An Otto cycle engine is analyzed using the cold air standard method. Given the conditions at state 1, compression ratio (r), and
pressure ratio (rp). The specific heats for air are Cp = 1.0045 kJ/kg-K and Cv = 0.7175 kJ/kg-K.
T1 = 300K
P1 = 100 kPa
r = 10
rp = 1.65
a) Determine the temperature (K) at state 2.
b) Determine the pressure (kPa) at state 2.
c) Determine the pressure (kPa) at state 3.
d) Determine the temperature (K) at state 3.
e) Determine the temperature (K) at state 4.
f) Determine the pressure (kPa) at state 4.
g) Determine the net work output (kJ/kg/cycle) for the engine.
h) Determine the heat addition (kJ/kg/cycle) for the engine.
i) Determine the efficiency (%) of the engine.
Points will be given to answer showing all the work done to get right answer.
T1 = 300K
P1 = 100 kPa
r = 10
rp = 1.65
a) Determine the temperature (K) at state 2.
b) Determine the pressure (kPa) at state 2.
c) Determine the pressure (kPa) at state 3.
d) Determine the temperature (K) at state 3.
e) Determine the temperature (K) at state 4.
f) Determine the pressure (kPa) at state 4.
g) Determine the net work output (kJ/kg/cycle) for the engine.
h) Determine the heat addition (kJ/kg/cycle) for the engine.
i) Determine the efficiency (%) of the engine.
Points will be given to answer showing all the work done to get right answer.
1 answer:
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The friction loss in the system is 3.480 kilowatts.
<h2>Procedure - Friction loss through a pump</h2><h2 /><h3>Pump model</h3><h3 />Let suppose that the pump within a distribution system is an open system at steady state, whose mass and energy balances are shown below:
<h3>Mass balance</h3> (1)
(2)
(3)
<h3>Energy balance</h3>
(4)
Where:
- Inlet mass flow, in kilograms per second.
- Outlet mass flow, in kilograms per second.
- Inlet volume flow, in cubic meters per second.
- Outlet volume flow, in cubic meters per second.
- Inlet specific volume, in cubic meters per kilogram.
- Outlet specific volume, in cubic meters per kilogram.
- Pump efficiency, no unit.
- Electric motor power, in kilowatts.
- Inlet specific enthalpy, in kilojoules per kilogram.
- Outlet specific enthalpy, in kilojoules per kilogram.
- Work losses due to friction, in kilowatts.
From steam tables we get the following water properties at inlet and outlet:
Inlet
,
,
,
, Subcooled liquid
Outlet
,
,
,
, Subcooled liquid
If we know that ,
,
,
,
and
, then the friction loss in the system is:
The friction loss in the system is 3.480 kilowatts.
To learn more on pumps, we kindly invite to check this verified question: brainly.com/question/544887
Answer:
Explanation:
Hello,
In this case, the coefficient of performance of this refrigerator is defined in terms of the removed heat and the work input as:
Moreover, the rate of heat rejection to the outside air tuns out:
Best regards.