A single-phase load is located 2800 ft from its source. The load draws a current of 86 A and operates on 480 V. The maximum volt
1 answer:
You might be interested in
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) (out)/2 + g Z(out)] - m(in) [h(in)
(in)/2 + g Z(in)]
Q - W = m2 [h2 2/2 + g Z2] - m1 [h1
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 2/2] - m1 [h1
1/2]
from mass balance m1 = m2
0 = [h2 2/2] - [h1
1/2]
Change in kinetic energy could be defined by
ΔKE = 2/2 -
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