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3 years ago

Which term defines the amount of mechanical work an engine can do per unit of heat energy it uses?

Engineering

1 answer:

skad [1K]3 years ago

7 0

Answer:

Explanation:

is the because that's the amount of work in making machine can do producing heat

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A bridge to be fabricated of steel girders is designed to be 500 m long and 12 m wide at ambient temperature (assumed 20°C). Exp

Volgvan

Answer:

a) 22.5number

b) 22.22 m length

Explanation:

Given data:

Bridge length = 500 m

width of bridge = 12 m

Maximum temperature = 40 degree C

minimum temperature = - 35 degree C

Maximum expansion can be determined as

$\Delta L = L \alpha (T_{max} - T_{min})$

where , \alpha is expansion coefficient $= 12\times 10^{-6}$ degree C

SO, $\Delta L = 500\times 12\times 10^{-6}\times ( 40 - (-35))$

$\Delta L = 0.45 m = 450 mm$

number of minimum expansion joints is calculated as

$n = \frac{450}{20} = 22.5$

b) length of each bridge

$Length = \frac{500}{22.5} = 22.22 m$

8 0

2 years ago

A barometer reads a height of 78 cmHg. Express this atmospheric pressure to:

Yakvenalex [24]

A barometer reads a height of 78 cmHg. Express this atmospheric pressure to $\large\mathsf{\red{\underline{Pascal(pa)}}}$

7 0

2 years ago

Read 2 more answers

A silicon diode has a saturation current of 6 nA at 25 degrees Celcius. What is the saturation current at 100 degrees Celsius?

Illusion [34]

Answer:

0.0659 A

Explanation:

Given that :

$I_{0} = 6nA$ ( saturation current )

at 25°c = 300 k ( room temperature )

n = 2 for silicon diode

Determine the saturation current at 100 degrees = 373 k

Diode equation at room temperature = I = Io $\frac{V}{e^{0.025*n} }$

next we have to determine the value of V at 373 k

q / kT = (1.6 * 10^-19) / (1.38 * 10^-23 * 373) = 31.08 V^-1

Given that I is constant

Io = $\frac{e^{0.025*2} }{31.08}$ = 0.0659 A

3 0

3 years ago

Does anybody know what plane this is? i saw it the other day doing a low pass through my community

Arlecino [84]

Answer:

Airbus A340-313

Explanation:

it is what it is

3 0

2 years ago

Calculate the convective heat-transfer coefficient for water flowing in a round pipe with an inner diameter of 3.0 cm. The water

olasank [31]

Answer:

h = 10,349.06 W/m^2 K

Explanation:

Given data:

Inner diameter = 3.0 cm

flow rate = 2 L/s

water temperature 30 degree celcius

$Q = A\times V$

$2\times 10^{-3} m^3 = \frac{\pi}{4} \times (3\times 10^{-2})^2 \times velocity$

$V = \frac{20\times 4}{9\times \pi} = 2.83 m/s$

$Re = \frac{\rho\times V\times D}{\mu}$

at 30 degree celcius $= \mu = 0.798\times 10^{-3}Pa-s , K = 0.6154$

$Re = \frac{10^3\times 2.83\times 3\times 10^{-2}}{0.798\times 10^{-3}}$

Re = 106390

So ,this is turbulent flow

$Nu = \frac{hL}{k} = 0.0029\times Re^{0.8}\times Pr^{0.3}$

$Pr= \frac{\mu Cp}{K} = \frac{0.798\times 10^{-3} \times 4180}{0.615} = 5.419$

$\frac{h\times 0.03}{0.615} = 0.0029\times (1.061\times 10^5)^{0.8}\times 5.419^{0.3}$

SOLVING FOR H

WE GET

h = 10,349.06 W/m^2 K

6 0

3 years ago