Nutrient Removal by Cows Grazing Corn Residue

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

Want to understand the numbers above? 

In the following scenario is 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 would be needed because the amount of residue available for grazing is less. 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 capability of a pregnant cow to maintain body weight without supplementation of protein or energy and suggest that you can graze a 1000 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. Thus, forcing the cow to consume cobs and stalks will result in body weight loss. 

Organic matter removal

Given the proportions available, a cow’s diet would be 80% Husk and 20% leaf. The OM digestibility of husk is 60% and leaf is 40% and multiplying the digestibility of the OM by the OM content of these plant parts will provide the digestible OM (DOM) content of 58% for husk and 34% for leaf. A cow eats about 2% of BW 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, 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 1 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 an 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 and results in 886 lbs. of OM/ac 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, 886 lbs. OM/ac seems excessive.  However, with 230 bu. corn/acre there would be 6290 lbs. OM/acre in the residue, so only about 14% of the total OM was removed.  For lower yielding fields this would be lower. 

An important question is “what percent of the OM on 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 lb OM/ac with no grazing and 540 lb 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, which are lost through excretion in the urine and manure (i.e. she is at balance). However, it is typical to graze cows that are pregnant. So they do retain some minerals for calf growth (i.e. mineral retention). Using fetal mineral concentrations the mineral removal rates can be predicted.

Based on estimates for the mineral retention due to the development of a gravis uterus (i.e. womb and calf) during the third trimester of pregnancy, 208 g P, 125 g Ca, and 38 g K are retained by the pregnant cow over the 90-day period. 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 that are grazing on corn residue because there is actually not enough P in the residue to meet her requirement (allow 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 5.7g P/day 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 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 considered one of the major potential sources of loss with at about 10%. With N content of residue at 1%, a cow would consume about 109 g N/day and retain, on average, about 3.2 g N/day of that. Thus she would retain 0.63 lbs. of N over the 90-day period.  She would excrete 106 g/d of N  resulting in an additional loss of 2.1 lbs. N to volatilization for a total of 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. 

In 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.