Team:SDU-Denmark/Model

Modeling

Heat loss of liquids in a closed cup environment

The PROSTATUS test uses the temperature sensitive RPA amplification, which is not functional below 25°C. In order to ensure that our final design for the test was feasible, we made a model in MATLAB that estimates how long it takes urine to reach 25°C in a closed cup. The model can estimate the heat conduction through a cup of any size and material. This could carry relevance for other projects, that have temperature sensitive procedures. If you wish to utilize the model you simply need to use the script below and fill in the following parameters:

Parameter Description Unit
r The top radius of the cup [cm]
R The bottom radius of the cup [cm]
S The edge length of the of the cup
(the model can also calculate this if necessary)		 [cm]
h The height of the cup
(only necessary to calculate S) 		[cm]
d The thickness of the cup walls [m]
k The thermal conductivity of the material [W m^-1 K^-1]
m The mass of the liquid [g]
c The specific heat of the liquid [J g^-1 C^-1]
T The starting temperature of the liquid [C]
T_r The temperature of the surroundings [C]
T_t The target temperature [C]

The above image shows the rate of diffusion for the PROSTATUS proof of concept cup, assuming a thermal conductivity of 0.13 W m^-1 K^-1

This model considers heat conduction through the walls of the container but does not take into account heat convection through the air. As such, the model will overestimate the rate of diffusion. Since the model was used as a rough estimate of the time interval in which the test can be conducted, this was not a concern for us. The rate of diffusion increases with temperature. Since urine is around 37°C there is a rather small temperature difference between the room temperature and urine, which further diminishes the effect of heat convection through air. For other teams, who intend to use this as a baseline, this might be necessary to consider, especially at higher temperatures. Please find the model below.

% This model is a part of iGEM SDU 2020 work.
% This model calculates the heat conduction through a closed cup
% It does not take into account the convection of air. If working with you
% are working with very warm liquids, this model will predict a greater
% rate of cooling than you will experimentally see, since it assumes the
% temperature around the cup is room temperature and does not take into
% account the fact that the cup walls become warm.
close all, clc, clear all
%Start by loading in the cup parameters
r = 4; %Top radius of the cup, cm
R = 3; %Bottom radius of the cup, cm
% S = 14; %Edge length of the cup (cm), if you want to calculate it, height is needed, see below

%
h = 10; %cm
S = sqrt((R-r)^2+h^2); %cm


% Calculate the total surface area of the cup and convert to meters
A = pi*(r^2+R^2)+pi*(r+R)*S; %cm^3
A = A*0.0001; %m^3
% Give the thickness of the cup walls
d = 2*10^-3; %m^3

% Thermal conductivity(k) is needed
% For polystyrene(styrofoam) it is given as 0.1-0.13 W m^-1 K^-1, at 23°c
% http://www.goodfellow.com/E/Polystyrene.html
k = 0.13;

% Mass of the liquid is needed in grams (if you use kg then you need to
% change the specific heat of the liquid, see below)
% If you know the value simply input it, but it can also be calculated from
% the volume

% %
% % s is the length of the side of the cup that touches the liquid. This
% % forms a right triangle with a side of d where rl (liquid radius at top)
% % is rl=x+d
% % assuming we know the height of the liquid, lets say half full, we can
% % solve for s, s = (h_l*S)/sqrt(S^2-(r1-r2)^2)
h_l = h/2; % assuming the cup is half full(or half empty)
s = (h_l*S)/sqrt(S^2-(r-R)^2);
% radius r_l can then be found
r_l = R+(s/S)*(r-R);
V_full =1/3*pi*h*(r^2+r*R+R^2); %cm^3
V_liquid = 1/3*pi*h_l*(r_l^2+r_l*R+R^2);
%

% Using the assumption that one cm^3 is equal to one g water
% % m=V_liquid;
% Using the assumption that one cm^3 is equal to 1.02 g urine
% % m = V_liquid*1.02

m = 250*1.020; %g

% specific heat of water is 4.184 Joules/gram*C
c = 4.184;

% Now imput your temperatures
T = 37; %starting temperature
T_r = 23; %room temperature
T_t = 25; %deg celsius that biomarkers become unusable
for i = 1:400
dT = -k*(A/(m*c*d))*(T-T_r);
T = T+(dT*(T-T_r));
temp(i)=T;
end


[minValue, closestIndex] = min(abs(T_t - temp.'))
closestValue = temp(closestIndex)
str = "RPA is possible for " + closestIndex + " Minutes";
loc = (max(temp)-min(temp))/2+min(temp)
figure
plot(temp,'LineWidth',3)
hold on
text(length(temp)/2,T_t+2,str,'FontSize',12)
text(closestIndex,T_t+0.5,"\downarrow",'FontSize',22)
grid off
baldurbeauty({'Estimated rate of temperature diffusion through', 'the PROSTATUS proof of concept cup'},'Time[m]','Temperature [C]','')

minValue =

   8.4539e-04


closestIndex =

   233


closestValue =

   24.9992


loc =

   30.4416


Published with MATLAB® R2020b

Note: This model uses a custom function to create the graph called "baldurbeauty". This is unrelated to the model itself and is only used to give the graph a specific style. Simply remove it for your use.