c Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 2 / 25 T145 = Q + W cv + i h 2 = h (V2 1 V 2 2)

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1 Thermodynamic Processes & Isentropic Efficiency Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & Technology (BUET Dhaka-1000, Bangladesh ME 6101: Classical Thermodynamics Assumptions: Control volume does not move relative to the coordinate frame. State of the mass at each point in the control volume does not vary with time. As for the mass that flows across the control surface, the mass flux and the state of this mass at each discrete area of flow on the control surface do not vary with time. The rates at which heat and work cross the control surface remain constant. For example, a centrifugal air compressor that operates with a constant mass rate of flow into and out it, constant properties at each point across the inlet and exit ducts, a constant rate of heat transfer to the surroundings, and a constant power input. At each point in the compressor the properties are constant with time. c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 1 / 5 Nozzles & Diffusers c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 / 5 [Moran Ex. 4.3]: Converging-diverging Steam Nozzle: Estimate A. T159 A nozzle is a flow passage of varying cross-sectional area in which the velocity of a gas or liquid increases in the direction of flow. In a diffuser, the gas or liquid decelerates in the direction of flow. For a nozzle or diffuser, the only work is flow work at locations where mass enters and exits the CV, so the term W cv drops out. PE 0 T145 h = h (V 1 V + gz i e Steady state / = 0 Z = Z 1 0 = (h 1 h + 1 (V 1 V ρ = ρ(steam,p = P,h = h = kg/m 3 = m = ρ A V = A = m W cv = 0 & Q = 0 = m e = kg/s ( c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 3 / 5 c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 4 / 5

2 Throttling Devices A significant reduction in pressure can be achieved simply by introducing a restriction into a line through which a gas or liquid flows. [Cengel Ex. 5.8]: R-134a enters the capillary tube of a refrigerator as saturated liquid at 0.8 MPa and is throttled to a pressure of 0.1 MPa. Determine x and T. Steady state / = 0 Z = Z 1 & V V 1 Q 0 & W cv = 0 T160 For a control volume enclosing a throttling device, the only work is flow work at locations where mass enters and exits the control volume, so the term W cv drops out. There is usually no significant heat transfer with the surroundings, and the change in potential energy from inlet to exit is negligible. c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 5 / 5 T148 h = h1 h 1 = enthapy(r134a,p 1 = 0.8 MPa,x 1 = 0.0 h 1 = h = h f, 0.1MPa + x h fg, 0.1MPa = x = 0.3 T 1 = T(R134a,P 1 = 0.8 MPa,x 1 = 0.0 = T 1 = 31.3 o C T = T(R134a,P = 0.1 MPa,sat. = T = 18.8 o C T = o C c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 6 / 5 Turbines [Moran Ex 4.4]: Heat Transfer from Steam Turbine: Determine heat loss, Q. A turbine is a device in which power is developed as a result of a gas or liquid passing through a set of blades attached to a shaft free to rotate. Steady state / = 0 Z = Z 1 = m e = (4600/3600 kg/s T161 Axial-flow steam or gas turbine. T16 Hydraulic turbine installed in a dam. T146 + gz i e [ W cv = Q + m h 1 = enthalpy(steam,p = P 1,T = T 1 h = enthalpy(steam,p = P,x = x = Q cv = kw (heat loss (h 1 h +( V 1 V ( ] c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 7 / 5 c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 8 / 5

