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3 years ago

If the operating control on a commercial water heater fails, the gas valve will be de-energized by?

Engineering

1 answer:

kogti [31]3 years ago

5 0

Answer:

Do not operate water heater if exposed to flooding or water damage. ... de-energize the 24-volt gas valve and end the current heating cycle. The control system

Explanation:

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2 years ago

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Think for a moment about the potential problems with big data that the speaker mentioned:

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Answer: 1. sadly yes, some people are treated unfairly for crimes people yet have commited. 2. no 3. yes

Explanation:

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3 years ago

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Air is saturated with water vapor at 35.0 oC and a total pressure of 1.50 atmospheres. If the molar flow rate of the dry air in

Marizza181 [45]

Answer:

11.541 mol/min

Explanation:

temperature = 35°C

Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa

note : partial pressure of water in mixture = saturation pressure of water at T = 35°c )

from steam table it is = 5.6291 Kpa

calculate the mole fraction of H $_{2}o$ ( YH $_{2}o$ )

= 5.6291 / 151.95

= 0.03704

calculate the mole fraction of air ( Yair )

= 1 - mole fraction of water

= 1 - 0.03704 = 0.9629

Now to determine the molar flow rate of water vapor in the stream

lets assume N = Total molar flow rate

NH $_{2}o$ = molar flow rate of water

Nair = molar flow rate of air = 300 moles /min

note : Yair * n = Nair

therefore n = 300 / 0.9629 = 311.541 moles /min

Molar flowrate of water

= n - Nair

= 311.541 - 300 = 11.541 mol/min

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3 years ago

A completely reversible heat pump produces heat at a rate of 300 kW to warm a house maintained at 24 °C. The exterior air, which

trapecia [35]

Answer:

No.

Explanation:

The Coefficient of Performance of the reversible heat pump is determined by the Carnot's cycle:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{297.15\,K}{297.15\,K-280.15\,K}$

$COP_{HP} = 3.339$

The power required to make the heat pump working is:

$\dot W = \frac{300\,kW}{3.339}$

$\dot W = 89.847\,kW$

The heat absorbed from the exterior air is:

$\dot Q_{L} = 300\,kW - 89.847\,kW$

$\dot Q_{L} = 210.153\,kW$

According to the Second Law of Thermodynamics, the entropy generation rate in a reversible cycle must be zero. The formula for the heat pump is:

$\frac{\dot Q_{L}}{T_{L}} - \frac{\dot Q_{H}}{T_{H}} + \dot S_{gen} = 0$

$\dot S_{gen} = \frac{\dot Q_{H}}{T_{H}} - \frac{\dot Q_{L}}{T_{L}}$

$\dot S_{gen} = \frac{300\,kW}{297.15\,K}-\frac{210.153\,kW}{280.15\,K}$

$\dot S_{gen} = 0.259\,\frac{kW}{K}$

Which contradicts the reversibility criterion according to the Second Law of Thermodynamics.

5 0

3 years ago

Air at 500 kPa and 400 K enters an adiabatic nozzle at a velocity of 30 m/s and leaves at 300 kPa and 350 K. Using variable spec

adoni [48]

Answer: a) Efficiency = 0.92

b) V = 319.19 m/s

c) S = 0.012 kj/kg.k

Explanation: please find the attached files for the solution

3 0

3 years ago