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2 years ago

Steam locomotives with a 4-6-2 wheel arrangement were usually classified as what?.

Engineering

1 answer:

Korvikt [17]2 years ago

7 0

Answer:

See Explanation

Explanation:

Locomotives with 4 leading wheels, 6 driving wheels and 2 trailing wheels knowns as "Pacific" type locomotives.

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How many different powerball combinations are there

svlad2 [7]

Answer:

There 11,238,513 different Powerball

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2 years ago

Which line from "On Becoming an Inventor" supports the idea that Dean's time at Worcester Polytechnic Institute was very useful

Stolb23 [73]

Answer:

Explanation:

The line from "On becoming an inventor" that says:

"I found that at college I could get help from my teachers with solving business problems and in learning new techniques for designing new things"

On Becoming an Inventor was by Dean Kamen an American Engineer, Inventor and Businessman

3 0

3 years ago

Read 2 more answers

2. The unthreaded part of a bolt or screw is called the

aksik [14]

Answer:

The grip

Explanation:

the head of all headed bolt (except countersunk head bolt)

3 0

2 years ago

Read 2 more answers

Initialize the tuple team_names with the strings 'Rockets', 'Raptors', 'Warriors', and 'Celtics' (The top-4 2018 NBA teams at th

Drupady [299]

Answer:

#Initialise a tuple

team_names = ('Rockets','Raptors','Warriors','Celtics')

print(team_names[0])

print(team_names[1])

print(team_names[2])

print(team_names[3])

Explanation:

The Python code illustrates or printed out the tuple team names at the end of a season.

The code displayed is a function that will display these teams as an output from the program.

4 0

3 years ago

Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, $T_{i} = 20^{\circ}C$ = 273 + 20 = 293 K

Temperature at the outlet, $T_{o} = 200{\circ}C$ = 273 + 200 = 473 K

Pressure at inlet, $P_{i} = 80 kPa = 80\times 10^{3} Pa$

Pressure at outlet, $P_{o} = 800 kPa = 800\times 10^{3} Pa$

Speed at the outlet, $v_{o} = 20 m/s$

Diameter of the tube, $D = 10 cm = 10\times 10^{- 2} m = 0.1 m$

Input power, $P_{i} = 400 kW = 400\times 10^{3} W$

Now,

To calculate the heat transfer, $Q$ , we make use of the steady flow eqn:

$h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}$

where

$h_{i}$ = specific enthalpy at inlet

$h_{o}$ = specific enthalpy at outlet

$v_{i}$ = air speed at inlet

$p_{s}$ = specific power input

H and H' = Elevation of inlet and outlet

Now, if

$v_{i} = 0$ and H = H'

Then the above eqn reduces to:

$h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}$

$Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}$ (1)

Also,

$p_{s} = \frac{P_{i}}{ mass, m}$

Area of cross-section, A = $\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}$

Specific Volume at outlet, $V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s$

From the eqn:

$P_{o}V_{o} = mRT_{o}$

$m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s$

Now,

$p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg$

Also,

$\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg$

Now, using these values in eqn (1):

$Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW$

Now, rate of heat transfer, q:

q = mQ = $0.925\times 813.33 = 752.33 kW$

4 0

3 years ago