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8/11/2019 Simulation of drilling riser disconnection http://slidepdf.com/reader/full/simulation-of-drilling-riser-disconnection 1/91 Simulation of drilling riser disconnection - Recoil analysis Arild Grønevik Marine Technology Supervisor: Carl Martin Larsen, IMT Department of Marine Technology Submission date: June 2013 Norwegian University of Science and Technology
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Simulation of drilling riser disconnection

- Recoil analysis

Arild Grønevik

Marine Technology

Supervisor: Carl Martin Larsen, IMT

Department of Marine Technology

Submission date: June 2013

Norwegian University of Science and Technology

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!"#$%&' )^ Q$$&+-0G%----------------------------------------------------------- % Recoil analysis of drilling riser% - % Master thesis by Arild Gr¯nevik % % script - muddischarge.m  %---------------------------------------------------------- % % Description: % Calculates muddischarge velocity and frictional % forces acting on the riser as the drilling mud flows % out of the riser at the LMRP. Results are input for % the recoil analysis in Riflex 

%---------------------------------------------------------- clear all clc

% Riser input data %---------------------------------------------------------- L = 500 ; %Water depth D = 0.489 ; %internal diameter e = 5e-5; %roughness parameter g = 9.81 ; %gravity rho_sw = 1025; %density sea water rho_mud = 1600 ; %density drilling mud v_mud = 1e-4 ; %viscosity mud (typical values: 3-30cP) v_sw = 1.15e-6 ; %viscosity seawatered = e/D ; %relativ roughness

area = pi*D^2/4 ; %internal area of riser M_mud = rho_mud*area; %mass per unit length of mud M_sw = rho_sw*area ; %mass per unit length of seawater 

%---------------------------------------------------------------------- % Dynamical equilibrium between mass, frictional forces, and% hydrostatic pressure.%---------------------------------------------------------------------- 

%initial values timestep = 0.05;i=2;vel(1,1)=0;time(1,1)=timestep;

acc(1,1)=0;L(1,1) = L;

%loop acting while there is mud in the riser  while L(i-1,1) > 0

time(i,1) = timestep*i ;

%velocity of discharge vel(i,1) = vel(i-1,1) + acc(i-1,1)*timestep ;

%length of mud and water column L(i,1) = L(i-1,1) - vel(i-1,1)*timestep ;L_water(i,1) = L(1,1) - L(i,1) ;

%friction coefficient for mud and seawater Re_mud(i,1) = vel(i,1)*D/v_mud ;haaland_mud = -1.8*log10(6.9/Re_mud(i,1) + (ed/3.7)^1.11);

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f_mud(i,1)=1/(haaland_mud^2);

Re_sw(i,1) = vel(i,1)*D/v_sw ;haaland_sw = -1.8*log10(6.9/Re_sw(i,1) + (ed/3.7)^1.11);f_sw(i,1)=1/(haaland_sw^2);

%hydrostatic pressure and pressure drop due to friction p_hydro(i,1) = (rho_mud-rho_sw)*g*L(i,1) ;p_mudfriction(i,1) = f_mud(i,1)*L(i,1)/D*vel(i,1)^2*rho_mud /2 ;p_waterfriction(i,1) = f_sw(i,1)*L_water(i,1)/D*vel(i,1)^2*rho_sw /2 ;p_friction(i,1) = p_mudfriction(i,1) + p_waterfriction(i,1) ;

%Gravity force on mud column G(i,1) = (rho_mud-rho_sw)*g*L(i,1)*area ;

%Newtons second law sum_forces(i,1) = p_hydro(i,1)*area + G(i,1) - p_mudfriction(i,1)*area -

p_waterfriction(i,1)*area ;

acc(i,1) = sum_forces(i,1) / ( (M_mud*L(i,1)) + M_sw*L_water(i,1) );

%resulting total friction force working on riser f_friction(i,1)= (p_mudfriction(i,1) + p_waterfriction(i,1) ) *area/ L(1,1) ;

i = i+1 ;end 

time_discharge = max(time)

figure(1)plot(time,vel) ;xlabel('time [s]')ylabel('velocity [m/s]')title('Mud discharge velocity')

figure(2)plot(time,p_friction) ;xlabel('time [s]')ylabel('pressure [Pa]')title('Frictional pressure loss')

figure(3)plot(time,L)xlabel('time [s]')ylabel('length [m]')title('Length of mud column')

figure(4)plot(time,f_friction);

xlabel('time')ylabel('friction [N/m]')title('Friction forces working on the riser')

