(1) Determine the power input
The energy balance for the oneinlet, oneexit steadyflow process is:
Introducing ,
and solving for
The specific volume of water v at 20 ^{o}C is 1.0018(10^{3})
m^{3}/kg (from water table). The mass flow rate can be evaluated
using the given volumetric flow rate.
=
V/v
= (0.1/1000)/(1.0018(10^{3})) = 0.1 kg/s
Since water is incompressible, the specific heat c is 4.18 kJ/kgK.
Hence, the enthalpy term can be determined.
h_{1}  h_{2} =
c(T_{1}  T_{2}) + P(v_{1} v_{2})
= 4.18(2023) + 0 = 12,540 J/kg
The second term of the right hand of the previous equation is zero because
water is modeled as incompressible.
The velocity at the inlet is:
v_{1} = V/A = 0.2 m/s
Hence the kinetic energy term can be evaluated as follow:
(v_{1}^{2}  v_{2}^{2})/2 = (0.2^{2
}  50^{2})/2 = 1,250 J/kg
Finally, the potential energy term is
g(z_{1}  z_{2}) = 9.81(0  5) = 49.05 J/kg
Substituting the above values to the energy equation yields
=
0.1/0.9(12,540+(1,250)+(49.05))
= 1,540 W
where the negative sign means the power is provided to the washer.
(2) Determine the dominate power consuming term
The table below shows the proportion of different terms in the
energy balance equation to the power input to the power washer.
