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Jobisdone [24]
2 years ago
6

In a power plant, pipes transporting superheated vapor are very common. Superheated vapor flows at a rate of 0.3 kg/s inside a p

ipe that is of diameter 5 cm and length10 m. The pipe is located in a power plant at a temperature of 22°C, and has a uniform pipe surface temperature of 100°C. If the temperature drop between the inlet and exit of the pipe is 30°C and the specific heat of the vapor is 2190 J/kg·K, determine the heat transfer coefficient as a result of convection between the pipe surface and the surroundings.
Physics
1 answer:
grigory [225]2 years ago
5 0

Answer:h=160.84 W/m^2-K

Explanation:

Given

mass flow rate=0.3 kg/s

diameter of pipe=5 cm

length of pipe=10 m

Inside temperature=22

Pipe surface =100

Temperature drop=30

specific heat of vapor(c)=2190 J/kg.k

heat supplied Q=mc\Delta T=0.3\times 2190\times (30)

Heat due to convection =hA(100-30)

A=\pi d\cdot L

A=\pi 0.05\times 10=1.571 m^2

Q_{convection}=h\times 1.571\times (100-22)=122.538 h

Q=Q_{convection}

19,710=122.538 h

h=160.84 W/m^2-K

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3 years ago
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 as \tau_{2} = I_{2} \alpha_{2}

                   = 1 kg m^{2} \times \alpha_{2}        

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3 years ago
the solubility product of lead fluoride is 3.6 x 10–8. what is its solubility in 0.10M NaF solution, in grams per liter
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Answer:

8.8 × 10⁻³ g/L

Explanation:

NaF is a strong electrolyte that ionizes according to the following reaction.

NaF(aq) → Na⁺(aq) + F⁻(aq)

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In order to find the solubility of PbF₂ (S), we will use an ICE Chart.

        PbF₂(s) ⇄ Pb²⁺(aq) + 2 F⁻(aq)

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E                          S              0.10 + 2S

The solubility product (Kps) is:

Kps = 3.6 × 10⁻⁸ = [Pb²⁺].[F⁻]² = S . (0.10 + 2S)²

In the term 0.10 + 2S, 2S is negligible in comparison with 0.10 and we can omit it to simplify calculations.

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