Fill in the blank to output the quotient of dividing 100 by 42. print (100______42)
1 answer:
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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
The load is placed at distance 0.4 L from the end of area.
<h3>What is meant by torque?</h3>
The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.
Let the beam is of length L
Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero
also, we know that stress at both ends are same
Now from two equations we have
solving the above equation we have
so the load is placed at distance 0.4 L from the end of area.
The complete question is:
47. the beam is supported by two rods ab and cd that have cross-sectional areas of and
, respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.
To learn more about torque refer to:
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