Steam enters a turbine with an enthalpy of 1292 Btu/lb and leaves with an enthal

Question

Steam enters a turbine with an enthalpy of 1292 Btu/lb and leaves with an enthalpy of 1098 Btu/lb. The transferred heat is 13 Btu/lb. What is the work in Btu/min and in hp for d flow of 2 lb/sec? Ans. 512.3 hp A thermodynamic steady flow system receives 4.56 kg per min of a fluid where p_1 = 137.90 kPa, v_1 = 0.0388 ms/kg, upsilon_1 = 122 m/s, and u_1 = 17.16 kJ/kg. The fluid leaves the system at a boundary where p_2 = 551.6 kPa, v_2 = 0.193 m^3/kg, upsilon_2 = 183 m/s and u_2 = 52.80 kJ/kg. During passage through the system the fluid receives 3, 000 J/s of heat. Determine the work. Ans. -486 kJ/min

Explanation / Answer

Ans 3) work, W = (1292-1098-13)Btu/lb(2lb/sec)(60sec/min) = 181Btu/lb* 120lb/min = 21720Btu/min

                  1 btu/min = 0.023580hp

so            W = 21720 * 0.023580 = 512.3 hp

Ans 4) enthalpy, h1 = u1 + p1*v1 = 22.51KJ/Kg,       h2 = u2 + p2*v2 = 159.2588KJ/Kg

           Q = 3000J/s = 3KJ/s ,             m = 4.56Kg/min = 0.076Kg/s

W = 3 + 0.076(22.51 + (1222 / 2000) -159.2588 - (1832 /2000)

W = -8.099KJ/s

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---