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3 years ago

Select the correct answer.

Engineering

2 answers:

Marysya12 [62]3 years ago

7 0

The answer would be C, a screw

olya-2409 [2.1K]3 years ago

6 0

Answer:

screw is the answer of the question

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Passband to baseband conversion: Consider the following passband signal,

defon

Answer:fr

Explanation:

5 0

3 years ago

Suppose that a wireless link layer using a CSMA-like protocol backs off 1ms on average. A packet’s link and physical layer heade

Liula [17]

Tell me why i got this question got it right and now won’t remember but i’ll get back at you when i remember

5 0

3 years ago

In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by

Stells [14]

Answer:

$\dfrac{\bar{h}}{h}=\dfrac{5}{4}$

Explanation:

Given that

$Nu_x=0.035Re_x^{0.8} Pr^{1/3}$

We know that

Rex=ρvx/μ

$Nu_x=0.035Re_x^{0.8} Pr^{1/3}$

$Nu_x=0.035\times\left(\dfrac{\rho vx}{\mu}\right)^{0.8}Pr^{1/3}$

All other quantities are constant only x is a variable in the above equation .so lets take all other quantities as a constant C

$Nu_x=C.x^{0.8}=C.x^{4/5}$

We also know that

Nux=hx/K

$C.x^{4/5}=\dfrac{hx}{k}$

m is the constant

$h=mx^{-1/5}$

This is local heat transfer coefficient

The average value of h given as

$\bar{h}=\dfrac{\int_{0}^{L}hdx}{L}$

$\bar{h}=\dfrac{5m}{4}\times\dfrac{L^{4/5}}{L}$

$\bar{h}=\dfrac{5m}{4}L^{-1/5}$ ---------1

The value of local heat transfer coefficient at x=L

$h=mx^{-1/5}$

$h=mL^{-1/5}$ -----------2

From 1 and 2 we can say that

$\dfrac{\bar{h}}{h}=\dfrac{5}{4}$

8 0

4 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

Water is stored in a tank which has vent open to the atmosphere. The water level is 1.0 m below the top the tank and the water i

sp2606 [1]

Answer:

6.99 x 10⁻³ m³ / s

Explanation:

Th e pressure difference at the two ends of the delivery pipe due to atmospheric pressure and water column will cause flow of water.

h = difference in the height of water column at two ends of delivery pipe

6 - 1 = 5 m

Velocity of flow of water

v = √2gh

= √ (2 x 9.8 x 5)

= 9.9 m /s

Volume of water flowing per unit time

velocity x cross sectional area

= 9.9 x 3.14 x .015²

= 6.99 x 10⁻³ m³ / s

7 0

3 years ago