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Lemur [1.5K]
3 years ago
9

It is suggested that the kitchen in your house could be cooled in the summer by closing the kitched from the rest of the house a

nd opening the door to the domestic electric refrigerator. Is it true or false? State the reason for your conclusion.
Engineering
1 answer:
jeka57 [31]3 years ago
8 0

Answer:

False

Explanation:

When we open the gate of refrigerator then the temperature of kitchen get increase.

But we think how it is possible that temperature of kitchen get increases.Actually the heat rejecting coil(condenser) is also present inside the room that is why it reject more heat as compare to heat absorbed by evaporator coil.This is the reason to increases the temperature of kitchen.

If heat rejecting coil put outside the kitchen then it will produces cooling effect even door of refrigerator open.This concept use in air conditioning.

You might be interested in
). A 50 mm diameter cylinder is subjected to an axial compressive load of 80 kN. The cylinder is partially
Delicious77 [7]

Answer:

\frac{e'_z}{e_z} = 0.87142

Explanation:

Given:-

- The diameter of the cylinder, d = 50 mm.

- The compressive load, F = 80 KN.

Solution:-

- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.

- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).

                          εx = - [ σx - ν( σy + σz ) ] / E

                          εy = - [ σy - ν( σx + σz ) ] / E

                          εz = - [ σz - ν( σy + σx ) ] / E

- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:

                          εx =  [ ν*σz ] / E

                          εy = [ ν*σz ] / E

                          εz = - [ σz ] / E   .... Eq 1

- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.

- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:

                         εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx

                         εx' = - [ σx' - ν( σy' + σz ) ] / E =  [ ν*σz ] / 2E

                         εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy

                         εx' = - [ σy' - ν( σx' + σz ) ] / E =  [ ν*σz ] / 2E

- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.

- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).

Note:  σx' = σy', The cylinder is radially enclosed around the entire surface.

Therefore,

                        - [ σx' - ν( σx'+ σz ) ] =  [ ν*σz ] / 2

                          σx' ( 1 - v ) = [ ν*σz ] / 2

                          σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]

- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):

                           εz' = - [ σz - ν( σy' + σx' ) ] / E

                           εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E

                          εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E  ... Eq2

- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:

                            \frac{e'_z}{e_z} = \frac{- \frac{s_z}{E} * [ 1 - \frac{v^2}{1 - v} ]  }{-\frac{s_z}{E}}  \\\\\frac{e'_z}{e_z} = [ 1 - \frac{v^2}{1 - v} ] = [ 1 - \frac{0.3^2}{1 - 0.3} ] \\\\\frac{e'_z}{e_z} = 0.87142... Answer

5 0
3 years ago
Which one of the following faults cause the coffee in a brewer to keep boiling after the brewing cycle is finished?
MrRa [10]

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.)

8 0
3 years ago
Are routers better for internet connection rather than a WiFi modem?
sleet_krkn [62]

Answer:

Depends

Explanation:

The modem connects you to the Internet via ISP. Without a modem, your router will only allow you to connect to a LAN. A modem will provide connections for just a single wired device. If you want to go wireless you need a router.

8 0
3 years ago
Read 2 more answers
Nearlyof all serious occupational injuries and illnesses stem from overexertion of repetitive motion.
Ostrovityanka [42]

Answer:

a) 1/2

Explanation:

Overexertion accounted for more than  half of all events that resulted in a disabling condition.

Furthermore, 30% of all overexertion cases were reported in the services industry, on the other hand, 25% of injuries resulting from contact with objects and  equipment occurred in the manufacturing industry.

The above piece of information is taken from the bureau of labor statistics, Survey of Occupational Injuries and Illnesses

"LOST-WORKTIME INJURIES AND ILLNESSES: CHARACTERISTICS  AND RESULTING DAYS AWAY FROM WORK, 2002"

8 0
3 years ago
Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E
White raven [17]

Answer:

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

Explanation:

To solve this problem we use the expression for the temperature film

T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5

Then, we have to compute the Reynolds number

Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77

but we also now that

Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

Pr=0.699

k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

8 0
3 years ago
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