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%------------------------------------------------------------------ % Recoil analysis of drilling riser% - % Master thesis by Arild Gr¯nevik % % script - w_dyn_force.m  %------------------------------------------------------------------ % % Description: % script that write user specified nodal forces to % to a text file. The code is then copied into the relevant % sima_dynmod.inp file.%------------------------------------------------------------------ 

% Main riser data %------------------------------------------------------------------ n_seg = 3; %number of segments in line 

n_elements1 = 40 ; %number of elements in segment n_elements2 = 304 ;n_elements3 = 40 ;n_elementstot = n_elements1 + n_elements2 + n_elements3 ;timeon = 67 ; %disconnection time l_elem = 1.25; %code is currently for elements with equal lengths 

%forces divided into working on every n_nodes%total forces (3 each written node) n_nodes=4 ;

%n_forces = n_elementstot/n_nodes*3+3;

%if mass loss included 

n_forces = n_elementstot/n_nodes*5+5;

%identifying the maximum force and time for it [f_max,time_max] = findpeaks(f_friction) ;time_max = time_max*timestep ;

%to replicate the force in a best possible way, 1 linear increasing%force, 1 constant force and 1 decreasing linear force starting at %different times are used, force is written in kN %---------------------------------------------------------- p1_up = f_max/time_max*n_nodes*l_elem/1000; %ramp force (linear) timemax = floor(timeon+time_max) ; %time for maximum force timeoff = floor(timeon+time_discharge); %end time p_max = f_max*n_nodes*l_elem/1000; %constant force (max) 

p1_down= f_max/(time_discharge-time_max)*l_elem *n_nodes/1000 ;%ramp force, linear decreasing 

% printing simplefied mass loss %----------------------------------------------------------------- time_m_max = 35 ; massloss= (M_mud-M_sw)*L(1,1) ;p1_m_ramp = massloss/L(1,1)*g *n_nodes*l_elem /time_m_max/1000 ;p1_m_const = massloss/L(1,1)*g *n_nodes*l_elem /1000 ;

% Prints to file (inlcuding dummy code) %------------------------------------------------------------------- 

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fid = fopen('dynamic_nodal_forces.txt','w+');fprintf(fid,'DYNAMIC NODAL FORCES') ;fprintf(fid,'\n ''----------------------------- ') ;fprintf(fid,'\n ''ndcomp \t cinput \t chfloa ') ;fprintf(fid,'\n %i \t NOFILE ',n_forces) ;fprintf(fid,['\n ''line-id \t ilseg \t ilnod \t ildof \t chicoo '... 

'\t iforty \t timeon \t timeoff \t p1 \t p2 \t p3 ']) ;

ilseg = 1;for j=1:n_seg

if j == 1n_elements = n_elements1 ;

elseif j == 2n_elements = n_elements2 ;

elseif j== 3n_elements = n_elements3 ;

end 

for i = 1:n_nodes:n_elementsilseg = j;%printing ramp force fprintf(fid,'\n %s \t %i \t %i \t %i \t %s \t %i \t %f \t %f \t %f\t %i

\t %i','line4', ilseg, i, 3, 'GLOBAL', 3, timeon, timemax, -p1_up, 0, 0 );

%printing ramp force mass loss fprintf(fid,'\n %s \t %i \t %i \t %i \t %s \t %i \t %f \t %f \t %f\t %i

\t %i','line4', ilseg, i, 3, 'GLOBAL', 3, timeon, (timeon+time_m_max),p1_m_ramp, 0, 0 );

%printing constant force fprintf(fid,'\n %s \t %i \t %i \t %i \t %s \t %i \t %f \t %f \t %f\t %i

\t %i','line4', ilseg, i, 3, 'GLOBAL', 1, timemax, timeoff, -p_max, 0, 0) ;

%printing constant force mass loss fprintf(fid,'\n %s \t %i \t %i \t %i \t %s \t %i \t %f \t %f \t %f\t %i\t %i','line4', ilseg, i, 3, 'GLOBAL', 1, (timeon+time_m_max), 220,p1_m_const, 0, 0) ;

%printing negative ramp force fprintf(fid,'\n %s \t %i \t %i \t %i \t %s \t %i \t %f \t %f \t %f\t %i

\t %i','line4', ilseg, i, 3, 'GLOBAL', 3, timemax, timeoff, p1_down, 0, 0) ;

end end 


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