Water enters a boiler at a temperature of 110°F. The boiler is to produce 2000 lb/hr of steam at a pressure of 130 psia. How man
1 answer:
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Answer:
a). Work transfer = 527.2 kJ
b). Heat Transfer = 197.7 kJ
Explanation:
Given:
= 5 Mpa
= 1623°C
= 1896 K
= 0.05
Also given
Therefore, = 1
R = 0.27 kJ / kg-K
= 0.8 kJ / kg-K
Also given :
Therefore, =
= 0.1182 MPa
a). Work transfer, δW =
= 527200 J
= 527.200 kJ
b). From 1st law of thermodynamics,
Heat transfer, δQ = ΔU+δW
=
=
=
= 197.7 kJ
Answer:
a) The mass flow rate of water is 14.683 kilograms per second.
b) The pressure difference across the pump is 245.175 kilopascals.
Explanation:
a) Let suppose that pump works at steady state. The mass flow rate of the water (), in kilograms per second, is determined by following formula:
(1)
Where:
- Pump power, in watts.
- Efficiency, no unit.
- Gravitational acceleration, in meters per square second.
- Hydrostatic column, in meters.
If we know that ,
,
and
, then the mass flow rate of water is:
The mass flow rate of water is 14.683 kilograms per second.
b) The pressure difference across the pump (), in pascals, is determined by this equation:
(2)
Where is the density of water, in kilograms per cubic meter.
If we know that ,
and
, then the pressure difference is:
The pressure difference across the pump is 245.175 kilopascals.