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3 years ago

I’ll give you 100 points

Engineering

2 answers:

Kaylis [27]3 years ago

6 0

Answer:

All are 4 to 5

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PLz dont delete my answer I know its right

Explanation:

its a square so each side is equal

Nikitich [7]3 years ago

4 0

Answer:

A. SS

B. Ss

C. SS

D. Ss

___________________________________________________________

Hope this helps :)

Pls brainliest...

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Air at 38°C and 97% relative humidity is to be cooled to 14°C and fed into a plant area at a rate of 510m3/min. (a) Calculate th

Katarina [22]

To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.

The gas ideal law is given as,

$PV=mRT$

Where,

P = Pressure

V = Volume

m = mass

R = Gas Constant

T = Temperature

Our data are given by

$T_1 = 38\°C$

$T_2 = 14\°C$

$\eta = 97\%$

$\dot{v} = 510m^3/kg$

Note that the pressure to 38°C is 0.06626 bar

PART A) Using the ideal gas equation to calculate the mass flow,

$PV = mRT$

$\dot{m} = \frac{PV}{RT}$

$\dot{m} = \frac{0.6626*10^{5}*510}{287*311}$

$\dot{m} = 37.85kg/min$

Therfore the mass flow rate at which water condenses, then

$\eta = \frac{\dot{m_v}}{\dot{m}}$

Re-arrange to find $\dot{m_v}$

$\dot{m_v} = \eta*\dot{m}$

$\dot{m_v} = 0.97*37.85$

$\dot{m_v} = 36.72 kg/min$

PART B) Enthalpy is given by definition as,

$H= H_a +H_v$

Where,

$H_a$ = Enthalpy of dry air

$H_v$ = Enthalpy of water vapor

Replacing with our values we have that

$H=m*0.0291(38-25)+2500m_v$

$H = 37.85*0.0291(38-25)-2500*36.72$

$H = 91814.318kJ/min$

In the conversion system 1 ton is equal to 210kJ / min

$H = 91814.318kJ/min(\frac{1ton}{210kJ/min})$

$H = 437.2tons$

The cooling requeriment in tons of cooling is 437.2.

3 0

3 years ago

Why is it important to review your plan when you write an algorithm?

user100 [1]

An algorithm is itself a general step-by-step solution of your problem. ... The most important point here is that you must use algorithms to solve problem, one way or the other. Most of the time it's better to think about your problem before you jump to coding - this phase is often called design.

7 0

3 years ago

The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm

aksik [14]

Answer:

Tmax=14.5MPa

Tmin=10.3MPa

Explanation:

T = 600 * 0.15 = 90N.m

$T_max =\frac{T_c}{j} = \frac{x}{y} = \frac{90 \times 0.0175}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}$

=14.5MPa

$T_{min} =\frac{T_c}{j} = \frac{x}{y} = \frac{90 \times 0.0125}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}$

=10.3MPa

7 0

3 years ago

In casting experiments performed using a certain alloy and type of sand mold, it took 155 sec for a cube-shaped casting to solid

Tems11 [23]

Answer:

A) Cm = 2.232 s/mm²

B) Time taken to solidify = 74.3 seconds

Explanation:

(A) Since a side is 50mm and all sides of a cube are equal, thus, Volume of the cube is;V = 50 x 50 x 50 = 125,000 mm³

There are 6 faces of the cube, thus Surface Area A = 6 x (50 x 50) = 15,000 mm²

So, Volume/Area = (V/A) = 125,000/15,000 = 8.333 mm

Cm is given by the formula; Cm =[Tts] /(V/A)² where Tts is time taken to solidify and it's 155 seconds in the question. Thus;

Cm = 155/(8.333)²= 2.232 s/mm²

(B) For;Cylindrical casting with D = 30 mm and L = 50 mm.;

Volume of cylinder is;

V = (πD²L) /4

So,V = (π x 30² x 50)/4 = 35,343mm³

Surface area of cylinder is;

A = (2πD²)/4 + (πDL)

Thus, A = ((π x 30²)/2) + (π x 30 x 50) = 6126 mm²

Volume/Area is;

V/A = 35,343/6126 = 5.77 mm

Same alloy and mold type was hsed as in a above, thus, Cm is still 2.232 s/mm²

Since Cm =[Tts] /(V/A)²

Making Tts the subject, we have;

Tts =Cm x (V/A)²

Tts = 2.232 x (5.77)² = 74.3 seconds

3 0

3 years ago

A small vehicle is powered by a pulsejet. The available net thrust is 6000 N and the traveling speed is 200 km/hr. The gases lea

kumpel [21]

Answer:

a) The mass flow rate is 19.71 kg/s

b) The inlet area is 0.41 m²

c) The thrust power is 333.31 kW

d) The propulsive efficiency is 26.7%

Explanation:

Please look at the solution in the attached Word file.

Download docx

4 0

3 years ago