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Scorpion4ik [409]

3 years ago

15

Hey, I have a question, I was thinking that if you have engineering skills or drawing skill you could help me to start a project

on scratch, which would be a life or WW3 simulation.
Comment or answer if you would like to help.

1 answer:

gayaneshka [121]3 years ago

7 0

Answer

Dont have one;(

Explanation:

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Select the correct answer from each drop-down menu.

Maurinko [17]

Answer:

Migration and indoor planting if too much rain outside.

Explanation:

If rain is a problem, migrating to another place is an option. If you can't leave your place, then I suggest planting inside. (No list was given)

5 0

4 years ago

If an internal combustion engine produces 12 W of power, how much power is this in a. hp and b. Ibf-ft/s. Show all unit cancella

AVprozaik [17]

Answer:

a)12 W = 0.01609 hp

b)12 W = 8.85 lbf-ft/s

Explanation:

Given that

Power produces by internal combustion engine = 12 W

a)As we know that

1 W = 0.001341 hp

So it means that

12 W = 0.001341 x 12 hp

12 W = 0.01609 hp

b)We also know that

1 W = 0.737 lbf-ft/s

So it means that

12 W = 0.737 x 12 lbf-ft/s

12 W = 8.85 lbf-ft/s

6 0

3 years ago

A smooth concrete pipe (1.5-ft diameter) carries water from a reservoir to an industrial treatment plant 1 mile away and dischar

Kamila [148]

ANSWER:

Q = 0.17ft3/s

EXPLANATION: since the water runs downhill on a 1:100 slope, that means the flow is laminar.

Using poiseuille equation:

Q = (π × D^4 × ∆P) ÷ (128 × U × ∆X)

Q is the volume flow rate.

π is pie constant value at 3.142

D is the diameter of the pipe

∆P is the pressure drop

U is the viscosity

∆X is the length of the pipe or distance of flow.

Form the question, we are to determine U then Find Q

Therefore;

D = 1.5ft

∆P = 1pa since the minor losses are negligible.

∆X = 1mile = 5280ft.

STEP1: FIND U

Viscosity is a function of the temperature of the liquid. An increase in temperature increases the viscosity of the liquid.

We know that at room temperature, which is 25°C the viscosity of water is 8.9×10^-4pa.s . We can find the viscosity of water at 4°C by cross multiplying.

Therefore;

25°C = 8.9×10^-4pa.s

4°C = U

Cross multiply

U25°C = 4°C × 8.9×10^-4pa.s

U25°C = 0.00356°C.pa.s

Therefore;

U = 0.00356°C.pa.s ÷ 25°C

U = 1.424×10^-4pa.s

Therefore at 4°C the viscosity of water in the pipe is 1.424×10^-4pa.s

STEP2: FIND Q

Imputing the values into poiseuille equation above.

Q = (3.142 × (1.5ft)^4 × 1pa) ÷ (128 × 1.424×10^-4pa.s × 5280ft)

Q = 15.906375pa.ft4 ÷ 96.239616pa.s.ft

Therefore;

Q = 0.16547887ft3/s

Approximately;

Q = 0.17ft3/s

6 0

3 years ago

Consider the 3.8-L engine used in the engine laboratory. At 2500 RPM the engine has a volumetric efficiency of 85%, air/fuel rat

AnnZ [28]

Answer:

A number of engine geometry properties and performance parameters obtained from dynamometer testing are given in Table below:

Please calculate:

i) Cylinder bore, bore to stroke ratio, connecting rod to crank radius ratio, and clearance volume in each engine cylinder.

ii) Brake power output at 3800 rpm and the torque output at 6000 rpm.

iii) Brake, indicated and friction mean effective pressures at both 3800 rpm and 6000 rpm.

iv) Brake specific fuel consumption in kg/(kW h) and brake thermal efficiency at 3800 rpm.

v) Volumetric efficiency at 3800 rpm.

vi) Air to fuel ratio and the relative air to fuel ratio at 3800 rpm. Is this a lean or rich mixture?

Engine type Displacement/ size Number of cylinders Stroke length Connecting rod length Compression ratio Maximum brake power6000 rpm Maximum [email protected] 3800 rpm Mass flow rate of fuel 3800 rpm Volumetric flow rate of air 3800 rpm Mechanical efficiency @ 3800 rpm Mechanical efficiency @ 6000 rpm Calorific value of fuel Ambient air temperature Ambient air pressure Specific gas constant for air Gasoline, naturally aspirated, four-stroke 1.329 L 4 (Straight) 85 mm 141 mm 11.5 73 kW 128 N m 3.2 g/s 37.5 L/s 0.88 0.72 46 MJ/kg 10 °C 101.3 kPa 287.1 J/(kg K)

Explanation:

3 0

4 years ago

A wind tunnel is used to study the flow around a car. The air is drawn at 60 mph into the tunnel. (a) Determine the pressure in

PSYCHO15rus [73]

Answer:

The answer is below

Explanation:

The complete question is attached.

a) Bernoulli equation is given as:

$P+\frac{1}{2}\rho V^2+ \rho gz=constant\\\\\frac{P}{\rho g} +\frac{V^2}{2g} +z=constant\\$

Where P = pressure, V = velocity, z = height, g = acceleration due to gravity and ρ = density.

$\frac{P}{\rho g} +\frac{V^2}{2g} +z=constant\\\\\frac{P}{\gamma} +\frac{V^2}{2g} +z=constant\\\\\frac{P_1}{\gamma} +\frac{V_1^2}{2g} +z_1=\frac{P_2}{\gamma} +\frac{V_2^2}{2g} +z_2\\\\but \ z_1=z_2,P_1=0,V_1=0,V_2=60\ mph=88\ ft/s. Hence:\\\\\frac{P_2}{\gamma} =-\frac{V_2^2}{2g} \\\\P_2=\gamma*-\frac{V_2^2}{2g} =\rho g*-\frac{V_2^2}{2g} \\\\P_2=-\frac{V_2^2}{2}*\rho=-\frac{(88.8\ ft/s)^2}{2} * 0.00238\ slug/ft^3=-9.22\ lb/ft^2\\\\P_2+\gamma_{H_2O}h-\gamma_{oil}(1/12 \ ft)=0\\\\$

$\gamma_{oil}=0.9\gamma_{H_2O}=0.9*62.4\ lb/ft^3=56.2\ lb/ft^3\\\\Therefore:\\\\-9.22\ lb/ft^2+62.4\ lb/ft^3(h)-56.2\ lb/ft^3(1/12\ ft)=0\\\\h=0.223\ ft$

b)

$\frac{P}{\gamma} +\frac{V^2}{2g} +z=constant\\\\\frac{P_2}{\gamma} +\frac{V_2^2}{2g} +z_2=\frac{P_3}{\gamma} +\frac{V_3^2}{2g} +z_3\\\\but \ z_3=z_2,V_3=0,V_2=60\ mph=88\ ft/s. \\\\\frac{P_2}{\gamma}+\frac{V_2^2}{2g} = \frac{P_3}{\gamma}\\\\\frac{P_3-P_2}{\gamma}=\frac{V_2^2}{2g} \\\\P_3-P_2=\frac{V_2}{2g}*\gamma=\frac{V_2^2}{2g}*\rho g\\\\P_3-P_2=\frac{V_2}{2}*\rho=\frac{(88\ ft/s^2)^2}{2}*0.00238\ slg\ft^3\\\\P_3-P_2=9.22\ lb/ft^2$

4 0

3 years ago