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3 years ago

7

Technician A says that 5W-30 would be better to use than 20W-50 in most vehicles in

2 answers:

shtirl [24]3 years ago

5 0

Technician is correct sorry if im wronghg

Ulleksa [173]3 years ago

5 0

And I don’t want to make it happen anymore I’m sorry for making

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g A food department is kept at -12oC by a refrigerator in an environment at 30oC. The total heat gain to the food department is

boyakko [2]

Answer:

a) $\dot W = 0.417\,kW$ , b) $COP_{R} = 2.198$ , c) Irreversible.

Explanation:

a) The power input required by the refrigerator is:

$\dot W = \dot Q_{H} - \dot Q_{L}$

$\dot W = \left(4800\,\frac{kJ}{h} - 3300\,\frac{kJ}{h}\right)\cdot \left(\frac{1}{3600} \,\frac{h}{s} \right)$

$\dot W = 0.417\,kW$

b) The Coefficient of Performance of the refrigerator is:

$COP_{R} = \frac{\dot Q_{L}}{\dot W}$

$COP_{R} = \frac{3300\,\frac{kJ}{h} }{(0.417\,kW)\cdot \left(3600\,\frac{s}{h} \right)}$

$COP_{R} = 2.198$

c) The maximum ideal Coefficient of Performance of the refrigeration is given by the inverse Carnot's Cycle:

$COP_{R,ideal} = \frac{T_{L}}{T_{H}-T_{L}}$

$COP_{R,ideal} = \frac{261.15\,K}{303.15\,K - 261.15\,K}$

$COP_{R,ideal} = 6.218$

The refrigeration cycle is irreversible, as $COP_{R} < COP_{R,ideal}$ .

3 0

3 years ago

19. A circuit contains four 100 S2 resistors connected in series. If you test the circuit with a digital VOM,

yanalaym [24]

Answer:

D

Explanation:

in series circuit the resistance is divided Total resistance is equal to the sum of resistances

6 0

3 years ago

A long, cylindrical, electrical heating element of diameter 10 mm, thermal conductivity 240 W/m·K, density 2700 kg/m3, and speci

Slav-nsk [51]

Answer:

Answer for the question is given in the attachment .

Explanation:

4 0

3 years ago

One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter

lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

V = 150× $10^{-3}$ $m^{3}$

m = 1 Kg

$P_{1}$ = 2 MPa

$T_{2}$ = 40°C

The waters specific volume is calculated:

$v_{1}$ = V/m

Here, the waters specific volume at initial condition is $v_{1}$ , the containers volume is V, waters mass is m.

$v_{1}$ = 150× $10^{-3}$ $m^{3}$ /1

$v_{1}$ = 0.15 $m^{3}$ / Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 $m^{3}$ / Kg and 0.13 $m^{3}$ / Kg.

$T_{1}$ = 350+(400-350) $\frac{0.15-0.13}{0.1522-0.1386}$

$T_{1}$ = 395.17°C

Hence, the initial temperature is 395.17°C.

The volume is constant in the rigid container.

$v_{2}$ = $v_{1}$ = 0.15 $m^{3}$ / Kg

In saturated water labels for $T_{2}$ = 40°C.

$v_{f}$ = 0.001008 $m^{3}$ / Kg

$v_{g}$ = 19.515 $m^{3}$ / Kg

The final state is two phase region $v_{f}$ < $v_{2}$ < $v_{g}$ .

In saturated water labels for $T_{2}$ = 40°C.

$P_{2}$ = $P_{Sat}$ = 7.3851 kPa

$P_{2}$ = 7.3851 kPa

7 0

3 years ago

Consider a plane composite wall that is composed of two materials of thermal conductivities kA = 0.1 W/m*K and kB = 0.04 W/m*K a

nadya68 [22]

Answer:

q=39.15 W/m²

Explanation:

We know that

Thermal resistance due to conductivity given as

R=L/KA

Thermal resistance due to heat transfer coefficient given as

R=1/hA

Total thermal resistance

$R_{th}=\dfrac{L_A}{AK_A}+\dfrac{L_B}{AK_B}+\dfrac{1}{Ah_1}+\dfrac{1}{Ah_2}+\dfrac{1}{Ah_3}$

Now by putting the values

$R_{th}=\dfrac{0.01}{0.1A}+\dfrac{0.02}{0.04A}+\dfrac{1}{10A}+\dfrac{1}{20A}+\dfrac{1}{0.3A}$

$R_{th}=4.083/A\ K/W$

We know that

Q=ΔT/R

$Q=\dfrac{\Delta T}{R_{th}}$

$Q=A\times \dfrac{200-40}{4.086}$

So heat transfer per unit volume is 39.15 W/m²

q=39.15 W/m²

4 0

3 years ago