School of Engineering & Computing Session 2015-16 Paisley Campus Trimester 1 MODULE CODE: ENGG08021 INTRODUCTION TO THERMOFLUIDS Date: 15 January 2016 Time: 10:00 12:00 Attempt FOUR QUESTIONS IN TOTAL Attempt AT LEAST ONE QUESTION FROM SECTION A Attempt AT LEAST ONE QUESTION FROM SECTION B Attempt AT LEAST ONE QUESTION FROM SECTION C Page 1 of 22
START OF PAPER SECTION A Attempt AT LEAST ONE QUESTION 1. (a) State the energy balance equation for a flow system and define all its terms with their units. (4) (b) A flow system delivers 5.0 MW of work to the surroundings and loses 250 kw of heat during the process. The velocity at the inlet is 48 m s -1 and at the outlet is 8 m s -1. The inlet is 20 m above the outlet. The fluid flows at a steady rate of 105 kg s -1. Calculate: (i) The rate of change in the system kinetic energy (3) (ii) The rate of change in the system potential energy (3) (iii) The change in the specific enthalpy per second. (5) Page 2 of 22
2. The turbine of a power cycle expands an intake of superheated steam at 56 bar and 450 C to 0.04 bar in an isentropic process according to the pressure volume relation pv 1.3 =C. Determine the following (basis 1 kg): (a) The initial specific volume, specific enthalpy, specific internal energy and specific entropy of the steam. (6) (b) The volume of steam after expansion. (2) (c) The dryness fraction of steam after expansion. (4) (d) The specific enthalpy of steam after expansion and the change in enthalpy. (3) Data for superheated steam: P(bar) T ( C) v (m 3 kg -1 ) u (kj kg -1 ) h (kj kg -1 ) s (kj kg -1 k -1 ) 50 450 0.0632 3000 3316 6.818 60 450 0.0521 2988 3301 6.719 Data for saturated steam: P(bar) Ts ( C) vg (m 3 kg -1 ) hf (kj kg -1 ) hg (kj kg -1 ) 0.04 29.0 34.80 121 2554 Page 3 of 22
SECTION B Attempt AT LEAST ONE QUESTION 3. (a) What is meant by: (i) Viscous flow (1) (ii) Inviscid flow (1) (iii) Drag force (1) (iv) Gauge pressure (1) (v) Dynamic pressure (1) (vi) Stagnation pressure (1) (b) An open tank contains a hydrocarbon oil of specific gravity 0.86 on top of water and mercury. The depth of oil is 3 m, the depth of water is 1.5 m and the depth of mercury is 0.12 m. Calculate the gauge and absolute pressures at the bottom of the tank when the atmospheric pressure is 1 bar. (9) Data: Density of water = 1000 kg m -3 Density of mercury = 13600 kg m -3 Acceleration due to gravity = 9.81 m -2 Page 4 of 22
4. (a) Define the following terms for a Newtonian fluid and identify their units: (i) Specific volume (2) (ii) Reynolds number (2) (b) Water is transferred at a rate of 5 kg s -1 from a large reservoir to a distribution point 1500 m away using a smooth pipe of 50 mm diameter. The transfer line includes two 90 large radius bends, two square elbow bends, 2 gate valves that are fully open during normal operation and four 45 elbows. (i) Is the flow laminar or turbulent? (3) (ii) Clearly stating any assumption made, calculate the total pressure drop due to friction in the pipe and the fittings. (6) Data: (iii) How can the pressure drop be reduced? (2) Average viscosity of water: 1.1 10-3 Ns m -2 Average density of water: 1000 kg m -3 Fanning Friction Factor: Blasius approximation for turbulent flow in smooth pipe: f = 0.079 N0.25 Re Laminar flow: f = 16 N Re Fitting Number of Equivalent Pipe Diameters Gate valve (fully open) 7 Globe valve (fully open) 60-300 45 elbow 15 90 square elbow 60 Large radius 90 bend 0.6 Entry into leg of tee junction 90 Entry from leg of tee junction 60 Page 5 of 22
