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3 years ago

10

A process involves the removal of oil and other liquid contaminants from metal parts using a heat-treat oven, which has a volume

of 15,000 ft3. The oven is free of solvent vapors. The ventilation rate of the oven is 2,100 cfm, and the safety factor (K) is 3. The solvent used in the process evaporates at a rate of 0.6 cfm (cubic feet per minute). The operator would like to know how long it would take the concentration to reach 425 ppm.

1 answer:

Aleks [24]3 years ago

7 0

Answer:

time = 4.89 min

Explanation:

given data

volume = 15,000 ft³

ventilation rate of oven = 2,100 cubic feet per minute

safety factor (K) = 3

evaporates at a rate = 0.6 cubic feet per minute

solution

we get here first solvent additional rate in oven that is

solvent additional rate = $\frac{0.6}{1500}$

solvent additional rate = 4 × $10^{-5}$ min

solvent additional rate == 4 × $10^{-5}$ × $10^{6}$

solvent additional rate = 40 ppm/min

and

solvent removal rate due to ventilation will be

removal rate = $\frac{2100}{1500}$ × concentration in ppm

removal rate = 0.14 C ppm/min

and

net additional rate is

net additional rate (c) = m - r

so net additional rate (c) is = 40 - 0.14C

so here

$\frac{dc}{dt}$ = 40 - 0.14C

so take integrate from 0 to t

$\int\limits^t_o {dt}$ = $\int\limits^{425/3}_0 \frac{dc}{40-0.14C}$ ....................1

here factor of safety is 3 so time taken is $\frac{425}{3}$

solve it we get

time = $[\frac{-50}{7} \ log(40-\frac{7c}{50} ]^{425/3} _0$

time = 4.89 min

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iVinArrow [24]

Answer:

Explanation:

Complete question:

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4 years ago

what is the transfer function of the loaded filter? express your answer in terms of the variables r , l , rl , and s .

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Air filters with depth loaded are made to achieve precisely that. They add particles gradually to create air passageways, reducing constriction. You may save time and money by using filters that last longer thanks to them. The bigger particles are caught at the filter's beginning, while the smaller particles are caught as it gets closer. This is intended to avoid rapid surface loading, hence facilitating more airflow. This enables longer-lasting filtration as well.

On the other hand, surface loading filters catch every particle that is on its surface. No matter how big or little the particles are, it doesn't care.

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1 year ago

A generator operating at 50 Hz delivers 1 pu power to an infinite bus through a transmission circuit in which resistance is igno

olga55 [171]

Answer:

critical clearing angle = 70.3°

Explanation:

Generator operating at = 50 Hz

power delivered = 1 pu

power transferable when there is a fault = 0.5 pu

power transferable before there is a fault = 2.0 pu

power transferable after fault clearance = 1.5 pu

using equal area criterion to determine the critical clearing angle

Attached is the power angle curve diagram and the remaining part of the solution.

The power angle curve is given as

= Pmax sinβ

therefore : 2sinβo = Pm

2sinβo = 1

sinβo = 0.5 pu

βo = $sin^{-1} (0.5) = 30$ ⁰

also ; 1.5sinβ1 = 1

sinβ1 = 1/1.5

β1 = $sin^{-1} (\frac{1}{1.5} )$ = 41.81⁰

∴ βmax = 180 - 41.81 = 138.19⁰

attached is the remaining solution

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3 0

3 years ago

A well-insulated, rigid vessel contains 3 kg of saturated liquid water at 40°C. The vessel also contains an electrical resistor

Olenka [21]

Answer:

The final temperature in the vessel after the resistor has been operating for 30 min is 111.67°C

Explanation:

given information:

mass, m = 3 kg

initial temperature, T₁ = 40°C

current, I = 10 A

voltage, V = 50 V

time, t = 30 min = 1800 s

Heat for the system because of the resistance is

Q = V I t

where

V = voltage (V)

I = current (A)

t = time (s)

Q = heat transfer to the system (J)

so,

Q = V x I x t

= 50 x 10 x 1800

= 900000

= 9 x 10⁵ J

the heat transfer in the closed system is

Q = ΔU + W

where

U = internal energy

W = work done by the system

thus,

Q = ΔU + W

9 x 10⁵ = ΔU + 0, W = 0 because the tank is a well-insulated and rigid.

ΔU = 9 x 10⁵ J = 900 kJ

then, the energy change in the system is

ΔU = m c ΔT

ΔT = ΔU / m c, c = 4.186 J/g°C

= 900 / (3 x 4.186)

= 71.67°C

so,the final temperature (T₂)

ΔT = T₂ - T₁

T₂ = ΔT + T₁

= 71.67°C + 40°C

= 111.67°C

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Answer:

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