Question

An ideal Stirling engine using helium as the working fluid operates between temperature limits of 300 and 2000 K and pressure limits of 150 kPa and 3 MPa. Assuming the mass of the helium used in the cycle is 0.12 kg, determine (a) the thermal efficiency of the cycle, (b) the amount of heat transfer in the regenerator, and (c) the work output per cycle.

Answer 1

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