Repetitive movements at work can lead to injuries. True or False

Two technicians are discussing cylinder honing technician a says a good cross hatch helps to trap the oil and retain it in the c

aleksley [76]

er:

Explanation:Technician A says that primary vibration is created by slight differences in the inertia of the pistons between top dead center and bottom dead center. Technician B says that secondary vibration is a strong low-frequency vibration caused by the movement of the piston traveling up and down the cylinder. Who is correct? O A. Neither Technician A nor B OB. Technician B O C. Both Technicians A and B D. Technician A

8 0

2 years ago

Find the following for an input of 120 VAC(RMS), 60 hertz, given a 10:1 stepdown transformer, and a full-wave bridge rectifier.

atroni [7]

Answer:

(i) 169.68 volt

(ii) 16.90 volt

(iii) 16.90 volt

(iv) 108.07 volt

(v) 2.161 A

Explanation:

Turn ratio is given as 10:1

We have given that input voltage $v_p=120volt$

(i) We know that peak voltage is give by $v_{peak}=\sqrt{2}v_p=\sqrt{2}\times 120=169.68volt$

(ii) We know that for transformer $\frac{v_p}{v_s}=\frac{n_p}{n_s}$

So $\frac{169.08}{v_s}=\frac{10}{1}$

$v_s=16.90volt$

So peak voltage in secondary will be 16.90 volt

(iii) Peak voltage of the rectifier will be equal to the peak voltage of the secondary

So peak voltage of the rectifier will be 16.90 volt

(iv) Dc voltage of the rectifier is given by $v_{dc}=\frac{2v_m}{\pi }=\frac{2\times 1.414\times 120}{3.14}=108.07volt$

(v) Now dc current is given by $i_{dc}=\frac{v_{dc}}{R}=\frac{108.07}{50}=2.1614A$

4 0

3 years ago

The weatherman states that the current temperature is 95 F and the dew point is 74 F. What is the current relative humidity?

spayn [35]

Answer:

Current Relative Humidity is 29.623

Given:

Current Temperature, $T_{c} = 95 F = 35^{\circ}C$

Dew point temperature, $T_{d} = 74 F = 23.34^{\circ}C$

Solution:

Now, in order to calculate the Relative Humidity, RH, we use the given formula:

$T_{d} =100( \frac{\frac{aT_{c}}{b + T}b + lnRH}{a - \frac{aT_{c}}{b + T} + lnRH})$

where

a = 17.625

b = 237.7

Now, using the above formula and given values:

$23.34 = (237.7\frac{\frac{17.625\times 35}{ 237.7 + 35} + lnRH}{17.625 - \frac{17.625\times 35}{237.7 + 35} + lnRH})$

$23.34(17.625 - 2.26 + lnRH) = (237.7\times 2.26 + lnRH)$

$358.62 + 23.34lnRH) = (537.20 + lnRH)$

On solving the above eqn, we get:

RH = 29.623

3 0

3 years ago

1 A long, uninsulated steam line with a diameter of 100 mm and a surface emissivity of 0.8 transports steam at 150°C and is expo

Alexeev081 [22]

Answer:

a) q' = 351.22 W/m

b) q'_total = 1845.56 W / m

c) q'_loss = 254.12 W/m

Explanation:

Given:-

- The diameter of the steam line, d = 100 mm

- The surface emissivity of steam line, ε = 0.8

- The temperature of the steam, Th = 150°C

- The ambient air temperature, T∞ = 20°C

Find:-

(a) Calculate the rate of heat loss per unit length for a calm day.

Solution:-

- Assuming a calm day the heat loss per unit length from the steam line ( q ' ) is only due to the net radiation of the heat from the steam line to the surroundings.

- We will assume that the thickness "t" of the pipe is significantly small and temperature gradients in the wall thickness are negligible. Hence, the temperature of the outside surface Ts = Th = 150°C.

- The net heat loss per unit length due to radiation is given by:

q' = ε*σ*( π*d )* [ Ts^4 - T∞^4 ]

Where,

σ: the stefan boltzmann constant = 5.6703 10-8 (W/m2K4)

Ts: The absolute pipe surface temperature = 150 + 273 = 423 K

T∞:The absolute ambient air temperature = 20 + 273 = 293 K

Therefore,

q' = 0.8*(5.6703 10-8)*( π*0.1 )* [ 423^4 - 293^4 ]

q' = (1.4251*10^-8)* [ 24645536240 ]

q' = 351.22 W / m ... Answer

Find:-

(b) Calculate the rate of heat loss on a breezy day when the wind speed is 8 m/s.

