3 years ago

Please help me on this it’s due now

Engineering

1 answer:

lbvjy [14]3 years ago

5 0

Answer:

road closures on deliver routes.. i think..

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How long should the shafts remain in the furnace to achieve a desired centerline temperature of 800K? 2) Determine the temperatu

Phantasy [73]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

6 0

4 years ago

A centrifugal pump is required to pump water to an open water link situated 4 km away from the location of the pump through a pi

11111nata11111 [884]

Answer:

P= 5.5 bar

Explanation:

Given that

L= 4000 m

d= 0.2 m

Friction factor(F) = 0.01

speed V= 2 m/s

Head = 5 m

Head loss due to friction

$h_f=\dfrac{FLV^2}{2gd}$

$h_f=\dfrac{0.01\times 4000\times 2^2}{2\times 9.81\times 0.2}$

$h_f=40.77m$

So the total head(H) = 5 + 40.77 + 10.3 =56.07

Where 10.3 m is the atmospheric head.

We know that

P=ρ g H

So total Pressure

P= 1000 x 9.81 x 56.07 Pa

$P=5.5\times 10^5\ Pa$

P= 5.5 bar

5 0

4 years ago

La iluminación de la superficie de un patio amplio es 1600 lx cuando el ángulo de elevación del sol 53°. Calcular la iluminación

gregori [183]

Answer:

I = 1205.69 Lx

Explanation:

The irradiation or intensity of the solar radiation on the earth is maximum for the vertical fire, with a value I₀

I = I₀ sin θ

in this case with the initial data we can calculate the initial irradiance

I₀ = $\frac{I}{sin \ \theta }$

I₀ = 1600 /sin 53

I₀ = 2003.42 lx

for when the angle is θ = 37º

I = 2003.42 sin 37

I = 1205.69 Lx

6 0

3 years ago

Water flows through a nozzle at the end of a fire hose. If the nozzle exit velocity must be 20 m/s and the exit diameter is 40 m

lara31 [8.8K]

Answer:

minimum flow rate provided by pump is 0.02513 m^3/s

Explanation:

Given data:

Exit velocity of nozzle = 20m/s

Exit diameter = 40 mm

We know that flow rate Q is given as

$Q = A \times V$

where A is Area

$A =\frac{\pi}{4} \times (40\times 10^{-3})^2 = 1.256\times 10^{-3} m^2$

$Q = 1.256\times 10^{-3} \times 20 = 0.02513 m^3/s$

minimum flow rate provided by pump is 0.02513 m^3/s

5 0

3 years ago

HELP PLEASE!!!!!!!!!!!

MAXImum [283]

C is your answers!!!!!$3&2)//

7 0

3 years ago

Read 2 more answers