# thrust at pipe outlet

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 June 29, 1998, 10:01 thrust at pipe outlet #1 A.H.Watson Guest   Posts: n/a I have a fluids application which uses a pressurised gas discharging to atmosphere from a pipe system. When calculating the force produced by the fluid, F=0.5 x K x density x area x (velocity)^2, books quote that 1 velocity head must be added to the system loss coefficient ,K, due to discharge into an infinite space. If a diffuser is used on the end of the pipe the system loss coefficient (for the diffuser discharging into an infinite space) is less than 1. This infers that a diffuser is more efficient at producing thrust for a given mass flow, is this correct? (the gas is air at @1 psi above atmospheric, and V=77 m/s)

 June 30, 1998, 04:32 Re: thrust at pipe outlet #2 Jonas Larsson Guest   Posts: n/a In the first case the force is F=m_dot*u [N] m_dot is the mass flow per second [kg/s] u is the velocity [m/s] m_dot [kg/s] = u [m/s] * density [kg/m3] * Area [m2] I'm not really sure of what you mean with a "diffuser" in the second case. If you mean that you let the air out through a divergent nozzle which is tuned so that it gives the same mass flow but has a larger exit area, then you will obtain less thrust. With a constant mass flow, m_dot, a larger discharge area will give you a smaller velocity, u, and thus a smaller thrust. The same formula as above also holds also for this case. This is only valid if both devices discharge into open space so that you can assume the outlet pressure to be equal to the surrounding pressure. Another requirement is that it is subsonic flow. For supersonic flow a divergent nozzle would accelerate the flow and produce a larger thrust (that's why you see those divergent nozzles on rocket engines). Hope I'm making sence and that I'm thinking correctly. Someone please correct me otherwise.

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