IndWW-HW#1-FlowCalc-answers
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Mechanical Engineering
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Apr 3, 2024
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ANSWER KEY
IndWW HW#1 - Spring 2023
The below columns will auto-populate from the spreadsheets.
pts
In the spreadsheets, the yellow data is required. The light orange blocks are for showing work (and will be used for partial credit).
12
Q1
Water Meter Supply
6,194 gpd-cal
Pts: 3 work
8,613 gpd-cal
Pts: 3 subtract vol
7,357 gpd-cal
Pts: 3 convert units
7,688 gpd-cal
Pts: 3 divide days
12,464 gpd-cal
11,345 gpd-cal
10,034 gpd-cal
13,645 gpd-cal
13,393 gpd-cal
13,367 gpd-cal
15,656 gpd-cal
15,600 gpd-cal
3
Q2
Design Basis
14,332 gpd-cal (if quick avg Feb-June
14,358 exact Feb-June
15,628 gpd-cal (if quick avg May-Jun
15,629 exact May-June
3
Explanation/judgement:
5
Q3 Sanitary Flow
3,946 gpd-cal
Would accept 3,946 gpd-cal or roundings to 3,900 gpd-cal (sig figure) or 4,000 gpd-cal (eng judgement of design basis).
4
Q4
WCT1 Evap Loss/Drift
504 gpd-cal
4
WCT1 Blowdown
576 gpd-cal
4
WCT4 Evap Loss/Drift
161 gpd-cal
Evap
Sewer
Haul
5
Q5
Source 1 - Spray Station
66 gpd-cal
5
Source 2 - Parts Washer
154 gpd-cal
5
Source 3 - Dip Tank
30
9 gpd-cal
5
Source 4 - Scrubber
400
720 gpd-cal
30
Q6
Flow balance
6 - 2 each, outfalls totaled
5
Q7
Supply (upstream-metered)
13,694
Usage (dnstream-estimated)
15,902
Ratio: dnstream to upstrea
116%
5
Q8
Agree?
Decent agreement, but would prefer tighter control to within 10%.
What to adjust?
5
Q9
Meter agreement?
2
Q10A
Bonus/Gas Meter Read
238,800 CF
2
Q10B
Bonus/Gas Meter Read
2,388 CCF
2
Q10C
Bonus/Gas BTU
1,030 BTU/CF
2
Q10D
Bonus/Evap Hours
20.6 hrs/day
2
Q10E
Bonus/Evap Flow
1,298 gpd
110
Total
Because production is going up, neglect 2013 and 2012 data. Could justify Feb-Jun or May/Jun avg. Rounded: 14,500 to 16,000, w/matching explanation.
Submit on sheet Q6-Q9; also submit a pdf of the flow balance Q6-
Q9 sheet (fit to page)
Adjust the sanitary flow. The other sources do not account for sufficient amounts of flow to cover a 15,902-13,694 = 2,208 gpd difference, when considered individually.
Adjust sanitary from 25 to 20 gpd-person; decreases sanitary from 9,500 gpd to 7,600 gpd, or a 1,900 gpd difference. Balance would then be: 15902-1900=14002. Ratio 14002/13694 = 102%
Conclusion: you did a very good job on your field measurements. Your field measurements showed 4,414 gpd and the meter is showing 4,387 gpd. Dnstream to upstream: 4414/4387 = 101%
IndWW HW#1 - Spring 2023
Question #1 & 2
Scenario: Your field visit is in July 2014 and you are preparing a flow balance for use in a WWTP upgrade
because production has expanded. The following water meter data is available.
Question 1: First, complete the table below (calculate yellow columns: TGL and gpd-cal). You may use the wh
After reviewing the data in Question 1, answer Question 2 (see below this table).
(if you use Column F for related calcs; label the header)
4/16/12
31
3449
3641
192
192,000
6,194
5/17/12
31
3641
3908
267
267,000
8,613
6/14/12
28
3908
4114
206
206,000
7,357
7/16/12
32
4114
4360
246
246,000
7,688
3/19/13
28
6649
6998
349
349,000
12,464
4/17/13
29
6998
7327
329
329,000
11,345
5/16/13
29
7327
7618
291
291,000
10,034
2/18/14
31
788
1211
423
423,000
13,645
3/18/14
28
1211
1586
375
375,000
13,393
4/17/14
30
1586
1987
401
401,000
13,367
5/19/14
32
1987
2488
501
501,000
15,656
6/18/14
30
2488
2956
468
468,000
15,600
Question 2: For the scenario described at top of page, and based on the data above, determine an appropriate
Hint: Engineering judgement is required. Answer:
14,332 gpd-cal
If avg Feb-June month
15,628 gpd-cal
If avg May/June mont
Explain your judgement:
Because production is going up, neglect 2013 and 2012 data. Could justify Fe
Read Date
Elapsed Time (Days)
Prior Reading (TGL)
Current Reading (TGL)
Consumption (TGL)
Consump
tion (gal)
Water Usage (gpd-cal)
hite columns as needed for intermediate formulas.
e design basis. hly
14,358 If correct/adjust for meter dates, the avg Feb-June
thly
15,629 If correct/adjust for meter dates, the avg May-June
eb-Jun or May/Jun avg. Rounded: 14,500 to 16,000, w/matching explanation.
