Simple differential equation modelling the smoke layer

Question

Q1 Here is a simple differential equation modelling the smoke layer depth (y) in a large atrium provided with dynamic smoke extraction system.

dy/ dt = 1/A (m/p - V_o)

Where

A - Cross-section area of the floor (m²)

V_o - Volume extraction rate at the top of the atrium (m³/s)

T_s - Average temperature of plume (K) and can be derived by T_s = T_o + Q/mc_p neglecting heat loss to walls and opening, T_o is the ambient temperature; c_p is the specify heat of the plume gases (kJ/kg K)

p Smoke density at smoke temperature of T_s (kg/m³) and can be derived by Ρ = 353/T_s

m- Mass flow rate of the axisymmetric plume (kg/s) at the specified smoke layer height (z); It can be calculated from

m = 0.0711^.2_c^1/3 z^s/3 + 0.0018Q_c

and Q_c is the convective heat release rate (kW)

       z is the height above the base of the fire (m)

Use numerical method to solve the differential equation and sketch the smoke layer depth and the clear smoke height as times goes from time zero to equilibrium status when the flows into and out of the smoke layer and the surrounding have reached equilibrium. Consider the following input parameters for the steady fire:

Ambient Temperature Ceiling Height Cross-section Area of Floor Total Heat Release Rate Convective Heat Transfer Fraction Fuel Height above Floor Smoke Extraction Rate

25 °C 8.5 m 150 m2 4500 kW 0.7 Om 30 m3/s