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3 years ago

Which is an obsolete technology

Engineering

2 answers:

zvonat [6]3 years ago

8 0

Answer:

your answer would be steam engines

Lemur [1.5K]3 years ago

3 0

Answer:

Steam Engine

Explanation:

It stopped being produced due to other types of engines being much more efficient.

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Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the

myrzilka [38]

Answer:

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

$\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}$

The head associated with the pump is cleared:

$h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})$

Inlet and outlet velocities are found:

$v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }$

$v_{1} \approx 4.278\,\frac{m}{s}$

$v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }$

$v_{2} \approx 11.573\,\frac{m}{s}$

Now, the head associated with the pump is finally computed:

$h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m$

$h_{pump} = 7.705\,m$

The power that pump adds to the fluid is:

$\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}$

$\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)$

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

4 0

3 years ago

An engine operates on gasoline (LHV=44 MJ/kg) with a brake thermal efficiency of 37.9 % What is the brake specific fuel consumpt

scZoUnD [109]

Answer:

$s =0.21\ kg/Kw.hr$

Explanation:

Given that

Calorific value (CV) = 44 MJ/Kg

CV= 44,000 KJ/kg

Brake thermal efficiency(η) = 37.9 %

We know that

$\eta =\dfrac{BP}{\dot{m_f}\times CV}$

Where BP is the brake power

$\eta =\dfrac{BP}{\dot{m_f}\times CV}$

$0.379 =\dfrac{BP}{\dot{m_f}\times 44000}$

$\dfrac{BP}{\dot{m_f}}=16676$

Brake specific fuel consumption (s)

$s =\dfrac{\dot{m_f}}{BP}$

$s =\dfrac{3600\times \dot{m_f}}{BP}$

$s =\dfrac{3600}{16,676}\ kg/Kw.hr$

$s =0.21\ kg/Kw.hr$

7 0

3 years ago

A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust end

hjlf

Answer:

a) $v=19.6 m/s$

b) $H=19.58 m$

c) $v_{f}=29.57 m/s$

Explanation:

a) Let's calculate the work done by the rocket until the thrust ends.

$W=F_{tot}h=(F_{thrust}-mg)h=(35-(2*9.81))*25=384.5 J$

But we know the work is equal to change of kinetic energy, so:

$W=\Delta K=\frac{1}{2}mv^{2}$

$v=\sqrt{\frac{2W}{m}}=19.6 m/s$

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

$v_{f}^{2}=v_{i}^{2}-2gh$

At the maximum height the velocity is 0, so v(f) = 0.

$0=v_{i}^{2}-2gH$

$H=\frac{19.6^{2}}{2*9.81}=19.58 m$

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)

$v_{f}^{2}=v_{i}^{2}-2gh$

$v_{f}=\sqrt{19.6^{2}-(2*9.81*(-25))}=29.57 m/s$

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

6 0

3 years ago

Consider a multiprocessor system and a multithreaded program written using the many-to-many threading model. Let the number of u

Montano1993 [528]

Answer:

At the point when the quantity of bit strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just part strings to processors and not client level strings to processors. At the point when the quantity of part strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used all the while. Be that as it may, when a part string obstructs inside the portion (because of a page flaw or while summoning framework calls), the comparing processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, in this way expanding the use of the multiprocessor system.When the quantity of part strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just bit strings to processors and not client level strings to processors. At the point when the quantity of bit strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used at the same time. Be that as it may, when a part string hinders inside the piece (because of a page flaw or while summoning framework calls), the relating processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, along these lines expanding the usage of the multiprocessor framework.

4 0

3 years ago

A piece of corroded steel plate was found in a submerged ocean vessel. It was estimated that the original area of the plate was

shepuryov [24]

Answer:

Time of submersion in years = 7.71 years

Explanation:

Area of plate (A)= 16in²

Mass corroded away = Weight Loss (W) = 3.2 kg = 3.2 x 106

Corrosion Penetration Rate (CPR) = 200mpy

Density of steel (D) = 7.9g/cm³

Constant = 534

The expression for the corrosion penetration rate is

Corrosion Penetration Rate = Constant x Total Weight Loss/Time taken for Weight Loss x Exposed Surface Area x Density of the Metal

Re- arrange the equation for time taken

T = k x W/ A x CPR x D

T = (534 x 3.2 x 106)/(16 x 7.9 x 200)

T = 67594.93 hours

Convert hours into years by

T = 67594.93 x (1year/365 days x 24 hours x 1 day)

T = 7.71 years

3 0

4 years ago