please show all work an steps! thank u A cylinder, fitted with a frictionless pi
ID: 1861557 • Letter: P
Question
please show all work an steps! thank u
A cylinder, fitted with a frictionless piston undergoes a process whereby energy is added to the water by heat transfer. The cylinder is 25 cm in diameter and contains 0.1 kg of water at state 1. When the piston has displaced 20 cm, it encounters a linear spring with a coefficient of 50 kN / m. Determine the work during the process. Determine the heat transfer. State 1 State 2 State 3 P1 = 150 kPa Spring engagement P3 = 1 MPa v1 = 1.0000 m3 / kg P2 = 150 kPaExplanation / Answer
Piston dia d = 25 cm = 0.25 m
Mass m = 0.1 kg
Spring stiffness k = 50 kN/m
Piston displacement from state 1 to 2, x = 20 cm = 0.2 m
Piston area A = 3.14/4 *0.25^2 = 0.0490625 m^2
v1 = 1 m3/kg
V1 = m*v1 = 0.1*1 = 0.1 m^3
V2 = V1 + A*x
V2 = 0.1 + 0.0490625*0.2
V2 = 0.1098125 m^3
v2 = V2/m
v2 = 0.1098125 / 0.1 = 1.098125 m^3/kg
From 1-2, process is constant pressure.
W1-2 = P*(V2 - V1) = 150*(0.1098125 - 0.1)
W1-2 = 1.471875 kJ
For 1-2:
Q - W = m*(u2 - u1)
From steam properties, at P1 = 150 kPa and v1 = 1 m3/kg, we get u1 = 2240 kJ/kg
From steam properties, at P2 = 150 kPa and v2 = 1.098125 m3/kg, we get u2 = 2410 kJ/kg
Q1-2 = 1.471875 + 0.1*(2410 - 2240)
Q1-2 = 18.471875 kJ
P3 = 1 MPa
Force on piston = spring force
(1*10^6)*0.0490625 = (50*10^3)*spring deflection
Spring deflection = 0.98125 m
Final volume V3 = V2 + 0.0490625*0.98125
V3 = 0.158 m^3
v3 = V3/m
v3 = 0.158 / 0.1 = 1.58 m^3/kg
Work done from 2-3 = Energy stored in spring
W2-3 = 1/2*kx^2
W2-3 = 1/2*50*0.98125^2
W2-3 = 24.07 kJ
From steam properties, at P3 = 1 MPa and v3 = 1.58 m3/kg, we get u3 = undefined!! (Some thing is wrong in inputs. I suspect 0.98 m spring deflection is too high. 50 kN/m spring stiffness might be 50 kN/cm.)
Here onwards, follow this process....
Get u3.
Then, Q2-3 - W2-3 = m*(u3 - u2)
This'll give you Q2-3.
Then W = W1-2 + W2-3
Q = Q1-2 + Q2-3
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