w2 example 1 answers

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1a Write an equation that can be used to work out the volume of water entering the basin without knowing the volume of water leaving the basin.

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Page 1: W2 Example 1 Answers

1a Write an equation that can be used to work out the

volume of water entering the basin without knowing the

volume of water leaving the basin.

Page 2: W2 Example 1 Answers

1a • Substitute V2 in the second equation with its equivalent from the first

equation:

Page 3: W2 Example 1 Answers

1a • Substitute V2 in the second equation with its equivalent from the first

equation:

V1S1 = (V1 +W)S2

Page 4: W2 Example 1 Answers

1a • Substitute V2 in the second equation with its equivalent from the first

equation:

V1S1 = (V1 +W)S2

• Then rearrange for V1:

Page 5: W2 Example 1 Answers

1a • Substitute V2 in the second equation with its equivalent from the first

equation:

V1S1 = (V1 +W)S2

• Then rearrange for V1:

V1S1 = V1S2 +WS2

Page 6: W2 Example 1 Answers

1a • Substitute V2 in the second equation with its equivalent from the first

equation:

V1S1 = (V1 +W)S2

• Then rearrange for V1:

V1S1 = V1S2 +WS2

V1S1 − V1S2 = WS2

Page 7: W2 Example 1 Answers

1a • Substitute V2 in the second equation with its equivalent from the first

equation:

V1S1 = (V1 +W)S2

• Then rearrange for V1:

V1S1 = V1S2 +WS2

V1S1 − V1S2 = WS2

V1 S1 − S2 = WS2

Page 8: W2 Example 1 Answers

1a • Substitute V2 in the second equation with its equivalent from the first

equation:

V1S1 = (V1 +W)S2

• Then rearrange for V1:

V1S1 = V1S2 +WS2

V1S1 − V1S2 = WS2

V1 S1 − S2 = WS2

𝐕𝟏 =𝐖𝐒𝟐

𝐒𝟏 − 𝐒𝟐

Page 9: W2 Example 1 Answers

1b It is important to be able to calculate the additions to the

water also without knowing the volume of water leaving

the basin. Write an equation to do this.

Page 10: W2 Example 1 Answers

1b • Rearrange your previous equation to make W the subject:

Page 11: W2 Example 1 Answers

1b • Rearrange your previous equation to make W the subject:

V1 =WS2

S1 − S2

Page 12: W2 Example 1 Answers

1b • Rearrange your previous equation to make W the subject:

V1 =WS2

S1 − S2

V1 S1 − S2 = WS2

Page 13: W2 Example 1 Answers

1b • Rearrange your previous equation to make W the subject:

V1 =WS2

S1 − S2

V1 S1 − S2 = WS2

𝐕𝟏 𝐒𝟏 − 𝐒𝟐𝐒𝟐

= 𝐖

Page 14: W2 Example 1 Answers

1c Over a year, a lagoon of surface area 50m2 gains 1.12m

of water by precipitation and 0.81m by run-off, but loses

1.26m by evaporation. What is W for the lake in m3s-1?

Page 15: W2 Example 1 Answers

1c • To convert m/year into m3s-1, you need to multiply it by the area if the

basin and divide by the number of seconds in a year:

Page 16: W2 Example 1 Answers

1c • To convert m/year into m3s-1, you need to multiply it by the area if the

basin and divide by the number of seconds in a year:

1 year = 356 x 24 x 60 x 60 seconds = 31536000s

Page 17: W2 Example 1 Answers

1c • To convert m/year into m3s-1, you need to multiply it by the area if the

basin and divide by the number of seconds in a year:

1 year = 356 x 24 x 60 x 60 seconds = 31536000s

P = 1.12my−1 =1.12 x 50

31536000= 1.78 x 10−6m3s−1

Page 18: W2 Example 1 Answers

1c • To convert m/year into m3s-1, you need to multiply it by the area if the

basin and divide by the number of seconds in a year:

1 year = 356 x 24 x 60 x 60 seconds = 31536000s

P = 1.12my−1 =1.12 x 50

31536000= 1.78 x 10−6m3s−1

R = 0.81my−1 =0.81 x 50

31536000= 1.28 x 10−6m3s−1

Page 19: W2 Example 1 Answers

1c • To convert m/year into m3s-1, you need to multiply it by the area if the

basin and divide by the number of seconds in a year:

1 year = 356 x 24 x 60 x 60 seconds = 31536000s

P = 1.12my−1 =1.12 x 50

31536000= 1.78 x 10−6m3s−1

R = 0.81my−1 =0.81 x 50

31536000= 1.28 x 10−6m3s−1

E = 1.26my−1 =1.26 x 50

31536000= 2.00 x 10−6m3s−1

Page 20: W2 Example 1 Answers

1c • To convert m/year into m3s-1, you need to multiply it by the area if the

basin and divide by the number of seconds in a year:

1 year = 356 x 24 x 60 x 60 seconds = 31536000s

P = 1.12my−1 =1.12 x 50

31536000= 1.78 x 10−6m3s−1

R = 0.81my−1 =0.81 x 50

31536000= 1.28 x 10−6m3s−1

E = 1.26my−1 =1.26 x 50

31536000= 2.00 x 10−6m3s−1

• Put the converted P, R and E back into the equation for W:

Page 21: W2 Example 1 Answers

1c • To convert m/year into m3s-1, you need to multiply it by the area if the

basin and divide by the number of seconds in a year:

1 year = 356 x 24 x 60 x 60 seconds = 31536000s

P = 1.12my−1 =1.12 x 50

31536000= 1.78 x 10−6m3s−1

R = 0.81my−1 =0.81 x 50

31536000= 1.28 x 10−6m3s−1

E = 1.26my−1 =1.26 x 50

31536000= 2.00 x 10−6m3s−1

• Put the converted P, R and E back into the equation for W:

W = 1.78 + 1.28 − 2 x 106m3s−1 = 𝟏. 𝟎𝟔 𝐱 𝟏𝟎−𝟔𝐦𝟑𝐬−𝟏

Page 22: W2 Example 1 Answers

1d Seawater of salinity 35 seeps into the lagoon at a rate of

5.4 x 10-5m3s-1 which has a salinity of exactly 35. What

will the salinity of the lagoon be after at least a year?

Page 23: W2 Example 1 Answers

1d • Substitute V2 in the second equation with its equivalent from

the first equation:

Page 24: W2 Example 1 Answers

1d • Substitute V2 in the second equation with its equivalent from

the first equation:

V1S1 = (V1 +W)S2

Page 25: W2 Example 1 Answers

1d • Substitute V2 in the second equation with its equivalent from

the first equation:

V1S1 = (V1 +W)S2

• Rearrange for S2 and substitute in the values you know:

Page 26: W2 Example 1 Answers

1d • Substitute V2 in the second equation with its equivalent from

the first equation:

V1S1 = (V1 +W)S2

• Rearrange for S2 and substitute in the values you know:

V1S1(V1 +W)

= S2

Page 27: W2 Example 1 Answers

1d • Substitute V2 in the second equation with its equivalent from

the first equation:

V1S1 = (V1 +W)S2

• Rearrange for S2 and substitute in the values you know:

V1S1(V1 +W)

= S2

S2 =5.4 x 10−5 x 35

5.4 x 10−5 + 1.06 x 10−6= 𝟑𝟒. 𝟑𝟑