All categories

3 years ago

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.)

You might be interested in

Products. of sheet metal

Blababa [14]

Automobiles and Airplanes is what I can think of

3 0

3 years ago

Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carb

Nezavi [6.7K]

Answer:

$y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02$

Explanation:

Hello,

a. On the attached document, you can see a brief scheme of the process. Thus, to know the degrees of freedom, we state the following unknowns:

- $\xi_1$ and $\xi_2$ : extent of the reactions (2).

- $F_{O_2}^2$ , $F_{CH_4}^2$ , $F_{H_2O}^2$ , $F_{HCHO}^2$ and $F_{CO_2}^2$ : Molar flows at the second stream (5).

On the other hand, we've got the following equations:

- $F_{O_2}^2=50mol/s-\xi_1-2\xi_2$ : oxygen mole balance.

- $F_{CH_4}^2=50mol/s-\xi_1-\xi_2$ : methane mole balance.

- $F_{H_2O}^2=\xi_1+2\xi_2$ : water mole balance.

- $F_{HCHO}^2=\xi_1$ : formaldehyde mole balance.

- $F_{CO_2}^2=\xi_2$ : carbon dioxide mole balance.

Thus, the degrees of freedom are:

$DF=7unknowns-5equations=2$

It means that we need two additional equations or data to solve the problem.

b. Here, the two missing data are given. For the fractional conversion of methane, we define:

$0.900=\frac{\xi_1+\xi_2}{50mol/s}$

And for the fractional yield of formaldehyde we can set it in terms of methane as the reagents are equimolar:

$0.860=\frac{F_{HCHO}^2}{50mol/s}$

In such a way, one realizes that the output formaldehyde's molar flow is:

$F_{HCHO}^2=0.860*50mol/s=43mol/s$

Which is equal to the first reaction extent $\xi_1$ , therefore, one computes the second one from the fractional conversion of methane as:

$\xi_2=0.900*50mol/s-\xi_1\\\xi_2=0.900*50mol/s-43mol/s\\\xi_2=2mol/s$

Now, one computes the rest of the output flows via:

- $F_{O_2}^2=50mol/s-43mol/s-2*2mol/s=3mol/s$

- $F_{CH_4}^2=50mol/s-43mol/s-2mol/s=5mol/s$

- $F_{H_2O}^2=43mol/s+2*2mol/s=47mol/s$

- $F_{HCHO}^2=43mol/s$

- $F_{CO_2}^2=2mol/s$

The total output molar flow is:

$F_{O_2}+F_{CH_4}+F_{H_2O}+F_{HCHO}+F_{CO_2}=100mol/s$

Therefore the output stream composition turns out into:

Best regards.

7 0

3 years ago

It is an important part of the differential maintenance which purpose is to make smoother the differential operation by lubricat

masha68 [24]

Answer:

lubrication

Explanation:

MARK ME BRAINLEIST

4 0

2 years ago

Technician A says that if fuel pump pressure is correct, fuel pump volume will be correct as well. Technician B says that a fuel

guajiro [1.7K]

Answer:

Technician B only

Explanation:

hope this helps :)

5 0

2 years ago

The volume of microbial culture is observed to increase according to the formula

saw5 [17]

The expression of V(m³)=e^(t(s)) to make V in in³ and t in minutes is;

V(in³) = (¹/₆₁₀₂₄)a $e^{\frac{1}{60}bt(h)$

We are given that;

Volume of microbial culture is observed to increase according to the formula;

V = e^(t)

where;

t is in seconds

V is in m³

We want to now express V in in³ and t in minutes.

Now, from conversions;

1 m³ = 61024 in³

Also; 1 second = 1/60 minutes

according to formula for exponential decay, we know that;

V = ae^(bt)

Thus, we have;

61024V = ae^(¹/₆₀b(t(h))

V(in³) = (¹/₆₁₀₂₄)a $e^{\frac{1}{60}bt(h)$

Read more about subject of formula at; brainly.com/question/790938

3 0

3 years ago