Respuesta :
Answer:
work transfer is - 120 kJ
heat transfer is - 76.25 kJ
constant K is 6.35 × [tex]10^{-3}[/tex] kW/min
Explanation:
given data
p1 = 0.9 MPa
u1 = 232.92 kJ/kg
transfers energy rate = 0.1 kW
time = 20 min
p2 = 1.2 MPa
u2 = 276.67 kJ/kg
to find out
the work and heat transfer and value of the constant K
solution
we will find work done of system as
work done W = [tex]\int\limits^a_b {W} \, dt[/tex]
here limit a to b is 0 to 20 min and W is 0.1 kW
so
W = [tex]\int\limits^0_20 {0.1} \, dt[/tex]
W = - 0.1 ( 20 ) × 60second
Work done W = - 120 kJ
here negative sign show that work done is on the system
so work transfer is - 120 kJ
and
for heat transfer we will apply first law of thermodynamic that is
dQ = dU + dW
here
Q = m ( u2-u1) + W
here Q is heat transfer and m = 1 and W is -120
so Q = 1 ( 276.67 - 232.92 ) - 120
Q = - 76.25 kJ
here negative sign show that heat transfer is system to surrounding
so
heat transfer is - 76.25 kJ
and
constant k is calculates as heat transfer formula that is
Q = [tex]\int\limits^a_b {Q} \, dt[/tex]
here a to b limit is 0 to t
and Q is Kt
Q = [tex]\int\limits^0_t {Kt} \, dt[/tex]
Q = [tex]\frac{-kt^2}{2}[/tex]
K = [tex]\frac{-2Q}{t^2}[/tex]
K = [tex]\frac{-2(-76.25)}{20^2}[/tex] × [tex]\frac{1min}{60 sec}[/tex]
K = 6.35 × [tex]10^{-3}[/tex] kW/min
so constant K is 6.35 × [tex]10^{-3}[/tex] kW/min