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Bas_tet [7]
3 years ago
7

Which one of the following faults cause the coffee in a brewer to keep boiling after the brewing cycle is finished?

Engineering
1 answer:
MrRa [10]3 years ago
8 0

Answer:

  C.  Welded contacts on the thermostat

Explanation:

Any fault that keeps the heating element heating when it should not is a fault that will cause the symptom described. The details <em>depend on the design of the brewer</em> (not given).

"A short at the terminals" depends on what terminals are being referenced. The device on-off switch terminals are normally connected together when the brewer is turned on, so a short there may not be observable.

"Welded contacts on the thermostat" will have the observed effect if the thermostat is the primary means of ending the brewing cycle. If the thermostat of interest is an overheat protective device not normally involved in ending the brewing cycle, then that fault may not cause the observed symptom.

__

If the heating element is open-circuit, no heating will occur. A gasket leak may cause a puddle, but may have nothing to do with the end of the brewing cycle. (Loss of water can be expected to end boiling, rather than prolong it.)

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a coil consists of 200 turns of copper wire and has a cross-sectional area of 0.8mm square . The mean length per turn is 80 cm a
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Answer:

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8 0
3 years ago
2. A well of 0.1 m radius is installed in the aquifer of the preceding exercise and is pumped at a rate averaging 80 liter/min.
hodyreva [135]

Question:

The question is not complete. See the complete question and the answer below.

A well that pumps at a constant rate of 0.5m3/s fully penetrates a confined aquifer of 34 m thickness. After a long period of pumping, near steady state conditions, the measured drawdowns at two observation wells 50m and 100m from the pumping well are 0.9 and 0.4 m respectively. (a) Calculate the hydraulic conductivity and transmissivity of the aquifer (b) estimate the radius of influence of the pumping well, and (c) calculate the expected drawdown in the pumping well if the radius of the well is 0.4m.

Answer:

T = 0.11029m²/sec

Radius of influence = 93.304m

expected drawdown = 3.9336m

Explanation:

See the attached file for the explanation.

8 0
3 years ago
A eutectic alloy is which of the following (two correct answers): (a) the composition in an alloy system at which the two elemen
NeTakaya

Answer:

Option (c) and option (d)

Explanation:

Eutectic system is one in which a solid and homogeneous mixture of two or more substances resulting in the formation of super lattice is formed which can melt or solidify at a temperature lower than the melting point of any individual metal.

Eutectic alloys are those which have its components mixed in a specific ratio.

It is the composition in an alloy system for which both the  liquidus and solidus temperatures are equal.

Eutectic alloys have the composition in which the melting point of the metal is lower than the other alloy composition.

3 0
3 years ago
WIL MARK U AS BRAINLIEST HELP PLZ
ss7ja [257]

Answer:

The 1st one:Your natural ability

4 0
3 years ago
A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given pressure drop per unit length of pip
JulsSmile [24]

Answer:

Part A: (d/D=0.1)

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV percent=21.7%

Explanation:

We are going to use the following volume flow rate equation:

DeltaV=\frac{\pi * DeltaP}{8*u*l}(R^{4}-r^{4} -\frac{(R^{2}-r^{2})}{ln\frac{R}{r}}^{2})

Above equation can be written as:

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{r}{R} )^{4}+\frac{(1-(\frac{r}{R} )^{2})}{ln\frac{r}{R}}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{d}{D} )^{4}+\frac{(1-(\frac{d}{D})^{2})}{ln\frac{d}{D}}^{2})

First Consider no wire i.e d/D=0

Above expression will become:

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0)^{4}+\frac{(1-(0)^{2})}{ln0}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}

Part A: (d/D=0.1)

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.1)^{4}+\frac{(1-(0.1)^{2})}{ln0.1}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.574

DeltaV percent=\frac{(\frac{\pi*R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.574}{\frac{\pi*R^{4}*DeltaP}{8*u*l} }*100

DeltaV percent=\frac{1-0.574}{1}*100

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.01)^{4}+\frac{(1-(0.01 )^{2})}{ln0.01}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.783

DeltaV percent=\frac{(\frac{\pi *R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.783}{\frac{\pi *R^{4}*DeltaP}{8*u*l} }*100

DeltaV percent=\frac{1-0.783}{1}*100

DeltaV percent=21.7%

5 0
3 years ago
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