Nutrient Removal by Cows Grazing Corn Residue

Cattle grazing a corn field
Cows grazing corn residue may not remove as much organic material and as many nutrients as might be expected, but they will likely redistribute some nutrients in the field due to their traffic patterns.

Nutrient Removal by Cows Grazing Corn Residue November 5, 2018

Key Points

Nutrient removal by cows grazing corn residue may not be as much as you might think. A pregnant cow grazing at suggested stocking rates for 90 days on a 230 bu/ac corn field will remove:

  • About ½ ton of residue.
  • About 2 lbs of nitrogen (N) per acre.
  • No phosphorus (P) or calcium (Ca). The amount of P available in residue is not sufficient to meet the nutritional needs. Therefore, cattle producers usually provide a free-choice mineral supplement that contains both P and Ca. There would actually be about ½ lb/acre of both P and Ca added in this situation.  
  • Essentially no potassium (K).

Want to Understand the Numbers Above?

Let’s look at a scenario for a field with a grain yield of 230 bu/ac. In this situation cows can graze for 90 days when stocked at 1.3 acres per cow.  In lower yielding fields, fewer cows or fewer days per acre should be applied because there is less residue for grazing. Thus, the nutrient removal estimates on a per acre basis would be lower (i.e., high yield fields can be stocked heavier and have greater potential removal than lower yielding fields).

Suggested Stocking Rates

Corn residue is about 10% husk and 34% leaf with the remaining residue being stalk and cob. Recommended stocking rates are based on the ability of a pregnant cow to maintain body weight without supplementation of protein or energy. The rates suggest that you can graze a 1200 lb cow for 30 days for every 100 bu. of corn grain produced. This would result in the cow consuming only about 12% to 15% of the corn residue in the field and nearly all would be husk with some leaf. Cob and stalk have less energy available. Forcing the cow to consume cobs and stalks will actually result in a loss of body weight.

Organic Matter Removal

Given the proportions available, a cow’s diet would be 80% husk and 20% leaf. The organic matter (OM) digestibility of the husk is 60% and of the leaves, 40%. Multiplying the OM digestibility by the OM content of these plant parts will provide a digestible OM (DOM) content of 58% for husk and 34% for leaf. A cow eats about 2% of its body weight per day (DM basis) when grazing corn residue. Thus, a 1200 lb cow would consume about 24 lbs of DM. Over a 90-day period, she would consume 2158 lbs of DM. The intake was on average 80% husk, so 1727 lbs DM from husk x 58% DOM = 1001 OM removed. For leaf matter, intake of 431 lbs DM x 34% DOM equals 147 lbs OM removed. Therefore, a total of 1148 lbs of OM was removed per cow over the 90-day period.

What is the organic matter removal per unit area of land? Remember that the target stocking rate was one cow per 100 bu of grain, so a cow could graze on a corn field that yielded 230 bu/acre for 69 days if stocked at one cow per acre. To get to 90 days, the stocking rate would need to be 1.3 acres per cow. Thus, the 1148 lbs of OM removed needs to be divided by 1.3. This would result in 886 lbs of OM/acre disappearance. In this example, no grain intake was assumed. Grain is more digestible, but typically less than 1 bu/acre of grain remains after harvest, which would be no more than 4% of the total intake. At a glance, that's 886 lbs. The OM/ac seems excessive; however, with 230 bu/acre corn, there would be 6290 lbs OM/acre in the residue. In this case, only about 14% of the total OM was removed.

An important question is “What percent of the OM on the soil surface becomes soil stable OM?” Not all of the OM from residue remaining on the soil surface will become soil organic matter. Some research suggests that in no-till systems about 10% of crop residue OM will become soil stable OM. With this in mind, the soil stable OM contribution from the residue would be 629 lbs OM/ac with no grazing and 540 lbs OM/ac in the grazed scenario.

Mineral Removal (P, Ca, K and N)

The mineral (N, P, K, etc.) requirements of a beef cow do not equal the amount that she retains in her body. A non-pregnant, mature cow is at maintenance (not gaining or losing weight) and would only require the minerals that are lost through excretion in the urine and manure (i.e., she is at balance). However, it is typical to graze pregnant cows, which do retain some minerals to support calf growth. Using fetal mineral concentrations, the mineral removal rates can be predicted.

It’s estimated that during the third trimester of pregnancy, a cow will retain 208 g P, 125 g Ca, and 38 g K over a 90-day period to support development of a gravis uterus (i.e., womb and calf). Using the above stocking rate assumption, this retention is equivalent to 0.35 lb P/acre, 0.21 lb Ca/acre, and 0.06 lb K/acre. However, we also need to consider that a typical production practice is to supplement a mineral with both P and Ca to cows grazing corn residue because there is not enough P in the residue to meet requirements (allowing her to recover what she excretes and retain what she needs for the pregnancy). For example, a 4 oz./day intake of free-choice mineral containing 5% P would contribute 5.7 g/day of P into the system or 513 g over the 90-day grazing period. Usually minerals containing P have an equal amount of Ca to make sure that the ratio in the diet remains in balance. The net result is actually more P and Ca being added than removed (in this case 0.51 lb P/acre and 0.65 lb Ca/acre).

The amount of N removed is more difficult to estimate. Not all N in plant residues will become plant available and there are many potential avenues for loss of N in the system, including volatilization, denitrification, leaching, and runoff. The amount of N retained by the pregnant cow would be estimated at 292 g over the 90-day period. In this situation, N loss from volatilization of N that is consumed and then excreted in urine and manure would be about 10% and considered one of the major potential sources of loss. With the N content of residue at 1%, a cow would consume about 109 g of N/day and retain, on average, about 3.2 g of N/day. Thus she would retain 0.63 lbs of N over the 90-day period. She would excrete 106 g of N per day, resulting in an additional loss of 2.1 lbs N to volatilization for a total loss of 2.7 lbs of N. Given the stocking rate of 1.3 ac per cow, we would assume a loss of 2.1 lbs of N/acre.

Nutrient Redistribution

It should also be noted that all nutrients will be redistributed across the landscape with an unequal distribution. It is also likely that more nutrients will be excreted in areas where cattle spend more time, such as around the water source and in sheltered areas.

Acknowledgment

The authors would like to thank Marty Schmer, research agronomist, and Virginia Jin, research soil scientist, both with USDA Agricultural Research Service, for their thoughts and review of this article.

Table 1. Nutrient removal for a pregnant cow grazing corn residue for 90 days in a field stocked at 1.3 acres per cow.
Daily intake (lbs)Intake (lbs/90d)Retention (lbs/90d)Excretion (returned) (lbs/90d)Loss of excreted N*
(lbs/90d)
Free choice mineral† (lbs/90d)Net removal (lb/ac)Amount in residue (lb/ac)Removal, % of residue content
Dry matter 24 2158 1252 906 966 7147 14
Organic matter 22 2007 1148 859 886 6290 14
N 0.24 21.6 0.64 20.9 2.1 2.11 71.5 3.0
P 0.02 1.8 0.46 1.3 1.1 -0.51 5.7 -8.9
Ca 0.09 8.0 0.28 7.7 1.1 -0.65 26.4 -2.4
K 0.27 24.0 0.08 23.9 0.06 79.3 0.1
* About 10% of nitrogen that is excreted (urine and manure) will be lost (volatilized).
† Corn residue does contain enough P to meet the cows' required intake needed to replace P turnover (excretion) of 0.03 lb/d plus retention for pregnancy thus the contribution from a free choice mineral containing 5% phosphorus and 5% calcium consumed at 4 oz. per day has been included.