Question

A power cycle operates between a lake’s surface water at a temperature of 25 oC and water at a depth whose temperature is 8oC. At steady state the cycle develops a power output of 12 kW, while rejecting energy by heat transfer to the lower temperature water at the rate 14,400 kJ/min.

Answer 1

4.76%

Step-by-step explanation:

The thermal efficiency of the cycle can be determined by using the formula

η = W / Qin. W is the work, and is equal to Qin - Qout.

Rearranging for Qin: Qin = W + Qout

Plugging in the given numbers: Qin = 12 + (14400/60) = 12 + 240 = 252

Plugging this value into the first equation: η = 12/252 = 0.0476

The thermal efficiency is 4.76%.

Answer 2

The thermal efficiency of the power cycle is 5 %.

Step-by-step explanation:

Given that,

Surface temperature = 25°C

Under water temperature = 80°C

Output power = 12 kW

Rate = 14400 kJ/min = 240 kJ/sec

Suppose we need to find the thermal efficiency of the power cycle

We need to calculate the thermal efficiency

Using formula of the thermal efficiency

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

Where, Q₁ = heat input

We know that,

`W=Q_(1)-Q_(2)`

`Q_(1)=W+Q_(2)`

`Q_(1)=10+240`

Now put the value of Q and W in to the formula of efficiency

`\eta=(12)/(240)`

`\eta=0.05`

`\eta=5\%`

Hence, The thermal efficiency of the power cycle is 5 %.