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IndWW HW#1 - Spring 2023
Question #3
The HR department provides the following info:
* 175 employees, split into 100 on 1st, 50 on 2nd, and 25 on 3rd shift. The 1st shift works M-F, and half also work on Saturdays. 10% of 2nd shift also comes on Saturdays.
* In addition to employees, the plant relies on 25 temp workers on 1st shift and 10 temp workers on 2nd shi
What is a reasonable initial estimate of sanitary flow?
Show your steps below in peach box, and enter final answer in yellow box:
3,946 gpd-cal
M-F: 175 + 35 = 210 people
Saturday: 50 + 5 = 55 people
25 gpd (210 people x 5 d/wk) + 25 gpd (55 people x 1 d/wk)= 27,625 gal/wk
3,946 gpd-cal
ift, M-F
IndWW HW#1 - Spring 2023
Question #4
The site has two cooling towers. Here are your field notes.
WCT1
WCT1
WCT4
WCT1 Evaporative Loss/Drift: Cooling tower 1 (WCT1) is used to cool a furnace that runs 24 hr/day, 7 d/wk. Open loop side is filled by a float. You have been unable to get info on temperature and pump rates. Tested by turning off make-up/blowdown and leaving the recirc pump (tower spray overpack) running. The base of cooling tower is 41" x 30" W. The water level dropped 1.25" in 19 minutes, and then you turned the make-up/blowdown back on. What is the evaporative+drift loss (gpd) to use in your flow balance?
WCT1 Blowdown
: The cooling tower blowdown rotameter reads 0.4 gpm. What is the blowdown (gpd) to use in your flow balance?
WCT4 Evaporative Loss/Drift:
Cooling tower 4 (WCT4) is used to cool a furnace that runs M-F, 12 hr/day. When the furnace is not running, the cooling tower is turned off. You have limited time on-site and cannot directly measure, but the set-up (tower size, equipment) is the same as a tower WCT3 that operates 24 hr/d, 7 d/wk and was estimated at 450 gpd evaporative+drift loss. What is the WCT4 evaporative + drift loss (gpd-cal).
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Define/Use as Needed For Calcs
Show notes or work as needed
CF
gpm
gpd
Answer (gpd-ca
0.8897569 0.3502833 504.40789
504
GPD: 0.4 gpm x 60 min/hr x 24 hr/d = 576 gpd.
0.4
576
576
160.71429
161
Drop: 41" x 30" x 1.25" in 19 min. CF: (41"x30"x1.25")/(12in/ft x 12in/ft x 12in/ft) = 0.889757 CF. GPM: (0.889757 CF) x (7.48 gal/CF)/19 min = 0.35 gpm. GPD: 0.35 gpm x 60 min/hr x 24 hr/d = 504 GPD.
WCT4 - assume same rate of loss as WCT3. Mental Math: For 12 hr/day instead of 24 hr/day, thus 225 gpd. Or, if more complicated and mental math not advised: 450 gpd WCT3 x d/24 hr = 18.75 gal/hr. Multiply by however many hours WCT4 runs: 18.75 gal/hr x 12 hr/day = 225 gpd WORK DAY basis. Weekly: 5 work days * 225 gpd-work = 1,125 gal/wk. Convert to calendar day basis: 1,125 gal/7 d = 160.7
al)
c6bcc2521885a94d4771305a25affc04127ebb3f.xlsx\Q5Misc
Page 9 of 15
Question #5
Evaporative Losses
Wastewater Generated
Description
Calc Notes
Hrs/ Wk
7.71008 2.03701
3.78
462
1,080
30.5
400
0.50
IndWW HW#1 - Calculate the yellow Wastewater Source
L (in)
W (in)
H (in)
Vol (Gal)
Evap (in)
Evap Loss/Refill (GPD-
work)
Evap Loss/Refill (GPD-Cal) Flow (L/min)
Flow (gpm)
Vol (Gal/ Wk)
Source 1 - Spray Station
Spray Station: Hand held rinse spray gun (bucket/stopwatch measurement): 2L/16 sec, 2L/14 sec, 2L/17 sec. Gun is on a timer cycle, so big parts are rinsed 4 minutes, small parts are rinsed 1 minute. Interviews: 1st shift: 48 big parts/wk, 8 small parts/wk. 2nd shift: 6 big parts/wk, 3 small parts/wk. 3rd shift: 0. Spray station operates 6 d/wk. What is the gpd-cal?
