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THERMODYNAMICS - CASE STUDY SOLUTION

John plans to wash his house using a power washer. The power input to the power unit needs to be determined.

Assumptions:

• The control volume enclosing the power unit and the delivery hose is at steady state
• The water is modeled as incompressible

Control Volume

(1) Determine the power input

The energy balance for the one-inlet, one-exit steady-flow process is:

Introducing , and solving for

The specific volume of water v at 20 oC is 1.0018(10-3) m3/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/kg-K. Hence, the enthalpy term can be determined.

h1 - h2 = c(T1 - T2) + P(v1 -v2)
= 4.18(20-23) + 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:

v1 = V/A = 0.2 m/s

Hence the kinetic energy term can be evaluated as follow:

(v12 - v22)/2 = (0.22 - 502)/2 = -1,250 J/kg

Finally, the potential energy term is

g(z1 - z2) = 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.

 Energy (J/kg) Enthalpy Kinetic Energy Potential Energy Heat Loss 12,540 1,250 49.05 1,540 Energy/power input (W/kg) 81.4% 8% 0.6% 10%