1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Mrac [35]
3 years ago
13

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 2

50 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?
Engineering
2 answers:
Furkat [3]3 years ago
7 0

Answer:

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

Explanation:

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

\frac{P_{1}}{\rho\cdot g} = \frac{P_{2}}{\rho \cdot g} + \frac{v^{2}}{2\cdot g}

The initial pressure is:

P_{1} = P_{2}+ \frac{1}{2}\cdot \rho v^{2}

The speed at outlet is:

v=\frac{\dot Q}{\frac{\pi}{4}\cdot D^{2}}

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} }

v\approx 24.592\,\frac{m}{s}\,(80.682\,\frac{ft}{s} )

The initial pressure is:

P_{1} = 101.325\times 10^{3}\,Pa+\frac{1}{2}\cdot (1000\,\frac{kg}{m^{3}} )\cdot (24.592\,\frac{m}{s} )^{2}

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

Contact [7]3 years ago
4 0

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

You might be interested in
A converging - diverging frictionless nozzle is used to accelerate an airstream emanating from a large chamber. The nozzle has a
vodomira [7]

Answer:

a. 617.958kpa

b. 1.351kg/sec

Explanation:

Please see attachment for step by step guide.

5 0
3 years ago
Read 2 more answers
A fluid has a dynamic viscosity of 0.048 Pa.s and a specific gravity of 0.913. For the flow of such a fluid over a flat solid su
sattari [20]

Answer:

Explanation:

First we should recall how Newton's laws relates shear stress to a fluid's velocity profile:

\tau = \mu \cfrac{\partial v}{\partial y}

where tau is the shear stress, mu is viscosity, v is the fluid's velocity and y is the direction perpendicular to flow.

Now, in this case we have a parabolic velocity profile, and also we know that the fluid's velocity is zero at the boundary (no-slip condition) and that the vertex (maximum) is at y=75 \, mm and the velocity at that point is 1.125 \, m/s

We can put that in mathematical terms as:

v(y)= A+ By +Cy^2 \\v(0) = 0\\v(75 \, mm) = 1.125 \, m/s\\v'(75 \, mm) = 0\\

From the no-slip condition, we can deduce that A=0 and so we are left with just two terms:

v(y) = By + C y ^2 \\

We know that the vertex is at y= 75 \, mm and so we can rewrite the last equation as:

v(y) = k(y-75 \, mm) ^2+h

where k and h are constants to be determined. First we check that v( 75 \, mm) = 1.125 \,  m/s :

v( 75 \, mm) = k(75 \, mm -75 \, mm) ^2+h = h = 1.125 \, m/s\\\\h= v_{max} = 1.125 \,  m/s

So we found that h was the maximum velocity for the fluid, now we have to determine k, for that we need to make use of the no-slip condition.

v( 0) = k( -75 \, mm) ^2+  1.125 \,  m/s= 0 \quad (no \, \textendash slip)  \\\\k= - \cfrac{ 1.125 \, m/s }{(75 \, mm ) ^2} = - \cfrac{ 1125 \, mm/s }{(75 \, mm ) ^2}\\\\k= -  \cfrac{0.2}{mm \times s}

And thus we find that the final expression for the fluid's velocity is:

v( y) = 1125-  0.2 ( y -75 ) ^2

where v is in mm/s and y is in mm.

In SI units it would be:

v( y) = 1.125-  200 ( y -0.075 ) ^2

To calculate the shear stress, we need to take the derivative of this expression and multiply by the fluid's viscosity:

\tau = \mu \cfrac{\partial v}{\partial y}

\tau =0.048\,   \cdot  (-400) ( y-0.075   )

for y= 0.050 \, m we have:

\tau =0.048\,   \cdot  (-400) ( 0.050 -0.075   ) = 0.48\, Pa

Which is our final result

5 0
3 years ago
A company specification calls for a steel component to have a minimum tensile strength of 1240 MPa. Tension tests are conducted
Pavlova-9 [17]

Answer:

a) The minimum acceptable value is 387.5 HV using Vickers hardness test.

b) The minimum acceptable value is 39.4 HRC using Rockwell C hardness test.

Explanation:

To get the tensile strength of a material from its hardness, we multiply it by an empirical constant that depends on things like yield strength, work-hardening, Poisson's ratio and geometrical factors. The incidence of cold-work varies this relationship.

According to DIN 50150 (a conversion table for hardness), the constant for Vickers hardness is ≈ 3.2 (an empirical approximate):

\mbox{Tensile strength}=HV*3.2\\\\HV  = \frac{\mbox{Tensile strength}}{3.2} =\frac{1240}{3.2}=387.5

According to DIN 50150, the constant for Rockwell C hardness test is ≈31.5 around this values of tensile strength:

\mbox{Tensile strength}=HRC*31.5\\\\HRC  = \frac{\mbox{Tensile strength}}{31.5} =\frac{1240}{31.5}=39.4

8 0
3 years ago
What are the different branches of engineering involved in manufacturing a general-purpose elevator?
Anestetic [448]

Answer:

technical engineering

8 0
3 years ago
Visual aids are useful for all of the following reasons except
Ksivusya [100]

Answer:

B

Explanation:

their presence allows the speaker to take a rest from talking

I couldnt see the question

4 0
2 years ago
Other questions:
  • A mysterious device found in a forgotten laboratory accumulates charge at a rate specified by the expression gm = 9 - 10tC from
    13·1 answer
  • Cng containers need to be inspected
    7·1 answer
  • In order to build a skyscraper Builders, Inc. hires 400 construction workers and 50 managers. Builders, Inc. represents A entrep
    8·1 answer
  • How does a carburetor work?
    7·1 answer
  • Are you able to text without looking at your phone?
    10·1 answer
  • Which of the following is not necessary a reason to machine a brake drum?
    8·1 answer
  • What are the benefits of using a multi view sketch to communicate a design
    14·1 answer
  • ───────────────────────────────
    7·1 answer
  • Which term describes the lowest of a foundation?
    10·1 answer
  • Jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!