Answer:
[tex]P_{1} = 403,708\,kPa\,(58.553\,psi)[/tex]
Explanation:
Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:
[tex]\frac{P_{1}}{\rho\cdot g} = \frac{P_{2}}{\rho \cdot g} + \frac{v^{2}}{2\cdot g}[/tex]
The initial pressure is:
[tex]P_{1} = P_{2}+ \frac{1}{2}\cdot \rho v^{2}[/tex]
The speed at outlet is:
[tex]v=\frac{\dot Q}{\frac{\pi}{4}\cdot D^{2}}[/tex]
[tex]v=\frac{(250\,\frac{gal}{min} )\cdot (\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot [(1.125\,in)\cdot(\frac{0.0254\,m}{1\,in} )]^{2} }[/tex]
[tex]v\approx 24.592\,\frac{m}{s}\,(80.682\,\frac{ft}{s} )[/tex]
The initial pressure is:
[tex]P_{1} = 101.325\times 10^{3}\,Pa+\frac{1}{2}\cdot (1000\,\frac{kg}{m^{3}} )\cdot (24.592\,\frac{m}{s} )^{2}[/tex]
[tex]P_{1} = 403,708\,kPa\,(58.553\,psi)[/tex]
Answer:
P1 = 42.93 psi
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
