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ladessa [460]
3 years ago
6

You get invited to a party at a 1,200 sf apartment, where there are a total of 23 people. The apartment has ceilings that are 8

ft above the floor. Three people begin smoking 2 cigarettes per hour each. The air handler system has a make-up air ventilation rate of 600 cfm. Assume an overhead fan ensures the air is well-mixed. Eye irritation from formaldehyde can start at concentrations of 0.01 ppm, with irritation of the lungs at 5 ppm. Will the concentrations become high enough to cause eye irritation
Engineering
1 answer:
yawa3891 [41]3 years ago
7 0

Answer:

concentration of air =  0.0000366 ppm is not sufficient to cause any eye irritation

Explanation:

Average amount of aldehyde generated by a cigarette = 1 mg = 10⁻⁶g

Room volume = Height * Base area

Room volume = 8ft * 1200 sq.ft = 9600 cubic feet

number of people in the room = 23

If 1 person smokes 2 cigarette in 1 hour,

23 people will smoke 23*2 cigarettes = 46 cigarettes in 1 hr

Mass of aldehyde generated by 1 cigarette = 10⁻⁶g

Mass of aldehyde generated by 46 cigarettes = 46 * 10⁻⁶g

Based on the law of mass conservation,

Mass of aldehyde coming in = mass of aldehyde going out = Rate of accumulation of aldehyde

Rate of accumulation of aldehyde in the steady state = 0

m*C*V= 0

m = 46 * 10⁻⁶g

C = concentration in the steady state

V = Volumetic flow of the fresh air

V = 600 ft³/min = 600 *60

V = 36000 ft³/hr

Density = Mass/volume

Density =  46 * 10⁻⁶g /36*10³

Density  = 1.28 * 10⁻⁹ g/ft³

If the density of air is taken as 36 g/ft³

concentration = (1.28 * 10⁻⁹ g/ft³)/36 g/ft³

concentration = 3.66 * 10⁻¹¹g/g

1*10⁻⁶g/g of air = 1ppm(part per million)

concentration of air =( 3.66 * 10⁻¹¹)(/1*10⁻⁶)

concentration of air = 3.66 * 10⁻⁵ppm = 0.0000366 ppm

concentration of air =  0.0000366 ppm is not sufficient to cause any eye irritation

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A motor driven water pump operates with an inlet pressure of 96 kPa (absolute) and mass flow rate of 120 kg/min. The motor consu
NeX [460]

Answer:

The maximum water pressure at the discharge of the pump (exit) = 496 kPa

Explanation:

The equation expressing the relationship of the power input of a pump can be computed as:

E _{pump,u} = \dfrac{m(P_2-P_1)}{\rho}

where;

m = mass flow rate = 120 kg/min

the pressure at the inlet P_1 = 96 kPa

the pressure at the exit P_2 = ???

the pressure \rho = 1000 kg/m³

∴

0.8 \times 10^{3} \ W = \dfrac{(120 \ kg/min * 1min/60 s)(P_2-96000)}{1000}

0.8 \times 10^{3}\times 1000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}

800000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}

\dfrac{800000}{2} = P_2-96000

400000 = P₂ - 96000

400000 +  96000 = P₂

P₂ = 496000 Pa

P₂ = 496 kPa

Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa

8 0
2 years ago
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Eva8 [605]
No you may not ask the question
3 0
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Show that for a linearly separable dataset, the maximum likelihood solution for the logisitic regression model is obtained by fi
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Answer for the question:

"Show that for a linearly separable dataset, the maximum likelihood solution for the logisitic regression model is obtained by finding a weight vector w whose decision boundary wx. "

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Explanation:

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6. During some actual expansion and compression processes in piston–cylinder devices, the gases have been
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During some actual expansion and compression processes in piston-cylinder devices, the gases have been are the P1= P2.

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During a few real enlargements and compression procedures in piston-cylinder devices, the gases were located to meet the connection PV n = C, wherein n and C are constants.

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