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3 years ago

QUESTIONS

Engineering

1 answer:

igomit [66]3 years ago

5 0

Answer:

According to many of the states' driving rules.

All vehicles of 3,000 pounds or more are required to have a brake system that makes them break as a response to the breaking of the vehicle's tow.

Explanation:

The reason behind this answer is that vehicles of more than 3,000 pounds are extremely dangerous and difficult to control. Therefore, when the tow breaks the automatic brake system is required. Because they are too big to be controlled, and if they are left without a brake system to reduce their damage they can destroy entire houses or other cars if this mechanism is not implemented.

You might be interested in

Consider the products you use and the activities you perform on a daily basis. Describe three examples that use both SI units an

Gwar [14]

Answer: Three examples of activities that I can perform on a daily basis that involves both metric units (SI units) and customary units include: measuring the length of a door using a tape measure, which includes both SI units and customary units (like feet, inches, and centimeters); baking a cake that requires one teaspoon (customary unit) of baking soda, which could also be converted into four grams (SI unit); weighing myself on a weighing scale, which can be measured by pounds (customary unit) or kilograms (metric unit).

Explanation: I big brain :) (Not Really I Just Wanted To Help) I hope this helped! ;)

4 0

3 years ago

Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra

gladu [14]

Answer:

a) $T_{2}=837.2K$

b) $e=91.3$ %

Explanation:

A) First, let's write the energy balance:

$W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})$ (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 $\frac{kJ}{kgK}$ And its specific R constant is 0.287 $\frac{kJ}{kgK}$ .

The only unknown from the energy balance is $T_{2}$ , so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

$T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K$

B) The isentropic efficiency (e) is defined as:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}$

Where ${h_{2s}$ is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because $h_{2}-h_{1}$ can be obtained from the energy balance $\frac{W}{m}=h_{2}-h_{1}$

$h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}$

An entropy change for an ideal gas with constant Cp is given by:

$s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})$

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

$0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})$

Applying logarithm properties:

$ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}$

Then,

$T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K$

So, now it is possible to calculate $h_{2s}-h_{1}$ :

$h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}$

Finally, the efficiency can be calculated:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3$ %

4 0

4 years ago

With thermodynamics, one cannot determine ________.

zloy xaker [14]

Answer:

Option B is correct.

Explanation:

With thermodynamics, one cannot determine the speed of the reaction.

Speed of the reaction is defined as the rate at which one component is changes to another component. With factor cannot be calculated from the thermodynamics branch of engineering.

With thermodynamics we can determine the value of the equilibrium constant,

the extent of a reaction, the temperature at which a reaction will be spontaneous & the direction of a spontaneous reaction. But we cannot determine the speed of a reaction.

Therefore option B is correct.

4 0

3 years ago

What are some advantages of making electronic components like transistors increasingly smaller?

Stells [14]

Answer:

Dr. Engelbart, who would later help develop the computer mouse and other personal computing technologies, theorized that as electronic circuits were made smaller, their components would get faster, require less power and become cheaper to produce — all at an accelerating pace

4 0

3 years ago

A steam power plant operating on a simple ideal Rankine cycle maintains the boiler at 6000 kPa, the turbine inlet at 600 °C, and

ohaa [14]

Answer:

ηa=0.349

ηb=0.345

Explanation:

The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:

$h_{3}=3658.8kJ/kg\\ s_{3}=7.1693kJ/kg$

The quality at state 4 is determined from the condition $s_{4} =s_{3}$ and the entropies of the components at the condenser pressure taken from table:

$q_{4} =\frac{s_{4}-s_{liq50} }{s_{evap,50} } \\=\frac{7.1693-1.0912}{6.5019}=0.935$

The enthalpy at state 4 then is:

$h_{4} =h_{liq50} +q_{4} h_{evap,50}\\ (340.54+0.935*2304.7)kJ/kg\\=2495.43kJ/kg\\$

Part A

In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:

$h_{1}=340.54kJ/kg\\ \alpha _{1}=0.00103m^3/kg$

The enthalpy at state 2 is determined from an energy balance on the pump:

$h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1} )$

=346.67 kJ/kg

The thermal efficiency is then determined from the heat input and output in the cycle:

$=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.349$

Part B

In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from and the saturated liquid values:

$h_{1}=293.7kJ/kg\\ \alpha _{1}=0.001023m^3/kg$

The enthalpy at state 2 is then determined from an energy balance on the pump:

$h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1} )$

=299.79 kJ/kg

The thermal efficiency in this case then is:

$=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.345$

8 0

3 years ago