Problem 1 Superheated steam flows into a turbine operating at steady state at 3

Question

Problem 1 Superheated steam flows into a turbine operating at steady state at 3 MPa, 370°C with a velocity of 100 m/s. The steam exits at 0.1 MPa and a velocity of 50 m/s. The elevation of the inlet is 9.84 ft higher than at the exit. The mass flow rate of the steam both at the entrance and exit ot turbine is 1985 Ibm/min, and the power developed is 7 MW. Let g9 (a) The area at the inlet, in m2 (b) The rate of heat transfer between the turbine and its surroundings, in kW Given the followings: Inlet specific enthalpy 2803.28 kJ/kg Exit specific enthalpy 2706.54 kJ/kg Inlet specific volume- 0.09089 m'/kg Exit specific volume 0.8803 m'kg

Explanation / Answer

given data-

at the inlet-

pressure = P1 = 3 MPa

velocity = V1 = 100 m/sec

temperature = T1 = 370+273 =643 K

specific enthalpy = E1= 2803.28 KJ/Kg

specific volume = V1 = 0.09089 m3/Kg

density of steam =d1= 1/V1 = 11.0023 Kg/m3

at the outlet-

pressure = P2 = 0.1 MPa

velocity = V2 = 50 m/sec

specific enthalpy = E2= 2706.54 KJ/Kg

specific volume = V2 = 0.8803 m3/Kg

density of steam = d2 = 1/V2 = 1.13597Kg/m3

calculation of area at the inlet -

mass flowrate = m = density x velocity x area of flow

= 11.0023 Kg/m3 x 100m/sec x A........................................(I)

but given mass flow rate =m = 1985 lbm/min

as 1lbm = 0.4536 kg

so, m = 19 85 x 0.4536 /60 kg /sec = 15.006 Kg/sec........................................(II)

from (I) and (II) we get

A = 0.013639 m2

(b) energy release to the surrounding-

energy balance -

energy In = energy converted to power + energy out + energy goes to surrounding

energy IN = Ei = inlet mass flow rate x enthalpy of the inlet steam

= density x velocity x area of flow x E1

= 11.0023 x 100 x 0.013639 x 2803.28

= 42066.123 KW

= 42.066 MW

energy OUT = E0 = outlet mass flow rate x enthalpy of the outlet steam

= density x velocity x area of flow x E2

= 1.13597 x 50 x 0.013639 x 2706.54

= 2096.688 KW

= 2.097 MW

energy converted to power = 7 MW (given)

energy released to the surroundings = we can calculate = Es

energy In = energy converted to power + energy out + energy goes to surrounding

42.066 MW = 7MW + 2.097 MW + Es

Es = 32.969 MW  

I hope you can understand the solution

Thank you!!!

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.