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
makvit [3.9K]
3 years ago
14

1:

Engineering
1 answer:
Nataly [62]3 years ago
7 0

Answer:

20.87 Pa

Explanation:

The formula for dynamic pressure is given as;

q= 1/2*ρ*v²

where ;

q=dynamic pressure

ρ = density of fluid

v = velocity of fluid

First find v by applying the formula for flow rate as;

Q = v*A   where ;

Q= fluid flow rate

v = flow velocity

A= cross-sectional area.

A= cross-sectional vector area of the pipe given by the formula;

A= πr² = 3.14 * 4² = 50.27 in²   where r=radius of pipe obtained from the diameter given divided by 2.

Q = fluid flow rate = 105 gpm----change to m³/s as

1 gpm = 0.00006309

105 gpm = 105 * 0.00006309 = 0.006624 m³/s

A= cross-sectional vector area = 50.27 in² -------change to m² as:

1 in² = 0.0006452 m²

50.27 in² = 50.27 * 0.0006452 = 0.03243 m²

Now calculate flow velocity as;

Q =v * A

Q/A = v

0.006624 m³/s / 0.03243 m² =v

0.2043 m/s = v

Now find the dynamic pressure q given as;

q= 1/2 * ρ*v²

q= 1/2 * 1000 * 0.2043² = 20.87 Pa

You might be interested in
Air is to be heated steadily by an 8-kW electric resistance heater as it flows through an insulated duct. If the air enters at 5
Furkat [3]

To solve this problem it is necessary to apply the concepts related to the heat exchange of a body.

By definition heat exchange in terms of mass flow can be expressed as

W = \dot{m}c_p \Delta T

Where

C_p = Specific heat

\dot{m}= Mass flow rate

\Delta T = Change in Temperature

Our values are given as

C_p = 1.005kJ/kgK \rightarrow Specific heat of air

T_1 = 50\°C

\dot{m} = 2kg/s

W = 8kW

From our equation we have that

W = \dot{m}c_p \Delta T

W = \dot{m}c_p (T_2-T_1)

Rearrange to find T_2

T_2 = \frac{W}{\dot{m}c_p}+T_1

Replacing

T_2 = \frac{8}{2*1.005}+(50+273)

T_2 = 326.98K \approx 53.98\°C

Therefore the exit temperature of air is 53.98°C

6 0
3 years ago
A fluid flows steadily through a pipe with a uniform cross sectional area. The density of the fluid decreases to half its initia
Vikentia [17]

Answer:

c. V2 equals V1

Explanation:

We can answer this question by using the continuity equation, which states that:

A_1 v_1 = A_2 v_2 (1)

where

A1 is the cross-sectional area in the first section of the pipe

A2 is the cross-sectional area in the second section of the pipe

v1 is the velocity of the fluid in the first section of the pipe

v2 is the velocity of the fluid in the second section of the pipe

In this problem, we are told that the pipe has a uniform cross sectional area, so:

A1 = A2

As a consequence, according to eq.(1), this means that

v1 = v2

so, the velocity of the fluid in the pipe does not change.

5 0
3 years ago
The cross-section of a rough, rectangular, concrete() channel measures . The channel slope is 0.02ft/ft. Using the Darcy-Weisbac
bazaltina [42]

Answer:

The following are the answer to this question:

Explanation:

In point a, Calculating the are of flow:

\bold{Area =B \times D_f}

         =6\times 5\\\\=30 \ ft^2

In point b, Calculating the wetter perimeter.

\bold{P_w =B+2\times D_f}

      = 6 +2\times (5)\\\\= 6 +10 \\\\=16 \ ft

In point c, Calculating the hydraulic radius:

\bold{R=\frac{A}{P_w}}

   =\frac{30}{16}\\\\= 1.875 \ ft

In point d, Calculating the value of Reynolds's number.

