Question

somebody can help me THERMODYNAMICS
*a thermodynamic steady flow system receives 4.56 kg per min of a fluid where p1=137.90 kPa, v1=0.0388m^3/kg, w1=122m/s, and u1=17.16kJ/kg. the fluid leaves the system at a boundary where p2=551.6kPa, v2=0.193m^3/kg, w2=183m/s and u2=52.80 kJ/kg. during passage through the system the fluid receives 3000J/s of heat. determine the work.

Answer 1

enthalpy: h1=u1+p1*v1=17.16+137.9*0.0388=22.51 KJ/Kg h2=u2+p2*v2=52.8+551.6*0.193=159.2588 KJ/Kg

Answer 2

Energy Balance for Open Systems without accumulation
\[Q-W+\sum m _{i}(h _{i}+\frac{w _{i}^2}{2}+gz _{i})-\sum m _{e}(h _{e}+\frac{w _{e}^2}{2}+gz _{e})=0\]

Answer 3

Q=3000J/s=3 KJ/s
m=4.56 Kg/min=0.076 Kg/s

Answer 4

then

\[W=3+0.076(22.51+\frac{122^2}{2000}-159.2588-\frac{183^2}{2000})\]

Answer 5

W is in KJ/s or KW
use a calculator to evaluate it

Answer 6

W=-8.099 KJ/s or KW

Answer 7

the answer was given and it was -486.0018KJ/s
...the difficult here is the conversion

Answer 8

let me check it again

Answer 9

my equation for W is rifht or not?

Answer 10

*right

Answer 11

there is a two work flow, given, so i think it is
wf1=mp1v1

Answer 12

thats work of flow
i think the problem is calculating work of shaft

Answer 13

@miat
the answer is given maybe is in KJ/min
because
my answer*60=answer is given
could u plz check the question again

Answer 14

the w1& w2 by the way is velocity

Answer 15

well thats right

Answer 16

W=-8.099 KJ/s or KW is correct but in order to have the final answer used this formula:
w=(KE1-KE2)+(Wf1-Wf2)+(V1-V2)+Q

Answer 17

oh yes thats correct

Answer 18

KE=m w^2/2
Wf=mPv
V=mu
so whats your problem with conversion?