1. AHD 3233
Advanced MNT
Case study 1: Question 5
Nurul Husna Abu Bakar 0912622
Nurfaraheen Mat Sahari 0913684
Safina Samsudin 0916686
Noor Syuhada Hj Mat 0913474
Nurul ‘Ain Abdul Hamid 0915774

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. ANTHROPOMETRY DATA
• Height : 1.63 m
• Weight : 80 kg
• BMI = 30.1 kg/m² (Obese Class 1)

4. BIOCHEMICAL DATA
Biomarkers Range Indication
Sodium 135 mmol/L Normal
Potassium 3.2 mmol/L Low
Urea 3.3 mmol/L Normal
Creatinine 42mmol/L Low
Albumin 22 g/L Low
Total protein 44 g/L Low
White blood cell 8.4 Normal
Platelet 179 Normal

5. NUTRITION DIAGNOSIS
• PES Statement
- Protein losses related to burns as evidence by low
total protein value which is 44g/L.

6. NUTRITION INTERVENTION
Objectives Principles
1. 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 balance
3. To treat macronutrients and 3. Provide enteral formula with
micronutrients deficiencies adequate amount of macronutrients
and micronutrients

7. NUTRITION INTERVENTION
1. ENERGY REQUIREMENT
1. Curreri formula 2. Harris Benedict
ER = (24 kcal X 80 kg) + (40 kcal X BEE = 66 + (13.7 X 80) + (5 X 16) – (6.8
17) 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. NUTRITION INTERVENTION
2. PROTEIN REQUIREMENT
1. Based on 17% TBSA:
= 1.5 g X 80 kg = 120 g = 19.2 N from protein
=120 g 6.25
2. Sutherland formula = (1g X 80 kg) + (3g X 17% TBSA)
= 131 g
Average = 120 g
% Protein = 120 g x 4 kcal x 100
2600
= 18%

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. Implementation
Name of products used:
1. Enercal Plus
1 scoop = 90 kcal + 45 ml water
= 3.5 g of protein
2. Myotein
1 scoop = 30.1 kcal
= 5 g of protein

11. Enteral Nutrition Regime
Steps 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. Enteral Nutrition Regime
Steps 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