- #106
- 23,514
- 5,785
In the earlier analysis I presented, if we include heat transfer between the tank gas and the surroundings, we obtain:
$$(F_3V-F_4m)\frac{dT}{dt}=\frac{dm}{dt}-\frac{UA}{(h_v-F_2)}(T-T_s)$$where A is the surface area of the tank, U is the overall heat transfer coefficient, and ##T_s## is the temperature of the surroundings; t is time.
$$(F_3V-F_4m)\frac{dT}{dt}=\frac{dm}{dt}-\frac{UA}{(h_v-F_2)}(T-T_s)$$where A is the surface area of the tank, U is the overall heat transfer coefficient, and ##T_s## is the temperature of the surroundings; t is time.