SECTION C Attempt AT LEAST ONE QUESTION 5. One of the product streams in a food processing facility producing soup is leaving the cooking area at the rate of 2000 kg h -1 and a temperature of 130 C. This stream is to be cooled to a maximum temperature of 35 C before it can be canned. A heat exchanger is used to cool the stream with water being the cooling fluid. Water enters the heat exchanger at 20 C and a rate of 7200 kg h -1. The average heat capacity of the soup for this temperature range is 3.8 kj kg -1 K -1 and that for water is 4.2 kj kg -1 K -1. The overall heat transfer coefficient is 720 W m -2 K -1. (a) Calculate the outlet water temperature. (7) (b) Calculate the total area if the flow is counter-current. (8) Page 6 of 22
6. Commercial cold storage rooms are normally built of concrete but in order to preserve energy they are usually insulated using suitable material that can increase resistance to heat flow. In one such insulation cork board is used as lagging for the concrete to reduce heat transfer from the surroundings to the room. The temperature at the wall surface inside the cold room is kept at 1 C to ensure the preservation of the food material being stored. (a) Calculate the heat loss per unit area of the concrete wall if there is no insulation and the wall thickness is 210mm. (6) Calculate the thickness of the cork necessary to ensure that the energy loss from the bare concrete is reduced by 95%. (9) Data: External wall surface temperature = 20 C Thermal conductivity of cork = 0.0433 W m -1 K -1 Thermal conductivity of concrete = 0.762 W m -1 K -1 END OF EXAM PAPER Fanning Friction Factor Chart Follows Dimensionless Numbers Definitions Follows Thermodynamics Formula Sheet Follows Fluid Flow Formula Sheet Follows Heat Transfer Formula Sheet Follows Page 7 of 22
Fanning Friction Factor Chart (C. Geankoplis, Prentice Hall, 4 th Edition, 2003) Page 8 of 22
Reynolds Number = Dimensionless Numbers Inertial Forces = Viscous Forces ρul μ Prandtl Number = Momentum Diffusion Thermal Diffusion = C P μ k fluid Nusselt Number = Convesctive Transfer Conductive Transfer = h L k fluid Peclet Number = Convective transfer Diffusive transfer = N Re N Pr = ρul μ C P μ k fluid = νl α Grashof Number = Buoyancy Forces Viscous Forces = L 3 ρ 2 gβ T μ 2 Fourier Number = Heat Conduction Thermal Energy Storage = k ρc P t L 2 = α t L 2 Biot Number = Internal Thermal Resistance BL Thermal Resistance = h L k solid Please note that L is the characteristic dimension and it is not necessary a length. For a circular pipe the characteristic dimension is the diameter or L d. Page 9 of 22
Thermodynamics Formula Sheet Kinetic Energy (E k ) = 1 2 mv2 Kinetic Energy (E k) = 1 2 m v2 (For flow systems) Specific Kinetic Energy (e k ) = 1 2 v2 Specific Kinetic Energy (e k) = 1 2 v2 (For flow systems) Potential energy(e P ) = mgz Potential energy(e P) = m gz (For flow systems) Specific Potential energy(e P ) = gz Specific Potential energy(e P) = gz (For flow systems) dq = du (Constant volume system) dq = dh = d(u + Pv) (Constant pressure system) Ideal Gas: h = u + Pv H = U + PV Q = U (Closed system) Q = H (Open system) C v (T) = lim T 0 u T Page 10 of 22
h C p (T) = lim T 0 T T 2 u = C v (T)dT T 1 T 2 h = C p (T)dT T 1 T 2 U = m u = m C v (T)dT T 1 T 2 H = m h = m C p (T)dT T 1 u = C T v (when Cv is approx. constant or the temperature range is narrow) h = C T p (when Cp is approx. constant or the temperature range is narrow) U = m u = mc T v (when Cv is approx. constant or the temperature range is narrow) H = m h = mc T p (when Cp is approx. constant or the temperature range is narrow) Entropy: δs = δq T Δs = dq T Work = Force Distance = F s 2 Work = F ds 1 (when force is not constant) Shaft work (Wsh) = 2πnT Page 11 of 22
Shaft Power: W sh = 2πn T 2 Pressure Volume Work: W p = pdv 1 (e.g. compressing gases) W p = PV (if V is approximately constant, e.g. pumping liquids) Q W = m {(h 2 h 1 ) + 1 2 (v 2 2 v 1 2 ) + g(z 2 z 1 )} = m e Q W = m{ h + ke + pe} = m e Q W = m h + m ke + m pe = m e Q W = H + KE + PE = E Q W = m {(h 2 h 1 ) + 1 2 (v 2 2 v 1 2 ) + g(z 2 z 1 )} (for flow systems) Q W = m { h + ke + pe} (for flow systems) Q W = m h + m ke + m pe (for flow systems) Q W = H + KE + PE (for flow systems) Ideal Gas Law: PV = nrt Isothermal Process: P 1V 1 = P 2V 2 = PV = Constant T 1 T 2 T Generalised Process: P 1 v 1 k = P 2 v 2 k = Pv k = Constant (k determined experimentally) Isentropic Process: P 1 v γ 1 = P 2 v γ 2 = Pv γ = Constant and γ = C p C v Avogadro s Principle: At STP (1 atm and 0 C) 1 kmol of an ideal gas will occupy 22.4m 3 Page 12 of 22
