Answer:
40.7 m
Explanation:
Let point 1 represent the surface of the water, point 2 be the top of the water trajectory and the reference be the bottom of the tank. Hence:
Answer:
symbolic machine code.
Explanation:
The instructions in the language are closely linked to the machine's architecture.
Answer:
ΔU =195 KJ
W=0 KJ
Explanation:
Given that
Mass of air m=3 kg
heat absorb by air =195 KJ
Cv=0.741 KJ/kgK
If we assume that air is as ideal gas so
Given that container is rigid it means that volume of system is constant so W=0
From first law of thermodynamics
Q=ΔU + W
⇒Q= ΔU (W=0)
So ΔU =195 KJ
And work done will be zero because because it is a constant volume process.
Answer:
y que cierren el orto si no demuestran nada
Answer:
POWER INPUT = 82.989 KW
Explanation:
For pressure P =0.24 MPa and T_1 = 0 DEGREE, Enthalapy _1 = 248.89 kJ/kg
For pressure P =1 MPa and T_1 = 50 DEGREE, Enthalapy _1 = 280.19 kJ/kg
Heat loss Q = 0.05w
Inlet diameter = 3 cm
exit diamter = 1.5 cm
volume of tank will be v = area * velocity
velocity at inlet
velocity at outlet
steady flow energy equation
solving wc = 1830.64 kJ/kg
wc in KWH
we know that