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2 years ago

A technician needs to check the heating operation of a heat pump that has no gauge access ports. The technician should start by:

Engineering

1 answer:

OverLord2011 [107]2 years ago

7 0

The technician should start by checking the temperature rise across the indoor coil.

<h3>Who is a technician?</h3>

This is a person who has skill in a particular area of job. A technician is responsible for repairs and also ensure different equipment and systems are working perfectly.

Hence, the technician should start by checking the temperature rise across the indoor coil.

Learn more about technician here : brainly.com/question/13315405

#SPJ1

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Joe is a chemical engineer whose plant discharges heavy metals into the local river. By the test authorized by the city governme

chubhunter [2.5K]

Answer:

B probably

Explanation:

Because the prompt doesn't specify what sort of violation it could be anything maybe when they release the metals during the day and so on.

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3 years ago

PLEASE HELP ME!!!!!! IM LOW ON POINTS BUT I NEED SOME HELP, QUICK!!! POINTS FOR HELPFUL ANSWERS + BRAINLIEST!!!!!

meriva

Answer:

Kitchen sinks can be clogged for a few reasons, food hair, and other things. Overtime, buildup of these things lead to clogging.

A few methods to fix clogging are, baking soda mixture, plunging, and boiling water.

The first method, baking soda mixture, what you do is you pour a cup of baking soda and vinegar down the drain, and place a rubber stopper to cover the drain opening. Wait 10 or more minutes, and take out the stopper, and run hot tap water until it's unclogged.

The second method is plunging. Make sure the sink has enough water to submerge the plunger, and start plunging. While you are plunging, maintain a good seal to get the best results. This usually works, but sometimes it doesn't.

The last method is boiling water. Pour boiling water into the drain, then after a while, the clogged drain should unclog, if it doesn't repeat the process until necessary.

Explanation:

8 0

3 years ago

Under normal conditions, gauge pressure is ______ absolute pressure.

levacccp [35]

Answer:

Option B Lower than

Explanation:

Gauge pressure is a relative measurement based on atmospheric pressure. Gauge pressure can be positive if it is above atmospheric pressure or it can also be negative it is below. On another hand, absolute pressure is an actual pressure in a space and its value has always to be zero or above. Basically absolute pressure is zero if it is in a perfect vacuum. So the measurement of absolute pressure is gauge pressure + atmospheric pressure. This is the reason in normal condition the gauge pressure = absolute pressure - atmospheric pressure and therefore is lower than absolute pressure

3 0

4 years ago

Sea A una matriz 3x3 con la propiedad de que la transformada lineal x → Ax mapea R³ sobre R³.

skelet666 [1.2K]

Answer:

Explanation:

7 0

3 years ago

Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra

gladu [14]

Answer:

a) $T_{2}=837.2K$

b) $e=91.3$ %

Explanation:

A) First, let's write the energy balance:

$W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})$ (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 $\frac{kJ}{kgK}$ And its specific R constant is 0.287 $\frac{kJ}{kgK}$ .

The only unknown from the energy balance is $T_{2}$ , so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

$T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K$

B) The isentropic efficiency (e) is defined as:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}$

Where ${h_{2s}$ is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because $h_{2}-h_{1}$ can be obtained from the energy balance $\frac{W}{m}=h_{2}-h_{1}$

$h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}$

An entropy change for an ideal gas with constant Cp is given by:

$s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})$

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

$0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})$

Applying logarithm properties:

$ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}$

Then,

$T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K$

So, now it is possible to calculate $h_{2s}-h_{1}$ :

$h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}$

Finally, the efficiency can be calculated:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3$ %

4 0

4 years ago