Find the gauge pressure at the pipe input

[solo]

A straight horizontal pipe with a diameter of 1.0 cm and a length of 50 m carries oil with a coefficient of viscosity of 0.12 N · s/m2. At the output of the pipe, the flow rate is 9.0 10-5 m3/s and the pressure is 1.0 atmosphere. Find the gauge pressure at the pipe input

I used the formula for flow rate: which is

Q = (pi)r^4(delta Pressure) / 8(coefficient of viscosity)(length)

I converted the diameter of 1.0 cm to .005 Meters radius, and the rest of the information is given

9.0 x 10^-5 = pi(6.25 x 10^-10)(Change in pressure) / 8(0.12)(50)

When i solve for pressure i get 21.7192 atm, and final pressure is 1,

so initial pressure is 1 + 21.7192 atm, which is 22.7192 ATM which is 2.30145 x 10^6 pascals

do you guys see anything wrong? thanks

 

[MexChemE]

First, there is not enough information to calculate the Reynolds number, but the diameter of the pipe is very small and the viscosity of oil is high, so we can safely assume laminar flow. (However, we could look at viscosity tables for different oils to check which ones coincide with our given value, then check the temperature where the values match, and then we can get the density, but that's probably not expected for this problem, and as I mentioned earlier, laminar flow is a good assumption.)

So, we have laminar flow in a pipe, so we can use the Hagen-Poiseuille equation
$$w = \frac{ \pi (p_0 - p_L) R^4 \rho}{8 \mu L}$$
$$Q = \frac{ \pi (p_0 - p_L) R^4}{8 \mu L}$$
Calculating pressure drop
$$p_0 - p_L = \frac{8 \mu L Q}{\pi R^4} = \frac{8 \left(0.12 \ \frac{kg}{m \cdot s} \right) (50 \ m) \left( 9 \times 10^{-5} \ \frac{m^3}{s} \right)}{\pi (0.005 m)^4} = 2.2 \ MPa$$
We're interested in gauge pressure, so at the end of the pipe, if the pressure is atmospheric, $p_L = 0$, then
$$p_0 = 2.2 \ MPa$$
Absolute pressures would be $P_0 = 2.3013 \ MPa$ and $P_L = 101.3 \ kPa$.