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3 years ago

A negative pressure respirator brings fresh air to you through a hoseA) TrueB)False

Engineering

2 answers:

madreJ [45]3 years ago

8 0

Answer:FALSE

Explanation: A negative pressure respirator is a respiratory system which is known to have a low air pressure inside the mask when compared to the air pressure on the outside during Inhalation.

Most of the personal protective equipment (PPE) which are in use in various industries are examples of Negative pressure respirator device,any leak or damage done to the device will allow the inflows of harmful and toxic Air into the person's respiratory system. AIR SUPPLY SYSTEMS ARE KNOWN TO SUPPLY FRESH UNCONTAMINATED AIR THROUGH AIR STORED INSIDE COMPRESSED CYLINDERS OR OTHER SOURCES AVAILABLE.

Elenna [48]3 years ago

8 0

Answer:

False

Explanation:

Any tight-fitting respirator in which the air pressure inside the face-piece is negative during inhalation with respect to the ambient air pressure outside the respirator is known to be a negative pressure respirator . An example of a negative pressure respirator is the disposable N95 filtering face-piece respirator.

Therefore, a negative pressure respirator brings fresh air to you through a face-piece, not through a hose.

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The mass flow rate in a 4.0-m wide, 2.0-m deep channel is 4000 kg/s of water. If the velocity distribution in the channel is lin

IceJOKER [234]

Answer:

V = 0.5 m/s

Explanation:

given data:

width of channel = 4 m

depth of channel = 2 m

mass flow rate = 4000 kg/s = 4 m3/s

we know that mass flow rate is given as

$\dot{m}=\rho AV$

Putting all the value to get the velocity of the flow

$\frac{\dot{m}}{\rho A} = V$

$V = \frac{4000}{1000*4*2}$

V = 0.5 m/s

4 0

3 years ago

Steam at 75 kPa and 8 percent quality is contained in a spring-loaded piston–cylinder device, as shown in Figure, with an initia

Rashid [163]

The heat transferred to and the work produced by the steam during this process is 13781.618 kJ/kg

<h3>How to calcultae the heat?</h3>

The Net Change in Enthalpy will be:

= m ( h2 - h1 ) = 11.216 ( 1755.405 - 566.78 ) = 13331.618 kJ/kg

Work Done (Area Under PV curve) = 1/2 x (P1 + P2) x ( V1 - V2)

= 1/2 x ( 75 + 225) x (5 - 2)

W = 450 KJ

From the First Law of Thermodynamics, Q = U + W

So, Heat Transfer = Change in Internal Energy + Work Done

= 13331.618 + 450

Q = 13781.618 kJ/kg

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brainly.com/question/13439286

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6 0

1 year ago

Steam enters a two-stage adiabatic turbine at 8 MPa and 5008C. It expands in the first stage to a state of 2 MPa and 3508C. Stea

Nataly [62]

Answer:

1) The exergy of destruction is approximately 456.93 kW

2) The reversible power output is approximately 5456.93 kW

Explanation:

1) The given parameters are;

P₁ = 8 MPa

T₁ = 500°C

From which we have;

s₁ = 6.727 kJ/(kg·K)

h₁ = 3399 kJ/kg

P₂ = 2 MPa

T₂ = 350°C

From which we have;

s₂ = 6.958 kJ/(kg·K)

h₂ = 3138 kJ/kg

P₃ = 2 MPa

T₃ = 500°C

From which we have;

s₃ = 7.434 kJ/(kg·K)

h₃ = 3468 kJ/kg

P₄ = 30 KPa

T₄ = 69.09 C (saturation temperature)

From which we have;

h₄ = $h_{f4}$ + x₄× $h_{fg}$ = 289.229 + 0.97*2335.32 = 2554.49 kJ/kg

s₄ = $s_{f4}$ + x₄× $s_{fg}$ = 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)

The exergy of destruction, $\dot X_{dest}$ , is given as follows;

$\dot X_{dest}$ = T₀ × $\dot S_{gen}$ = T₀ × $\dot m$ × (s₄ + s₂ - s₁ - s₃)

$\dot X_{dest}$ = T₀ × $\dot W$ ×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)

∴ $\dot X_{dest}$ = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138 - 2554.49) ≈ 456.93 kW

The exergy of destruction ≈ 456.93 kW

2) The reversible power output, $\dot W_{rev}$ = $\dot W_{}$ + $\dot X_{dest}$ ≈ 5000 + 456.93 kW = 5456.93 kW

The reversible power output ≈ 5456.93 kW.

6 0

3 years ago

Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia

nekit [7.7K]

Answer:

98°C

Explanation:

Total surface area of cylindrical fin = πr² + 2πrl , r = 0.015m; l= 0.1m; π =22/7

22/7*(0.015)² + 22/7*0.015*0.1 = 7.07 X 10∧-4 + 47.1 X 10∧-4 = (54.17 X 10∧-4)m²

Temperature change, t = (50 - 25)°C = 25°C = 298K

Hence, Temperature = 150 X (54.17 X 10∧-4) X 298/123 = 242.14/124 = 2.00K =

∴ Temperature change = 2.00K

But temperature, T= (373 - 2)K = 371 K

In °C = (371 - 273)K = 98°C

7 0

3 years ago

What is the governing ratio for thin walled cylinders?

Ann [662]

Answer:

The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words:

if $\frac{radius}{thickness} >10$ then you have a thin walled cylinder

or using the diameter:

if $\frac{diameter}{thickness} >20$ then you have a thin walled cylinder

3 0

3 years ago

A negative pressure respirator brings fresh air to you through a hoseA) TrueB)False​

A negative pressure respirator brings fresh air to you through a hoseA) TrueB)False