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1 year ago

Write down the tracking error such that the adaptive cruise control objective is satisfied.

1 answer:

5 0

Answer:
The most common reason a cruise control stops working is due to a blown fuse or a defective brake pedal switch. It can also be caused by issues with the throttle control system or the ABS. In older cruise control systems it can be caused by a broken vacuum line.

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A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a

bazaltina [42]

Answer:

$Q_{cv}=-339.347kJ$

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

$P_iV_i=m_iRT_i$ (i could be 1 for initial and 2 for the end)

State1

$1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295$

$m_1=232kg$

State2

$8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350$

$m_2=11.946$

So, the total mass of the aire entered is

$m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg$

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

$u_1=210.49kJ/kg$

For Specific enthalpy

$h_1=295.17kJ/kg$

For the second state the Specific internal Energy (6bar, 350K)

$u_2=250.02kJ/kg$

At the end we make a Energy balance, so

$U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e$

No work done there is here, so clearing the equation for Q

$Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)$

$Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)$

$Q_{cv}=-339.347kJ$

The sign indicates that the tank transferred heat<em> to</em> the surroundings.

8 0

2 years ago

True or false tensile forces are smaller in arch bridges

Ivan

Answer:

True

Explanation:

The tensile forces are small in most arches and usually negligible.

4 0

2 years ago

Calculate the volume of a hydraulic accumulator capable of delivering 5 liters of oil between 180 and 80 bar, using as a preload

Vinil7 [7]

Answer:

1) $V_o = 10 liters$

2) $V_o = 12.26 liters$

Explanation:

For isothermal process n =1

$V_o =\frac{\Delta V}{(\frac{p_o}{p_1})^{1/n} -(\frac{p_o}{p_2})^{1/n}}$

$V_o = \frac{5}{[\frac{72}{80}]^{1/1} -[\frac{72}{180}]^{1/1}}$

$V_o = 10 liters$

calculate pressure ratio to determine correction factor

$\frac{p_2}{p_1} =\frac{180}{80} = 2.25$

correction factor for calculate dpressure ration for isothermal process is

c1 = 1.03

$actual \ volume = c1\times 10 = 10.3 liters$

b) for adiabatic process

n =1.4

volume of hydraulic accumulator is given as

$V_o =\frac{\Delta V}{[\frac{p_o}{p_1}]^{1/n} -[\frac{p_o}{p_2}]^{1/n}}$

$V_o = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}$

$V_o = 12.26 liters$

calculate pressure ratio to determine correction factor

$\frac{p_2}{p_1} =\frac{180}{80} = 2.25$

correction factor for calculate dpressure ration for isothermal process is

c1 = 1.15

$actual \volume = c1\times 10 = 11.5 liters$

8 0

3 years ago

In a certain chemical plant, a closed tank contains ethyl alcohol to a depth of 71 ft. Air at a pressure of 17 psi fills the gap

Yuliya22 [10]

Answer:

the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Explanation:

Given that;

depth 1 = 71 ft

depth 2 = 10 ft

pressure p = 17 psi = 2448 lb/ft²

depth h = 71 ft - 10 ft = 61 ft

we know that;

p = P_air + yh

where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )

so we substitute;

p = 2448 + ( 49.3 × 61 )

= 2448 + 3007.3

= 5455.3 lb/ft³

= 37.88 psi

Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

5 0

3 years ago

The Role of Fuel Cells in Renewable Energy Solutions

gogolik [260]

Answer:

chemical energy directly

4 0

2 years ago