Answer:
Explanation:
Given that,
Temperature limit are 300 K and 2000K
Then,
Hot Temperature TH = 2000K
Cold Temperature TC = 300K
pressure limits of 150 kPa and 3 MPa.
P1 = 150kPa
P2 = 3MPa = 3000 kPa
Mass of helium given
M = 0.12kg
The gas constant and specific heat of helium at room temperature are
R = 2.0769 kJ/KgK
Cv = 3.1156 kJ/KgK
Cp = 5.1926 kJ/KgK
A. Thermal efficiency?
Thermal efficiency is given as
η = 1 — TC / TH
η = 1 — 300 / 2000
η = 1 —0.15
η = 0.85
B. Heat transfer in the generator?
The heat transfer in the generator can be determined from the energy balance
Q(req) = Q⁴~¹
Q(req) = m∆u
Q(req) = m•cv•∆T
Q(req) = m•cv•(TH - TC)
Q(req) = 0.12 × 3.1156 × (2000-300)
Q(req) = 0.12 × 3.1156 × 1700
Q(req) = 635.6 kJ
C. Work output?
The work output can be determine from the efficiency and the heat input.
W = ηQ(in)
W = ηQ¹~²
W = η•m•TH•∆s
∆s = Cp•In(TH/TC) — R•In(P2/P1)
Since from 1 to 2, the temperature did not change
∆s = —R•In(P1•TH/P2•TC)
W = η•m•TH•(—R•In(P1•TH/P2•TC))
W = 0.85 × 0.12 × 2000 ×(-2.0176 × In(150 × 2000/3000×300)
W = -423.69 In(⅓)
W = 465.47 kJ
