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sertanlavr [38]
3 years ago
13

Lars is a medical coder. He works for a hospital in Wisconsin but currently lives in Georgia. Lars does his work online and over

conference calls. Lars is an example of a remote worker.
True or False
Engineering
1 answer:
Brums [2.3K]3 years ago
6 0
The answer is True because he works in one location while on another
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1 A power transmission includes a belt drive, a chain drive and a gear drive. Which of the following is the best arrangement bet
Setler79 [48]

Answer:d

Explanation:

(d) chain drive belt drive gear drive

3 0
3 years ago
Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes
loris [4]

Answer:

a). 8.67 x 10^{-3} m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L  = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X 10^{-5} Pa-s

We know density of air is ρ = 1.21 kg /m^{3}

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = \frac{\rho .U.x}{\mu }

Re = \frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = \frac{5.x}{\sqrt{Re}}

   = \frac{5\times 1}{\sqrt{332052.6}}

   = 8.67 x 10^{-3} m

a). Boundary layer thickness at x = 1 is δ = 8.67 X 10^{-3} m

b). Given Re = 100000

    Therefore the critical distance from the leading edge can be found by,

     Re = \frac{\rho .U.x}{\mu }

     100000 = \frac{1.21\times5\times x}{1.822 \times10^{-5}}

     x = 0.3011 m

c). Given x = 3 m from the leading edge

    The Reyonld no at x = 3 m from the leading edge

     Re = \frac{\rho .U.x}{\mu }

     Re = \frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }

     Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

    δ = \frac{0.38\times x}{Re^{\frac{1}{5}}}

       = \frac{0.38\times 3}{996158.06^{\frac{1}{5}}}

       = 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

  Re = \frac{\rho \times U\times x}{\mu }

5 X 10^{5} = \frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}

 x = 1.505 m

We know that the force acting on the plate is

F_{D} = \frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}

and C_{D} at x= 1.505 for a laminar flow is = \frac{1.328}{\sqrt{Re}}

                                                                         = \frac{1.328}{\sqrt{5\times10 ^{5}}}

                                                                       = 1.878 x 10^{-3}

Therefore, F_{D} =  \frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}

                                          = \frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}

                                         = 0.2137 N

e). The flow is turbulent at the end of the plate.

  Re = \frac{\rho \times U\times x}{\mu }

       = \frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }

       = 1992316

Therefore C_{D} = \frac{0.072}{Re^{\frac{1}{5}}}

                                           = \frac{0.072}{1992316^{\frac{1}{5}}}

                                           = 3.95 x 10^{-3}

Therefore F_{D} = \frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}

                                           = \frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}

                                          = 1.792 N

3 0
3 years ago
Examine a process whereby air at 300 K, 100 kPa is compressed in a piston/cylinder arrangement to 600 kPa. Assume the process is
professor190 [17]

Answer:

See attachment and explanation.

Explanation:

- The following question can be solved better with the help of a MATLAB program as follows. The code is given in the attachment.

- The plot of the graph is given in attachment.

- The code covers the entire spectrum of the poly-tropic range ( 1.2 - 1.6 ) and 20 steps ( cases ) have been plotted and compared in the attached plot.

3 0
3 years ago
An air conditioning system operating on reversed carnot cycle is required to remove heat from the house at a rate of 32kj/s to m
Brilliant_brown [7]

Answer:

(e) 1.64 kW

Explanation:

The Coefficient of Performance of the Reverse Carnot's Cycle is:

COP = \frac{T_{L}}{T_{H}-T_{L}}

COP = \frac{293.15\,K}{308.15\,K-293.15\,K}

COP = 19.543

Lastly, the power required to operate the air conditioning system is:

\dot W = \frac{\dot Q_{L}}{COP}

\dot W = \frac{32\,kW}{19.543}

\dot W = 1.637\,kW

Hence, the answer is E.

3 0
3 years ago
The mathematical relationship between
11Alexandr11 [23.1K]
4) Ohms law thats the answer
7 0
3 years ago
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