The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The height difference from the upper to the lower channels is about 2 m and the distance between them is about 100 m. How large a pipe is needed to carry the flow? State any assumptions you make

2738.5 cubic metres

Explanation: Given that

Flow rate Q = 350m^3/s

Height h = 2m

Distance x = 100m

Using pythagorean theorem to find the length L of the pipe

L^2 = 100^2 + 2^2

L^2 = 10004

L = 100.02 m

Let assume that the pipe is uniform of same diameter at both ends and water flows through the pipe

Let also consider the atmospheric pressure at the upper channel

Using bernoulli equation

P1 = P2 + 1/2pV^2

Where P2 = phg

P1 = atmospheric pressure = 101325pa

V = velocity

p = density = 1000kg/m3

101325 = 1000×9.81×2 + (0.5×1000V^2)

101325 - 19620 = 500V^2

81705 = 500V^2

V^2 = 163.41

V = 12.8 m/s

Q = V × A

Where A = area of the pipe

A = Q/V

A = 350/12.8 = 27.4 square metre

The volume of the pipe = A × L

Volume = 27.4 × 100.02

Volume = 2738.5 cubic metres

The volume of the pipe determines how large a pipe is needed to carry the flow

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