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3 years ago

List at least two things that scientists and explorers have in common.

Engineering

2 answers:

Tresset [83]3 years ago

8 0

• educated
•very curious
•discover new things
•Investigate
•search unknown

madam [21]3 years ago

8 0

Curiosity,
education,
proof,
discovery

You might be interested in

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

List six possible valve defects that should be included in the inspection of a used valve?

olchik [2.2K]

Answer:

Valvular stenosis , Valvular prolapse , Regurgitation,

Explanation:

8 0

3 years ago

Design a plate and frame heat exchanger for the following problem:

qwelly [4]

Answer:

See explaination and attachment.

Explanation:

Iteration method is a repetitive method applied until the desired result is achieved.

Let the given equation be f(x) = 0 and the value of x to be determined. By using the Iteration method you can find the roots of the equation. To find the root of the equation first we have to write equation like below

x = pi(x)

Let x=x0 be an initial approximation of the required root α then the first approximation x1 is given by x1 = pi(x0).

Similarly for second, thrid and so on. approximation

x2 = pi(x1)

x3 = pi(x2)

x4 = pi(x3)

xn = pi(xn-1).

please go to attachment for the step by step solution.

8 0

3 years ago

A 1000 kg turbine has a rotating unbalance of 0.1 kg.m. The turbine operates at a speed between 500 to 750 rpm. What is the maxi

raketka [301]

Answer:

maximum isolator stiffness k =1764 kN-m

Explanation:

mean speed of rotation $=\frac{N_1 +N_2}{2}$

$Nm = \frac{500+750}{2} = 625 rpm$

$w =\frac{2\pi Nm}{60}$

=65.44 rad/sec

$F_T = mw^2 e$

$F_T = mew^2$

= 0.1*(65.44)^2

F_T =428.36 N

Transmission ratio $=\frac{300}{428.36} = 0.7$

also

transmission ratio $= \frac{1}{[\frac{w}{w_n}]^{2} -1}$

$0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}$

SOLVING FOR Wn

Wn = 42 rad/sec

$Wn = \sqrt {\frac{k}{m}$

k = m*W^2_n

k = 1000*42^2 = 1764 kN-m

k =1764 kN-m

3 0

3 years ago

If the car passes point A with a speed of 20 m>s and begins to increase its speed at a constant rate of at = 0.5 m>s 2 , d

sattari [20]

Answer:

a = 1.68m/ $S^{2}$

Explanation:

Please kindly find the attached file for explanations

3 0

3 years ago