Question

Liquid water enters a valve at 300 kPa and exits at 275 kPa. As water flows through the valve, the change in its temperature, stray heat transfer with the surroundings, and potential energy effects are negli-gible. Operation is at steady state. Modeling the water as incompress-ible with constant rho= 1000 kg/m3, determine the change in kinetic energy per unit mass of water flowing through the valve, in kJ/kg

Answer 1

Answer:

The change in kinetic energy per unit mass of water flowing through the valve is - ΔKE = 0.025 KJ/Kg

Explanation:

Knowing

-Fluid is air

-inlet 1: P1 = 300 kPa

-exit 2: P2 = 275 kPa

density - rho= 1000 kg/m3

Using the formula

Δh = cΔT + Δp/rho

as change in temperature is neglected then change in enthalpy becomes

Δh = Δp/rho

energy equation could be defined by

Q - W = m(out) [h(out) $V^{2}$(out)/2 + g Z(out)] - m(in) [h(in) $V^{2}$(in)/2 + g Z(in)]

Q - W = m2 [h2 $V^{2}$2/2 + g Z2] - m1 [h1 $V^{2}$1/2 + g Z1]

as for neglecting potential energy effects

Q - W = m2(h2) - m1(h1)

as the system is adiabatic and has no work done

0 = m2 [h2 $V^{2}$2/2] - m1 [h1 $V^{2}$1/2]

from mass balance m1 = m2

0 = [h2 $V^{2}$2/2] - [h1 $V^{2}$1/2]

Change in kinetic energy could be defined by

ΔKE = $V^{2}$2/2 - $V^{2}$1/2

Change in specific enthalpy could be defined by

Δh = h2 - h1

Then the change in kinetic energy per unit mass of water flowing through the valve could be calculated as following

ΔKE = -Δh = ΔP/rho

-(275 - 300)/1000 = 0.025 KJ/Kg

- ΔKE = 0.025 KJ/Kg