We will use Poiseuille's law to solve this problem:

\(\displaystyle \Delta P=\frac{8\mu LQ}{\pi r^4}\)  

We can determine the volumetric flow rate from the mass flow rate:

\(\displaystyle Q = \frac{\dot{m}}{\rho_w} = \frac{1,500\frac{kg}{s}}{1,000\frac{kg}{m^3}} = 1.5\frac{m^3}{s}\)  

Now to determine an expression for radius:

\(\displaystyle A = \pi r^2\)

\(\displaystyle r^2 = \frac{A}{\pi}\)

\(\displaystyle r^4 = \frac{A^2}{\pi^2}\)  

Plugging this into the first expression, we get:

\(\displaystyle \Delta P = \frac{8\pi \mu L Q}{A^2}\)

Now plugging in all of our values:

\(\displaystyle \Delta P = \frac{8\pi (1x10^{-3}Pa\cdot s)(10m)\left ( 1.5\frac{m^3}s{}\right )}{(0.5m^2)^2}\)

\(\displaystyle \Delta P = 1.5Pa\)

Let's begin with the generic form of Poiseulle's law:

\(\displaystyle \Delta P = \frac{8\mu_wLQ}{\pi r^4}\)

Where:

\(\displaystyle Q = vA_c = v\pi r^2\)

Substituting this in:

\(\displaystyle \Delta P = \frac{8\mu_wLv\pi r^2}{\pi r^4} = \frac{8\mu_wLv}{r^2}\)

And:

\(\displaystyle r = \frac{d}{2}\)

Substituting this in:

\(\displaystyle \Delta P = \frac{8\mu_wLv}{\left (\frac{d}{2} \right )^2}\)

\(\displaystyle \Delta P = \frac{32\mu_wLv}{d^2}\)

Rearranging for diameter:

\(\displaystyle d = \sqrt{\frac{32\mu_wLv}{\Delta P}}\)

Plugging in values:

\(\displaystyle d = \sqrt{\frac{32(1.002x10^{-3}Pa\cdot s)(2,000m)\left ( 20\frac{m}{s}\right )}{4,000Pa}}\)

\(\displaystyle d = 57cm\)

Using

\(\displaystyle \Delta P=\frac{32\mu*L*v}{d^2}\)

Where \(\displaystyle \mu\) is the dynamic viscosity

\(\displaystyle L\) is the length of the pipe

\(\displaystyle v\) is the velocity of the fluid

\(\displaystyle d\) is the diameter of the pipe

Plugging in values:

\(\displaystyle \Delta P=\frac{32*8.9*10^{-4}*15*3}{.20^2}\)

\(\displaystyle \Delta P= 32.04 Pa\)

Using

\(\displaystyle \Delta P=\frac{32\mu*L*v}{d^2}\)

Where \(\displaystyle \mu\) is the dynamic viscosity

\(\displaystyle L\) is the length of the pipe

\(\displaystyle v\) is the velocity of the fluid

\(\displaystyle d\) is the diameter of the pipe

Plugging in values:

\(\displaystyle \Delta P=\frac{32*8.9*10^{-4}*25*9}{.36^2}\)

\(\displaystyle \Delta P= 49.44 Pa\)

For this question, we're asked to consider two pipes. Each pipe has the same length, is made out of the same material, and has the same fluid moving through it. The only difference is the cross-sectional diameter of these pipes. We're asked to find how the flow rate will differ between the two pipes.

In order to solve this question, we'll need to use Poiseuille's equation.

\(\displaystyle Q=\frac{\pi \Delta P r^{4}}{8\eta l}\)

This equation tells us that the volume flow rate is directly proportional to two things: the pressure gradient between the ends of the pipe and the radius of the pipe raised to the fourth power. Moreover, the volume flow rate is inversely proportional to the viscosity of the fluid and also to the length of the pipe.

Now we need to ask the question - which of the variables in this equation is different in pipes A and B? Both pipes have the same length. Since each pipe has the same fluid moving through it, the viscosity will also be the same. Furthermore, we can assume that the pressure gradient at the end of each pipe is the same. The only thing that's left is the radius.

We're told that the diameter of pipe A is twice as great as pipe B. Since the radius is just half of the diameter, this also means that the radius of pipe A is twice as great as pipe B. Because the volume flow rate is dependent on the radius of the pipe raised to the fourth power, we can see that doubling the radius will result in a \(\displaystyle 16\) fold difference in the volume flow rate. Thus, the flow rate in pipe A will be \(\displaystyle 16\) times the flow rate in pipe B.

Turbulent flow is caused by sufficiently high flow rates, and is characterized by chaotic and irregular flow patterns. Streamlines represent the curves tangent to the point of the direction of flow; thus, if a flow is turbulent the streamlines will be chaotic as well. 

The wisps of smoke from a blown out candle start out a smooth, but eventually swirl and become chaotic and turbulent. The other examples can be intuitively thought of as smooth flow, and thus are laminar. 

We will use the expression for Reynold's number for this problem:

\(\displaystyle Re = \frac{\rho_m vD_H}{\mu_m}\)

Where \(\displaystyle D_H\) is the hydraulic diameter, and for a cylindrical tube, \(\displaystyle D_H = D\)

Rearranging for velocity:

\(\displaystyle v = \frac{Re\mu_m}{\rho_m D}\)

For laminar flow: \(\displaystyle Re \leq 10\)

Plugging in our values into the expression, we get:

\(\displaystyle v = \frac{10(.611x10^{-3}Pa\cdot s)}{\left ( 792\frac{kg}{m^3}\right )(0.02m)}\)

\(\displaystyle v = 3.9x10^{-4}\frac{m}{s}\)

Methanol has a relatively low viscosity, so it only has a very small range for laminar flow and quickly becomes turbulent

We will use the expression for Reynold's number for this problem:

\(\displaystyle Re = \frac{\rho_w vD_H}{\mu_w}\)

Where \(\displaystyle D_H\) is the hydraulic diameter, and for a cylindrical tube, \(\displaystyle D_H = D\)

Plugging in our values, we get:

\(\displaystyle Re = \frac{\left ( 1,000\frac{kg}{m^3}\right )\left ( 6\frac{m}{s}\right )(0.08m)}{\left ( 1.002x10^{-3}Pa\cdot s\right )}\)

\(\displaystyle Re = 480,000\)

We will use the expression for Reynold's number for this problem:

\(\displaystyle Re = \frac{\rho_w vD_H}{\mu_w}\)

For a fully filled rectangular duct, the hydraulic diameter is:

\(\displaystyle D_H = \frac{2wh}{w+h}\)

Plugging in values:

\(\displaystyle D_H = \frac{2(0.5m\cdot 0.3m)}{0.5m + 0.3m} = 0.375m\)

Now we can plug our values into the original expression:

\(\displaystyle Re = \frac{\left ( 1,000\frac{kg}{m^3}\right )(2.5\frac{m}{s})(0.375m)}{\left ( 1.002x10^{-3}Pa\cdot s\right )}\)

\(\displaystyle Re = 940,000\)