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3 years ago

The air loss rate for a straight truck or bus with the engine off and the brakes

Engineering

1 answer:

vesna_86 [32]3 years ago

7 0

Answer:

Explanation:

The air loss rate for a straight truck or bus with the engine off and the brakes applied should not be more than (D) 3 psi in one minute. An air loss rate greater than 3 psi per minute can cause damage to air tank so the to avoid such condition the air leak rate cannot be more than 3 psi per minute.

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Charging method .Constant current method

mina [271]

Answer:

There are three common methods of charging a battery; constant voltage, constant current and a combination of constant voltage/constant current with or without a smart charging circuit.

Constant voltage allows the full current of the charger to flow into the battery until the power supply reaches its pre-set voltage. The current will then taper down to a minimum value once that voltage level is reached. The battery can be left connected to the charger until ready for use and will remain at that “float voltage”, trickle charging to compensate for normal battery self-discharge.

Constant current is a simple form of charging batteries, with the current level set at approximately 10% of the maximum battery rating. Charge times are relatively long with the disadvantage that the battery may overheat if it is over-charged, leading to premature battery replacement. This method is suitable for Ni-MH type of batteries. The battery must be disconnected, or a timer function used once charged.

Constant voltage / constant current (CVCC) is a combination of the above two methods. The charger limits the amount of current to a pre-set level until the battery reaches a pre-set voltage level. The current then reduces as the battery becomes fully charged. The lead acid battery uses the constant current constant voltage (CC/CV) charge method. A regulated current raises the terminal voltage until the upper charge voltage limit is reached, at which point the current drops due to saturation.

4 0

2 years ago

An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and ind

Misha Larkins [42]

Answer:

The unknown element is Capacitor.

Explanation:

The sinusoidal voltage is given as:

v(t) = V Sin (ωt + Ф)

Where:

V = Amplitude of Voltage

ω = 2π / T = > Time period (T)

Ф = phase shift

Considering no horizontal phase shift in the wave form, the equation can be written as:

v(t) = V Sin (ωt)--------(1)

Since, current in the capacitor can be given as:

i(t) = C dv(t)/dt = ωCV Cos (ωt)--------(2)

Now, checking all conditions:

At t=0 :

Equation (1) implies:

v(t) = V Sin [(2π/T)(0)] = V Sin (0)

v(t) = 0

The above finding satisfies the condition in the question. Now checking other conditions.

At t = T/4:

Equation (2) implies:

i(t) = ωCV Cos [(2π/T)(T/4)] = ωCV Cos [(π/2)] = ωCV (0)

i(t) = 0

At t = T/2 :

Equation (2) implies:

i(t) = ωCV Cos [(2π/T)(T/2)] = ωCV Cos [(π)] = ωCV (-1)

i(t) = - ωCV = max amplitude of current in negative direction

All three conditions of voltage and currents of question are satisfied with equations of capacitor hence, the unknown element is capacitor.

4 0

3 years ago

What is the key objective of data analysis

IceJOKER [234]

Answer: The process of data analysis uses analytical and logical reasoning to gain information from the data. The main purpose of data analysis is to find meaning in data so that the derived knowledge can be used to make informed decisions.

3 0

3 years ago

Consider a 1.5-m-high and 2.4-m-wide glass window whose thickness is 6 mm and thermal conductivity is k = 0.78 W/m⋅K. Determine

Bess [88]

Answer:

The steady rate of heat transfer through the glass window is 707.317 watts.

Explanation:

A figure describing the problem is included below as attachment. From First Law of Thermodynamics we get that steady rate of heat transfer through the glass window is the sum of thermal conductive and convective heat rates, all measured in watts:

$\dot Q_{total} = \dot Q_{cond} + \dot Q_{conv, in} + \dot Q_{conv, out}$ (Eq. 1)

Given that window is represented as a flat element, we can expand (Eq. 1) as follows:

$\dot Q_{total} = \frac{T_{i}-T_{o}}{R}$ (Eq. 2)

Where:

$T_{i}$ , $T_{o}$ - Indoor and outdoor temperatures, measured in Celsius.

$R$ - Overall thermal resistance, measured in Celsius per watt.

Now, we know that glass window is configurated in series and overall thermal resistance is:

$R = R_{cond} + R_{conv, in}+R_{conv, out}$ (Eq. 3)

Where:

$R_{cond}$ - Conductive thermal resistance, measured in Celsius per watt.

$R_{conv, in}$ , $R_{conv, out}$ - Indoor and outdoor convective thermal resistances, measured in Celsius per watt.

And we expand the expression as follows:

$R = \frac{l}{k\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d}$

$R = \frac{1}{w\cdot d}\cdot \left(\frac{l}{k}+\frac{1}{h_{i}}+\frac{1}{h_{o}} \right)$ (Eq. 4)

Where:

$w$ - Width of the glass window, measured in meters.

$d$ - Length of the glass window, measured in meters.

$l$ - Thickness of the glass window, measured in meters.

$k$ - Thermal conductivity, measured in watts per meter-Celsius.

$h_{i}$ , $h_{o}$ - Indoor and outdoor convection coefficients, measured in watts per square meter-Celsius.

If we know that $w = 2.4\,m$ , $d = 1.5\,m$ , $l = 0.006\,m$ , $k = 0.78\,\frac{W}{m\cdot ^{\circ}C}$ , $h_{i} = 10\,\frac{W}{m^{2}\cdot ^{\circ}C}$ and $h_{o} = 25\,\frac{W}{m^{2}\cdot ^{\circ}C}$ , the overall thermal resistance is:

$R = \left[\frac{1}{(2.4\,m)\cdot (1.5\,m)}\right] \cdot \left(\frac{0.006\,m}{0.78\,\frac{W}{m\cdot ^{\circ}C} }+\frac{1}{10\,\frac{W}{m^{2}\cdot ^{\circ}C} }+\frac{1}{25\,\frac{W}{m^{2}\cdot ^{\circ}C} } \right)$

$R = 0.041\,\frac{^{\circ}C}{W}$

Now, we obtain the steady rate of heat transfer from (Eq. 2): ( $R = 0.041\,\frac{^{\circ}C}{W}$ , $T_{i} = -5\,^{\circ}C$ , $T_{o} = 24\,^{\circ}C$ )