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wariber [46]
3 years ago
11

The period-luminosity relation is critical in finding distances with

Physics
1 answer:
muminat3 years ago
5 0

D hopefully this helps

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An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,
insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

Where

t_{hf} = Temperature at the inside of the pipe = 300 °C

t_{f} = Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

k_A = 15 W/m·K

k_B = 0.038 W/m·K

h_{hf} = 75 W/m²·K

h_{cf} = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

Q = \frac{t_A-t_B}{R_T} Where R_T for the air film on the pipe outer surface is given by

R_T= \frac{1}{\alpha A}

where A =area of the outside of the pipe

= \frac{1}{10*2\pi*0.07*1 } = 0.227 K/W

Therefore

131.32 W = \frac{t_A-20}{0.227} which gives

t_A = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

8 0
3 years ago
A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s
Paraphin [41]

Answer:

\omega = 0.016\,\frac{rad}{s}

Explanation:

The rotation rate of the man is:

\omega = \frac{v}{R}

\omega = \frac{0.80\,\frac{m}{s} }{5\,m}

\omega = 0.16\,\frac{rad}{s}

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega

The final angular speed is:

\omega = 0.016\,\frac{rad}{s}

3 0
3 years ago
Which three statements about electromagnetic radiation are true?
grandymaker [24]
I think that numbers one, three, and four are true
3 0
3 years ago
Read 2 more answers
A train’s mass is 18181.81 kg. What is its weight?
wel

Answer:

The trains mass in pounds would be 40084.029 if you would round it to the hundreths

Explanation:

7 0
3 years ago
Read 2 more answers
22) A driver traveling at 80 km/h brakes her 2000 kg truck to stop for a red light. How much internal energy is produced
blsea [12.9K]

Answer:

E=4.9\times 10^5\ J

Explanation:

Given that,

The speed of a driver, v = 80 km/h = 22.22 m/s

The mass of the truck, m = 2000 kg

We need to find how much internal energy is produced. The internal energy produced by a truck is its kinetic energy and it is given by :

E=\dfrac{1}{2}mv^2\\\\E=\dfrac{1}{2}\cdot22.22^{2}\cdot2000\\\\E=493728.4\ J

or

E=4.9\times 10^5\ J

So, 4.9\times 10^5\ J of internal energy is produced.

3 0
3 years ago
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