Do not ionize in solutions
Poor conductors of electricity/heat
Low melting/boiling points
gases or liquids at room temperature
Complete Question
The complete question is shown on the first question
Answer:
a) The duty of the heat exchanger is given as 6.8658 KJ /sec
b) The temperature of the water leaving the exchanger is TOUT = 29.84 ⁰C
c) The log mean difference is given as TZ = 47.317 ⁰ C
d) the UA value is UA = 145.10
Explanation:
The explanation is uploaded on the first and second ,third and fourth image
Answer:
Final temperature: 659.8ºC
Expansion work: 3*75=225 kJ
Internal energy change: 275 kJ
Explanation:
First, considering both initial and final states, write the energy balance:
Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:
The pressure is constant, so:
(There is a multiplication by 100 due to the conversion of bar to kPa)
So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):
On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:
Subtracting the first from the second:

Isolating
:

Assuming that it is water steam, n=0.1666 kmol

ºC