Solution:-

- We have an added heat loss due to the convection current of air with free stream velocity of U∞ = 8 m/s.

- We will first evaluate the following properties of air at T∞ = 20°C = 293 K

Kinematic viscosity ( v ) = 1.5111*10^-5 m^2/s

Thermal conductivity ( k ) = 0.025596

Prandtl number ( Pr ) = 0.71559

- Determine the flow conditions by evaluating the Reynold's number:

Re = U∞*d / v

= ( 8 ) * ( 0.1 ) / ( 1.5111*10^-5 )

= 52941.56574 ... ( Turbulent conditions )

- We will use Churchill - Bernstein equation to determine the surface averaged Nusselt number ( Nu_D ):

$Nu_D = 0.3 + \frac{0.62*Re_D^\frac{1}{2}*Pr^\frac{1}{3} }{[ 1 + (\frac{0.4}{Pr})^\frac{2}{3} ]^\frac{1}{4} }*[ 1 + (\frac{Re_D}{282,000})^\frac{5}{8} ]^\frac{4}{5} \\\\Nu_D = 0.3 + \frac{0.62*(52941.56574)^\frac{1}{2}*(0.71559)^\frac{1}{3} }{[ 1 + (\frac{0.4}{0.71559})^\frac{2}{3} ]^\frac{1}{4} }*[ 1 + (\frac{52941.56574}{282,000})^\frac{5}{8} ]^\frac{4}{5} \\\\$

$Nu_D = 0.3 + \frac{127.59828 }{ 1.13824 }*1.27251 = 142.95013$

- The averaged heat transfer coefficient ( h ) for the flow of air would be:

$h = Nu_D*\frac{k}{d} \\\\h = 143*\frac{0.025596}{0.1} \\\\h = 36.58951 W/m^2K$

- The heat loss per unit length due to convection heat transfer is given by:

q'_convec = h*( π*d )* [ Ts - T∞ ]

q'_convec = 36.58951*( π*0.1 )* [ 150 - 20 ]

q'_convec = 11.49493* 130

q'_convec = 1494.3409 W / m

- The total heat loss per unit length ( q'_total ) owes to both radiation heat loss calculated in part a and convection heat loss ( q_convec ):

q'_total = q_a + q_convec

q'_total = 351.22 + 1494.34009

q'_total = 1845.56 W / m ... Answer

Find:-

For the conditions of part (a), calculate the rate of heat loss with a 20-mm-thick layer of insulation (k = 0.08 W/m ⋅ K)

Solution:-

- To reduce the heat loss from steam line an insulation is wrapped around the line which contains a proportion of lost heat within.

- A material with thermal conductivity ( km = 0.08 W/m.K of thickness t = 20 mm ) was wrapped along the steam line.

- The heat loss through the lamination would be due to conduction " q'_t " and radiation " q_rad":

$q'_t = 2*\pi*k \frac{T_h - T_o}{Ln ( \frac{r_2}{r_1} )}$

q' = ε*σ*( π*( d + 2t) )* [ Ts^4 - T∞^4 ]

Where,

T_o = T∞ = 20°C

T_s = Film temperature = ( Th + T∞ ) / 2 = ( 150 + 20 ) / 2 = 85°C

r_2 = d/2 + t = 0.1 / 2 + 0.02 = 0.07 m

r_1 = d/2 = 0.1 / 2 = 0.05 m

- The heat loss per unit length would be:

q'_loss = q'_rad - q'_cond

- Compute the individual heat losses:

$q'_t = 2*\pi*0.08 \frac{150 - 85}{Ln ( \frac{0.07}{0.05} )}\\\\q'_t = 0.50265* \frac{65}{0.33647}\\\\q'_t = 97.10 W/m$

Therefore,

q'_loss = 351.22 - 97.10

q'_loss = 254.12 W / m .... Answer

- If the wind speed is appreciable the heat loss ( q'_loss ) would increase and the insulation would become ineffective.

6 0

3 years ago

Help plsss fast!!!!!!

Semmy [17]

Answer:

1. Cast iron or aluminum

2. aluminum (first blank) iron(second blank)

3. aluminum

4. dry sleeve

5. Wet sleeve

6. matching operation that cuts a series of holes through the block for crankshaft bearing

8 0

2 years ago