One good way to improve your gas Milage is to __

a circular pile, 19 m long is driven into a homogeneous sand layer. The piles width is 0.5 m. The standard penetration resistanc

Elena L [17]

Answer:

Point force (Qp) = 704 kn/m²

Explanation:

Given:

length = 19 m

Width = 0.5 m

fs = 4

Vicinity of the pile = 25

Find:

Point force (Qp)

Computation:

Point force (Qp) = fs²(l+v)

Point force (Qp) = 4²(25+19)

Point force (Qp) = 16(44)

Point force (Qp) = 704 kn/m²

5 0

3 years ago

Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa

Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air, R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes 2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0

4 years ago

A generator is to be driven by a small Pelton wheel with a head of 91.5m at inlet to the nozzle and discharge of 0.04m^3/s. The

maria [59]

Answer:

ratio = 412

speed of wheel = 13.63

Explanation:

given data

head h = 91.5 m

discharge Q = 0.04 m³/s

wheel rotates N = 720 rpm

velocity coefficient Cv = 0.98

efficiency of the wheel η = 80 %

ratio of bucket speed to jet speed $\frac{u}{v}$ = 0.46

to find out

wheel to jet diameter ratio and power specific speed of the wheel

solution

we know that discharge = area × velocity .............1

here velocity = Cv√(2gh)

velocity = 0.98√(2×9.81×91.5) = 41.52 m/s

and area = $\frac{\pi }{4} d^2$

so from equation 1

0.04 = $\frac{\pi }{4} d^2$ × 41.52

d = 0.00123 m = 1.23 mm

we know bucket speed to jet speed = 0.46

so bucket speed u = 0.46 v

bucket speed u = 0.46 × 41.52 = 19.1 m/s

and bucket speed = $\frac{\pi D N}{60}$

so 19.1 = $\frac{\pi D 720}{60}$

so D = 0.506 m = 506.62 mm

so ratio is $\frac{D}{d} = \frac{506.62}{1.23}$

ratio = 412

and

we know efficiency = $\frac{outputpower}{inputpower}$

0.80 × ρQgh = output power

power output = 0.80 ×1000×0.04×9.81×91.5

power output P = 28.72 kW

so

speed of wheel is

speed of wheel = $\frac{N\sqrt{P} }{h^{5/4}}$

speed of wheel = $\frac{720\sqrt{28.72} }{91.5^{5/4}}$

speed of wheel = 13.63

7 0

4 years ago

A two-bus power system is interconnected by one transmission line. Bus 1 is a generator bus with specified terminal voltage magn

mixas84 [53]

Answer:

a) | v2 | , β2 ( load, bus voltage at bus 2 )

p1 , q1 ( slack, bus power at bus 1 )

b) q2 , β2

p1 and q1 ( slack, bus power at bus 1 )

Explanation:

Attached below is a schematic representation of the solution

a) Identify the variables in the solution vector assume Bus 2 is a load bus

The specified parameters are ; P2 and q2

while | v2 | and β2 are not specified

given that bus 2 is a load bus, bus 1 is a slack bus with ; | v1 | and β1 been specified while p1 and q1 are not specified

Hence the variables in the solution

= | v2 | , β2 ( load, bus voltage at bus 2 )

p1 , q1 ( slack, bus power at bus 1 )

b) Identify the variables in the solution vector ( assume Bus 2 is a PV bus )

specified at Bus 2 are ; | p2 | , | v2 |

unspecified : q2 , β2

Bus 1 ( still a slack bus )

specified parameter : | v1 | and β1

unspecified : p1 and q1

Hence the variables in the solution

= q2 , β2

p1 and q1 ( slack, bus power at bus 1 )

7 0

3 years ago

A jar made of 3/16-inch-thick glass has an inside radius of 3.00 inches and a total height of 6.00 inches (including the bottom

swat32

Answer:

1. Volume of the glass shell (Vg) is simply volume of the empty part of the jar (Ve) subtracted from volume of the entire jar (Vj):

Vg = Vj - Ve

Volume is calculated as base (B) multiplied with height (h). Base of the jar is circle, so its surface is πr^2 (r being the radius).

However radius is different depending on the part of the jar; for empty part of the jar, inner radius is d = 3 in, for the whole jar it is inner radius plus thickness of the glass a = 3 + 3/16 = 3.1875 in.

We are also given height of the whole jar, h = 6 in, but height of the empty part is entire height minus thickness of the jar h' = 6 - 0.1875 = 5.8125 in.

Now, let's calculate:

Vj = πa^2 • h = 191.42 in^3

Ve = πd^2 • h' = 164.26 in^3

So, volume of the glass shell is Vj - Ve which is 27.16 in^3.

2. Mass of the glass jar is density of the glass multiplied with volume:

m = ρ • Vg

Density of the glass is given here in cubic feet so, first, we need to convert it to cubic inches, dividing it by 1728:

ρ = 165 lb/ft^3 / 1728 = 0.095 lb/in^3

So, mass of the jar is:

m = 0.095 lb/in^3 • 27.16 in^3 = 2.59 lb

5. To find weight and volume of the water displaced we first need to find how deep the jar sinks (H), because volume of the displaced water is equal to the volume of the jar submerged. Jar will sink until gravity force (pulling it down) and buoyancy force (pushing it up) become equal. Displaced water is πa^2 • H and the buoyancy is ρw • g • Vd (ρw is density of water which is 62.5 lb/ft^3 / 1728 = 0.036 lb/in^3, and Vd is displaced water).

So, buoyancy is:

B = ρw • g • πa^2 • H

We said that buoyancy must be equal to gravity:

B = m • g (m being mass of the jar). So:

ρw • g πa^2 • H = m • g

ρw • πa^2 • H = m

From this, we can find H:

H = m / ρw•πa^2

H = 2.25 inches

That means that the jar will sink 2.25 inches in the water.

3. Now, it's easy to find volume of displaced water. It's the same as the volume of the jar submerged:

Vd = πa^2 • H

Vd = 71.94 in^3

4. And finally, the weight of water is:

m = ρw • Vd

m = 0.036 lb/in^3 • 71.94 in^3

m = 2.59 lb

Of course, we see that the mass of the jar equals the mass of the displaced water. Taking this as a rule, this question could have been solved easier However I wanted to do it more detailed, to explain it more clearly

6 0

3 years ago

One good way to improve your gas Milage is to ___.