I need all of the parts so that I can check what I have so far and figure out wh

Question

I need all of the parts so that I can check what I have so far and figure out where to go from there.  You'll need to use a table to find out the properties (density, specific heat, etc.)  Thanks!

Cycles Consider 1 kg of liquid water at T = 50 degree C and P - 2atin. You may use saturated liquid properties. A device compresses the water up to 20atm but docs not change its temperature. How much did the enthalpy of the system change? Heat is now added at constant pressure until the water reaches a temperature of 500 degree C. How much did the enthalpy of the system change? Next, we have a clever (and ideal) device which allows an isentropic (no-entropy- change) process to extract work by expanding the steam down to a lower temperature and pressure. The pressure at the end of this process is again 2aim. How much did the enthalpy of the system change? Finally, we reject heat at a constant pressure of 2atm until the water is back to 50 degree C, where it started. How much did the enthalpy of the system change? Consider this question as a whole: how much work was extracted (in part c) per heat input (in part b)? I need all of the parts so that I can check what I have so far and figure out where to go from there. You'll need to use a table to find out the properties (density, specific heat, etc.) Thanks!

Explanation / Answer

a)

From property tables, at T = 50 deg C and P = 2 atm, we get h1 = 210 kJ/kg

and, at 20 atm, T = 50 deg C, we get h2 = 211 kJ/kg

Enthalpy chnage = m(h2 - h1) = 1*(211-210) = 1 kJ

b)

At 20 atm and 500 deg C, we get h3 = 3470 kJ/kg and s3 = 7430 J/kg-K

Enthalpy change = m(h3 - h2) = 1*(3470 - 211) = 3259 kJ

c)

For s4 = s3 = 7430 J/kg-K and 2 atm, we get h4 = 2840 kJ/kg

Enthalpy change = m*(h4 - h3) = 1*(2840 - 3470) = -630 kJ

d)

Enthalpy change = m*(h1 - h4) = 1*(210 - 2840) = -2630 kJ

Net enthalpy change of system = 1 + 3259 + (-630) + (-2630) = 0 kJ

e)

Work extracted per heat input = (h3 - h4) / (h3 - h2) = 630 / 3259 = 0.193 kJ (or 19.3 %)

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