*Sample Problems*

* *

1. What is q, the volume flow rate of vapor formed, if 0.5 gallons of
toluene are evaporated uniformly over a 7-hr shift? What volume flow rate Q_{d }is required for dilution to 10ppm, if
K_{mixing }= 2? (Assume STP; d=1.0)

What is the average face velocity of
air in a room 10’ * 8’ * 40’ for these conditions?

**Solution:**

** **

SG=0.866, MW=92.1

Q = 387 * lbs evaporated/MW * minutes * d

Note: lbs evaporated = gal * 8.31 * SG

Q = 387 * 0.5gal * 8.31lbs/gal * 0.866/9.21 * 480
minutes * 1.0 = 0.0315 scfm

Q_{d }= q * 10^{6 }* K_{mixing}/Ca
(ppm)

Q_{d }= 0.0315 * 10^{6 }* 2/10 =
6,300 scfm

V_{face }= Qd/A = 6300/80 = about 80 fpm

2.
What
is q, the volume flow rate of vapor formed, if 500 grams of xylene are
evaporated uniformly over a 4-hr shift? What volume flow rate Q_{d }is
required for dilution to 25 ppm, if K_{mixing }= 1.5? (Assume STP; d=1.0)

**Solution:**

** **

q = 0.0241 * grams evaporated/MW * seconds * d

q = 0.0241 * 500/106.2 * 4 * 60 * 60 * 1.0 = 7.88 *
10^{-6} acms at STP (i.e. scms)

Q_{d }= q * 10^{6 }* K_{mixing}/Ca
(ppm)

Q_{d }= 7.88 * 10^{-6 }* 10^{6 }*
1.5/25 = 0.47 acms at STP

3.
Calculate
N (number of air changes per hour)

**Solution:**

** **

(i) Q = 6,300 scfm, Volume, V = 10’ * 8’ * 40’ =
3,200 cubic feet [3,200 * 0.0283 = 90.5 cubic meters)

N = Q *
60/Vol = 6,300 * 60/3200 = 120 ac/hr

(i) Q = 0.47 acms, Volume, V = 10’ * 8’ * 40’ =
3,200 cubic feet [3,200 * 0.0283 = 90.5 cubic meters)

N =Q * 3600/Vol = 0.47 * 3600/90.5 = 19 ac/hr

4. What is the estimated time required for
the buildup in an acetone concentration to 750 ppm, given the following
conditions?

STP K_{mix
}= 2

Co=0, start of shift Ct=750 ppm (after time t has elapsed)

US Units SI units

Volume 100,000 cubic feet 2832 cubic meters

q 2.25 scfm 0.00106 scms

Q_{d} 2,000 scfm 0.9440 scms

**Solution:**

** **

In US units

t = -2 * (10000/2000) * Ln [((2.25 * 10^{6}/2000)
- 750)/((2.25 * 10^{6}/2000) - 0)]

=
-2 * 50 * Ln (375/1125) = 110 minutes

In SI units

t = -2 * (2832/0.944) * Ln[((0.00106 * 10^{6}/0.944)
- 750)/( (0.00106 * 10^{6}/0.944) - 0)]

=-2 * 3000 * Ln (373/1123) = 6600 seconds =
110 minutes.

5. An automobile garage was contaminated with carbon monoxide. If
dilution air is provided for 30 minutes, what will the CO concentration be?
Assume STP, q=0; C_{o}=500 ppm; K_{mix}=2.0

US Units SI Units

Volume
10,000 cubic feet 283.2 cubic meters

Q_{d} 200 scfm 0.0944 scms

t 30
min 1800 seconds

**Solution:**

** **

US units:

C_{t}=500 * EXP [-200/(2 * 10000) * 30
minutes]

C_{t}=500 * EXP [-0.30] = 500 * 0.74 = 370
ppm

In SI units:

C_{t}=500 * EXP [-0.0944/(2 * 283.2) * 1800
seconds] = 370 ppm

6. Methyl chloroform is lost by evaporation from a tank at a rate
of 1.5 pints per 60 minutes. What is the effective ventilation rate

Q’ and the actual ventilation rate (Q) required to maintain the vapor concentration at the TLV?

**Solution:**

** **

TLV = 350 ppm, SG = 1.32, MW = 133.4, Assume K = 5

Assuming perfect dilution, the effective ventilation
rate (Q’) is given by

Q’ = 403 * 10^{6 }* 1.32 * (1.5/60)/(133.4 *
350) = 285 cfm

Due to incomplete mixing the actual ventilation rate
(Q) is

Q = 403 * 10^{6 }* 1.32 * (1.5/60) *
5/(133.4 * 350) = 1425 cfm

7.
A
cleaning and gluing operation is being performed; methyl ethyl ketone (MEK) and
toluene are both being released. Both have narcotic properties and the effects
are considered additive. Air samples disclose concentrations of 150 ppm MEK and
50 ppm toluene. The volumetric flow rate at standard conditions required for
dilution of the mixture to the TLV would be as follows.

**Solution:**

** **

Assume 2 pints of each is being released each 60 min. Select a K value for MEK and a K value of 5 for toluene; Sp gr for MEK=0.805, for toluene = 0.866; MW for MEK = 72.1, for toluene = 92.13

Q for MEK =403 * 0.805 * 10^{6 }* 4 *
(2/60)/(72.1 * 200) = 3000 cfm

Q for toluene = 403 * 0.866 * 10^{6 }* 5 *
(2/60)/(92.13 * 100) = 6313 cfm

8.
A
batch of enamel-dipped shelves is baked in a recirculating oven at 350 F for 60
minutes. Volatiles in the enamel applied to the shelves consist of two pints of
xylene. What oven ventilation rate, in cfm, is required to dilute the xylene
vapor concentration within the oven to a safe limit at all times?

**Solution:**

** **

For xylene, the LEL = 1.0%; sp g = 0.88; MW = 106; S_{f
}= 10; B = 0.7

Q = 403 * 0.88* (2/60) * 100 * 10/(106 * 1.0 * 0.7)
= 159 cfm

Since the above equation is at standard conditions,
the airflow rate must be converted from 70 F to 350 F.

Q_{A }= (cfm_{STP}) (Ratio of
Absolute Temperature)

=( cfm_{STP}) (460 F +
350 F)/(460 F + 70 F)

Q_{A }= 159(810/530)
= 243 cfm

9.
Air
contains 400 ppm of acetone (TLV = 750 ppm), 150 ppm of ethyl acetate (TLV =
200 ppm) and 100 ppm of methyl ethyl
ketone (TLV = 200 ppm). What is the TLV of the mixture?

**Solution:**

** **

C_{1}/ T_{1}
+ C_{2}/ T_{2} + C_{3}/ T_{3 }+ ------------------+
C_{n}/ T_{n} = 1

Hence 400/750 + 150/200 +
100/200 = 0.53 + 0.75 + 0.5 >1

Hence threshold limit is
exceeded.