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4 years ago

A __________ is an added note showing additional or more specific information.

Engineering

1 answer:

xxMikexx [17]4 years ago

8 0

Answer:

awnsers should be added to know to show additional

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Please. my brain isn’t working right now

Alex787 [66]

Answer:

10kQ is to 36......... D2 _ D1 D4

7 0

3 years ago

How much power is needed to operate a Carnot heat pump if the pump receives heat 10°C and delivers 50 kW of heat at 40°C? at A)

mariarad [96]

Answer:

Power needed to pump=4.79 KW.

Explanation:

Given that: $T_{1}=283K,T_{2}=313K,Q_{H}=50KW$

We know that coefficient of performance of heat pump

COP= $\dfrac{T_{H}}{T_{H}-T_{L}}$

So COP= $\dfrac{313}{313-283}$

COP=10.43

COP= $\frac{Q_{H}}{W_{in}}$

10.43 = $\frac{50}{W_{in}}$

$W_{in}$ =4.79 KW

So power needed to pump=4.79 KW.

3 0

4 years ago

Atmospheric air at 25 °C and 8 m/s flows over both surfaces of an isothermal (179C) flat plate that is 2.75m long. Determine the

vekshin1

Answer:

Re=100,000⇒Q=275.25 $\frac{W}{m^2}$

Re=500,000⇒Q=1,757.77 $\frac{W}{m^2}$

Re=1,000,000⇒Q=3060.36 $\frac{W}{m^2}$

Explanation:

Given:

For air $T_∞$ =25°C ,V=8 m/s

For surface $T_s$ =179°C

L=2.75 m ,b=3 m

We know that for flat plate

$Re$ ⇒Laminar flow

$Re>30\times10^5$ ⇒Turbulent flow

Take Re=100,000:

So this is case of laminar flow

$Nu=0.664Re^{\frac{1}{2}}Pr^{\frac{1}{3}}$

From standard air property table at 25°C

Pr= is 0.71 ,K=26.24 $\times 10^{-3}$

So $Nu=0.664\times 100,000^{\frac{1}{2}}\times 0.71^{\frac{1}{3}}$

Nu=187.32 ( $\dfrac{hL}{K_{air}}$ )

187.32= $\dfrac{h\times2.75}{26.24\times 10^{-3}}$

⇒h=1.78 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=275.25 $\frac{W}{m^2}$

Take Re=500,000:

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 500,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=1196.18 ⇒h=11.14 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=11.14(179-25)

= 1,757.77 $\frac{W}{m^2}$

Take Re=1,000,000:

So this is case of turbulent flow

$Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}$

$Nu=0.037\times 1,000,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}$

Nu=2082.6 ⇒h=19.87 $\frac{W}{m^2-K}$

heat transfer rate =h $(T_∞-T_s)$

=19.87(179-25)

= 3060.36 $\frac{W}{m^2}$

7 0

3 years ago

Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface of th

True [87]

Um I think the answer for this is 10l

5 0

3 years ago

A race car is travelling around a circular track. The velocity of the car is 20 m/s, the radius of the track is 300 m, and the m

Zielflug [23.3K]

Answer:

μ = 0.136

Explanation:

given,

velocity of the car = 20 m/s

radius of the track = 300 m

mass of the car = 2000 kg

centrifugal force

$F_c = \dfrac{mv^2}{r}$

$F_c = \dfrac{2000\times 20^2}{300}$

F c = 2666. 67 N

F f= μ N

F f = μ m g

2666.67 = μ × 2000 × 9.8

μ = 0.136

so, the minimum coefficient of friction between road surface and car tyre is equal to μ = 0.136

5 0

4 years ago