Question

slader An electrically driven pump operating at steady state draws water from a pond at a pressure of 1 bar and a rate of 20 kg/s and delivers the water at a pressure of 4 bar. There is no significant heat transfer with the surroundings, and changes in kinetic and potential energy can be neglected. The isentropic pump efficiency is 75%. Evaluating electricity at 8.5 cents per kW · h, estimate the hourly cost of running the pump.

Answer 1

70.9 cents.

Step-by-step explanation:

Let assume that water at inlet is a saturated liquid. The process within pump is modelled after the First Law of Thermodynamics:

`\dot W_(in) + \dot m \cdot (h_(in)-h_(out)) = 0`

The power consumed by the pump is:

`\dot W_(in) = \dot m \cdot (h_(out)-h_(in))`

The isentropic effciency of the pump is:

`\eta_(s) = (h_(out,s)-h_(in))/(h_(out)-h_(in))`

Where,

`h_(out) - h_(in) = (h_(out,s)-h_(in))/(\eta_(s))`

The power equation is modified into the following form:

`\dot W_(in) = \dot m \cdot ((h_(out,s)-h_(in)))/(\eta_(s))`

Specific enthalpies and entropies at inlet and outlet are obtained from steam tables and included below:

State 1 (Saturated Liquid)

`h = 417.51\,(kJ)/(kg)`

`s = 1.3028\,(kJ)/(kg\cdot K)`

State 2s (Subcooled Liquid)

`h = 417.823\,(kJ)/(kg)`

`s = 1.3028\,(kJ)/(kg\cdot K)`

The power consume by the pump is:

`\dot W_(in) = (20\,(kg)/(s) )\cdot \left((417.823\,(kJ)/(kg)-417.51\,(kJ)/(kg) )/(0.75) \right)`

`\dot W_(in) = 8.347\,kW`

The hourly energy consumption is:

`E_(hourly) = (8.347\,kW)\cdot (3600\,s)\cdot \left((1\,kWh)/(3600\,kJ) \right)`

`E_(hourly) = 8.347\,kWh`

The hourly cost is:

`C_(hourly) = (0.085\,(USD)/(kWh) )\cdot (8.347\,kWh)`

`C_(hourly) = 0.709\,USD`

Answer 2

Answer: The electricity cost = 70938 Cents

Explanation: Please find the attached file for the solution