A municipal solis waste (MSW) mixture has the following composition: a) Calculat

Question

A municipal solis waste (MSW) mixture has the following composition:

a) Calculate air requirements for combustion of 100 pounds of MSW for the folowing conditions:

- with theorical air

- with 50% excess air

b) Calculate the composition of stack gases when 50% excess air is used (assume water is in gas form)

c) Calculate the total volume (cubic feet/lb of MSW) of stack gases if these gases are 8000F and 50% excess air is used. Assume is 100% complete (at STP 1 mole of gas is 22.4liters, 1 cubic ft =28.3 liters).

COMPONENT Weight% atomic/molecular weight C 36 12 H 6 1 O 32 16 Ash 8 - H2O 18 18

Explanation / Answer

a) Carbon (C):

C+O2 CO2

12Kg C + 32Kg O2 44Kg CO2

Oxygen required = 0.36 X (32/12) = 0.96 Kg/Kg MSW

Hydrogen (H):

H2 + 1/2 O2 H2O

2Kg H2 + 16Kg O2 18Kg H2O

1Kg H2 + 8Kg O2 9Kg H2O

Oxygen required = 0.06 X (8/1) = 0.48 Kg/Kg MSW

- with theoritical air,

Oxygen required per Kilogram of MSW = 1.12 Kg

Air required per Kilogram of MSW = 1.12 / 0.233 = 4.81 Kg (Where air is assumed to contain 23.3% O2 by mass)

1 pound = 0.45 kg, 100 pounds = 45 kg

- with theoritical air, Air required for 45 Kg or 100 pound of MSW = 45 X 4.81 = 216.45 Kg

- with 50% excess air,

Air required per Kilogram of MSW = 1.12 / [0.233 + (0.233 X 0.5] = 1.12 / 0.3495 = 3.2 Kg

- with 50% excess air, Air required for 45 Kg or 100 pound of MSW = 45 X3.2 = 144 Kg

Component Weight fraction Oxygen required (Kg/Kg MSW) C 0.36 0.96 H 0.06 0.48 O 0.32 -0.32 Ash 0.08 - H2O 0.18 - Total 1 1.12

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.