Steady flow energy eq....by Bilal Ashraf

sfeee

1. Group members: Muhammad Bilal Hanif ( ME1510 04 ) Hasnain Ahmad (ME1510 09 ) Muhammad Bilal Ashraf ( ME1510 26 ) Nadeem Iqbal ( ME1510 49 ) Sufian Arshad ( ME1510 50 ) 2 GROUP # 04

2. Steady Flow Energy Equation Presentation of Thermodynamics-I Topic on Assigned by: Dr. Ijaz Khan

3. What is SFEE? A steady flow process is one in which matter and energy flow steadily. ṁ in = ṁ out Unsteady flow is: ṁ in ≠ ṁ out

4. Derivation of Steady Flow Equation By the conservation of energy states: Energy (in) = Energy (out)

5. Then; Dividing the equation by yields

6. Uniform-Flow Processes At any instant during the process, the state of control volume is uniform The fluid properties may differ from one inlet or exit to another, But the fluid flow at an inlet or exit is uniform and steady

7. Applications of SFEE: Nozzles and Diffusers: Nozzles and diffusers are properly shaped ducts used to increase or decrease the speed of the fluid q = Δh + Δ V 2 / 2 + gΔz + w q = 0 Δz = 0 w = 0

8. Steam Turbine: from q = Δh + Δ V 2 / 2 + gΔz + w s if q = 0; Δc = 0; Δz = 0 then w s = -Δh = h 1 - h 2 A turbine is a device W ith rows of blades mounted on a shaft Could be rotated about its axis

9. Throttling Valves: From q = Δh + Δ V 2 / 2 + gΔz + w s as q = 0; w s = 0 ; z = 0 ; Δ V 2 / 2 = then Δh = That is h 1 = h 2 A throttling valve is a device U sed to cause a pressure drop in a flowing fluid. It does not involve any work.

10. Heat exchangers: From q = Δh + Δ V 2 / 2 + gΔz + w s as q = 0; w s = 0 ; Δ z = 0; Δ V 2 / 2 = Δh = That is ∑h in = ∑ h exit It is a device H ot fluid stream exchanges heat with a cold fluid stream W ithout mixing with each other

11. SFEE for More Than One Outlet: Energy (in) = Energy (out ) See Blow Pic . . . . . 1 2 3 1 2 3

12. Non-Flow Energy Equation:- When the fluid in closed system It is not passing through the system boundary The flow terms in SFEE will not apply In derivation of it u neglect PV and velocity

13. Equation of non-Flow Equation:- U 1 + Q = U 2 + W Q = (U 2 –U 1 ) + W Q = ∆U + W Hence , Heat Transferred the boundary of system = change of internal energy + work done

14. Problem # 01 :- Steam enter a turbaine with velocity of 16m/sec and sp. enthalpy 2990kj/kg. The steam leaves the turbine with velocity of 37m/s and sp. enthalpy 2530kJ/kg . The heat lost to the surrounding as the steam passes through the turbine is 25kJ/kg. The steam flow rate is 324000kg/hour. Determine the work out in kw (power). Given; ṁ = 32400kg/hr P = w/m x ṁ By equation of SFFEE , we will find w/m and in above equation we get power

15. Bernoullie’s Equation: By the help SFEE we can derive B ernoullie’s equation If there is no change in internal energy Q = 0 , W = 0 Then Equation will remain of Bernoullie’s Eq.

16. Problem# 02 Steam flows along a horizontal duct . At one point in the duct the pressure of steam is 1 bar and the temperature is 400 . A t a second point some distance from first , the pressure is 1.5 bar and temp. is 500 . A ssuming flow is be friction less and adiabatic . Determine whether the flow is accelerating or decelerating. Solve: In SFEE we put the values like w = 0, Q = 0, P E = 0 Then we will get ; h2-h1 = - So flow is decelerating as we negative value.  

17. Problem # 03 Steam is expanded isentropically in a turbine from 30 bar 400 Calculate the work done per unit mass flow of steam. Neglect changes as K.E and P.E. Solve: Q = 0, K.E = P.E = 0, Hence ; W = m [h 2 -h 1 ] By table , we get h 1 = 3230.9kJ/kg, h 2 = 2750kJ/kg & m = 1 Hence we get; Answer by putting values in above Eq.  

18. Problem# 04 A compressor takes in air at 1 bar and 20 .and discharge into line. The average air velocity in the line at appoint close the discharge is 0.7m/s and discharge pressure is 3.5 bar. Assuming isentropically , calculate the work input to compressor .Assume that air inlet velocity is very small. Slove : By putting the value Q=0 & P.E=0 in SFEE Eq. we get this relation: - W= } then - W= } T 1 and P 1 and P 2 is given by this we get T 2 : T2=T1 So put this values in above Eq. and Required Answer.  

19. Problem #05 Air is expanded isentropically in a nozzle from 13.8 bar and 150 To a presume of 6.9 bar. The inlet velocity to the nozzle is very small and process occur under steady flow. Calculate the exit velocity from the nozzle knowing that the nozzle is laid in a horizontal plane and that the inlet is 10m/s. Solve: By putting these Q = 0 , P.E = 0, w = 0 .So SFEE is: 0= m[ Cp (T2-T1)+ ] By the calculating T2 from T2=T1 Then putting values in above Eq. Get Answer  

20. THANKS for Bearing Us. From our side it is enough.