\bold{Re =\frac{4VR}{v}}

     =\frac{4V \times 1.875}{1 \times 10^{-5} \frac{ft^2}{s}}\\\\

     =750,000 V

Calculating the velocity:

V= \sqrt{\frac{8gRS}{f}}

   = \sqrt{\frac{8\times 32.2 \times 1.875 \times 0.02}{f}}\\\\=\frac{3.108}{\sqrt{f}}\\\\

\sqrt{f}=\frac{3.108}{V}\\\\

calculating the Cole-brook-White value:

\frac{1}{\sqrt{f}}= -2 \log (\frac{K}{12 R} +\frac{2.51}{R_e \sqrt{f}})\\\\ \frac{1}{\frac{3.108}{V}}= -2 \log (\frac{2 \times 10^{-2}}{12 \times 1.875} +\frac{2.51}{750,000V\sqrt{f}})\\

\frac{V}{3.108} =-2\log(8.88 \times 10^{-5} + \frac{3.346 \times 10^{-6}}{750,000(3.108)})

After calculating the value of V it will give:

V= 25.18 \ \frac{ft}{s^2}\\

In point a, Calculating the value of Froude:

F= \frac{V}{\sqrt{gD}}

= \frac{V}{\sqrt{g\frac{A}{\text{Width flow}}}}\\

= \frac{25.18}{\sqrt{32.2\frac{30}{6}}}\\\\= \frac{25.18}{\sqrt{32.2 \times 5}}\\\\= \frac{25.18}{\sqrt{161}}\\\\=  \frac{25.18}{12.68}\\\\= 1.98

The flow is supercritical because the amount of Froude is greater than 1.  

Calculating the channel flow rate.

Q= AV

   =30x 25.18\\\\= 755.4 \ \frac{ft^3}{s}\\

4 0
3 years ago
When cutting a FBD through an axial member, assume that the internal force is tension and draw the force arrow _______ the cut s
shepuryov [24]

Problem-Solving Tip: When cutting an FBD through an axial member, assume that the internal force is tension and draw the force arrow directed away from the cut surface. If the computed internal force value turns out to be a positive number, then the assumption of tension is confirmed.

3 0
2 years ago
the velocity of a particle is given by v=16t^2i+4t^3j+(5t+2k) m/s, where t is in seconds. if the particle is at the origin when
pshichka [43]

Answer:

80.16 m/s^2

at t=2 s

x=42.3 m

y=16 m

z=14 m

Explanation:

solution

The x,y,z components of the velocity are donated by the i,j,k vectors.

v_{x}=16t^{2}  \\v_{y}=4t^{3}\\v_{z}=5t+2

acceleration is a derivative of velocity with respect to time.

a_{x}=\frac{d}{dt} v_{x}=\frac{d}{dt}[16t^{2}]=32t\\a_{y}=\frac{d}{dt} v_{y}=\frac{d}{dt}[4t^{3}]=12t^{2} \\a_{z}=\frac{d}{dt} v_{z}=\frac{d}{dt}[5t+2]=5

evaluate acceleration at 2 seconds

a_{x} =32*2=64m/s^{2}\\ a_{y} =12*2^{2} =48m/s^{2}\\a_{z} =5m/s^{2}

the magnitude of the acceleration is the square root of the sum of the square of each component of the acceleration.

=\sqrt{a_{x}^2 +a_{y}^2+a_{z} ^2 } \\=\sqrt{64^2 +48^2+5 ^2 }\\=80.16m/s^2

position is the integral of velocity with respect to time position at a time can be found by taking by taking the definite intergral of each component.

x=\int\limits {v_{x} } \, dx=\int\limits^2_0 {{16t^2} \, dt=42.7m\\\\y=\int\limits {v_{y} } \, dx=\int\limits^2_0 {{4t^3} \, dt=16m\\\\\\\\\\z=\int\limits {v_{z} } \, dx=\int\limits^2_0 {{5t+2} \, dt=14m\\\\

3 0
3 years ago
Other questions:
  • A corroded metal gusset plate was found on a bridge. It was estimated that the original area of the plate was 750 cm2 and that a
    11·1 answer
  • 11 Notează, în caiet, trăsăturile personajelor ce se pot
    13·1 answer
  • Carbon dioxide flows at a rate of 1.5 ft3 /s from a 3-in. pipe in which the pressure and temperature are 20 psi (gage) and 120 °
    8·1 answer
  • Determine the Thevenin/Norton Equivalent Circuit with respect to the terminalsa,bas shown in the figure. (Here 1A is an independ
    11·1 answer
  • In unguided medium (free space), the electromagnetic (EM) signal wave spreads as it leaves the transmit antenna. Since the power
    10·1 answer
  • 12 times the square root of 8737
    13·1 answer
  • Consider a sphere made of stainless steel with diameter of 25 cm. It is heated to temperature of 300°C for some chemical tests.
    12·1 answer
  • What forced induction device is more efficient?
    8·2 answers
  • What is the maximum fine for knowingly refilling a disposable refrigerant drum?
    11·1 answer
  • Do better then me......................................
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!