a 800 MW nuclear power plant
    is 32 percent.
  • Assume no heat is lost to the atmosphere directly
    from the system.
  • Calculate the minimum rate at which cooling water
    must be pumped through the condensers of the
    plant if the maximum cooling water temperature
    rise is 15 0F.
  • Assuming the plant is designed with five
    identical pipes in parallel to carry this cooling
    water, calculate the diameter of each pipe if the
    water velocity is 6.0 ft/sec.
  • Give your answer in pounds/hr, and ft3/sec.

20
HW 3.10
  • Coal input
  • heat in800/0.32 2500MW
  • heat out 2500-8001700MW

21
HW 3.11
  • A new regulation requires that the wastewater
    from a tannery be routed to a municipal
    wastewater treatment plant for final treatment.
  • The tannery wastewater must be pretreated to
    remove certain organic pollutants.
  • The pretreatment process is a biological
    anaerobic process that has a limit on total
    dissolved solids (TDS) of no more than 20,000
    mg/L.

22
HW 3.11
23
HW 3.11
Rearrange the schematics as below
QM
24
HW 3.11
  • Overall Balance

25
HW 3.11
Point I balance
QM
26
HW 3.11
  • Mixing Point 1 Balance

27
HW 3.13
G1
G2
Straight-run gasoline
High-Octane gasoline
G3
Overall balance ?G1 ?G2 ?G3 10,000 gal
28
HW 3.13
  • Iso-Octane balance
  • 0.92 G1 0.63 G2 0.83 (10,000)
  • G2 900/0.29 3,103 gal of straight run
  • G1 10,000 3,103 6,897 gal of Iso-Octane

29
HW 3.15
(a)
  • (b) 2A?B, so for each B created , 2A are
    destroyed
  • Destruction rate of A is difference between input
    of A its output.
  • 100(0.10)-100(0.05)5moles A/min being destroyed
Write a Comment
User Comments (0)
About PowerShow.com