Flowrate: 60 sec/min * (2L/16 sec + 2L/14 sec + 2L/17 sec)/3 = 7.71 L/min. Also: 7.71 L/min x gal/3.785 L = 2.037 gpm. Total time running: (54 big parts/wk * 4 min/big parts + 11 small parts/wk * 1 min/small part)/(60 min/hr) = 3.78 hr/wk. Vol (Gal/Wk): 2.037 gpm * 60 min/hr x 3.78 hr/wk = 462 gal/wk. Flow GPD-CAL: (462 gal/wk)/7 days/wk = 66 gpd. The info that the operation is 6 d/wk is extraneous info, because you can get the minutes of operation directly from the parts and timer info.
Source 2 - Parts Washer
Parts washer: 6-7 cycles/day, 5 d/wk, 1st shift. 4-5 cycles/day, 5 d/wk, 2nd shift. 0 on 3rd. Triggered a cycle, no water came out till end of cycle. Drain sump: 3 ft x 3 ft sump, water level raised 3.5" from one cycle. What is the gpd-cal?
10-12 cycles/day, 5 d/wk. Use 11 cycles/day x 5 d/wk = 55 cycles/wk. Flow: 3 ft x 3 ft x 3.5"/12" = 2.625 CF/cycle x 7.48 gal/CF = 19.635 gal/cycle. Vol (gal/wk): 19.365 gal/cycle x 55 cyles/wk = 1,080 gal/wk. Note: I used 11 cycles/day because that is the midpt of the interviews. When doing a flow balance, if you always pick max from interviews, you will normally not get a good balance with the flow meter. It is better to get a flow balance that is in agreement, and then apply safety factors (as/if needed) to your balance as far as a design basis. That way, you can use your balance for multiple purposes (e.g., permit, reporting, etc. as well as designs). Therefore, suggest using 11 cycles. However, I also accepted answers that used max cycles (12 cycles).
Source 3 - Dip Tank
Hot dip tank. 3.5'Wx3'Lx4'H sidewall (6" freeboard; freeboard = distance from operating water level to top of sidewall). Once/mo hauled. Daily top-offs with tap water to make-up for evaporative loss. Operator snap-shots of top-offs: 11/21/14 3" water added Friday morning, 11/21/14 2" added Friday afternoon, 11/22/14 2" added Saturday morning. Provide estimates of evap loss and hauled volumes.
EVAP (GPD-CAL): 3+2+2=7 inches/1.5 days = 4.66 inch/day. (3.5'x3'x4.66"/day)/(12 in/ft)= 4.08 CF @ 7.48 gal/CF = 30.5 gal/day. Hauled: 3.5'x3'x(4-0.5')x 7.48 gal/CF x 12 hauls/yr = 3,299 gal/yr hauled. GPD hauled: 3,299 gal/yr x (yr/365 day) = 9 gpd-cal. Tip: the tanks were topped off in morning, afternoon, morning. That first morning, the topoff was due to evap loss overnight. So, these 3 refills really represented 1.5 days of evap time.
