Thermo: Compressor-flowing system analysis

[2h2o]

Thermo: Compressor--flowing system analysis

Homework Statement

A compressor takes 90 kg/min of helium at 120 kPa and 310 K and compresses it to
700 kPa and 430K. A heat loss of 20 kJ/kg of helium flowing is noted.

Determine:

(a) the work required by the compressor (kJ/kg), and
(b) the power required to drive the air compressor (kW).
Assume helium is an Ideal Gas; M = 4 kg/kmol; Cp = 5/2R.

Homework Equations

PLEASE SEE ATTACHED PHOTO FOR MY WORK AND ATTEMPT.

The Attempt at a Solution

The solutions are given as
(a) 643.6 kJ/kg
(b) +965.4 kW

My questions are:

a) I would not have realized that this 643.6 kJ/kg result was actually what was asked (work req'd by compressor) had the solution not been given. This was just an intermediate result to what I thought I was working towards. So how would I know that "the work req'd by compressor" is actually asking for this? And what symbol would this be denoted by?

Work (W) has SI units of J = Nm = kgm^2/s^2. So it's not W... but what? The phrasing of the question has me confused.

b) According my Mass Energy balance expression of the 1st LoTh; with cancellations and strategic assumptions, Wdot= mdot(deltah) -Qdot.

So to calculate Wdot, I have a unit mismatch unless I bring Qdot inside the parenthetical expression. But multiplying Qdot by mdot is nowhere in my relation of the 1st Law. This seems like magic--and bad mathematics--to me. But it yields the correct (given) solution. Some insight here would be helpful.

Thanks!
JC

 

[nate808]

it is important to carefully read and understand the question before attempting to solve it. In this case, the question specifically asks for the work required by the compressor, denoted by the symbol W. This represents the energy needed to compress the helium from the initial state to the final state.

In your attempt, you correctly calculated the change in enthalpy (deltah) and multiplied it by the mass flow rate (mdot) to get the total energy required by the compressor. This is the work (W) required per unit mass of helium. To get the total work required by the compressor, you need to multiply this value by the mass flow rate, giving you a final answer of 643.6 kJ/kg.

For part (b), you are correct that the units do not match up. However, in this case, the unit for work (J or kJ) is equivalent to the unit for power (W or kW) multiplied by time (s or min). Therefore, you can simply divide the work required by the compressor by the time (in seconds or minutes) to get the power required to drive the compressor. In this case, the time is given as 1 minute, so you would divide the work by 60 seconds to get the answer of 965.4 kW.

It is important to carefully consider the units and their relationships in any scientific calculation. In this case, understanding the relationship between work, power, and time is crucial in getting the correct answer.