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3 years ago

Dan is applying for a job as an engineering technician. What are the tasks that he may be asked to perform?

Engineering

1 answer:

OleMash [197]3 years ago

8 0

Answer:

Engineering technicians help solve technical problems in many ways. They build or set up equipment, conduct experiments, and collect data and calculate results. They might also help to make a model of new equipment. Some technicians work in quality control, where they check products, do tests, and collect data.

Short Answer:

He may be asked to build or set up equitment, Go through experiments, Collect Data and calculate results. may be in charge of making new models or equitment and may be asked to work in quality control(collect data, do tests and check products).

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Challenge:

goldenfox [79]

Answer:

what?

Explanation:

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4 years ago

/ Air enters a 20-cm-diameter 12-m-long underwater duct at 50°C and 1 atm at a

Digiron [165]

Answer:

A) EXIT TEMPERATURE = 14⁰C

b) rate of heat transfer of air = - 13475.78 = - 13.5 kw

Explanation:

Given data :

diameter of duct = 20-cm = 0.2 m

length of duct = 12-m

temperature of air at inlet= 50⁰c

pressure = 1 atm

mean velocity = 7 m/s

average heat transfer coefficient = 85 w/m^2⁰c

water temperature = 5⁰c

surface temperature ( Ts) = 5⁰c

properties of air at 50⁰c and at 1 atm

= 1.092 kg/m^3

Cp = 1007 j/kg⁰c

k = 0.02735 W/m⁰c

Pr = 0.7228

v = 1.798 * 10^-5 m^2/s

determine the exit temperature of air and the rate of heat transfer

attached below is the detailed solution

Calculate the mass flow rate

= p*Ac*Vmean

= 1.092 * 0.0314 * 7 = 0.24 kg/s

4 0

3 years ago

What is the highest possible theoretical efficiency of a heat engine operating with a hot reservoir of furnace gases at 727°C wh

hichkok12 [17]

Answer:

$\eta_{max} = 70%$

Explanation:

given data:

maximum temperature = 727 degree C =727 +273 = 1000 K

Minimum temperature = 27 degree C = 27 +273 = 300 K

Maximum efficiency is given as $\eta$

$\eta = \frac{T_{max} - T_{min}}{T_{max}} *100%$

plugging all value from the above to get maximum efficiency

$\eta = \frac{1000-300}{1000} *100%$

$\eta_{max} = 70%$

8 0

3 years ago

A ________ is a condition or capability needed by a user to solve a problem or achieve an objective that satisfies a standard or

Aliun [14]

A successful brain to help

7 0

3 years ago

Ammonia gas is diffusing at a constant rate through a layer of stagnant air 1 mm thick. Conditions are such that the gas contain

fiasKO [112]

Answer:

The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)

That is the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

Explanation:

The diffusion through a stagnant layer is given by

$N_{A} = \frac{D_{AB} }{RT} \frac{P_{T} }{z_{2} - z_{1} } ln(\frac{P_{T} -P_{A2} }{P_{T} -P_{A1} })$

Where

$D_{AB}$ = Diffusion coefficient or diffusivity

z = Thickness in layer of transfer

R = universal gas constant

$P_{A1}$ = Pressure at first boundary

$P_{A2}$ = Pressure at the destination boundary

T = System temperature

$P_{T}$ = System pressure

Where $P_{T}$ = 101.3 kPa $P_{A2} =0$ , $P_{A1} =y_{A}$ , $P_{T} =$ 0.5×101.3 = 50.65 kPa

Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m

R = $\frac{kJ}{(kmol)(K)} ,$ T = 298 K and $D_{AB}$ = 1.18 $\frac{cm^{2} }{s}$ = 1.8×10⁻⁵ $\frac{m^{2} }{s}$

$N_{A} = \frac{1.8*10^{-5} }{8.314*295} *\frac{101.3}{1*10^{-3} }* ln(\frac{101.3-0}{101.3-50.65})$ = 5.153×10⁻⁴ $\frac{kmol}{m^{2}s }$

Hence the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

5 0

3 years ago