Source 4 - Scrubber
Scrubber info: 14,000 cfm air flow; 30 gpm recirc rate (pump recirculation from scrubber sump to scrubber packing); 0.5 gpm blowdown; assume 400 gpd evap loss; 7 d/wk, 24 hr/d operation. The 14,000 cfm and the 30 gpm are extraneous info. They could be used to estimate if no other info was available, but info is directly given. Given 400 gpd evap loss (directly enters, as this is shown as a 7 d/wk operation). Flow: 0.5 gpm x 60 min/hr x 24 hr/d = 720 gpd
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c6bcc2521885a94d4771305a25affc04127ebb3f.xlsx\Q5Misc
Page 10 of 15
d
Off-Site Haul
66 154 3,299
9
720 GPD-
Calendar
Haul (Gal/ Yr)
Haul (GPD - Cal)
ANSWER KEY
SOURCES
FLOW GPD (CALENDAR)
FINAL DISPOSAL
Water Meter Data
gpd
Water Use
Evap/Loss
Discharge
13,694 gpd
Supply (upstream-metered)
13,694
4,716 gpd
Process WW Sources
Usage (dnstream-estimated)
15,902
to on-site WWTP
Ratio: dnstream to upstream
116%
Clean/Strip Acid Line
2,530
0
2,530
X-Ray Room
115
0
115
Washing Area
512
0
512
Scrubber System
1,022
302
720
Wet Tumblers
537
0
537
4,414 gpd
Subtotal WWTP =
4,716
302
4,414
WWTP Outfall 001
823 gpd
RO Reject
823 gpd
823
0
823
RO Reject
5,237 gpd
WWTP+RO Flow
9,500 gpd
Restrooms
9,500 gpd
15,160 gpd
9,500
0
9,500
Restrooms
Outfall 003 to City
863 gpd
Cooling Towers
423 gpd
863
440
423
Cooling Towers
9,923 gpd
Sewer Outfall 002
ANSWER KEY (ANOTHER OPTION)
SOURCES
FLOW GPD (CALENDAR)
FINAL DISPOSAL
Water Meter Data
gpd
Water Use
Evap/Loss
Discharge
13,694 gpd
Supply (upstream-metered)
13,694
4,716 gpd
Process WW Sources
Usage (dnstream-estimated)
15,902
to on-site WWTP
Ratio: dnstream to upstream
116%
Clean/Strip Acid Line
2,530
0
2,530
X-Ray Room
115
0
115
Washing Area
512
0
512
Scrubber System
1,022
302
720
Wet Tumblers
537
0
537
4,414 gpd
Subtotal WWTP =
4,716
302
4,414
WWTP Outfall 001
823 gpd
RO Reject
823 gpd
823
0
823
RO Reject
5,237 gpd
WWTP+RO Flow
Points
10,363 gpd
Restrooms & 9
9 pt - recognizable flow balance
Cooling Towers
4
4 pt - meter data, supply, usage, ratio
Restrooms
9,500
0
9,500
4
4 pt - group WWTP flows in one category; omit RO from WWTP
Cooling Towers
863
440
423
9,923 gpd
3
3 pt - correctly depict the water use as the use not the discharge #s
Subtotal RR/WCT =
10,363
440
9,923
Sewer Outfall 002
3
3 pt - Outfall 001 depicted & with correct flow
3
3 pt - Outfall 002 depicted & with correct flow
15,160 gpd
3
3 pt - Outfall 003 depicted & with correct flow
Outfall 003 to City
IndWW HW#1 - Spring 2023
For use in answering Questions #6-9 (for preparation of your flow balance on Q6-Q9FlowBalance sheet).
Given the below plan view and field notes (line 27-43 below), construct a flow balance
(this flow balance is not related to questions Q1-Q5 on the homework; all info for the flow balance is on this sheet).
Transparent facility background is generic/solely for pictorial purposes.
Meter Info
* The site's water bill shows a base supply of water charged at $2 per CCF, which has an avg of 11,238 gpd-cal.
Additional water is charged at $4 per CCF, and shows 2,456 gpd-cal supplied at that rate.
* This site operates 7 d/wk.
From interviews and a field visit, you have gathered the following initial info.
Item
Description
Flow info (all gpd-cal)
WW Source 1
RO Reject
823
WW Source 2
Clean/Strip Acid Line
2,530
WW Source 3
X-ray room
115
WW Source 4
Washing Area
512
WW Source 5
Scrubber System
302 gpd evaporative/drift loss, 720 gpd blowdown to WWTP
WW Source 6
Wet Tumblers
537
Restrooms
380 people @ 25 gpd =
9,500
Cooling Towers
WCT to sanitary sewer
440 gpd evaporative/drift loss, 423 gpd blowdown to WWTP
When your purchases were > downstream use, check what sanitary adjustments would do:
People
GPD
Flow (gpd)
380
24
9,120
380
23
8,740
380
22
8,360
380
21
7,980
380
20
7,600
realistic adjustment
380
19
7,220 can justify others as well
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A copy of Client B flow balance, with notes removed for clarity is provided below. Provide a similar format (sum upstream vs. downstream, and calculate the ratio) when completing your balance on question 6 (see the question 6 sheet).
Tip: Place your cursor on the green fields below to see how they are calculated.
Simplified Version of Client B Flow Balance
WATER SUPPLY SOURCES
WATER DEMAND/USAGE
FLOW GPD (CALENDAR)
FINAL DISPOSAL
Sprinklers
563 gpd (calendar)
Lawn Irrigation
WATER USE
EVAP/LOSS
DISCHARGE
Meter # 12345678
563
563
0
Lost to Atmosphere
Sprinkler Meter data based on dd-mm-yy to dd-mm-yy records.
