Steady Flow Energy Problems PDF

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SOLVED EXAMPLES ON STEADY FLOW PROCESSES Example 1. A chilled water of 15 kg/s enters the system for air conditioning a tall building with a velocity of 60 m/s at an height of 40m from the ground. The water leaves the system with a velocity of 20 m/s at an height of 70 m. The enthalpies of water entering in and leaving out are 30 KJ/kg and 50 KJ/kg respectively. The rate of workdone by a pump in the line is 40 KW. Find out the rate at which heat is removed from the building. Solution. Given m = 15 kg/ s W= – 40KW From eqn. (5.8) we have Q + m ( h + v2 / 2 + gz) in = m ( h+ v2 / 2 + gz) out + W Q = m( h2 –h1) + m ( v2 – v2 /2 + m g ( z2 – z1 ) + W = 150 ( 50 – 30 ) + 15 ( 202 – 602 / 2) 1/1000 + 15 x 9.81 ( 70 -40 / 1000) – 40 = 240.41 kw = heat removed rate. Example 2. A steam turbine receives steam at the rate of 0.42 kg/s. The inlet and outlet conditions of steam are as follows. Inlet Pressure Temperature Enthalpy Velocity Elevation

1.2 MPa 1880C 2785 KJ/kg 33.3 m/s 3m

Outlet 20 KPa – 2512 KJ/kg 100 m/s 0m

If the heat lost to the surroundings is 0.29 KJ/s, find out the power output of the turbine under steady flow conditions. Solution. Given Q = – 0.29 KJ/s;m = 0.42 kg/s, h1 = 2785 KJ/ kg, V1 = 33.3 m/s, z1 =3m, h2 = 2512 KJ/kg, V2 = 100 m/s, z2 = 0 m we have -0.29 + 0.42 ( 2785 + 33.3 2 / 2000 + 9.81 x 3 / 1000) = 0.42 ( 2512 + 1002 / 2000+9.81 x0/ 1000 ) + W Example 3. An adiabatic horizontal nozzle receives steam at velocity of 60 m/s and enthalpy of 3000 KJ/kg. It discharges it at an enthalpy 2762 KJ/kg, (a) find the velocity at exist from the nozzle (b) if the inlet area is

0.1 m2 and the specific volume at inlet is 0.187 m /kg, find the mass flow rate (c) if the specific volume at the nozzle exit is 0.498 m3 /kg, find the exit area of the nozzle. Solution. For adiabatic (Q = 0), horizontal (z1 = z2) nozzle; we have h2 – h2 = v1 – v2/2 2762 – 3000 = 602 – v2/2000 V2 = 692.5315 m/s. m = A1 V1 / v1 =0.1 x 60 / 0.187 = 32.08 kg / s Example 4. In an oil cooler, oil flows steadily through a bundle of metal tubes submerged in a steady stream of cooling water. Under steady flow conditions, the oil enters at 90°C and leaves at 30°C, while water enters at 25°C and leaves at 70°C. The enthalpy of oil at t°C is given by h = 1.68 t + 10.5 X 10– 4 t2 KJ/kg. What is the cooling water flow required for cooling 2.78 kg/s of oil ? Solution. h = 1.68 t + 10.5 x 10-4 t2 h A in = 1.68 x 90 + 10.5 X 10– 4 x 902 = 159.7 KJ/kg H A out = 1.68 x 30 + 10.5 x 10– 4 x 302 = 51.345 KJ/kg we have mA (h A in –h A out) = ma (h Bout-h Bin) or 2.78 (159.7 – 51.345) = Mb x Cw (tBsout – Tb in) [A = hot liquid= oil, B =Water, i.e.; cold liquid] Or 2.78 (159.7- 51.345) = m B X 4.1868 (70 – 25) Or MB = 1.59 kg/ s. Ans....