Question

A Carnot engine operates between 1350 °F and 125 °F. If it rejects 55 Btu as heat, determine the work output.

a. W =68.54 Btu
b. W=145.19 Btu
c. W = 115.12 Btu
d. W=235.7 Blu

Answer 1

C.
`W = 115.12\,Btu`

Step-by-step explanation:

Thermodynamically speaking, a Carnot engine represents an entirely reversible thermal process and its energy efficiency represents the maximum theoretical efficiency that thermal machines can reach. The efficiency of the ideal thermal process (
`\eta`), no unit, is:

`\eta = \left(1-(T_(L))/(T_(H)) \right)` (1)

Where:

`T_(L)` - Temperature of the cold reservoir, measured in Rankine.

`T_(H)` - Temperature of the hot reservoir, measured in Rankine.

If we know that
`T_(H) = 1809.67\,R` and
`T_(L) = 584.67\,R`, then the energy efficiency of the ideal thermal process is:

`\eta = 0.678`

By First Law of Thermodynamics, we calculate the work output:

`W = Q_(H)-Q_(L)`

`W = (W)/(\eta) -Q_(L)` (By definition of efficiency)

`Q_(L) = (W)/(\eta)-W`

`Q_(L) = \left((1)/(\eta)-1 \right)\cdot W`(2)

Where:

`Q_(H)` - Heat received by the engine, measured in Btu.

`Q_(L)` - Heat rejected by the engine, measured in Btu.

`W` - Work output, measured in Btu.

If we know that
`\eta = 0.678` and
`Q_(L) = 55\,Btu`, then the work output of the Carnot engine is:

`W = (Q_(L))/((1)/(\eta)-1 )`

`W = 115.807\,Btu`

The work output of the Carnot engine is 115.807 Btu. (Answer: C)