A horizontal pipe has an abrupt expansion from D1 5 8 cm to D2 5 16 cm. The water velocity in the smaller section is 10 m/s and
the flow is turbulent. The pressure in the smaller section is P1 5 410 kPa. Taking the kinetic energy correction factor to be 1.06 at both the inlet and the outlet, determine the downstream pressure P2, and estimate the error that would have occurred if Bernoulli’s equation had been used.
1 answer:
- Answer: Explanation: Application of the bernoulli's equation comes in from conservation of mass flow. The cross sectional area of the two pipes are calculated. from A = πD²/4 The velocity of water from conservation of mass flow is also calculated ; V2 = Ac1V1/Ac2 The Loss coefficient is then calculated from KL = (1 - Ac1²/Ac2²)² Then the head Loss (hL) is calculated The detailed calculated and appropriate steps is as shown in the attached files.
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Answer:
I)E= 40.95 GPa
II)E=5.29 GPa
Explanation:
I)
Given that
E₁ = 2.41 GPa ,V₁=1-0.55 = 0.45
E₂ = 72.5 GPa ,V₂=0.55
Longitudinal moduli given as ;
E= E₁V₁+E₂V₂
E= 2.41 x 0.45 + 72.5 x 0.55 GPa
E= 40.95 GPa
II)
E₁ = 2.41 GPa ,V₁=1-0.55 = 0.45
E₂ =230 GPa ,V₂=0.55
Transverse moduli given as:
E=5.29 GPa
To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are
Real power in 3 phase
Now the Phase Voltage is,
The current phase would be,
Rearranging,
Replacing,
Therefore the current per phase is 2.26kA