How are eras different from decades?

Assume a steel pipe of inner radius r1= 20 mm and outer radius r2= 25 mm, which is exposed to natural convection at h = 50 W/m2.

Mekhanik [1.2K]

Answer:

98,614.82 W/m²

Explanation:

$Q = 2\pi hL(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})$

Where;

Q = the amount of heat loss from the pipe

h = the heat transfer coefficient of the pipe = 50 W/m².K

T₁ = the ambient temperature of the pipe = 30⁰C

T₂ = the outside temperature of the pipe = 100⁰C

L= the length of pipe

r₁ = inner radius of the pipe = 20mm

r₂ = outer radius of the pipe = 25mm

To determine the amount of heat loss from the pipe per unit length

From the equation above

$\frac{Q}{L} = 2\pi h(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})$

$\frac{Q}{L} = 2\pi* 50(\frac{100-30}{Ln\frac{25}{20}})$

$\frac{Q}{L} = 314.159(\frac{70}{0.223})$

$\frac{Q}{L} = 314.159(313.901)$ = 98,614.82 W/m²

3 0

3 years ago

Analyze the example of this band saw wheel and axle. The diameter of the wheel is 14 inches. The diameter of the axle that drive

Kazeer [188]

The answer for the ideal mechanical advantage and actual mechanical advantage for the different scenarios are;

A) Ideal Mechanical Advantage = 18.67

B) Actual Mechanical Advantage = 4.1067

We are given;

Input distance; The diameter of the wheel; d_w = 14 inches

Output distance; The diameter of the axle that drives the wheel; d_a = 3/4 inches

The force needed to cut a one-inch-thick softwood board; F = 1.75 pounds

The efficiency of the band saw; η = 22% = 0.22

A) Formula for Mechanical advantage is;

M.A = Force output/Force input = (Input distance)/(Output distance)

Thus;

Ideal mechanical advantage = 14/(3/4)

Ideal mechanical advantage = 18.67

B) Now, we are given that efficiency of the band saw is η = 22% = 0.22.

Thus using the mechanical advantage formula above;

Actual mechanical advantage = 0.22 × Expected output

Actual mechanical advantage = 0.22 × 18.67

Actual mechanical advantage ≈ 4.1067

Read more about Mechanical Advantage at; brainly.com/question/18345299

5 0

2 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

Ninety-five percent of the acetone vapor in an 85 vol.% air stream is to be absorbed by countercurrent contact with pure water i

Sati [7]

Answer:

Explanation:

.......................................................................................................................

5 0

3 years ago

wo companies, Ajax Co. and Boho Inc., were negotiating a merger. In the course of the negotiations, an Ajax representative told

boyakko [2]

Answer:

Yes

Explanation:

If the Ajax representative fails to correct the previous statement this can cause misrepresentation.

4 0

3 years ago