Question

5 kg of saturated liquid water at 10 bars is contained in a piston-cylinder assembly and undergoes two processes. process 1-2: heated at constant pressure until it is saturated vapor. process 2-3: cooled as it is compressed at constant volume until it reaches 110 degrees celsius. determine the overall work and heat transfer for these processes. Hint: Must used linear interpolation from values in the correct Water tables.

Answer 1

Mass of sat liq water M = 5 kg

P1= 10 bar

process 1-2 heating at constant pressure, isobaric process

process2-3 cooled and compressed at const. volume ; isochoric process

T2= 110 deg C

from saturated water table at P1= 10 bar = 10 x 10⁵ Pa = 1000 KPa : Temp saturation = 179.88 deg C:V_f= specific volume of saturated liquid =0.001127 m3/ kg

and Vg = sp. volume of saturated vapour= 0.19436 m3/kg ; (enthaly change of water at 1000 kPa hfg = 2014.6KJ/kg)

So work done in process 1-2 at const pressure conversion of 5kg water to vapour (saturated)

W1-2 = P1 (Vg-Vf) KJ/kg (per kg I am taking here) = 1000 x (0.19436-0.001127) = 193.23 KJ/kg

Now, process 2-3 water is already converted to sat. vapour at T_saturated = 110 deg C

Since this is constant volume process W2-3 = P (V2-V1) = 0 as V2 = V1

So total overall work during process = m W 1-2 = 5 x 193.23 = 966.17 KJ (ANSWER)

Now heat transfer during process 1-2 constant pressure process is just change in enthalpy of water

Q 1-2 = m x hfg = 5 x 2014.6= 10073 KJ

Now heat trasnfer during process 2-3 (compression) which is constant volume process T g = 100 deg C

At this temp. ufg = internal energy change of vapour at temp 100 deg C from saturation water table = 2087 KJ/kg

Q2-3 = m x ufg = 5 x 2087 = - 10435 KJ (negative because of compression)

Total heat transfer = Q = Q1-2 - Q2-3 = 2087-10435 = -8348 KJ (ANSWER)