4-12

1. m = 1/v1V1A1 = 1/0.06857 m3/kg *(80 m/s)(50 X 10-4 m 2) = 5.833 kg/s
2. The energy balance becomes Q = m(h2 - h1 + (V22 - V12)/2). So -90 kJ/s = (5.833 kg/s(3247.6 - 3433.8 + ((V22- (80m/s)2)/2(1kJ/kg/1000m2/s2) and V2 = 585.2. Watch out for the conversion.
3. A2 = mv2/V2 = (5.833 kg/s)(0.1512 m3/kg)/585.2 m/s = 15.1 X 10-4 m2

4.32

The energy balance becomes -W = m(h2 - h1). -200 kJ/s = (3 kg/s)(h2 - 3373.7 kJ/kg) and h2 = 707 kJ/kg. The exit temperature at P2 = 20 kPa and h2 = 2707 kJ/kg is 110.8 °C.

4.45
The enthalpy at P = 0.8 Mpa and T = 25 °C is compressed so it's hf @ 25 °C = 84.33 kJ/kg. This is also the enthalpy at the final conditions since h1 = h2 for a throttling valve. At T = -20 ° C, since the h is between hf and hg, it's a saturated mixture and the P = Psat at T = -20 ° C or 0.13299 MPa. Find x2 using the h: x = (h - h-f-)/hfg = (84.33 - 24.26)/211.05 = 0.285. Then u2 = uf + xufg = 24.17 + 0.85*(215.84-24.17) = 78.8 kJ/kg.

4.53
We can find all the enthalpies at the conditions stated. The h1 = hf @ 50 ° C since it's compressed liquid = 209.33 kJ/kg. The h2 @ P = 800 kPa and T = 200 °C = 2839.3 kJ/kg. The h3 = hf @ 800 kPa = 721.22 kJ/kg. Combining the energy balance and the mass balance gives us: m1h1 + m2 h2 = (m1 + m2)h3 since m3 = m1 + m2. Dividing through by m2 and letting m1/m2 = y gives us: yh1 + h2 = (y+1)h3. Then y = (h3-h2)/(h1 - h3) and y = 4.14.

4.63
As most of you heard, there was a typo in this problem and the ethylene glycol was cooled to 40 °C. First find the rate of heat transfer using the ethylene glycol. Q = mC(Tout - Tin) = (2 kg/s)(2.56 kJ/kg-°C)(40 - 80 °C) = -04.8 kW. The negative of this must equal the heat gained by the water. So 204.8 kW = mC(Tout - Tin) and m = Q/C(Tout-Tin) = 204.8/(4.18 kj/kg-°C)(55 - 20 )°C = 1.4 kg/s.

4.88
a) The m = 1/v1A1V1 = 1/.2327 m3/kg [p(0.06 m)2](2 m/s) = 0.0972 m/s b) The energy balance for this pipe simplifies to Q = m(h2 - h1) = (0.0972 kg/s)(2942.6-2839.3)kJ/kg = -10.4 kJ/s.