Finding Density Constant for Part (c) of Question

[WY]

Hey I am completely stumped as to how to find the density in this following question so I can answer part (c)
Q: At a bottling plant soft drink flows in a pipe at a rate that would fill 200 375mL cans per minute. At point 1 in the pupe the gauge pressure is 152 kPa and the crossectional area is 8cm^2. At point 2, 1.35 m above point 1, the cross sectional area is 2cm^2. Find
(a) the volume flow rate - which I have found
(b) the flow speeds at points 1 and 2 - which I also have found
(c) the gauge pressure at point 2 -this one has got me in a spin

Can someone help me out?
Thanks in advance!

 

[Doc Al]

Hey I am completely stumped as to how to find the density in this following question so I can answer part (c)

If they didn't give you the density of the soft drink, you'll have to make some assumptions. I'd take the density to be that of water. (Of course, that's not quite right. Sugar-laden sodas probably have a density greater than water; diet sodas, less.)

 

[thepatientmental]

To find the density in this question, we can use the continuity equation, which states that the volume flow rate at one point in a pipe is equal to the volume flow rate at another point in the pipe. In other words, the amount of fluid entering a pipe must be equal to the amount of fluid leaving the pipe.

Since we already have the volume flow rate and flow speeds at points 1 and 2, we can use the continuity equation to find the density. The equation is:

ρ = m/V = ρ1A1v1/ A2v2

Where:
ρ = density
m = mass
V = volume
A1 = cross sectional area at point 1
v1 = flow speed at point 1
A2 = cross sectional area at point 2
v2 = flow speed at point 2

Plugging in the values from the question, we get:

ρ = ρ1(8cm^2)(v1)/(2cm^2)(v2)

Since we don't have the density at point 1, we can use the ideal gas law to find it. The ideal gas law states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is the temperature. Since we are dealing with a liquid, we can assume that n and R are constant and cancel out. Therefore, we can rearrange the equation to solve for density:

ρ = P/RT

Plugging in the values for pressure, temperature, and the gas constant, we get:

ρ1 = (152 kPa)/(8.31 J/molK)(298 K) = 0.059 kg/m^3

Now, we can plug this value into our original equation to solve for the density at point 2:

ρ = (0.059 kg/m^3)(8cm^2)(0.5 m/s)/(2cm^2)(1.5 m/s) = 0.01475 kg/m^3

Therefore, the density at point 2 is approximately 0.01475 kg/m^3. This can then be used to find the gauge pressure at point 2 using the ideal gas law:

P2 = ρ2RT = (0.01475 kg/m^3)(8.31 J/molK)(298 K) = 36.9 kPa