Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J
Answer:
The mass rate of the cooling water required is:
Explanation:
First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.
Where w refers to the cooling water and s to the steam flow. Reorganizing,
Write the difference of enthalpy for water as Cp (Tout-Tin):
This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, and
can be calculated as:
.
The Cp of water at 25ºC (which is the expected average temperature for water) is: 4.176 . If the average temperature is actually different, it won't mean a considerable mistake. Also we know that
, so let's work with the limit case, which is
to calculate the minimum cooling water mass rate required (A higher one will give a lower temperature difference as a result). Finally, replace data: