Concept explainers
(a)
Interpretation:
The water temperate, the liquid and head space volume (L), and the mass of water vapor in the head space (kg) at time t1 should be determined.
Concept introduction:
All the properties of liquid and vapor of water, enthalpy, specific volume, and specific entropy vary with change with saturation pressure.
(b)
Interpretation:
The water temperature, the liquid and head space volume (L), and the mass of water vapor in(g) that evaporates between t1 and t2 should be determined.
Concept introduction:
The energy transfer properties of saturated water can be calculated by saturated steam table. If pressure is given then we can find value of temperature and same with if temperature given we can find other properties such as enthalpy, specific volume etc.
(c)
Interpretation:
The amount of heat (kJ) transferred to the tank contains between t1 and t2 should be calculated and two resins why the actual heat input to the tank must have been greater than the calculated value should be given.
Concept introduction:
Sum of the heat of liquid and vapor at t1 gives total heat contains in the tank similarly at t2 we can calculate total heat contains. And difference of heat contains at t2 and t1 gives heat transferred to the tank between t1 and t2.
(d)
Interpretation:
The three different factors responsible for the increase in pressure resulting from the transfer of heat to the tank should be listed.
Concept introduction:
If increase the temperature vaporization increases and pressure increases, density of liquid decrease, because system is closed system.
(e)
Interpretation:
The ways in which this accident could have been avoided should be listed.
Concept introduction:
Sensors play an important role in such situation because if processes cross the set point processes automatic open safety valve.
(f)
Interpretation:
The temperature at which the valve is open and the rate at which the needed to release steam (kg/KJ of added heat) in order to keep the tank pressure from rising should be determined.
Concept introduction:
From stem table find the corresponding value at 10 bars. In this case saturated steam table give the other properties such as enthalpy, specific volume etc.
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- step by steparrow_forwardThe power out of an adiabatic steam turbine is 5 MW and the steam enters turbine at 2 MPa and velocity of 50 m/s, specific enthalpy (h) of 3248 kJ/kg. The elevation of the inlet is 10 m higher than at the datum. The vapor mixture exits at 15 kPa and a velocity of 180 m/s, specific enthalpy (h) of 2361.01 kJ/kg. The elevation of the exit is 6 m higher than at the datum. Let g = 9.81 m/s². Assuming the ideal gas model and R = 0.462 KJ/(kg.K). The steam specific heat ratio is 1.283. Calculate:arrow_forwardThe power out of an adiabatic steam turbine is 5 MW and the steam enters turbine at 2 MPa and velocity of 50 m/s, specific enthalpy (h) of 3248 kJ/kg. The elevation of the inlet is 10 m higher than at the datum. The vapor mixture exits at 15 kPa and a velocity of 180 m/s, specific enthalpy (h) of 2361.01 kJ/kg. The elevation of the exit is 6 m higher than at the datum. Let g = 9.81 m/s². Assuming the ideal gas model and R = 0.462 KJ/(kg.K). The steam specific heat ratio is 1.283. Calculate:arrow_forward
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