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3 years ago

What is valve overlap?

Engineering

1 answer:

Thepotemich [5.8K]3 years ago

6 0

Answer:

Effects of Valve Overlap. As discussed previously, valve overlap is the time when both intake and exhaust valves are open. In simple terms, this provides an opportunity for the exhaust gas flow and intake flow to influence each other.

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The use of seatbelts in a car has significantly reduced the number of crash fatalities. Which statement best explains how societ

satela [25.4K]

as the public need for preventing injuries and deaths from car crashes became known,laws were enacted to mandate the inclusion of seatbelts in almost all passengers vehicles

8 0

2 years ago

Liquid water is fed to a boiler at 24°C and 10 bar is converted at a constant pressure to saturated steam.

zepelin [54]

We can find the change in the enthalpy through the tables A5 for Saturated water, pressure table.

For 1bar=1000kPa:

$T_{sat}=179.88\°c$

$H_{fg} = 2014.6kJ/kg$

$c_p=4.18 kJkg^{-1}{K^{-1}$

$\nu_g = 0.19436m^3/kg$

Replacing,

$\Delta h = h_{fg}+c_p(T_{sat}-T_{inlet})$

$\Delta h = 2014.6+4.18(179.88-24)$

$\Delta h=2666.17kJ/kg$

With the specific volume we know can calculate the mass flow, that is

$\dot{m}=\frac{\frac{15000}{3600}}{0.19436}$

$\dot{m} = 21.4378kg/s$

Then the heat required in input is,

$Q=\dot{m}\Delta h$

$Q=21.4378*2666.17$

$Q=57157.036kW$

With the same value required of 15000m^3/h, we can calculate the velocity of the water, that is given by,

$V= \frac{\dotV}{A}$

$V = \frac{\frac{15000}{3600}}{\pi /4 *(0.15)^2}$

$V=235.79m/s$

Finally we can apply the steady flow energy equation, that is

$\dot{m}(h_1+\frac{V^2}{2000})+Q = \dot{m}h_2$

Re-arrange for Q,

$Q=\dot{m}(h_2-h_1-\frac{V^2}{2000})$

$Q=\dot{m}(\Delta h-\frac{V^2}{2000})$

$Q= (21.4378)(2666.17-\frac{235.79^2}{2000})$

$Q= 56560.88kW$

We can note that consider the Kinetic Energy will decrease the heat input.

4 0

3 years ago

Acquisition of resources from an external source is called?

densk [106]

Answer:

subcontracting

Explanation:

I hope this is right

6 0

3 years ago

A steam pipe passes through a chemical plant, where wind passes in cross-flow over the outside of the pipe. The steam is saturat

valina [46]

Answer:

a) the rate of heat transfer from the pipe to the air is 23.866 watts

b) YES, the rate of heat transfer changes to 3518.61 watt

Explanation:

Given that:

steam is saturated at 17.90 bar.

the pipe is stainless steel and has an outside diameter of 6.75 cm

length = 34.7 m

Air flows over the pipe at 7.6 m/s

Bulk fluid temperature of 27°C

we know that

hD/k = 0.028 (Re)^0.8 (Pr)^0.33

Outside diameter of pipe = 6.75 cm

length of the pipe = 34.7 m

velocity of air = 7.6 m/s

Cp of air = 1.005 kJ/Kgk

viscosity of air = 1.81 × 10⁻⁵ kg/(m.sec)

thermal conductivity of air = 2.624 × 10⁻⁵ kw/m.k

so as

hD/k = 0.028 (Re)^0.8 (Pr)^0.33

hD/k = 0.028 (Dvp / u)^0.8 (Cpu / k)^0.33

(h × 0.0675 / 2.624 × 10⁻⁵) = (0.028 ([0.0675 × 7.6 × 1.225] / [1.81 ×10⁻⁵])^0.8) (((1.005 × 1.81 × 10⁻⁵) / (2.624 × 10⁻⁵))^0.33))

h = 0.0414 w/m².k

Now to find the rate of heat transfer Q

Q = hAΔT

Q = 0.0414 × (2π × 0.03375 × 34.7) × (105.383 - 27)

Q = 23.866 watts

therefore the rate of heat transfer from the pipe to the air is 23.866 watts

Now the flow direction changes to parallel flow, then

(h × 0.0675 / 2.624 × 10⁻⁵) = (0.028 ([34.7 × 7.6 × 1.225] / [1.81 ×10⁻⁵])^0.8) (((1.005 × 1.81 × 10⁻⁵) / (2.624 × 10⁻⁵))^0.33))

h = 6.1036 w/m².k

so from the steam table, saturated steam at 17.70 bar, temperature of steam will be 105.383°C

so to find the rate of heat transfer Q

Q = hAΔT

Q = 6.1036 × (2π × 0.03375 × 34.7) × (105.383 - 27)

Q = 3518.61 watt

Therefore the rate of heat transfer changes to 3518.61 watt

4 0

3 years ago

An op-amp is connected in an inverting configuration with R1 = 1kW and R2 = 10kW, and a load resistor connected at the output, R

Svetllana [295]

Answer:

View Image

Explanation:

You didn't provide me a picture of the opamp.

I'm gonna assume that this is an ideal opamp, therefore the input impedance can be assumed to be ∞ . This basically implies that...

no current will go in the inverting(-) and noninverting(+) side of the opamp
V₊ = V₋ , so whatever voltage is at the noninverting side will also be the voltage at the inverting side

Since no current is going into the + and - side of the opamp, then

i₁ = i₂

Since V₊ is connected to ground (0V) then V₋ must also be 0V.

V₊ = V₋ = 0

Use whatever method you want to solve for v_out and v_in then divide them. There's so many different ways of solving this circuit.

You didn't give me what the input voltage was so I can't give you the entire answer. I'll just give you the equations needed to plug in your values to get your answers.

8 0

4 years ago