Subtotal Irrigation =
563
563
0
or Irrigation
gpd (cal)
Main Water Supply
Supply (upstream-metered)
28,196
13,842 gpd (calendar)
Non-Process WW
Meter # 101010101
28,196 gpd (calendar)
Usage (dnstream-estimated)
26,227
Sources to 001
mm/yyyy to mm/yyyy
Ratio: dnstream to upstream
93%
Sanitary & Cooling WW Includes:
Sanitary RR (Front, South, Middle)
6,188
0
6,188
Main Plant RO Reject/Backwash
2,353
0
2,353
Drinking Water RO Reject/Backwash
103
0
103
LEGEND:
gpd = gallons per day
FIC Furnace #1 and #2 Cooling Water
5,193
0
5,193
calendar = calendar day basis (average over 7 calendar days)
FIC Furnace #2 Heat Exchanger (neglible)
0
0
0
Handwash Sink
5
0
5
13,872 gpd
HVAC Condensate (nominal)
0
0
30
To Outfall 1
Subtotal Sanitary/Cooling Water =
13,842
0
13,872
8,311 gpd (calendar)
Non-Process WW
Sources to 002
Sanitary RR (NE Upstairs/Downstairs)
2,063
0
2,063
Exercise Area Showers (NE Upstairs)
200
0
200
5,317 +
Vacuum Furnace #2 Hoses (negligible)
0
0
0
604 =
HVAC Condensate (nominal)
0
0
30
5,921 gpd
WCT-1, -2, -3, -4, -5 Cooling Towers
6,048
3,024
3,024
To Outfall 2
Subtotal Sanitary/Cooling Water =
8,311
3,024
5,317
Local Sewer Ordinance
Pt of Compliance
1,662 gpd (calendar)
Metals WWTP
Categorical to 003
Process Wastewater Includes:
FIC Scrubbers 1 - 2 see below
40
see below
Tumblers, Met Lab
see below
0
see below
Chem Line Acids & Rinses see below
48
see below
Etch/Titration Hood/Sink
see below
0
see below
Coating Furnace Scrubbers 1 - 4
see below
400
approx 71
Chem Line C/S Scrubber
see below
400
288 to 432
Platinum Line Scrubber
see below
20
approx 4
Sludge Dryer Scrubber
see below
150
approx 192
Treated via Metals Precipitation.
Lime, Polymer Solutions
see below
0
see below
Through Outfall 3
Subtotal of Metals WWTP (T11/T12) =
1,662
1,058
604
Categorical Pt of Compliance.
48 gpd (calendar)
Series 700 Evaporator
Metal /Chloride WW
Includes: Mop Buckets, Barrel Tumbler, Chem Line Alkalines
X-Ray Wastewater
Typical 0 gpd (optional route here; typical to Series 800 Evaporator)
T11/T12 Final Hold WW
Typical 0 gpd (optional here; typical to 003 City Discharge)
Water Lost to Evaporation.
Furnace Scrubbers
Typical 0 gpd (optional here; typical to Metals WWTP)
Residues Disposed Off-Site.
Subtotal of Series 700 Evaporator =
48
48
0
Zero Discharge to Sewer.
1,602 gpd (calendar)
Series 800 Evaporator
Dilute Oily/X-Ray WW
Air Compressor Blowdown
0
80
0
Parts Washers, FPI Lines, X-Ray WW
1,602
1,602
0
Water Lost to Evaporation.
Subtotal of Series 800 Evaporator =
1,602
1,682
0
Residues Disposed Off-Site.
Zero Discharge to Sewer.
650 gpd (calendar)
Non-Process WW
Cafeteria to 005
Cafeteria - sink, food vats, tray washing
650
0
650
660 gpd
HVAC Condensate (nominal)
0
0
10
To Outfall 5
Subtotal Cafeteria =
650
0
660
Grease Trap
112 gpd (calendar)
Equipment Refills
112
112
0
Water for Equipment Top-Offs & & Hauled Wastes
112
112
0
Residues Disposed Off-Site.
Sewer Tee
IndWW HW#1 - Spring 2023
Question #10 (BONUS)
Q10A. Read this gas meter. Neglect the dials labeled 1/2 and 2. Provide answer in CF and CCF.
238,800 CF
Remember - gas meters arrows both clockwise and CCW
Read whatever # you have crossed; do not read next # up.
2388 for the #s. Then, how many 0s? Look at far R
dial. One full rev is 1,000. You are at 8 on that dial.
Note that the dial itself says: 1,000 per REV
So, that 8 is 800. Meter reading is: 238800.
