10. Which of these requires a wheel alignment after replacement?

The best way to identify common masonry problems is to call the engineer.<br> True or False

Daniel [21]

Answer:

True

Explanation:

5 0

2 years ago

Air exits a compressor operating at steady-state, steady-flow conditions at 150 oC, 825 kPa, with a velocity of 10 m/s through a

ioda

Answer:

a) Qe = 0.01963 m^3 / s , mass flow rate m^ = 0.1334 kg/s

b) Inlet cross sectional area = Ai = 0.11217 m^2 , Qi = 0.11217 m^3 / s

Explanation:

Given:-

- The compressor exit conditions are given as follows:

Pressure ( Pe ) = 825 KPa

Temperature ( Te ) = 150°C

Velocity ( Ve ) = 10 m/s

Diameter ( de ) = 5.0 cm

Solution:-

- Define inlet parameters:

Pressure = Pi = 100 KPa

Temperature = Ti = 20.0

Velocity = Vi = 1.0 m/s

Area = Ai

- From definition the volumetric flow rate at outlet ( Qe ) is determined by the following equation:

Qe = Ae*Ve

Where,

Ae: The exit cross sectional area

Ae = π*de^2 / 4

Therefore,

Qe = Ve*π*de^2 / 4

Qe = 10*π*0.05^2 / 4

Qe = 0.01963 m^3 / s

- To determine the mass flow rate ( m^ ) through the compressor we need to determine the density of air at exit using exit conditions.

- We will assume air to be an ideal gas. Thus using the ideal gas state equation we have:

Pe / ρe = R*Te

Where,

Te: The absolute temperature at exit

ρe: The density of air at exit

R: the specific gas constant for air = 0.287 KJ /kg.K

ρe = Pe / (R*Te)

ρe = 825 / (0.287*( 273 + 150 ) )

ρe = 6.79566 kg/m^3

- The mass flow rate ( m^ ) is given:

m^ = ρe*Qe

= ( 6.79566 )*( 0.01963 )

= 0.1334 kg/s

- We will use the "continuity equation " for steady state flow inside the compressor i.e mass flow rate remains constant:

m^ = ρe*Ae*Ve = ρi*Ai*Vi

- Density of air at inlet using inlet conditions. Again, using the ideal gas state equation:

Pi / ρi = R*Ti

Where,

Ti: The absolute temperature at inlet

ρi: The density of air at inlet

R: the specific gas constant for air = 0.287 KJ /kg.K

ρi = Pi / (R*Ti)

ρi = 100 / (0.287*( 273 + 20 ) )

ρi = 1.18918 kg/m^3

Using continuity expression:

Ai = m^ / ρi*Vi

Ai = 0.1334 / 1.18918*1

Ai = 0.11217 m^2

- From definition the volumetric flow rate at inlet ( Qi ) is determined by the following equation:

Qi = Ai*Vi

Where,

Ai: The inlet cross sectional area

Qi = 0.11217*1

Qi = 0.11217 m^3 / s

- The equations that will help us with required plots are:

Inlet cross section area ( Ai )

Ai = m^ / ρi*Vi

Ai = 0.1334 / 1.18918*Vi

Ai ( V ) = 0.11217 / Vi .... Eq 1

Inlet flow rate ( Qi ):

Qi = 0.11217 m^3 / s ... constant Eq 2

6 0

3 years ago

The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam

lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

$T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy$

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

$T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy$

$T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy$

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

6 0

2 years ago

The simply supported beam in the Figure has a rectangular cross-section 150 mm wide and 240 mm high.

8_murik_8 [283]

Same question idea but different values... I hope I helped you... Don't forget to put a heart mark

4 0

3 years ago

Lydia is the CEO for a large pharmaceutical manufacturer. Her company is in the final stages of FDA

weqwewe [10]

OSHA inspections are generally unannounced. In fact, except in four exceptional circumstances when advance notice may be given.

It is a criminal offense for any person to give unauthorized advance notice of an OSHA inspection.

5 0

3 years ago