Question

Problem 5.024 A power cycle receives 1000 Btu by heat transfer from a reservoir at 1500°F and discharges energy by heat transfer to a reservoir at 300°F. The thermal efficiency of the cycle is 75% of that for a reversible power cycle operating between the same reservoirs. (a) For the actual cycle, determine the thermal efficiency and the energy discharged to the cold reservoir, in Btu b. Repeat for the reversible cycle.

Answer 1

Answer:

a) $\eta_{actual} = 0.4592 = 45.92%$

$Q_2 = 540.8 Btu$

b) $\eta_{ideal} = 0.61235 = 61.23%$

$Q_2 = 387.645 Btu$

Explanation:

Given data:

$Q_1 = 1000Btu$

$T_1 = 1500 degree F = 1088.70 K$

$T_2 = 300 F = 422.03 k$

$\eta_{actual} = 0.75 \times \eta_[ideal}$

$\eta_{actual} = 0.75 \times [1 -\frac{T_2}{T_1}]$

$\eta_{actual} = 0.75 \times [1 -\frac{422.03}{1088.70}]$

$\eta_{actual} = 0.4592 = 45.92%$

$\eta_{actual} = 1 - \frac{Q_2}{Q_1}$

$0.4592 = 1 -\frac{Q_2}{1000}$

$0.5408 =\frac{Q_2}{1000}$

$Q_2 = 540.8 Btu$

$\eta_{ideal} = 1 - \frac{T_2}{T_1}]$

$\eta_{ideal} = 1 -\frac{422.03}{1088.70}]$

$\eta_{ideal} = 0.61235 = 61.23%$

$\eta_{ideal} = 1 - \frac{Q_2}{Q_1}$

$1 - 0.6123 = \frac{Q_2}{Q_1}$

solving for Q2

$Q_2 = 387.645 Btu$