Q10B
2,388 CCF
CCF is 100 CF
Q10C. What BTU content per CF can I use for natural gas (hint: lecture slide 50)
1,030 BTU/CF
Q10D. I installed an evaporator 30 days ago. The gas meter was installed at same time (started at 0) and now reads 450,000 CF.
The evaporator is rated at 63 gph and 750,000 BTU/hr.
How many hours per day has the evaporator been running on average?
Show Work:
450,000 CF x 1030 BTU
x
hr
=
618 hrs
CF
750,000 BTU
Answer: 20.6 hrs/day
=618 hrs/30 days ago installation
Q10E. What is the average gpd of wastewater processed by this evaporator?
Show Work:
20.6 hr/d x 63 gph = Answer: 1,298 gpd
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Question 3
The activity relation chart for NOFRILLS departmental store in Richmond Hill is shown in the figure below.
The space requirements are in square feet. Construct the relationship diagram and develop a block layout
using the systematic layout planning (SLP) method. Your solution must include flow-between chart,
activity relationship chart, value chart, nodal relationship, block calculations, and grid representation, as
covered in class.
Dept A
400 (sq. ft)
0
Dept B
1,000
12
Dept C
2,600
Dept D
176 212
136 132
240 216 16
400
140 180
36
184
Dept E
144
172
2,400
220
188
108
20
Dept F
1,000
192
168
24
248
204
124
Dept G
3,600
228
112 156
120
Dept H
1,200
160
224
148
Dept I
152
400
108
200
Dept J
2,400
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R5 please answer all problems and also write out all steps and reasoning. The second image shows the correct answers.
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Row indexing: Snow fall records
stateSnowRecord contains the amount of snow fall (in cm) for various cities. Rows represent a city and columns represent the snow fall
for a given month. Assign citySnow Record with all elements of row rowNum.
Ex: If stateSnowRecord is [17.3, 20.3; 4.8, 6.2] and rowNum is 2, then citySnowRecord is [4.8, 6.2].
Function >
6 %
7 %
8
9
10
11
12
13 end
function citySnowRecord = Get CitySnowFall (stateSnowRecord, rowNum)
2% Get CitySnow Fall: Returns all snow fall values for city in row rowNum
%
4 %
Inputs: stateSnowRecord matrix of snow fall records for cities for various months
rowNum specified row of stateSnowRecord to return
stateSnowRecord = [17.3, 20.3; 4.8, 6.2]
Outputs: citySnowRecord - all snow fall values for a given city
% Assign citySnowRecord with all elements of a row rowNum.
citySnowRecord = stateSnowRecord (rowNum, :);
Code to call your function >
1 GetCitySnow Fall([17.3, 20.3; 4.8, 6.2], 2)
Previous Assessment: 1 of 2 Tests Passed (50%)
Save…
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Hello I’m trying to make the graph that you see in the picture but I am having trouble on make the two lines that you see on the graph to over lap each other, I don’t want the two lines to be separate I want the two lines on top of eachother like you see in the picture please fix the code so that the Diesel line is on top of the Petrol line as seen on the picture.
% Sample data for Diesel and Petrol cars
carPosition = linspace(1, 60, 50); % Assumed positions of cars
% Fix the random seed for reproducibility
rng(45);
% Assumed positions of cars
CO2Diesel = 25 + 5*cos(carPosition/60*2*pi) + randn(1, 50)*5; % Random data for Diesel
CO2Petrol = 20 + 5*sin(carPosition/60*2*pi) + randn(1, 50)*5; % Random data for Petrol
% Fit polynomial curves
pDiesel = polyfit(carPosition, CO2Diesel, 3);
pPetrol = polyfit(carPosition, CO2Petrol, 3);
% Generate points for best fit lines
fitDiesel = polyval(pDiesel, carPosition);
fitPetrol = polyval(pPetrol, carPosition);
% Plotting the data
figure;
hold on;…
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PART A Using the graph to the right, what would Air Pressure be (mb) at an altitude of 10 km? What is the approximate altitude where the air pressure is 400 mb?
PART B At an altitude of 0 km, what is the air pressure in mb?
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You own a wheat warehouse with a capacity of 20,000bushels. At the beginning of month 1, you have 6,000 bushels of wheat. Each month, wheat can be bought and sold at theprice per 1000 bushels given in Table 46.The sequence of events during each month is as follows:a You observe your initial stock of wheat.b You can sell any amount of wheat up to your initialstock at the current month’s selling price.c You can buy (at the current month’s buying price) asmuch wheat as you want, subject to the warehouse sizelimitation.Your goal is to formulate an LP that can be used to determine how to maximize the profit earned over the next 10months
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This question about Sanitary Engineering
Note: please solve the question using Incremental Increase Method.