3 Compressors & Pumps Compressors and pumps are devices in which work is done on the substance flowing through them in order to increase the pressure and/or elevation. Compressor is used to compress a gas (vapour and the term pump is used when the substance is a liquid. T147 [Moran Ex. 4.5]: Air Compressor Power: Determine power required, W cv. Steady state / = 0 Z = Z 1 + gz i e [ W cv = Q + m Q cv = 180kJ/min = 3.0 kw m = ρav ρ = P/RT (h 1 h +( V 1 V ( ] T163 c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 9 / 5 T164 h 1 h = C p (T 1 T = kj/kg ρ 1 = P 1 /RT 1 = 1.0 m 3 /kg m = ρ 1 A 1 V 1 = 0.7 kg/s = W cv = kw (work input required c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 ( / 5 [Borgnakke Ex. 4.6] : A water pump is located 15 m down in a well, taking water in at 10 o C, 90 kpa at a rate of 1.5 kg/s. The exit line is a pipe of diameter 0.04 m that goes up to a receiver tank maintaining a gauge pressure of 400 kpa. Assume that the process is adiabatic, with the same inlet and exit velocities, and the water stays at 10 o C. Find the required pump work. Heat Exchangers Steady state / = 0 Q = 0, V i = V e, T i = T e P i = 90 kpa, P e = kpa. z e z i = 15.0 m T44 [ W cv = m + gz i (h 1 h +( V 1 V m = 1.5 kg/s e ( ] + g(z 1 z W cv = 0.8 kw (work input required : h 1 h = P 1 P, if ρ & T are constant. ρ c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 ( / 5 T166 Common heat exchanger types. (a Direct contact heat exchanger. (b Tube-within a-tube counterflow heat exchanger. (c Tube-within-a-tube parallel flow heat exchanger. (d Cross-flow heat exchanger. c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 1 / 5

[Cengel P5-85]: Condenser Cooling Water: Determine m water. Isentropic (First Law Efficiency [Borgnakke Ex. 7.]: Reversible adiabatic flow of steam through nozzle, V e =?

4 kg/s h 1 = h(h O, P = 100 kpa, T = 0 h = h(h O, P = 100 kpa, T = 30 h 3 = h(h O, P = 0 kpa, x = 0.95 h 4 = h(h O, P = 0 kpa, x = 0 0 = ( mh i ( mh e c Dr. Md.

4 [Cengel P5-85]: Condenser Cooling Water: Determine m water. Isentropic (First Law Efficiency [Borgnakke Ex. 7.]: Reversible adiabatic flow of steam through nozzle, V e =? W = 0, Q = 0, (KE = 0, (PE = 0 m 3 = (0000/3600 kg/s m 1 = m & m 3 = m 4 T149 ( m 3 h 3 + m 1 h 1 = ( m 4 h 4 + m h = m water = m 1 = 97.4 kg/s h 1 = h(h O, P = 100 kpa, T = 0 h = h(h O, P = 100 kpa, T = 30 h 3 = h(h O, P = 0 kpa, x = 0.95 h 4 = h(h O, P = 0 kpa, x = 0 0 = ( mh i ( mh e c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 ( / 5 T185 Continuity equation: = m e = m First law: 0 = 0+0+(h i h e + V i V e + 0 Second law: s i = s e. h i = h(steam, P i = 1 MPa, T i = 300 o C = s i = s(steam, P i = 1 MPa, T i = 300 o C = s e = s i & P e = 0.3 MPa: state e defined. h e = h(p e = 0.3 MPa, s e = = = V e = (h i h e + V i = V e = m/s c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 ( / 5 Isentropic (First Law Efficiency Isentropic Nozzle Efficiency Isentropic (First Law Efficiency [Moran Ex. 6.6]: Entropy production in a steam turbine T178 T181 For nozzle: w = 0, (PE = 0, V 1 0 h 1 = h + V = η N = Actual KE at nozzle exit Isentropic KE at nozzle exit = V V s h1 h h 1 h s c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 ( / 5 Continuity equation: = m e = m First law: 0 = q + w +(h i h e + V Second law: (s e s i = q T b + σcv m. i V e + 0 P 1 = 30 bar, T 1 = 400 o C := h 1 =, s 1 = T = 100 o C, x = 1.0 := h =, s = = q = 3.9 kj/kg = σcv m = kj/kg.k. c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 ( / 5

