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Anna [14]
3 years ago
12

If the power to a condensing unit has been turned off for an extended period of time, the _________________________ should be en

ergized first for the recommended time to protect the compressor from liquid floodback. A. blower motor B. thermostat C. crankcase heater D. condenser fan motor
Engineering
1 answer:
Mama L [17]3 years ago
7 0

Answer:

C. Crankcase heater

Explanation:

The crankcase heater of the compressor was developed to mitigate the problematic variation of temperature on the in and out sides of the crankcase that has the capacity to change the physical states of the refrigerant and oil, cause them migrate and possibly mix together. It does so by allowing the regulation of the temperature on the inside of the crankcase

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Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following
torisob [31]

Given:

Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following for air at 300 K and engine oil at 380 K. Assume the free stream velocity is 3 m/s.

To Find:

a. The distance from the leading edge at which the transition will occur.

b. Expressions for the momentum and thermal boundary layer thicknesses as a function of x for a laminar boundary layer

c. Which fluid has a higher heat transfer

Calculation:

The transition from the lamina to turbulent begins when the critical Reynolds

number reaches 5\times 10^5

(a).  \;\text{Rex}_{cr}=5 \times 10^5\\\\\frac{\rho\;vx}{\mu}=5 \times 10^5\\\text{density of of air at}\;300K=1.16  \frac{kg}{m\cdot s}\\\text{viscosity of of air at}\;300K=1.846 \times 10^{-5} \frac{kg}{m\cdot s} \\v=3m/s\\\Rightarrow x=\frac{5\times 10^5 \times 1.846 \times 10^{-5} }{1.16 \times 3} =2.652 \;m \;\text{for air}\\(\text{similarly for engine oil at 380 K for given}\; \rho \;\text{and} \;\mu)\\

(b).\; \text{For the lamina boundary layer momentum boundary layer thickness is given by}:\\\frac{\delta}{x} =\frac{5}{\sqrt{R_e}}\;\;\;\;\quad\text{for}\; R_e(c). \frac{\delta}{\delta_t}={P_r}^{\frac{r}{3}}\\\text{For air} \;P_r \;\text{equivalent 1 hence both momentum and heat dissipate with the same rate for oil}\; \\P_r >>1 \text{heat diffuse very slowly}\\\text{So heat transfer rate will be high for air.}\\\text{Convective heat transfer coefficient will be high for engine oil.}

7 0
3 years ago
A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.
denpristay [2]

Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is 8.621 \times 10^{-4} in/in

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

\frac{M}{I}=\frac{F}{y}

Here

  • M is the moment which is to be calculated
  • I is the moment of inertia given as

                         I=\frac{bd^3}{12}

Here

  • b is the breath given as 0.75"
  • d is the depth which is given as 8"

                     I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4

  • y is given as

                     y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\

  • Force is 50 ksi

\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in

The yield moment is 400 k.in.

Part b:

The strain is given as

Strain=\frac{Stress}{Elastic Modulus}

The stress at the station 2" down from the top is estimated by ratio of triangles as

                        F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi

Now the steel has the elastic modulus of E=29000 ksi

Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in

So the strain is 8.621 \times 10^{-4} in/in

Part c:

For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi

The plastic moment is 600 ksi.

3 0
3 years ago
Both model building codes and NFPA __________ can be used to determine the type of construction used in a building.
Alex777 [14]

Answer:

Both model building codes and NFPA 220 can be used to determine the type of construction used in a building.

8 0
2 years ago
Some organizations prohibit workers from bringing certain kinds of devices into the workplace, such as cameras, cell phones, and
gulaghasi [49]

Answer:

Personal computers:

Personal computers may be useful and lead to productivity as using a computer an employee familiars with is a good thing. However, the disadvantages in some facilities especially ones dealing with customer and information security can include data theft, unauthorized data sharing, uses of internet connection for personal purposes, as this can slow down internet connection at the facility, distraction at work place etc.

Hard drive:

Due to large amount of data that can be stored on a hard drive, it might not be allowed in some facilities to avoid data theft and unauthorized transfer.

Music players:

This might be restricted to avoid distraction at work. Noice in places such as libraries would cause abnormality and poor service delivery. An employee with loud speaker at work would not only distracts himself but also other staffs and customers.

PSP Game Device and other game devices:

Playing games during working hour may jeopardize the productivity and therefore might be resctrited in some facilities and working places.

Electronic digital notepad:

Carrying a handheld electronic digital notepad to the work place can cause lack of concentration and division of attention on work and other personal activities. These can harm working harmony and business productivity.

Video recorder:

In some facilities, this device might not be allowed due to facility privacy and protection from unwanted navigation.

Explanation:

6 0
3 years ago
21.Why are throttling devices commonly used in refrigeration and air-conditioning<br> applications?
Sloan [31]

Answer is given below

Explanation:

we know that some common types of throttling devices are

  • Hard -throttling devices
  • Capillary valve
  • Constant pressure throttling devices
  • Thermostatic expansion valve
  • Float expansion valve

so here throttling devices commonly used in refrigeration and air-conditioning because

  • To reduce the coolant pressure, the high pressure of the refrigerant from the condenser is necessary to reduce the evaporation to obtain evaporation at the right temperature  
  • To meet the refrigerated load, the throttling valve flows through the coolant to cool the load at high temperatures.
5 0
3 years ago
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