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## Homework Statement

As shown in Fig. P5.39, a system undergoing a power cycle develops a net power output of 1 MW while receiving energy by heat transfer from steam condensing from saturated vapor to saturated liquid at a pressure of 100 kPa. Energy is discharged from the cycle by heat transfer to a nearby lake at 17°C. These are the only significant heat transfers. Kinetic and potential energy effects can be ignored. For operation at steady state, determine the minimum theoretical steam mass flow rate, in kg/s, required by any such cycle.

Fig. 5.39: https://gyazo.com/49a3ca702a5fa239633f8f054618345e

(Can't get it to embed)

## Homework Equations

η = W

_{cycle}/ Q

_{H}= 1 - T

_{C}/ T

_{H}

W

_{cycle}= Q

_{H}- Q

_{C}

mass flow: dm

_{cv}/ dt = ∑ m

_{i}- ∑ m

_{e}

energy balance: dEcv / dt = Q - W + ∑ m

_{i}(h

_{i}+ V

_{i}

^{2}/ 2 + gz

_{i})- ∑ m

_{e}(h

_{e}+ V

_{e}

^{2}/ 2 + gz

_{e})

## The Attempt at a Solution

Knowing that the system operates at steady state, I know that the entry mass flow and exit mass flow are the same. I also know that since kinetic energy and potential energy can be ignored, the energy balance equation simplifies to:

0 = Q - W + ∑ m

_{i}(h

_{i})- ∑ m

_{e}(h

_{e}) =>

W = Q + m(h

_{i}- h

_{e})

Knowing that the initial and final states are at saturated vapor and saturated liquid, respectively, using the tables in the back of the book, I found the specific enthalpy values to be:

h

_{1}= 2675.5 kJ/kg

h

_{2}= 417.46 kJ/kg

I know that I can use the thermal efficiency relation to find the Q

_{H}value, but I don't know how exactly to find the thermal efficiency since there's no listed T

_{H}value, unless I'm missing something in the problem statement. I feel that the energy balance equation will be cake once the heat transfer value is found, but I just can't figure out how to get it. If anyone could steer me in the right direction, that would be appreciated.

Edit: The temperature of the lake is 17°C, not 178C. Copy-and-paste didn't recognize the degree symbol.

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