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2 years ago

6

calculate force and moment reactions at bolted base O of overhead traffic signal assembly. each traffic signal has a mass 36kg,

while the masses of member OC and AC are 50Kg and 55kg, respectively. The mass center of mmber AC at G.

1 answer:

Igoryamba2 years ago

7 0

Answer:

The free body diagram of the system is, 558 368 368 508 O ?? O, Consider the equilibrium of horizontal forces. F

Explanation:

I hope this helps you but I think and hope this is the right answer sorry if it’s wrong.

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Ai r is compressed by a 30-kW compressor from P1 to P2. The air t emperature i s maintained constant at 25oC during thi s proces

RUDIKE [14]

Answer:

The rate of entropy change of the air is -0.10067kW/K

Explanation:

We'll assume the following

1. It is a steady-flow process;

2. The changes in the kinetic energy and the potential energy are negligible;

3. Lastly, the air is an ideal gas

Energy balance will be required to calculate heat loss;

mh1 + W = mh2 + Q where W = Q.

Also note that the rate of entropy change of the air is calculated by calculating the rate of heat transfer and temperature of the air, as follows;

Rate of Entropy Change = -Q/T

Where Q = 30Kw

T = Temperature of air = 25°C = 298K

Rate = -30/298

Rate = -0.100671140939597 KW/K

Rate = -0.10067kW/K

Hence, the rate of entropy change of the air is -0.10067kW/K

3 0

3 years ago

A hydrauliic jack is rated at 5000 pound capacity. The area of the large piston on the jack is 4.45 Square inches. Calculate the

sergeinik [125]

Answer:

1123.6 pounds/ square inch.

Explanation:

Fluid pressure is the ratio of force or weight applied by the fluid per unit area.

i.e Fluid pressure = $\frac{weight}{area}$

The maximum load of the jack is obtained at its maximum capacity = 5000 pounds

Area of the large piston on the jack = 4.45 square inches

Thus,

Fluid pressure = $\frac{5000}{4.45}$

= 1123.5955

Fluid pressure = 1123.6 pounds/ square inch

Thu, the fluid pressure in the jack at maximum load is 1123.6 pounds/ square inch.

7 0

2 years ago

The alignment readings for the front of a vehicle are shown above. Camber and toe are within specification, caster is not. Techn

dlinn [17]

Answer:

B. B only

Given Information:

1. Camber and toe are within specification

2. Caster is not within specification

Technician A says that with the current settings, the left front tire tread may wear on the inside edge.

Technician B says that with the current settings, the vehicle may pull to the left

Explanation:

Lets discuss the effects of Camber, toe and caster misalignment

Effects of Camber and Toe misalignment:

Camber is the inward or outward tilt of the fron tires and is used to distribute load across the tread. Any misalignment causes uneven loading on the tires which results in tire wear on one edge.

The most common cause of tire wear on the inside edge is due to the camber misalignment which results in premature tire wear.

Another reason is of tire wear is vehicle’s toe. A slight misalignment of the toe reduces the life of the tire.

Since it is given that camber settings and toe settings are within specification therefore, tire tread wear on the inside edge cannot happen if camber and toe are within specification.

Technician A cannot be right.

Effects of Caster misalignment:

Whenever there is a misalignment of the castor then the vehicle will not be able to go in straight line rather it will pull to either left or right side. Caster misalignment also causes heavy or light steering depending upon the positive or negative misalignment of caster.

Since it is given that caster settings are not within specification therefore, the vehicle may pull to the left due to the caster misalignment.

Technician B must be right.

4 0

3 years ago

Steam enters a turbine operating at steady state with a mass flow of 10 kg/min, a specific enthalpy of 3100 kJ/kg, and a velocit

Natali [406]

Answer:

$\dot W_{out} = 133.327\,kW$

Explanation:

The model for the turbine can be derived by means of the First Law of Thermodynamics:

$-\dot Q_{out}-\dot W_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right] =0$

The work produced by the turbine is:

$\dot W_{out}=-\dot Q_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right]$

The mass flow and heat transfer rates are, respectively:

$\dot m = (10\frac{kg}{min})\cdot (\frac{1\,min}{60\,s} )$

$\dot m = 0.167\,\frac{kg}{s}$

$\dot Q_{out} = (0.167\,\frac{kg}{s} )\cdot (1.1\times 10^{3}\,\frac{J}{kg} )$

$\dot Q_{out} = 183.7\,W$

Finally:

$\dot W_{out} = -183.7\,W + (0.167\,\frac{kg}{s} )\cdot \left(8\times 10^{5}\,\frac{J}{kg} -562,5\,\frac{J}{kg} +29.43\,\frac{J}{kg} \right)$

$\dot W_{out} = 133.327\,kW$

3 0

2 years ago

A dryer is shaped like a long semi-cylindrical duct of diameter 1.5 m. The base of the dryer is occupied with water-soaked mater

Anestetic [448]

Answer:

0.0371 kg/s.m

Explanation:

From the given information, let's have an imaginative view of the semi-cylinder; (The image is shown below)

Assuming the base surface of both ends of the cylinder is denoted by:

$A_1 \ and \ A_2$

Thus, using the summation rule, the view factor $F_{11$ and $F_{12$ is as follows:

$F_{11}+F_{12}=1$

Let assume the surface (1) is flat, the $F_{11} = 0$

Now:

$0+F_{12}=1$

$F_{12}=1$

However, using the reciprocity rule to determine the view factor from the dome-shaped cylinder $A_2$ to the flat base surface $A_1$ ; we have:

$A_2F_{21} = A_{1}F_{12} \\ \\ F_{21} = \dfrac{A_1}{A_2}F_{12}$

Suppose, we replace DL for $A_1$ and

$A_2$ = $\dfrac{\pi D}{2}$

Then:

$F_{21} = \dfrac{DL}{(\dfrac{\pi D}{2}) L} \times 1 \\ \\ =\dfrac{2}{\pi} \\ \\ =0.64$

Now, we need to employ the use of energy balance formula to the dryer.

i.e.

$Q_{21} = Q_{evaporation}$

But, before that; let's find the radian heat exchange occurring among the dome and the flat base surface:

$Q_{21}= F_{21} A_2 \sigma (T_2^4-T_1^4) \\ \\ Q_{21} = F_{21} \times \dfrac{\pi D}{2} \sigma (T_2^4 -T_1^4)$

where;

$\sigma = Stefan \ Boltzmann's \ constant$

$T_1 = base \ temperature$

$T_2 = temperature \ of \ the \ dome$

∴

$Q_{21} = 0.64 \times (\dfrac{\pi}{2}\times 1.5) \times 5.67 \times 10^4 \times (1000^4 -370^4)\\ \\ Q_{21} = 83899.15 \ W/m$

Recall the energy balance formula;

$Q_{21} = Q_{evaporation}$

where;

$Q_{evaporation} = mh_{fg}$

here;

$h_{fg}$ = enthalpy of vaporization

m = the water mass flow rate

∴

$83899.15 = m \times 2257 \times 10^3 \\ \\ m = \dfrac{83899.15}{ 2257 \times 10^3 }\\ \\ \mathbf{m = 0.0371 \ kg/s.m}$

6 0

2 years ago