Volume of 1 kg of an ideal gas at STP is: 22.4 Average Relative Molecular Mass(M Wt) (m3 kg ) Volume of x kg of an ideal gas at STP is: 22.4 Average Relative Molecular Mass(M Wt) x(m3 ) The volume of x kg of an ideal gas at any temperature and pressure is given by: V(m 3 ) = x (22.4) (101.325 M Wt P (kpa) ) ( T(K) 273.15 ) Average Relative Molecular Mass (M Wt) of Air is 29 kg/kmol Average Relative Molecular Mass (M Wt) of Water (and steam) is 18 kg/kmol Average Relative Molecular Mass (M Wt) of Carbon Dioxide is 44 kg/kmol All symbols have their usual meanings and units (SI) Page 13 of 22
Fluid Flow Formula Sheet (All symbols have their usual meaning) Pressure = Force Exerted Area of Boundary or p = F A Mean Pressure, p = δf δa δf Pressure at a point, p = lim δa 0 δa = df da Absolute Pressure = Gauge Pressure + Atm Pressure P = ρg h For Newtonian Fluid: τ yx = μ ( dv x dy ) Power law for non-newtonian fluids: τ yx = m(γ yx) n The Bingham Plastic Model: τ yx = τ B B o + μ B (γ yx) for τ yx > τ o B γ yx = 0 for τ yx < τ o Page 14 of 22
Hydraulic Diameter (Equivalent Diameter) D h = 4 A c p 4 Flow Area = Wetted Perimeter The velocity profile in a circular pipe for laminar flow: u(r) = R2 4μ (dp r2 ) (1 dx R 2) Average velocity in a circular pipe for laminar flow: V avg = V = R2 8μ (dp dx ) u(r) = 2V avg (1 r2 R 2) u max = 2V avg Pressure drop for laminar flow in a circular pipe: P = P 1 P 2 = 8μLV avg R 2 = 32μLV avg D 2 Average velocity for laminar flow in an inclined pipe: Q = ( P ρgl sinθ)πd4 128μL Page 15 of 22
Volumetric flow rate for laminar flow in an inclined pipe: Q = ( P ρgl sinθ)πd4 128μL Pressure drop due to friction in a circular pipe: P f = 2f L D ρu 2 avg Head loss due to friction in a circular pipe: h f = P ρg = 2f L D 2 u avg g Pump work: W pump,l = Q P L = Qρgh f = m gh f The mechanical energy balance equation between two points 1 and 2: (P 2 P 1 ) ρ + (u 2 2 u 1 2 ) 2 + g(z 2 z 1 ) + F = W s m The mechanical energy balance equation between two points (1 and 2) both open to the atmosphere (in terms of head): P 1 ρg + u 2 1 2g + z 1 + h pump = P 2 ρg + u 2 2 2g + z 2 + h friction + h fittings Page 16 of 22
Fourier s Law of Heat Conduction: Heat Transfer Formula Sheet q x A = k dt dx Integrated Fourier Equation (Flat Slab): q A = k (T x 2 x 1 T 2 ) 1 q = T 1 T 2 x ka = T 1 T 2 R = Driving Force Resistance Integrated Fourier Equation (Hollow Cylinder): or where and 2πL q = k ln(r 2 r 1 ) (T 1 T 2 ) T 1 T 2 T 1 T 2 q = ka lm = = T 1 T 2 r 2 r 1 (r 2 r 1 ) (ka lm ) R A lm = (2πLr 2) (2πLr 1 ) ln ( 2πLr 2 2πlr 1 ) R = r 2 r ln ( r 2 1 r1 ) = ka lm 2πkL = A 2 A 1 ln ( A 2 A 1 ) Page 17 of 22
Integrated Fourier Equation (Hollow Sphere): q = 4πk(T 1 T 2 ) 1 1 = r 1 r 2 (1 r 1 1 r 2 T 1 T 2 ) (4πk) Integrated Fourier Equation (Plane Walls in Series): q = T 1 T 4 x A k A A + x B k B A + x C k C A T 1 T 4 = R A + R B + R C Integrated Fourier Equation (Multi-layer Cyclinders): Or where T 1 T 4 q = r 2 r 1 + r 3 r 2 + r 4 r 3 k A A Alm k B A Blm k C A Clm q = T 1 T 4 = T 1 T 4 R A + R B + R C R A Alm = A 2 A 1 ln ( A 2 A1 ), A Blm = A 3 A 2 ln ( A 3 A2 ), A Clm = A 4 A 3 ln ( A 4 A3 ) Page 18 of 22
Log Mean Temperature ( T lm ): 1 2 T c1 T h1 T c1 Cold Fluid Hot Fluid Cold Fluid T c2 T h2 T c2 T lm = T 2 T 1 ln ( T 2 T 1 ) = (T h 2 T c2 ) (T h1 T c1 ) ln ( (T h 2 T c2 ) (T h1 T c1 ) ) q = UA T lm = m cc Pc (T c2 T c1 ) = m hc Ph (T h2 T h1 ) Page 19 of 22
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Critical Thickness of insulation: Page 22 of 22