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4 years ago

7

Technician a says multiple discharge ignition system fires the spark plug during each of the engine's four cycle strokes. Techni

an B says that a dual spark plug ignition system ensures better ignition and combustion of the air- fuek mixture that a sigle spark plug per cylinder system who is right?

1 answer:

patriot [66]4 years ago

7 0

Answer:

Technician B only is correct

Explanation:

Here we have that the multiple discharge ignition system is a system that fires the spark plug multiple times during each power stroke to provide ample spark to complete the ignition. Therefore, technician A is not correct.

The dual spark plug ignition system is a method if obtaining the ideal combustion and improved fuel consumption than a single spark plug ignition system.

Therefore, only technician B is correct.

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a closed systems internal energy changes by 178 j as a result of being heated with 658 j of energy. the energy used to do work b

Akimi4 [234]

<h3><u>Answer;</u></h3>

= 480 Joules

<h3><u>Explanation;</u></h3>

We use the formula, Q - W = ΔU

Where, Q = Heat transferred to the system

W = Work done by the system

ΔU = Change of internal energy.

As per the question, Q = 658 J

ΔU = 178 J

Thus, W = Q - ΔU = (658 - 178) J = 480 J.

The energy used to do work by the system is 480 J.

7 0

3 years ago

A car traveling 28 mi/h accelerates uniformly for 8.9 s, covering 599 ft in this time. what was its acceleration? round your ans

almond37 [142]

Answer:

5.90 ft/s^2

Explanation:

There are mixed units in this question....convert everything to miles or feet

and hr to s

28 mi / hr = 41.066 ft/s

Displacement = vo t + 1/2 at^2

599 = 41.066 (8.9) + 1/2 a (8.9^2)

solve for a = ~ 5.90 ft/s^2

5 0

2 years ago

Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of

marusya05 [52]

Answer:

a) $\dot m = 6.878\,\frac{kg}{s}$ , b) $T = 104.3^{\textdegree}C$ , c) $\dot S_{gen} = 11.8\,\frac{kW}{K}$

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

$-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0$

The mass flow rate is:

$\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}$

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

$P = 7000\,kPa$

$T = 500^{\textdegree}C$

$h = 3411.4\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

State 2s - Liquid-Vapor Mixture

$P = 100\,kPa$

$h = 2467.32\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

$x = 0.908$

The isentropic efficiency is given by the following expression:

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

The real specific enthalpy at outlet is:

$h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})$

$h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )$

$h_{2} = 2684.46\,\frac{kJ}{kg}$

State 2 - Superheated Vapor

$P = 100\,kPa$

$T = 104.3^{\textdegree}C$

$h = 2684.46\,\frac{kJ}{kg}$

$s = 7.3829\,\frac{kJ}{kg\cdot K}$

The mass flow rate is:

$\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}$

$\dot m = 6.878\,\frac{kg}{s}$

b) The temperature at the turbine exit is:

$T = 104.3^{\textdegree}C$

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

$\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0$

$\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})$

$\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )$

$\dot S_{gen} = 11.8\,\frac{kW}{K}$

4 0

4 years ago

In an open system, such as a campfire, matter can?

-Dominant- [34]

I am positive with my and answer and i think yes

7 0

3 years ago

Is a 3D parallel flow necessarily irrotational?

satela [25.4K]

Answer:

No

Explanation:

We know that

Rationality in 3D flow is given as follows

$\omega =\omega _xi+\omega _yj+\omega _zk$

If ω is equal to zero then flow will be irrotational and if ω is not equal to zero then flow will be rotational .

So if flow is ir-rotational it means that the component of ω in x-direction,in y-direction and in z-direction direction should be zero.

So it is not necessarily that 3D parallel flow will be irrotaional.

3 0

4 years ago