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3 years ago

What separates the work of technology transfer research from implementation of the products of such research?

Engineering

1 answer:

tatuchka [14]3 years ago

4 0

Answer:

The thing that separates the work of technology transfer research from implementation of the products of such research is:

Technology transfer research looks at how technology may be transferred but does not actually make the transfer.

Explanation:

This suggests that technology transfer research is different from the technology transfer (implementation) itself. The first stops at making scientific investigations into technology transfer activities while the next step performs the actual transfer or implementation. In other words, technology transfer conveys the results of scientific and technological research to the marketplace and to the wider society. It is the bridge-builder between the research and the implementation.

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25°C is flowing in a covered irrigation ditch below ground. Every 100 ft, there is a vent line with a 1.0 in. inside diameter an

Gnesinka [82]

The total evaporation loss of water is 87.873 × $10^{-5}$ lbm /day.

<u>Explanation:</u>

Assume A is the water and B is air.

A is diffusing to non diffusing B.

$N_{A}=\frac{D_{A B} P}{R T Z P_{B/ M}}\left(P_{A_{1}} \ - P_{B_{2}}\right)$

By the table 6.2 - 1 at 25°C, the diffusivity of air and water system is $0.260 \times 10^{-4} \mathrm{m}^{2} / \mathrm{s}$ .

Total pressure P = 1 atm = 101.325 KPa

$P_{A_{1} }$ = 23.76 mm Hg

$P_{A_{1} }$ = $\frac{23.76}{760}$

$P_{A_{1} }$ = 0.03126 atm

$P_{A_{1} }$ = 3.167 K Pa

When air surrounded is dry air, then $PA_{2}$ = 0 mm Hg

R = 8.314 $\frac{K P a \cdot m^{3}}{mol \cdot k}$

$P_{B/M}$ = $\frac{P_{B_{1}}-P_{B_{2}}}{\ln \left(P_{B_{1}} / P_{B_{2}}\right)}$

$P_{B_{1}}=P_{T}-P_{A1_{}}$

= 101.325 - 3.167

$P_{B_{1} }$ = 98.158 K Pa

$P_{B_{2}}=P_{T}-P_{A_{2}}$

$P_{B_{2} }$ = 101.325 - 0

$P_{B_{2} }$ = 101.325 K Pa

$P_{B / M}=\frac{98.158-101.325}{\ln (98.158 / 101 \cdot 325)}$

$P_{B/M}$ = 99.733 K Pa

Z = 1 ft = 0.3048 m

T = 298 K

$N_{A}$ = $\frac{(0.206*10^{-4})(101.325)(3.167 - 0) }{(8.314) ( 298 )( 0.3048 )( 99.733 ) }$

$N_{A}$ = 0.11077 × $10^{-6}$ mol/ $m^{2}$ .s

$N_{A}$ = 0.11077 × $10^{-6}$ × 18 × (60×60×24)

$N_{A}$ = 0.1723 $lb_{m}$ / $m^{2}$ .day

$\tilde{N}_{A}=N_{A} \times Area$

Area of individual pipe is

$A=\frac{\pi}{4}(0.0254)^{2}$

A = 0.00051 $m^{2}$

$\bar{N}_{\boldsymbol{A}}$ = 0.1723 × 0.00051

$\bar{N}_{\boldsymbol{A}}$ = 0.000087873 lbm/day

In 1000 ft length of ditch,there will be a 10 pipes. The amount of evaporation water is

= 10 × 0.000087873 = 0.00087873 lbm /day

The total evaporation loss of water is 87.873 × $10^{-5}$ lbm /day.

5 0

3 years ago

In some companies, workers who increase the quantity or quality of their work receive a. benefits. c. performance bonuses. b. pe

Greeley [361]

Answer:

performance bonuses

Explanation:

3 0

3 years ago

You have an ac voltage of 10 sin(60t). Design a full wave bridge rectifier which will give you an output voltage that varies a m

Masteriza [31]

As there are 10 V, for Vp1, that is the peak-voltage of the source:

$Vp1=10*\sqrt{2}=14.14 V$

Then, transformer's theory says that the relation of transformations is:

V1/V2=a

Where V1 is the voltage in the primary and V2 in the secondary.

V1=14.14 V

V2=8.55 V

a=1.65

Then, with the 8.5 V, we find the real peak-voltage, taking in account that in the diodes we have a drop of 0.7 V each, so:

8.5 -1.4=7.1 V

And this will be called VpL

Now we proceed to calculate the mean voltage:

$V_{mean}=VpL-(\frac{Vr}{2} )$

Where Vr is the ripple voltage, we asume that is 1 V

So, Vmean = 6.6 V

Then we have

Vmean/R= I mean

We have that R=1000 Ohm

Imedia=6.6 V/1000 Ohm

Imedia=6.6 mAmps

Finally, we can calculate the capacitor:

C=Q/Vr

C=Imean/(Vr*2f)

Where f is 60Hz

C=6.6mA/(1V*120)

C=5.5 uFarads

Therefore:

C=5.5 uFarads that works at 12 V

6 0

3 years ago

A very large plate is placed equidistant between two vertical walls. The 10-mm spacing between the plate and each wall is filled

Vikentia [17]

Answer:

Force per unit plate area is 0.1344 $N/m^{2}$

Solution:

As per the question:

The spacing between each wall and the plate, d = 10 mm = 0.01 m

Absolute viscosity of the liquid, $\mu =1.92\times 10^{- 3} Pa-s$

Speed, v = 35 mm/s = 0.035 m/s

Now,

Suppose the drag force that exist between each wall and plate is F and F' respectively:

Net Drag Force = F' + F''

$F = \tau A$

where

$\tau$ = shear stress

A = Cross - sectional Area

Therefore,

Net Drag Force, F = $(\tau ' +\tau '')A$

$\frac{F}{A} = \tau ' +\tau ''$

Also

F = $\frac{\mu v}{d}$

where

$\mu$ = dynamic coefficient of viscosity

Pressure, P = $\frac{F}{A}$

Therefore,

$\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}$

$\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}$

8 0

3 years ago

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the str

Murljashka [212]

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

$J=\frac{\pi}{32}d^4$