1. On page 0-34 of the course notes, we compared dairy manure table values and a

Question

1. On page 0-34 of the course notes, we compared dairy manure table values and an actual analysis (given on page O-35). Assume the difference in estimated and actual total N contents is due to the fact that newspaper (0.10% N on dry basis, 8% moisture content wet basis) was used as bedding material. a) How many pounds of dry newspaper were added per pound of dry manure solids to b) How many pounds of newspaper ("as is", i.e., including the moisture) have been added c) If we want to compost the manure by adding more newspaper to achieve a mixture dilute the total N content of the as-voided manure (3.94% total N) to the value in the analysis (3.05% total N). per cow per day. Use the manure production data for 1000-lb dairy cattle on page O-32 of the notes. (manure plus newly-added newspaper) with 55% moisture content (ideal for composting), how much newspaper needs to be added per pound of wet manure? 2. Estimate the C:N ratio for the dairy manure described on the attached analysis report. Does your value fall within the range reported for 29 dairy manure samples in the University of California report: http://manuremanagement.ucdavis.edu/files/134369.pdf?

Explanation / Answer

a)N-content of the voided manure=3.94% of N

N-content of dry news paper=0.1%

Let the mass of voided manure be x lbs

So, N content by mass=0.0394x

Amount of drynews paper=y

N-content by mass=0.001y

After addition the total mass=x+y

and total N-content=0.0305(x+y)

Equating the given values,0.0394x+0.001y=0.0305(x+y)

solving this linear equation you get,

0.0089x=0.0295y

or,x=3.315y

or y/x=0.0302

If 1 lbs of x, pounds of newspaper=[0.0302/(0.0302+1)]*1lbs=0.232 lbs

So 0.232 lbs dry newspaper per lb dry manure is to be added=0.232 lbs/lb manure

b) for 1000-lb dairy cattle,

total manure produced per day=82 lbs/d

Or,So pounds of newspaper added=(0.232 lbs/lb manure)*(82 lbs manure/d)=19.024 lbs

c)mass of wet manure=1lbs

Let z lbs of newspaper be added (newly)with moisture content =8%

So,moisture content added(newly)=0.08z

Moisture content added previously=0.232 lbs/lb manure

mass of manure after addition=1+z

net moisture content=55% of (1+z)=0.55(1+z)

Equating the above equations,

0.55(1+z)=0.08(z+0.232)

solving for z,

0.55+0.55z=0.08z+0.0186

0.47z=0.531 lbs

z=1.130 lbs

So 1.130 lbs of dry news paper per lb of manure needs to be added

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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