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Could you please fix my code it’s supposed to look like the graph that’s on the picture. But the lines do not cross eachother at the beginning. Could you make the lines look like the lines on the graph?
Use this code in MATLAB and fix it.
% Sample data for Diesel and Petrol cars
carPosition = linspace(1, 60, 50); % Assumed positions of cars
% Define your seed here
seed = 50;
rand('seed',seed); % Set the seed for reproducibility
% Assumed CO2 emissions for Diesel and Petrol
CO2Diesel = 25 + 5*cos(carPosition/60*2*pi) + randn(1, 50)*5; % Random data for Diesel
CO2Petrol = 20 + 5*sin(carPosition/60*2*pi) + randn(1, 50)*5; % Random data for Petrol
% Fit polynomial curves with a reduced degree of 2
pDiesel = polyfit(carPosition, CO2Diesel, 2);
pPetrol = polyfit(carPosition, CO2Petrol, 2);
% Generate points for best fit lines
fitDiesel = polyval(pDiesel, carPosition);
fitPetrol = polyval(pPetrol, carPosition);
% Plotting the data
figure;
hold on;
% Plot Diesel best fit line…
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Don't Use Chat GPT Will Upvote And Give Handwritten Solution Please
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Note: Round your final answer to 2 decimal places if it is not a whole number.
Note:-
Do not provide handwritten solution. Maintain accuracy and quality in your answer. Take care of plagiarism.
Answer completely.
You will get up vote for sure.
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Please show all work and highlight your answers
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I am sorry but your values when plugging in are not the correct values. Not too sure where your values are from but can you please redo the question using the given values in this problem and not from a different problem please? (example, height is not 29, it is 22) all values do not correlate to this problem. I was also told when finding the area of triangle to use (1/2)*b*h*Ad^2 and not 1/36*b*h*Ad^2
Find the moment of inertia and radius of gyration of the section of this bar about an axis parallel to x-axis going through the center of gravity of the bar.
The bar is symmetrical about the axis parallel to y-axis and going through the center of gravity of the bar and about the axis parallel to z-axis and going through the center of gravity of the bar.
The dimensions of the section are:
l=55 mm, h=22 mm
The triangle: hT=12 mm, lT=19 mm
and the 2 circles: diameter=8 mm, hC=6 mm, dC=8 mm.
A is the origin of the referential axis.
Provide an organized table and explain all your steps…
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sample calculations
目
File Edit View Insert Format Data To
100%
$ % .0
.00 12:
fx |
A
1
Variable
Value
Units
diameter
height
m
4
volume
m^3
3
The image above shows a section of a Google sheet. What formula should be written in cell
B4 to calculate the volume of a cylinder with a diameter value in B2 and height in B3?
a) =0.25*PI()*B2^2 * B3
b) =PI()*B2^2 * B3
c) =pi*diameter^2 * height
d) =PI*B2^2 * B3
3 5
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Please solve the following parts a) b) c) and show the detail work for each part.
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I already have the answer and provide it. Show work please
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Please make the exact graph that you see in the picture, along with the graph are numbers that are data. Please make the exact graph do not make anything thing different, the blue and orange circles should be there and the lines should be the same including the titles. Please make sure everything is exactly the same. Use MATLAB, and send the code and please make sure no error signs comes up. Take your time please I need help.
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I need help with the empty table, you can use the solved table to answer it.
The images are below.
Thank you very much!
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Problem 3:
Insulation
To=1
Toowwww
Steam
Tx2
T₂ T3
www www
R₁ R₁ R₂
www.T
R₂
Steam at T1 = 320 °C flows in a cast iron pipe (k= 80 W/m. °C) whose inner and outer diameters are
5 cm = 0.05 m and D₂ = 5.5 cm = 0.055 m, respectively. The pipe is covered with 3-cm-thick glass wool
insulation with k = 0.05 W/m. °C. Heat is lost to surroundings at T2 = 5 °C by natural convection and
radiation, with a combined heat transfer coefficient of h₂ = 18 W/m². °C. Taking the heat transfer coefficient
inside the pipe to be h₁ = 60 W/m². °C, determine the temperature drops across the pipe and the insulation.
The determination is based on a unit length of the pipe (L = 1 m).
Assumptions
1. Heat transfer is one-dimensional since there is no indication of any change with time.
2.
Heat transfer is one-dimensional since there is thermal symmetry about the centreline and no
variation in the axial direction.
3. Thermal conductivities are constant.
4. The thermal contact resistant at the interface is…
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Can someone please help me to solve both of the following parts of this question showing all work. Thank you!
