A 8.0 cm-diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through this pipe at a certain rate, the gauge p
1 answer:
Answer:
7.11x10^-3
Explanation:
We are to get the volume rate of flows
1/2pv1² + pl = 1/2pv2²
Such that A1V1 = A2V2
V1 = A2V2/A1
From the attachment I uploaded, we have a formula named equation 1 from which I have plugged in these values
P2 = 33000
P2 = 24000
P = 1000
r2 = 2.25
r1 = 4
When we put these values into the equation,
V2 = 4.47
A2V2 = pi(0.0225)²x4.47
= 7.7x19^-3m³/s
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Explanation:
For point a:
Energy balance equation:
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Since the tank was initially empty and the inlet is constant hence,
Interpolate the enthalpy between . The surrounding air
temperature:
Substituting the value from ideal gas:
Follow the ideal gas table.
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Substitute values from the table.
For point b:
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For point c:
Entropy is given by the following formula: