A compressed air drill requires 0.25 kg / s of air at 650 kPa (gage) at the drill. The hose from

A compressed air drill requires 0.25 kg / s of air at 650...

A compressed air drill requires $0.25 \mathrm{kg} / \mathrm{s}$ of air at 650 kPa (gage) at the drill. The hose from the air compressor to the drill is $40 \mathrm{mm}$ inside diameter. The maximum compressor discharge gage pressure is 670 kPa; air leaves the compressor at $40^{\circ} \mathrm{C}$. Neglect changes in density and any effects of hose curvature. Calculate the longest hose that may be used.

Key Concepts

Pressure Drop and System Design

Understanding how friction and geometry affect pressure drop is key to ensuring that a fluid system meets its operational requirements. In system design, the available pressure head from a compressor must overcome these losses so that sufficient pressure is maintained at the point of use, influencing decisions such as pipe length and diameter.

Continuity Equation

The continuity equation expresses the principle of mass conservation, relating the mass flow rate to the fluid’s density, flow velocity, and the cross-sectional area of the pipe. This relationship is essential for converting between mass flow rate and volumetric flow rate when designing or analyzing fluid transport systems.

Darcy-Weisbach Equation

This equation is fundamental in fluid mechanics for calculating pressure loss due to friction in a pipe. It relates the pressure drop to the fluid’s velocity, the length and diameter of the pipe, and the friction factor, providing a framework for designing piping systems to meet pressure requirements.

Friction Factor

The friction factor is a dimensionless quantity that encapsulates the effect of the pipe’s roughness and flow turbulence on the resistance to flow. It is critical for accurately estimating frictional losses using the Darcy-Weisbach equation and is often determined based on the Reynolds number and the relative roughness of the pipe.

Transcript

00:01 Applying the energy equation between compressor discharge and at the drill bit, we can write by bernoulli equation p1 divided by rho plus v1 squared divided by 2 plus gz1 minus is equal to p2 divided by row plus v2 squared divided by 2 plus gz2 plus head loss due to friction.

00:32 It allows due to friction so so here in this in our question we the velocity there's no change in velocity so these two terms will cancel out each other v1 divided by 2 or and v2 divided by 2 moreover the day there is no change in the date term level so this is also going to cancel each other hence our final equation will be p1 divided by row is equal to p2 divided by row plus plus hf or we can write or we can write p1 minus p2 is equal to rho hf or we can further write it as rowe multiplied by hf is equal to flv flv squared divided by 2d.

01:24 Let this equation be equation number one.

01:27 This is d.

01:29 Now, now, now we can find the pressure p1.

01:33 Pressure p1.

01:34 P1 absolute is equal to p1 gauge p1 gauge plus p atmosphere p atmosphere so p1 gauge is given as 670 plus p atm is 101 kioskle so it will be equals to 700 and 71 7771 kilo pascal 7771 kilo pascal now we will now we can find the density density intensity can be written as r is equal to p divided by r t so p we have already found as 771 multiplied by 10 to the power of 3 pascal divided by the value of r r is 0 .287 0 .287 kilo kilo kilo per kg kelvin and the value of temperature t is given as 40 degrees centigrade which is equal to 40 plus 273 kelvin so from here so from here we will get our answer.

02:39 So this is kilo pascal.

02:40 The upper upper part should also be in a kilo pascal...

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