CASE STUDY (Q5)

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CASE STUDY (Q5)

  1. 1. AHD 3233 Advanced MNTCase study 1: Question 5Nurul Husna Abu Bakar 0912622Nurfaraheen Mat Sahari 0913684Safina Samsudin 0916686Noor Syuhada Hj Mat 0913474Nurul ‘Ain Abdul Hamid 0915774
  2. 2. Question 5• Mr RK age 49,Indian male, was admitted to ICU ward due to burn with hot oil and scalded wound over left side of face and trunk (face involving nose and mouth). He then was diagnosed with 17% mixed thickness burn (43% superficial burn). Pt was put on RT (Ryle’s Tube). Plan for this patient’s EN regime.
  3. 3. ANTHROPOMETRY DATA• Height : 1.63 m• Weight : 80 kg• BMI = 30.1 kg/m² (Obese Class 1)
  4. 4. BIOCHEMICAL DATA Biomarkers Range IndicationSodium 135 mmol/L NormalPotassium 3.2 mmol/L LowUrea 3.3 mmol/L NormalCreatinine 42mmol/L LowAlbumin 22 g/L LowTotal protein 44 g/L LowWhite blood cell 8.4 NormalPlatelet 179 Normal
  5. 5. NUTRITION DIAGNOSIS• PES Statement - Protein losses related to burns as evidence by low total protein value which is 44g/L.
  6. 6. NUTRITION INTERVENTION Objectives Principles1. To provide nutrient support 1. Introduce suitable enteral formula consistent with patient’s that meets energy and nutrient medical condition requirement.2. To improve wound healing 2. Provide sufficient protein intake for nutrient balance3. To treat macronutrients and 3. Provide enteral formula withmicronutrients deficiencies adequate amount of macronutrients and micronutrients
  7. 7. NUTRITION INTERVENTION 1. ENERGY REQUIREMENT 1. Curreri formula 2. Harris BenedictER = (24 kcal X 80 kg) + (40 kcal X BEE = 66 + (13.7 X 80) + (5 X 16) – (6.817) X 49) = 2600 kcal = 1643.8 kcal TEE: = 1643.8 X 1.05 (ventilated) X 1.5 (burns) = 2589 kcal Average = 2600 + 2589 2 = 2594.5 kcal ~ 2600 kcal
  8. 8. NUTRITION INTERVENTION2. PROTEIN REQUIREMENT1. Based on 17% TBSA:= 1.5 g X 80 kg = 120 g = 19.2 N from protein=120 g 6.252. Sutherland formula = (1g X 80 kg) + (3g X 17% TBSA) = 131 gAverage = 120 g% Protein = 120 g x 4 kcal x 100 2600 = 18%
  9. 9. Macronutrient distribution CHO = 55 % Protein = 18 % Fat = 27 %= 55 X 2600 kcal = 18 X 2600 = 27 X 2600 100 100 100= 1430 kcal / 4 = 468 kcal / 4 = 702kcal / 9= 357.5 g = 117 g ~ 120 g = 78 g NPE: N ratio = CHO (1430 kcal) + Fat (702 kcal) 19.2 = 2132 kcal 19.2 N in Protein = 111: 1
  10. 10. ImplementationName of products used:1. Enercal Plus 1 scoop = 90 kcal + 45 ml water = 3.5 g of protein2. Myotein 1 scoop = 30.1 kcal = 5 g of protein
  11. 11. Enteral Nutrition RegimeSteps Formula Calories Protein 1 2 scoops Enercal Plus = 90 kcal x 2 scoops x 6 times = 3.5 x 2 x 6 + 90 ml water = 1080 kcal = 42 g 2 4 scoops Enercal Plus = 90 kcal x 4 x 6 = 3.5 x 4 x 6 + 180 ml water = 2160 kcal = 84 g 3 4 ½ scoops Enercal Plus = 90 kcal x 4 ½ x 6 = 3.5 x 4 ½ x 6 + 203 ml water = 2430 kcal = 94.5 g 2520.3 109.5 3 scoops Myotein = 30.1 kcal x 3 =5x3 = 90.3 kcal = 15 g 4 4 ½ scoops Enercal Plus 90 kcal x 4 ½ x 6 = 3.5 x 4 ½ x 6 + 203 ml water = 2430 kcal = 94.5 g 2580.5 119.5 5 scoops Myotein = 30.1 kcal x 5 =5x5 = 150.5 = 25 g
  12. 12. Enteral Nutrition RegimeSteps Formula Calories Protein 1 2 scoops Enercal Plus = 90 kcal x 2 scoops x 6 times = 3.5 x 2 x 6 + 90 ml water = 1080 kcal = 42 g 2 4 scoops Enercal Plus = 90 kcal x 4 x 6 = 3.5 x 4 x 6 + 180 ml water = 2160 kcal = 84 g 3 4 ½ scoops Enercal Plus = 90 kcal x 4 ½ x 6 = 3.5 x 4 ½ x 6 + 203 ml water = 2430 kcal = 94.5 g 2580.5 119.5 2 ½ scoops Myotein = 30.1 kcal x 2 ½ x 2 =5x2½x2 = 150.5 kcal = 25

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