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sashaice [31]
3 years ago
14

. (20 pts) A horizontal cylindrical pipe (k = 10 W/m·K) has an outer diameter of 15 cm and a wall thickness of 5 cm. The pipe is

situated in a stationary air, where the air and surrounding temperature is 27°C. The outer surface temperature of the pipe is 127°C, and the pipe surface has an emissivity of 0.5. Determine the inner surface temperature of the pipe. Use the following air properties for the analysis: k = 0.03 W/m∙K, ν = 20.92 × 10−6 m2 /s, α = 29.90 × 10−6 m2 /s, Pr = 0.70

Engineering
1 answer:
nata0808 [166]3 years ago
7 0

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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expeople1 [14]

It is study of the relationships between heat, temprature, work and energy

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3 years ago
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Water from the Earth's surface turns into a gas, or water vapor, when it is warmed and
bagirrra123 [75]

Answer:

Evaporation.

Explanation:

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4 years ago
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Gas is kept in a 0.1 m diameter cylinder under the weight of a 100 kg piston that is held down by a spring with a stiffness k =
Artyom0805 [142]

Answer:

The spring is compressed by 0.275 meters.

Explanation:

For equilibrium of the gas and the piston the pressure exerted by the gas on the piston should be equal to the sum of  weight of the piston and the force the spring exerts on the piston

Mathematically we can write

Force_{pressure}=Force_{spring}+Weight_{piston}

we know that

Force_{pressure}=Pressure\times Area=300\times 10^{3}\times \frac{\pi \times 0.1^2}{4}=750\pi Newtons

Weight_{piston}=mass\times g=100\times 9.81=981Newtons

Now the force exerted by an spring compressed by a distance 'x' is given by Force_{spring}=k\cdot x=5\times 10^{3}\times x

Using the above quatities in the above relation we get

5\times 10^{3}\times x+981=750\pi \\\\\therefore x=\frac{750\pi -981}{5\times 10^{3}}=0.275meters

5 0
3 years ago
Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank is fitted with a paddle whee
Juli2301 [7.4K]

Answer:

a) 1 m^3/Kg  

b) 504 kJ

c) 514 kJ

Explanation:

<u>Given  </u>

-The mass of C_o2 = 1 kg  

-The volume of the tank V_tank = 1 m^3  

-The added energy E = 14 W  

-The time of adding energy t = 10 s  

-The increase in specific internal energy Δu = +10 kJ/kg  

-The change in kinetic energy ΔKE = 0 and The change in potential energy  

ΔPE =0  

<u>Required  </u>

(a)Specific volume at the final state v_2

(b)The energy transferred by the work W in kJ.  

(c)The energy transferred by the heat transfer W in kJ and the direction of  

the heat transfer.  

Assumption  

-Quasi-equilibrium process.  

<u>Solution</u>  

(a) The volume and the mass doesn't change then, the specific volume is constant.

 v= V_tank/m ---> 1/1= 1 m^3/Kg  

(b) The added work is defined by.  

W =E * t --->  14 x 10 x 3600 x 10^-3 = 504 kJ  

(c) From the first law of thermodynamics.  

Q - W = m * Δu

Q = (m * Δu) + W--> (1 x 10) + 504 = 514 kJ

The heat have (+) sign the n it is added to the system.

7 0
3 years ago
The substance is steam (H2O). NOTE: The purpose of this problem is to illustrate that there are conditions where water vapor is
Gennadij [26K]

Answer:

See the attached pictures for detailed answer.

Explanation:

See the attached pictures for step by step explanation.

7 0
3 years ago
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