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katrin [286]
3 years ago
7

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

Engineering
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
You might be interested in
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
2 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
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