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maw [93]
3 years ago
6

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.

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Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−
sergeinik [125]

Answer:

Power output, P_{out} = 178.56 kW

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, T = 350^{\circ}C

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, \dot{m} = 0.1 kg.s^{- 1}

Diameter of exhaust pipe, d_{h} = 15 cm = 0.15 m

Pressure at exhaust, P' = 50 kPa

temperature, T' =  100^{\circ}C

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

\dot{m} = \frac{Av_{i}}{V_{1}}

0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}

0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}

v_{i} = 3.986 m/s

Now, eqn for compressible fluid:

\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}

Now,

\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}

\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}

\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}

v_{e} = 19.33 m/s

Now, the power output can be calculated from the energy balance eqn:

P_{out} = -\dot{m}W_{s}

P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}

P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW

4 0
3 years ago
A kernel-level thread wishes to acquire a mutex lock declared as global in the process. True or False: the function call used be
Alik [6]

Answer:

  1. A kernel-level thread wishes to acquire a mutex lock declared as global in the process.  <u>"</u><u>True"</u>  
  2. The function call used below causes the thread to block at this point, if the mutex lock is already acquired by a different thread. int resultcode -pthread mutex_trylock (myMutex). <u>"False"</u>                              

7 0
3 years ago
How would you expect an increase in the austenite grain size to affect the hardenability of a steel alloy? Why?
seraphim [82]

Answer:

The hardenability increases with increasing austenite grain size, because the grain boundary area is decreasing. This means that the sites for the nucleation of ferrite and pearlite are being reduced in number, with the result that these transformations are slowed down, and the hardenability is therefore increased.

3 0
3 years ago
Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°
cricket20 [7]

Answer:

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

m\cdot (h_{in} - h_{out}) = 0

h_{in} = h_{out}

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

P = 1500\,kPa

T = 198.29\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 6.3068\,\frac{kJ}{kg\cdot K}

x = 0.967

State 2 - Outlet (Superheated Vapor)

P = 200\,kPa

T = 130\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 7.1776\,\frac{kJ}{kg\cdot K}

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

6 0
3 years ago
Assume that a p+ - n diode is built with an n region width l smaller than a hole diffusion length (l
RSB [31]

Answer: hiiiiii

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Explanation:

7 0
3 years ago
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