00:01
From the nlt balance equation given, we can simplify the terms as follows.
00:07
Du divided by dt equals to 0, ma equals to 100, kg, q equals to 0 and w equals to 0.
00:25
To calculate the mass of steam to be added, we need to use the following equation.
00:30
M steam equals to m water multiplied by t final minus t initial divided by h steam minus h water.
00:55
Where m water equals to 100 kg, t initial is 25 degree celsius, t final is 80 degree celsius, h water, enthalpy of water at 25 degree celsius and h steam is the enthalpy of steam that is 33 .0 bar and 300 degree celsius.
01:31
We can obtain the values of h water and h steam from the steam tables.
01:36
Using the steam tables, we find the h water equals to 104 .8 kg per kg and h steam equals to 3129 .1 kg per kg.
01:57
Substituting the values in the equation, we get m steam equals to 100 multiplied by 80 minus 25 divided by 3129 .1 minus 104 .8.
02:17
The value will be 3 .23 kg.
02:21
Therefore, 3 .23 kg of steam should be added...