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- R5 please answer all problems and also write out all steps and reasoning. The second image shows the correct answers.arrow_forwardRow indexing: Snow fall records stateSnowRecord contains the amount of snow fall (in cm) for various cities. Rows represent a city and columns represent the snow fall for a given month. Assign citySnow Record with all elements of row rowNum. Ex: If stateSnowRecord is [17.3, 20.3; 4.8, 6.2] and rowNum is 2, then citySnowRecord is [4.8, 6.2]. Function > 6 % 7 % 8 9 10 11 12 13 end function citySnowRecord = Get CitySnowFall (stateSnowRecord, rowNum) 2% Get CitySnow Fall: Returns all snow fall values for city in row rowNum % 4 % Inputs: stateSnowRecord matrix of snow fall records for cities for various months rowNum specified row of stateSnowRecord to return stateSnowRecord = [17.3, 20.3; 4.8, 6.2] Outputs: citySnowRecord - all snow fall values for a given city % Assign citySnowRecord with all elements of a row rowNum. citySnowRecord = stateSnowRecord (rowNum, :); Code to call your function > 1 GetCitySnow Fall([17.3, 20.3; 4.8, 6.2], 2) Previous Assessment: 1 of 2 Tests Passed (50%) Save…arrow_forwardHello I’m trying to make the graph that you see in the picture but I am having trouble on make the two lines that you see on the graph to over lap each other, I don’t want the two lines to be separate I want the two lines on top of eachother like you see in the picture please fix the code so that the Diesel line is on top of the Petrol line as seen on the picture. % Sample data for Diesel and Petrol cars carPosition = linspace(1, 60, 50); % Assumed positions of cars % Fix the random seed for reproducibility rng(45); % Assumed positions of cars CO2Diesel = 25 + 5*cos(carPosition/60*2*pi) + randn(1, 50)*5; % Random data for Diesel CO2Petrol = 20 + 5*sin(carPosition/60*2*pi) + randn(1, 50)*5; % Random data for Petrol % Fit polynomial curves pDiesel = polyfit(carPosition, CO2Diesel, 3); pPetrol = polyfit(carPosition, CO2Petrol, 3); % Generate points for best fit lines fitDiesel = polyval(pDiesel, carPosition); fitPetrol = polyval(pPetrol, carPosition); % Plotting the data figure; hold on;…arrow_forward
- PART A Using the graph to the right, what would Air Pressure be (mb) at an altitude of 10 km? What is the approximate altitude where the air pressure is 400 mb? PART B At an altitude of 0 km, what is the air pressure in mb?arrow_forwardYou own a wheat warehouse with a capacity of 20,000bushels. At the beginning of month 1, you have 6,000 bushels of wheat. Each month, wheat can be bought and sold at theprice per 1000 bushels given in Table 46.The sequence of events during each month is as follows:a You observe your initial stock of wheat.b You can sell any amount of wheat up to your initialstock at the current month’s selling price.c You can buy (at the current month’s buying price) asmuch wheat as you want, subject to the warehouse sizelimitation.Your goal is to formulate an LP that can be used to determine how to maximize the profit earned over the next 10monthsarrow_forwardThis question about Sanitary Engineering Note: please solve the question using Incremental Increase Method.arrow_forward
- Could you please fix my code it’s supposed to look like the graph that’s on the picture. But the lines do not cross eachother at the beginning. Could you make the lines look like the lines on the graph? Use this code in MATLAB and fix it. % Sample data for Diesel and Petrol cars carPosition = linspace(1, 60, 50); % Assumed positions of cars % Define your seed here seed = 50; rand('seed',seed); % Set the seed for reproducibility % Assumed CO2 emissions for Diesel and Petrol CO2Diesel = 25 + 5*cos(carPosition/60*2*pi) + randn(1, 50)*5; % Random data for Diesel CO2Petrol = 20 + 5*sin(carPosition/60*2*pi) + randn(1, 50)*5; % Random data for Petrol % Fit polynomial curves with a reduced degree of 2 pDiesel = polyfit(carPosition, CO2Diesel, 2); pPetrol = polyfit(carPosition, CO2Petrol, 2); % Generate points for best fit lines fitDiesel = polyval(pDiesel, carPosition); fitPetrol = polyval(pPetrol, carPosition); % Plotting the data figure; hold on; % Plot Diesel best fit line…arrow_forwardDon't Use Chat GPT Will Upvote And Give Handwritten Solution Pleasearrow_forwardNote: Round your final answer to 2 decimal places if it is not a whole number. Note:- Do not provide handwritten solution. Maintain accuracy and quality in your answer. Take care of plagiarism. Answer completely. You will get up vote for sure.arrow_forward
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