Answer:
Enthalpy, hsteam = 2663.7 kJ/kg
Volume, Vsteam = 0.3598613 m^3 / kg
Density = 2.67 kg/ m^3
Explanation:
Mass of steam, m = 1 kg
Pressure of the steam, P = 0.5 MN/m^2
Dryness fraction, x = 0.96
At P = 0.5 MPa:
Tsat = 151.831°C
Vf = 0.00109255 m^3 / kg
Vg = 0.37481 m^3 / kg
hf = 640.09 kJ/kg
hg = 2748.1 kJ/kg
hfg = 2108 kJ/kg
The enthalpy can be given by the formula:
hsteam = hf + x * hfg
hsteam = 640.09 + ( 0.96 * 2108)
hsteam = 2663.7 kJ/kg
The volume of the steam can be given as:
Vsteam = Vf + x(Vg - Vf)
Vsteam = 0.00109255 + 0.96(0.37481 - 640.09)
Vsteam = 0.3598613 m^3 / kg
From the steam table, the density of the steam at a pressure of 0.5 MPa is 2.67 kg/ m^3
Answer:
"Macro Instruction"
Explanation:
A macro definition is a rule or pattern that specifies how a certain input sequence should be mapped to a replacement output sequence according to a defined procedure. The mapping process that instantiates a macro use into a specific sequence is known as macro expansion.
It is a series of commands and actions that can be stored and run whenever you need to perform the task. You can record or build a macro and then run it to automatically repeat that series of steps or actions.
Answer:
The atmospheric pressure is found to be 
Explanation:
We know that pressure exerted by a standing column of fluid is calculated using the equation
In our case the pressure of the standing column of mercury is equal to the atmospheric pressure.
According to the given data we have
Using the values in the equation above we calculate atmospheric pressure to be
Answer:
ΔV = 
Explanation:
Distance of earth from sun = 
Spacecraft perihelion = 
gravitational parameters are now given as
radius of earth = 6378 Km
Heliocentric spacecraft velocity at earth sphere of influence =
Heliocentric velocity of earth = 
assume
Geometric spacecraft velocity of spacecraft at perigee of departure hyperbola
geometric space craft velocity in its circular parking orbit
ΔV =
