Steam initially at 0.3 MPa, 2500 C is cooled at constant volume. (a) At what temperature will steam become saturated vapour? [12
3.90 C] (b) What is the quality at 800 C? [0.234]. What is the heat transfer per kg of steam in cooling from 2500 C to 800 C? [-1890.2 kJ/Kg]
1 answer:
Answer:
a. 123.9°C
b.
c.
Explanation:
Hello, I'm attaching a picture with the numerical development of this exercise.
a. Since the steam is overheated vapour, the specific volume is gotten from the corresponding table. Then, as it became a saturated vapour, we look for the interval in which the same volume of state 1 is, then we interpolate and get the temperature.
b. Now, at 80°C, since it is about a rigid tank (constant volume for every thermodynamic process), the specific volume of the mixture is 0.79645 m^3/kg as well, so the specific volume for the liquid and the vapour are taken into account to get the quality of 0.234.
c. Now,since this is an isocoric process, the heat transfer per kg of steam is computed as the difference in the internal energy, considering the initial condition (showed in a. part) and the final one computed here.
** The thermodynamic data were obtained from Cengel's thermodynamics book 7th edition.
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Answer:
ν = 7.04 × 10¹³ s⁻¹
λ = 426 nm
It falls in the visible range
Explanation:
The relation between the energy of the radiation and its frequency is given by Planck-Einstein equation:
E = h × ν
where,
E is the energy
h is the Planck constant (6.63 × 10⁻³⁴ J.s)
ν is the frequency
Then, we can find frequency,
Frequency and wavelength are related through the following equation:
c = λ × ν
where,
c is the speed of light (3.00 × 10⁸ m/s)
λ is the wavelength
A 426 nm wavelength falls in the visible range (≈380-740 nm)