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1 month ago

8

Which of the following types of protective equipment protects workers who are passing by from stray sparks or metal while anothe

r worker is welding?

1 answer:

choli [191]1 month ago

7 0

Answer:

Flame-resistant clothing and aprons

Explanation:

Workers involved with welding are generally mandated to wear flame-resistant clothing and aprons to shield them from various hazards, including heat, flames, burns, and potential radiation. In the context of welding, this gear protects individuals from flying sparks that can ignite and cause fires. Hence, such clothing helps to prevent accidents in these situations.

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Consider a rectangular fin that is used to cool a motorcycle engine. The fin is 0.15m long and at a temperature of 250C, while t

Kisachek [217]

Answer:

q' = 5826 W/m

Explanation:

Given:-

- The length of the fin in question, L = 0.15 m

- The fin's surface temperature, Ts = 250°C

- The velocity of free stream air, U = 80 km/h

- The air temperature, Ta = 27°C

- The flow is parallel over both sides of the fin, assuming turbulent flow conditions throughout.

Find:-

What is the heat removal rate per unit width of the fin?

Solution:-

- Steady state conditions are assumed, along with negligible radiation and turbulent flow conditions.

- From Table A-4, we gather air properties (T = 412 K, P = 1 atm ):

Dynamic viscosity, v = 27.85 * 10^-6 m²/s

Thermal conductivity, k = 0.0346 W / m.K

Prandtl number Pr = 0.69

- Compute the Nusselt Number (Nu) corresponding to - turbulent conditions - using the relevant relationship as follows:

$Nu = 0.037*Re_L^\frac{4}{5} * Pr^\frac{1}{3}$

Where, Re_L: Average Reynolds number for the entire length of fin:

$Re_L = \frac{U*L}{v} \\\\Re_L = \frac{80*\frac{1000}{3600} * 0.15}{27.85*10^-^6} \\\\Re_L = 119688.80909$

Consequently,

$Nu = 0.037*(119688.80909)^\frac{4}{5} * 0.69^\frac{1}{3}\\\\Nu = 378$

- The convective heat transfer coefficient (h) can now be derived from:

$h = \frac{k*Nu}{L} \\\\h = \frac{0.0346*378}{0.15} \\\\h = 87 \frac{W}{m^2K}$

- The heat loss rate q' per unit width can be established using the convection heat transfer formula and should be multiplied by (x2) since the airflow is present on both sides of the fin:

$q' = 2*[h*L*(T_s - T_a)]\\\\q' = 2*[87*0.15*(250 - 27)]\\\\q' = 5826\frac{W}{m}$

- Ultimately, the heat loss per unit width from the rectangular fin is q' = 5826 W/m

- The thermal loss per unit width (q') attributed to radiation:

$q' = 2*a*T_s^4*L$

Where, a signifies the Stefan-Boltzmann constant = 5.67*10^-8

$q' = 2*5.67*10^-^8*(523)^4*0.15\\\\q' = 1273 \frac{W}{m}$

- It is observed that radiation losses are not insignificant, accounting for 20% of thermal loss by convection. As the emissivity (e) of the fin is unspecified, this value is dismissed from the calculations as it pertains to the provided information.

7 0

4 days ago

A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7

mote1985 [204]

Answer:

r=0.31

Ф=18.03°

Explanation:

Provided:

Original diameter of bar = 75 mm

Diameter post-cutting = 73 mm

Average diameter of the bar d= (75+73)/2=74 mm

Average length of uncut chip = πd

Average length of uncut chip = π x 74 =232.45 mm

Thus, cutting ratio r

$Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}$

$r=\dfrac{73.5}{232.45}$ r=0.31

Therefore, the cutting ratio equals 0.31.

Now, the shearing angle is given as

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

Next by substituting the values

$tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }$

$tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }$ \

Ф=18.03°

Concluding, the shearing angle is 18.03°.

4 0

1 month ago

6. You are evaluating flow through an airway. The current flow rate is 10 liters per minute with a fixed driving pressure (P1) o

iogann1982 [279]

Answer:

B) P1 would have to increase to sustain the flow rate (correct)

C) Resistance would rise (correct)

Explanation:

Flow rate is measured at 10 liters per minute

Driving pressure (P1) stands at 20 cm H2O

Fixed downstream pressure (P2) is 5 cm H2O

The accurate statements when the lumen is pinched in the center of the tube are: P1 will increase to maintain the flow rate, and resistance will rise. This occurs because pinching the lumen decreases its diameter, leading to higher resistance, which is linearly related to pressure, thus P1 will also increase.

The incorrect statement is: the flow would decrease.

6 0

16 days ago

You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an

pantera1 [220]

Answer:

$t'_{1\2} = 6.6 sec$

Explanation:

The half-life for the specified RC circuit can be expressed as

$t_{1\2} =\tau ln2$

where [/tex]\tau = RC[/tex]

$t_{1\2} = RCln2$

Given $t_{1\2} = 3 sec$

The circuit has a resistance of 40 ohms, and by adding a new resistor of 48 ohms, the total resistance becomes 40 + 48 = 88 ohms.

Thus, the new half-life is

$t'_{1\2} =R'Cln2$

Now, divide equation 2 by 1

$\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}$

$t'_{1\2} = t'_{1\2}\frac{R'}{R}$

After substituting all values, we can calculate the revised half-life

$t'_{1\2} = 3 * \frac{88}{40} = 6.6 sec$

$t'_{1\2} = 6.6 sec$

7 0

25 days ago

An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

Mrrafil [253]

Answer:

Volume change percentage is 2.60%

Water level increase is 4.138 mm

Explanation:

Provided data

Water volume V = 500 L

Initial temperature T1 = 20°C

Final temperature T2 = 80°C

Diameter of the vat = 2 m

Objective

We aim to determine percentage change in volume and the rise in water level.

Solution

We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.

It can similarly relate to density changes.

Thus,

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

Here, ρ denotes density. The density at 20°C = 998 kg/m³.

The density at 80°C = 972 kg/m³.

Plugging in these values into equation 2 gives

$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

dV = 0.0130 m³

Therefore, the percentage change in volume will be

dV % = $-\frac{dV}{V}$ × 100

dV % = $-\frac{0.0130}{500*10^{-3} }$ × 100

dV % = 2.60 %

Hence, the percentage change in volume is 2.60%

Initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

Final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

From equations 3 and 4, subtract v1 from v2.

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

dV = $\frac{\pi }{4} *d^2*dl$

Substituting all values yields

0.0130 = $\frac{\pi }{4} *2^2*dl$

Thus, dl = 0.004138 m.

Consequently, the water level rises by 4.138 mm.

8 0

25 days ago