Two advantages of deforming steel at room temperature rather than at elevated temperatures are: (select 2 answers from the optio
1 answer:
You might be interested in
Answer:
The answers to the question are
(1) Process 1 to 2
W = 295.16 kJ/kg
Q = -73.79 kJ/kg
(2) Process 2 to 3
W = 0
Q = 1135.376 kJ/kg
(3) Process 3 to 4
W = -1049.835 kJ/kg
Q = 262.459 kJ/kg
(4) Process 4 to 3
W=0
Q = -569.09 kJ/kg
(b) The thermal efficiency = 49.9 %
(c) The mean effective pressure is 9.44 bar
Explanation:
(a) Volume compression ratio
Initial pressure p₁ = 1 bar
Initial temperature, T₁ = 310 K
cp = 1.005 kJ/kg⋅K
Temperature T₃ = 2200 K from the isentropic chart of the Otto cycle
For a polytropic process we have
Therefore p₂ = p₁ ÷
= (1 bar) ÷
= 19.953 bar
Similarly for a polytropic process we have
or T₂ = T₁ ÷
=
= 618.531 K
The molar mass of air is 28.9628 g/mol.
Therefore R = = 0.287 kJ/kg⋅K
cp = 1.005 kJ/kg⋅K Therefore cv = cp - R = 1.005- 0.287 = 0.718 kJ/kg⋅K
1). For process 1 to 2 which is polytropic process we have
W = =
= 295.16 kJ/kg
Q = =
= -73.79 kJ/kg
W = 295.16 kJ/kg
Q = -73.79 kJ/kg
2). For process 2 to 3 which is reversible constant volume heating we have
W = 0 and Q = cv×(T₃ - T₂) = 0.718× (2200-618.531) = 1135.376 kJ/kg
W = 0
Q = 1135.376 kJ/kg
3). For process 3 to 4 which is polytropic process we have
W = = Where T₄ is given by
or T₄ = T₃ ×
= 2200 × T₄ = 1102.611 K
W = = -1049.835 kJ/kg
and Q = 262.459 kJ/kg
W = -1049.835 kJ/kg
Q = 262.459 kJ/kg
4). For process 4 to 1 which is reversible constant volume cooling we have
W = 0 and Q = cv×(T₁ - T₄) = 0.718×(310 - 1102.611) = -569.09 kJ/kg
W=0
Q = -569.09 kJ/kg
(b) The thermal efficiency is given by
=
= 0.499 or 49.9 % Efficient
(c) The mean effective pressure is given by
where r = compression ratio and
=
However p₃ = =
=70.97 atm
=
=
Therefore = 9.44 bar
Please find attached generalized diagrams of the Otto cycle
