Answer:
Explanation:
Given:
Steam Mass rate, ms = 1.5 kg/min
= 1.5 kg/min × 1 min/60 sec
= 0.025 kg/s
Air Mass rate, ma = 100 kg/min
= 100 kg/min × 1 min/60 sec
= 1.67 kg/s
A.
Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.
xf, quality = 0.9.
Tsat = 89.9°C
hf = 376.57 kJ/kg
hfg = 2283.38 kJ/kg
Using the equation for specific enthalpy,
hi = hf + (hfg × xf)
= 376.57 + (2283.38 × 0.9)
= 2431.552 kJ/kg
The specific enthalpy of the outlet, h2 = hf
= 376.57 kJ/kg
B.
Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy
= ms × (hi - h2)
= 0.025 × (2431.552 - 376.57)
= 0.025 × 2055.042
= 51.37455 kW
= 51.38 kW.
Answer:the pressure depends on gas and it will be half as much underwater
Explanation:
Water pressure increases with the depth of the water. This is because the weight of the column of water above the object increases. But a large, shallow pond may have more water in it than a small, deep pond.
This is due to an increase in hydrostatic pressure, the force per unit area exerted by a liquid on an object. The deeper you go under the sea, the greater the pressure of the water pushing down on you. For every 33 feet (10.06 meters) you go down, the pressure increases by one atmosphere .
The freezing point is the same as the melting point.
If it freezes at -58°C, hence the melting point is also <span>-58°C.</span>
