Question

A fire hose nozzle has a diameter of 1.125 in. According to some fire codes, the nozzle must be capable of delivering at least 250 gal/min. If the nozzle is attached to a 3-in.-diameter hose, what pressure must be maintained just upstream of the nozzle to deliver this flowrate?

Answer 1

P1 = 42.93 psi

Step-by-step explanation:

For incompressible fluid, we know that;

A1V1 = A2V2

Making V1 the subject, we obtain;

V1 = A2V2/A1

Now A2V2 is the volumetric flow rate (V') .

Thus; V1 = V'/A1

A1 = πD²/4

Thus, V1 = 4V'/πD²

V' = 250 gal/min

But the diameter is in inches, let's convert to inches³/seconds.

Thus, V' = 250 x 3.85 = 962.5 in³/s

Substituting the relevant values to obtain,

V1 = (4 x 962.5)/(π x 3²) = 136.166 in/s.

Now let's convert to ft/s;

V1 = 136.166 x 0.0833 = 11.34 ft/s

Also for V2;

V2 = (4 x 962.5)/(π x 1.125²) = 968.29 in/s.

Now let's convert to ft/s;

V2 = 968.29 x 0.0833 = 80.66 ft/s

Setting bernoulli equation between the hose and the exit, we obtain;

(p1/γ) + (V1²/2g) = V2²/2g

Where V1 and V2 are intial and final velocities and γ is specific weight of water which is 62.43 lb/ft³ and g i acceleration due to gravity which is 32.2 ft/s²

Making p1 the subject, we obtain;

p1 = (γ/2g)(V2² - V1²)

p1 = (62.43/(2x32.2))(80.66² - 11.34²)

p1 = 6182.35 lb/ft²

So Converting to psi, we have;

p1 = 6182.35/144 = 42.93 psi

Answer 2

`P_(1) = 403,708\,kPa\,(58.553\,psi)`

Step-by-step explanation:

Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:

`(P_(1))/(\rho\cdot g) = (P_(2))/(\rho \cdot g) + (v^(2))/(2\cdot g)`

The initial pressure is:

`P_(1) = P_(2)+ (1)/(2)\cdot \rho v^(2)`

The speed at outlet is:

`v=(\dot Q)/((\pi)/(4)\cdot D^(2))`

`v=((250\,(gal)/(min) )\cdot ((3.785* 10^(-3)\,m^(3))/(1\,gal) )\cdot((1\,min)/(60\,s) ))/((\pi)/(4)\cdot [(1.125\,in)\cdot((0.0254\,m)/(1\,in) )]^(2) )`

`v\approx 24.592\,(m)/(s)\,(80.682\,(ft)/(s) )`

The initial pressure is:

`P_(1) = 101.325* 10^(3)\,Pa+(1)/(2)\cdot (1000\,(kg)/(m^(3)) )\cdot (24.592\,(m)/(s) )^(2)`

`P_(1) = 403,708\,kPa\,(58.553\,psi)`