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Aleonysh [2.5K]
3 years ago
6

Consider a fully developed laminar flow in a circular pipe. The velocity at R/2 (midway between the wall surface and the centerl

ine) is given by _____ provided that Vmax is the maximum velocity in the tube.
Engineering
1 answer:
ryzh [129]3 years ago
4 0

Answer:

The velocity at R/2 (midway between the wall surface and the centerline) is given by (3/4)(Vmax) provided that Vmax is the maximum velocity in the tube.

Explanation:

Starting from the shell momentum balance equation, it can be proved that the velocity profile for fully developedblaminar low in a circular pipe of internal radius R and a radial axis starting from the centre of the pipe at r=0 to r=R is given as

v = (ΔPR²/4μL) [1 - (r²/R²)]

where v = fluid velocity at any point in the radial direction

ΔP = Pressure drop across the pipe

μ = fluid viscosity

L = pipe length

But the maximum velocity of the fluid occurs at the middle of the pipe when r=0

Hence, maximum veloxity is

v(max) = (ΔPR²/4μL)

So, velocity at any point in the radial direction is

v = v(max) [1 - (r²/R²)]

At the point r = (R/2)

r² = (R²/4)

(r²/R²) = r² ÷ R² = (R²/4) ÷ (R²) = (1/4)

So,

1 - (r²/R²) = 1 - (1/4) = (3/4)

Hence, v at r = (R/2) is given as

v = v(max) × (3/4)

Hope this Helps!!!

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An asbestos pad is square in cross section, measuring 5 cm on a side at its small end increasing linearly to 10 cm on a side at
polet [3.4K]

Answer:

q = 1.73 W

Explanation:

given data

small end  = 5 cm

large end = 10 cm

high = 15 cm

small end is held = 600 K

large end at = 300 K

thermal conductivity of asbestos  = 0.173 W/mK

solution

first we will get here side of cross section that is express as

S = S1 + \frac{S2-S1}{L} x     ...............1

here x is distance from small end and S1 is side of square at small end

and S2 is side of square of large end and L is length

put here value and we get

S = 5 + \frac{10-5}{15} x

S = \frac{0.15 + x}{3}    m

and  

now we get here Area of section at distance x is

area A = S²    ...............2

area A = (\frac{0.15 + x}{3})^2    m²

and

now we take here small length dx and temperature difference is dt

so as per fourier law

heat conduction is express as

heat conduction q = \frac{-k\times A\  dt}{dx}      ...............3

put here value and we get

heat conduction q = -k\times (\frac{0.15 + x}{3})^2 \   \frac{dt}{dx}  

it will be express as

q \times \frac{dx}{(\frac{0.15 + x}{3})^2} = -k (dt)  

now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K

q \int\limits^{0.15}_0 {\frac{dx}{(\frac{0.15 + x}{3})^2 } = -0.173 \int\limits^{300}_{600} {dt}          

solve it and we get

q (30)  = (0.173) × (600 - 300)

q = 1.73 W

5 0
2 years ago
Select four geometric shapes that are commonly used by engineers in their designs.
stealth61 [152]

Answer:

Arc ,sphere ,line ,prism

Explanation:

  • Arc is used for semicircular or short term projects
  • Sphere is used for design of spherical surfaces or projects
  • Prism is used for creating buliding designs
  • Lines are especially used for creating roads, flyover design
3 0
2 years ago
With a very precise volumetric measuring device, the volume of a liquid sample is determined to be 6.321 L (liters). Three stude
zheka24 [161]

Answer:

See explanation

Explanation:

Solution:-

- Three students measure the volume of a liquid sample which is 6.321 L.

- Each student measured the liquid sample 4 times. The data is provided for each measurement taken by each student as follows:

                                                 Students

                      Trial          A            B               C

                         1            6.35        6.31          6.38

                        2            6.32        6.31          6.32

                        3            6.33        6.32         6.36

                        4            6.36        6.35         6.36

- We will define the two terms stated in the question " precision " and "accuracy"

- Precision refers to how close the values are to the sample mean. The dense cluster of data is termed to be more precise. We will use the knowledge of statistics and determine the sample standard deviation for each student.

- The mean measurement taken by each student would be as follows:

                       E ( A ) = \frac{6.35 +6.32+6.33+6.36}{4} \\\\E ( A ) = 6.34\\\\E ( B ) = \frac{6.31 +6.31+6.32+6.35}{4} \\\\E ( B ) = 6.3225\\\\E ( C ) = \frac{6.38 +6.32+6.36+6.36}{4} \\\\E ( C ) = 6.355\\

- The precision can be quantize in terms of variance or standard deviation of data. Therefore, we will calculate the variance of each data:

 

                        Var ( A ) = \frac{6.35^2+6.32^2+6.33^2+6.36^2}{4} - 6.34^2\\\\Var ( A ) = 0.00025\\\\Var ( B ) = \frac{6.31^2+6.31^2+6.32^2+6.35^2}{4} - 6.3225^2\\\\Var ( B ) = 0.00026875\\\\Var ( C ) = \frac{6.38^2+6.32^2+6.36^2+6.36^2}{4} - 6.355^2\\\\Var ( C ) = 0.000475\\

- We will rank each student sample data in term sof precision by using the values of variance. The smallest spread or variance corresponds to highest precision. So we have:

                   Var ( A )          <          Var ( B )        <    Var ( C )

                   most precise                                      Least precise

- Accuracy refers to how close the sample mean is to the actual data value. Where the actual volume of the liquid specimen was given to be 6.321 L. We will evaluate the percentage difference of sample values obtained by each student .

                       P ( A ) = \frac{6.34-6.321}{6.321}*100= 0.30058\\\\P ( B ) = \frac{6.3225-6.321}{6.321}*100= 0.02373\\\\P ( C ) = \frac{6.355-6.321}{6.321}*100= 0.53788\\

- Now we will rank the sample means values obtained by each student relative to the actual value of the volume of liquid specimen with the help of percentage difference calculated above. The least percentage difference corresponds to the highest accuracy as follows:

                   P ( B )         <       P ( A )         <      P ( C )

            most accurate                                least accurate

7 0
3 years ago
The constant A in Equation 17.2 is 12π4 R/5(θD) 3 where R is the gas constant and θD is the Debye temperature (K). Estimate θD f
kirill115 [55]

Answer:

The Debye temperature for aluminum is 375.2361 K

Explanation:

Molecular weight of aluminum=26.98 g/mol

T=15 K

The mathematical equation for the specific heat and the absolute temperature is:

C_{v} =AT^{3}

Substituting in the expression of the question:

C_{v} =(\frac{12\pi ^{4}R }{5\theta _{D}^{3}  } )T^{3}

\theta _{D} =(\frac{12\pi ^{4}RT^{3}  }{5C_{v}   } ) ^{1/3}

Here

C_{v} =4.6\frac{J}{kg-K} *\frac{1kg}{1000g} *\frac{26.98g}{1mol} =0.1241J/mol-K

Replacing:

\theta _{D}  =(\frac{12\pi ^{4}*8.31*15^{3}  }{5*0.1241} )^{1/3} =375.2361K

3 0
2 years ago
Consider a fully developed flow in a circular pipe (negligible entrance effects), how does the convective heat transfer coeffici
Alexeev081 [22]

Answer: A)  Gradually decrease

Explanation:

  The convection value of heat transfer rate are gradually decreasing with the flow of the heat. Flow in a circular pipe, flow direction does not change in the velocity path. The average of the coefficient of heat transfer and the number of pipes are needed and the effects are get neglected so that is why the flow are fully developed.

8 0
3 years ago
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