5 Isentropic (First Law Efficiency Isentropic Turbine Efficiency Isentropic (First Law Efficiency [Cengel Ex.7.14]: Isentropic Efficiency of a Steam Turbine T176 w t = h 1 h : for steady-state, adiabatic expansion. σ cv m = s s 1 0: states with s < s 1 is not attainable with adiabatic expansion. w t s = h 1 h s : state s is for internally reversible expansion. Isentropic turbine efficiency, η t = wt w t s = h1 h h 1 h s c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 ( / 5 T179 (P 1 = 3 MPa,T 1 = 400 o C : h 1 =,s 1 = (P = 50 kpa,t = 100 o C : h = (P s = 50 kpa,s s = s 1 : h s = = η t = h1 h h 1 h s = 66.7± = m = Power h 1 h s = 3.64 kg/s c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 ( / 5 Isentropic (First Law Efficiency Isentropic Compressor and Pump Efficiencies Isentropic (First Law Efficiency [Cengel Ex. 7.15]: Effect of efficiency on compressor power, if η c = 80± T177 w c = (h h 1 : for steady-state, adiabatic compression. σ cv m = s s 1 0: states with s < s 1 is not attainable with adiabatic compression. w c s = (h s h 1 Isentropic compressor/pump efficiency, η c = wc s w c = hs h1 h h 1 c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 ( / 5 T180 T s = T 1 (P /P 1 (k 1/k = K η c = hs h1 h h 1 = cp(ts T1 c P(T T 1 T = K W c s = m(h s h 1 = mc P (T s T 1 = 46.5 kw W c = m(h h 1 = mc P (T T 1 = 58.1 kw c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 0 / 5

6 Isentropic (First Law Efficiency Steady-state Flow Process First Law: CM system & reversible process: δq = du δw = d(h Pv+Pdv = dh vdp Second Law: δq = Tds Tds = dh vdp 1 Tds = 1 (dh vdp = (h h 1 First Law: CV system & reversible process: 0 = q 1 + w 1 +(h 1 h + (ke+ (pe = w 1 = q 1 +(h 1 h +0+0 = 1 Tds +(h 1 h = T754 w sf = w 1 = W1 = 1 VdP c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 1 / 5 W1 = 1 VdP Isentropic (First Law Efficiency w 1 : Pv n = constant w sf = w 1 = { = n (P v P 1 v 1 = nr(t T1 = RT ÐÒ( v v 1 : n 1 : n = 1 = RT ÐÒ( P P 1 T754 q 1 = h h 1 + w 1 Example: A compressor operates at steady state with nitrogen entering at 100 kpa and 0 o C and leaving at 500 kpa. During this compression process, the relation between pressure and volume is Pv 1.3 =constant. R = (8.314/8= kj/kg.k, T /T 1 = (P /P 1 (n 1/n = T = 45 K. w 1 = nr(t T1 = kj/kg q 1 = c P (T T 1 +w 1 = 37.5 kj/kg c P = 1.0 kj/kg.k c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 / 5 Isentropic (First Law Efficiency Isentropic (First Law Efficiency W Pv n = constant w b = w 1 = 1 Pdv [Cengel Ex.7.1]: Compression & Pumping Works W 1 = 1 PdV w 1 : { = P v P 1v 1 = RT ÐÒ( v v 1 = R(T T1 = RT ÐÒ( P P 1 : n 1 : n = 1 T753 q 1 = u u 1 + w 1 Example: In a reversible process, nitrogen is compressed in a cylinder from 100 kpa and 0 o C to 500 kpa. During this compression process, the relation between pressure and volume is Pv 1.3 =constant. R = (8.314/8= kj/kg.k, T /T 1 = (P /P 1 (n 1/n = T = 45 K. w 1 = R(T T1 = kj/kg q 1 = c V (T T 1 +w 1 = 3. kj/kg c v = R/(1.4 1 = 0.74 kj/kg.k T191 w P = v f(p P 1 = kj/kg w C = h h 1 = kj/kg = w c >> w p c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 3 / 5 c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 4 / 5

7 Isentropic (First Law Efficiency Isentropic Efficiencies of Some Devices Device First-Law Efficiency Typical Range Turbine Compressor Pump Nozzle Boiler Heat-exchanger Throttling-device η t = wa w s = wa w s = h h1 η c = ws w a = ws w a = hs h1 η p = ws w a = ws w a = hs h1 η n = (kea (ke s = h1 h h 1 h s V V s h s h 1 80 to 90% h h 1 70 to 85% h h 1 50 to 90% 90% c Dr. Md. Zahurul Haq (BUET Thermodynamic Processes & Efficiency ME 6101 (017 5 / 5

Consequences of Second Law of Thermodynamics. Entropy. Clausius Inequity

Consequences of Second Law of Thermodynamics. Entropy. Clausius Inequity onsequences of Second Law of hermodynamics Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & echnology BUE Dhaka-000, Bangladesh zahurul@me.buet.ac.bd