CN3135 PYP Sem 1, 2010/2011 Question 1.

(a) From table 2-8 (P58), TLV-TWA for nitrogen dioxide in the unit of ppm ppm ppm

Source term model: vapor flow from an orifice From P133, for nitrogen dioxide, which is a triatomic gas,

. Assume the flow is choked,

choked .

From question, P0 = 105 bar = . Pa Thus,

. . . Pa. An external pressure less than 805.28 psia will result in choked flow through the leak. Because the external pressure is atmospheric in this case, choked flow is expected and Equation (4-50) applies.

The discharge coefficient is assumed to be 1.0. Also, . mm

.

. ,

.

.

..

. J/g-molK

kgm/s2 /N g/g-mol kg/g-mol

. kg/s Box model Use equation 3-9

Therefore,

ppm

From Table (3-12), for excellent ventilation and a vapour concentration of 10ppm,

.

g/g-mol

. J/g-molK Therefore,

. m3/s (b) From Probit model, to get a 1% probability of death, Y=2.67

Probit variable Y

For nitrogen dioxide deaths . , .

Therefore, ∑ . .

when the ventilation fan only delivers 10% of the required ventilation rate,

We could notice the QV is 1/10 as before, and k decrease from 1/6 to 1/8 ppm

Therefore, . mins

(c) A. Suggest workers to wear proper PPE. B. Install an alarm to remind technicians when the ventilation fan is faulty. C. Install spare ventilation fan to guarantee the ventilation system runs well. D. Re-design the laboratory layout and arrange the equipment properly so that the condition of ventilation stay excellent.

(d) The nitrogen oxides and hydrocarbons will form photochemical smog under the influence of sunlight, NOx + uv + VOC NO + ozone, PAN, acrolein, which would cause severe eye irritants and it also destroys lung tissue and chlorophyll in plants. SO2 and NOx could react with water vapor in the air in the presence of oxidising agents to form sulphuric and nitric acids. This would cause acid rain. This is corrosive to structures and highly damaging to forest and fresh water ecosystems.

Question 2 (a) A gaseous mixture of CH4 and CO leaks from the storage tanks suddenly, puff model

is chosen. From P190, The puff concentration directly downind along the cloud centerline is given by equation (5-41)

⟨ ⟩ , , ,∗

√ /

∗ kg as given in the question This happened on a clear night with 2m/s wind speed, stability class is F from table 5-1 to be more conservative. For stability class E puff, from Table 5-3,

. . . .

The mole fractions on a fuel-only basis are calculated in the following table. The UFL data are obtained from appendix B. Mass Mole Mole fraction UFLi (vol. %) CH4 160 10 66.7% 15.0 CO 140 5 33.3% 74 Equation (6-2) is used to determin the LFL of the mixture:

∑ . . . %byvolumetotalcombustibles

Convert to concentration units (see example 7-10, P341, but note the molar volume is 24.0 m3/kmol for 25˚C, instead of 22.4 for 0˚C)

⟨ ⟩

.m3mixturem3air

g-molmixturem3mixture

. % . % gmixtureg-molmixture

. g/m3 . kg/m3

Thus,

∗

√ / ⟨ ⟩ , , ,

Solve for x, . m

Therefore, .

(b) The mole fractions on a fuel-only basis are calculated in the following table. The LFL data are obtained from appendix B. Mass Mole Mole fraction LFLi (vol. %) CH4 160 10 66.7% 5.3 CO 140 5 33.3% 12.5 Equation (6-2) is used to determin the LFL of the mixture:

∑ ..

..

. %byvolumetotalcombustibles

Convert to concentration units (see example 7-10, P341, but note the molar volume is 24.0 m3/kmol for 25˚C, instead of 22.4 for 0˚C)

⟨ ⟩

.m3mixturem3air

g-molmixturem3mixture

. % . % gmixtureg-molmixture

. g/m3 . kg/m3

For the isopleth,

⟨ ⟩ , , ,⟨ ⟩ , , ,

when . m, . . . m

Therefore, the radius of the cloud

...

. m

Therefore, the diameter of the cloud is 25.42 m.

(c) When the concentration equals LFLmix ∗

√ / ⟨ ⟩ , , ,

Solve for x, . m

Therefore, ∆x = 1431.28 – 889.9 = 541.38 m. No. The toxic effects of this cloud must also be considered. The cloud contains CO, which is a gas would cause people to have respiratory problems. To be more conservative, all the people should stay further away where the concentration of CO is smaller than its TLV value.

Question 3 (a) Refer to the following figure

(b) Control system failures contains B, C; Protective system failures contain A, D, and E.

Fault tree:

(c) (d) T = (0.6 + 0.1 + 0.2)(0.05 + 0.01) = 0.054. (e) There are several considerations we need to take account when we proof-test. (1)

What is the failure rate? If it is very high, we should consider to test. (2) What is the repair time? If it is very difficult to repair, i.e. repair time is long, we should consider to test. (3) What is the expenditure of a new component? If it is very expensive, we should consider to test to reduce the costs. Since we don’t have enough information, I would recommend testing on the agitator motor and temperature sensor/transmitter.