Yes, there is. The pressure drop over the tube is known (pressure in the bladder/vessel minus ambient pressure at the end of the hose). Knowing that pressure drop, you know is has to be equal to the pressure drop due to viscous forces acting inside the tube.
Note, it changes whether you are in the laminar or turbulent regime. It is very difficult to calculate if the length is equal to or shorter than the development length.
Re=\frac{\rho D V_{avg}}{\mu}
Laminar: Re < 2300
L_developing = 0.03*Re*D
Q=\frac{-\pi R^{4}}{8 \mu}\frac{\Delta P}{L}
(ΔP is pressure drop, L is length of tube, Q is volume flow rate, or V_avg*A_cross sectional).
Source: http://faculty.poly.edu/~rlevicky/Handout12_6333.pdf
Turbulent: Re > 2300
L_developing ≈ 10*D
\frac{1}{\sqrt{\lambda}}=-2LOG\left(\frac{2.51}{Re\sqrt{\lambda}}+0.269\frac{k}{D}\right)
\Delta P=\lambda \frac{L}{D}\frac{\rho}{2}V_{avg}^{2}
(ΔP is pressure drop, L is length of tube, λ is the friction factor, k is the absolute roughness of the tube where 0 is smooth).
Source:
http://www.engineersedge.com/fluid_flow/pressure_drop/pressure_drop.htm
Note: It is iterative, and you will have to guess a value for Re, calculate the friction factor, calculate a new velocity, calculate a new Re, until your solution has converged.
