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
Maksim231197 [3]
3 years ago
8

In a certain chemical plant, a closed tank contains ethyl alcohol to a depth of 71 ft. Air at a pressure of 17 psi fills the gap

at the top of the tank. Determine the pressure at a closed valve attached to the tank 10 ft above its bottom.
Engineering
1 answer:
Yuliya22 [10]3 years ago
5 0

Answer:

the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Explanation:

Given that;

depth 1 = 71 ft

depth 2 = 10 ft

pressure p = 17 psi = 2448 lb/ft²

depth h = 71 ft - 10 ft = 61 ft

we know that;

p = P_air + yh

where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )

so we substitute;

p = 2448 + ( 49.3 × 61 )

= 2448 + 3007.3

= 5455.3 lb/ft³

= 37.88 psi

Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

You might be interested in
A. Briefly describe the microstructural difference between spheroidite and tempered martensite. Explain why tempered martensite
masha68 [24]

Answer:

Answered below.

Explanation:

A) Both spheroidite & tempered martensite possess sphere - like cementite particles within their microstructure known as a ferrite matrix. However, the difference is that these particles are much larger for spheroidite than tempered.

B) Tempered martensite is much harder and stronger than spheroidite primarily because there is much more ferrite - cementite phase boundary area for its sphere - like cementite particles.

This is because the greater the boundary area, the more the hardness.

4 0
3 years ago
Which type of modeling can create virtual designs that can save clients thousands of dollars?
swat32

Answer:

VR Prototyping

VR Prototyping Can Save you Thousands of Dollars.

Explanation:

there you go lad

8 0
3 years ago
Firefighters are holding a nozzle at the end of a hose while trying to extinguish a fire. The nozzle exit diameter is 8 cm, and
ivanzaharov [21]

Question

Determine the average water exit velocity

Answer:

53.05 m/s

Explanation:

Given information

Volume flow rate, Q=16 m^{3}/min

Diameter d= 8cm= 0.08 m

Assumptions

  • The flow is jet flow hence momentum-flux correction factor is unity
  • Gravitational force is not considered
  • The flow is steady, frictionless and incompressible
  • Water is discharged to the atmosphere hence pressure is ignored

We know that Q=AV and making v the subject then

V=\frac {Q}{A} where V is the exit velocity and A is area

Area, A=\frac {\pi d^{2}{4} where d is the diameter

By substitution

V=\frac {16\times 4}{\pi 0.08^{2}}=3183.098862 m/min

To convert v to m/s from m/s, we simply divide it by 60 hence

V=\frac {3183.098862  m/min}{60 s}=53.0516477 m/s\approx 53.05 m/s

3 0
3 years ago
An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40MPa. It has been dete
Nataly [62]

Answer:

Yes, fracture will occur

Explanation:

Half length of internal crack will be 4mm/2=2mm=0.002m

To find the dimensionless parameter, we use critical stress crack propagation equation

\sigma_c=\frac {K}{Y \sqrt {a\pi}} and making Y the subject

Y=\frac {K}{\sigma_c \sqrt {a\pi}}

Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, \sigma_c  is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain

Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682

When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m

\sigma_c=\frac {K}{Y \sqrt {a\pi}}  and making K the subject

K=\sigma_c Y \sqrt {a\pi}  and substituting 260 MPa for \sigma_c  while a is taken as 0.003m and Y is already known

K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa

Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material

6 0
3 years ago
Only answer this if your name is riley
Sati [7]

Answer:

hey im like kinda riley

Explanation:

y u wanna talk to moi

3 0
3 years ago
Read 2 more answers
Other questions:
  • Given int variables k and total that have already been declared, use a do...while loop to compute the sum of the squares of the
    15·1 answer
  • Where Does a Solar Engineer Work? <br> (2 sentences or more please)
    14·2 answers
  • For ceramic-matrix composites, high interfacial strength is desirable. ( True , False )
    8·1 answer
  • For what two reasons do countries specialize? Countries specialize so that opportunity costs can be increased. Countries special
    13·1 answer
  • 10. True or False: You should select your mechanic before you experience vehicle failure.
    6·2 answers
  • Plz give solutions..... ​
    6·1 answer
  • Contrast the electron and hole drift velocities through a 10 um (micro meter) layer of intrinsic silicon across which a voltage
    11·1 answer
  • Christopher has designed a fluid power system that repeatedly gets clogs. Which of the following objects should he choose to add
    13·1 answer
  • Is there a project idea, or invention that would be good for<br> my class.
    6·2 answers
  • In order to live and grow, bacteria need moisture, food, the right temperature, and ______? Fill in the blank
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!