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

User Mchr
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2 Answers

3 votes

Final answer:

The maximum theoretical efficiency for a heat engine operating between a high temperature of 300°C (573.15 K) and a low temperature of 27°C (300.15 K) is 47.63%, calculated using the Carnot efficiency formula.

Step-by-step explanation:

To calculate the maximum theoretical efficiency for a heat engine operating between two temperatures, we can use the efficiency formula derived from the Carnot cycle, which is given by:

\(\eta = 1 - \frac{T_{cold}}{T_{hot}}\)

Where \(\eta\) is the efficiency, \(T_{cold}\) is the cold reservoir temperature, and \(T_{hot}\) is the hot reservoir temperature. Temperatures must be in Kelvin.

First, convert the temperatures from Celsius to Kelvin:

\(T_{cold} = 27 \degree C + 273.15 = 300.15 K\)

\(T_{hot} = 300 \degree C + 273.15 = 573.15 K\)

Now, substitute these values into the efficiency formula:

\(\eta = 1 - \frac{300.15}{573.15}\)

\(\eta = 1 - 0.5237\)

So, the maximum theoretical efficiency is:

\(\eta = 0.4763\)

Or in percentage:

\(\eta = 47.63\%\)

This calculation assumes an ideal Carnot engine, which is a theoretical limit and cannot be achieved in practical engines; the actual efficiency will be lower due to various inefficiencies.

User Jeffbyrnes
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6 votes

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

Step-by-step 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* 2190* (30)

Heat due to convection =hA(100-30)


A=\pi d\cdot L


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


Q_(convection)=h* 1.571* (100-22)=122.538 h


Q=Q_(convection)

19,710=122.538 h


h=160.84 W/m^2-K

User DBoyer
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