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spin [16.1K]
3 years ago
5

Assume the triac of an AC discrete output module fails in the shorted state. How would this affect the device connected to this

output ?
Engineering
1 answer:
Aloiza [94]3 years ago
3 0

Answer:

The device connected to the output would behave the same as if it had been connected directly to the ac power supply. It would run at full power and it would not be controllable.

Explanation:

If the triac of an ac output module fails in the shorted state, the device connected to the output would behave the same as if it had been connected directly to the ac power supply. It would run at full power and it would not be controllable.

This implies that there is some sort of power storage device with the discrete output module which makes the system to run as if it is been connected to a source of power supply.

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Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of
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Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant

Since the losses are neglected thus applying this theorm between upper and lower porion we have

\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}

Now by continuity equation we have

A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s

Applying the values in the Bernoulli's equation we get

\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars

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What should be your strongest tool be for gulding your ethical decisions making process
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R_{conduction}=\frac{ln(\frac{r_{2}}{r_{1}})}{2\pi LK}

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