(b) 1. Look up c p for air in Table A.6. c p = 1004 J/kg K 2. Use equation (1) and given and looked up values to find s 2 s 1.

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1 Problem 1 Given: Air cooled where: T 1 = 858K, P 1 = P = 4.5 MPa gage, T = 15 o C = 88K Find: (a) Show process on a T-s diagram (b) Calculate change in specific entropy if air is an ideal gas (c) Evaluate heat transferred per unit mass if process is reversible Assumptions: (1) Constant pressure (1) For an ideal gas: δq () For a reversible process: Tds = = δq m (3) Gibbs equation form : Tds = dh vdp (4) For an ideal gas: dh = c dt (a) p (b) 1. Look up c p for air in Table A.6. c p = 1004 J/kg K. Use equation (1) and given and looked up values to find s s 1. T 88K s s1 = c p ln R ln(1) = (1004J / kgk)ln = 1.1kJ / kgk T 858K 1 (c) 3. Plug equation (3) into equation () and find: δ q = Tds = dh vdp 4. Plug equation (4) into this new equation to find: δ q = c pdt vdp 5. Plug in values to find q. q = c p T T ) v( p p ) = 1004J / kgk (88 858) K = 57kJ / kg ( 1 1

2 Problem Given: A closed system: gas undergoes a cycle made up of the following processes: 1- reversible isothermal compression, -3 reversible constant volume heating, 3-4 reversible constant pressure expansion, and 4-1 reversible adiabatic expansion. Find: (a) Sketch p-v and T-s diagrams of the cycle (b) State whether each of the following quantities is positive, zero, negative, or δ W, δq, ds, du, indeterminate in sign: dh Assumptions: (1) Ideal gas (1) δ W + δq = du (a) (b) 1. Use equation (1) and find: δ W δq + = du. So: 1- doing work on system: +work, and -3, 3-4 system doing work: -work. So Net W is negative, and net Q is positive. δ W dh = negative, δq = positive, ds = 0, du = 0, = 0

3 Problem 3 Given: Jet transport aircraft cruising with: M = 0.85, and z = 1.5 km Find: (a) The stagnation pressure sensed by a probe on the aircraft (b) Speed calculated from the incompressible Bernoulli equation (c) Percentage error in this speed compared to the true speed Assumptions: (1) Air is ideal gas (1) Local isentropic stagnation conditions: p V () Bernoulli: + + gz = cons tant ρ (3) For ideal gas, speed of sound: c = V / M = (krt) 1/ (a) 1. Look up properties of air at z = 1.5 km in Table A.3. k = 1.4, T = 16.7 K (b) p p SL ρ = where psl = 101.3kPa, = where ρ SL = 1.5kg / m ρ. Calculate pressure and density values: p = ρ = SL (101.3kPa) = 17.99kPa, = 0.361(1.5kg / m ) 0.89kg / m 3. Plug these numbers and given into equation (1) and rearrange to find p 0. k /( k 1) 1.4 /(0.4) k p0 = p 1+ M = 19.99kPa 1 (0.85) = kPa 4. Use equation () between known pressure and stagnation conditions: 3 (c) p V 0 0 p V gz + + = + ρ ρ V = 74.1m / s + gz V = ( p 0 ρ p) ( ) kpa = kg / m 5. Use a rearranged equation (3) so calculate actual plane velocity for compressible flow: V = M (krt) 1/ = (0.85) [(1.4)(87 J / kgk)(16.7 K)] 1/ = 50.8 m/s Vactual Vcalculated Percent error = x100 % = x100% = 9% error V 50.8 actual

4 Problem 4 Given: Air enters a long, insulated duct as shown below. Find: (a) Local isentropic stagnation conditions at the inlet section, T 0,1, p 0,1 (b) Local isentropic stagnation conditions at the outlet section, T 0,, p 0, (c) Change in specific entropy along the duct, s s 1 (d) Plot static and stagnation state points on a T-s diagram Assumptions: (1) Air is ideal gas (1) Local isentropic stagnation conditions: () For ideal gas: (a) 1. Plug in values at inlet to the temperature equation from equations (1). (b) k T0,1 = T1 1 + M 1 = 86K 1 (0.) = 88. 3K +. Plug in values at inlet to the pressure equation from equations (1). k /( k 1) 1.4 /(0.4) k p0,1 = p1 1+ M 1 = 98.5kPa 1 (0.) = 101.3kPa + 3. Plug in values at outlet to the temperature equation from equations (1). k T0, = T 1 + M = 68.9K 1 (0.6) = 88. 3K + 4. Plug in values at outlet to the pressure equation from equations (1). p k /( k 1) 1.4 /(0.4) 0, = = k 1 p 1+ M 0.4 = 31.3kPa 1 + (0.6) 39.9kPa

5 (c) 5. Look up c p for air in Table A.6. c p = 1004 J/kg K 6. Use values at inlet and outlet and equation (3) to find change in entropy s s1 = (1004J / kgk) ln (87J / kgk) ln = 67J / kgk (d)

6 Problem 5 Given: Bicycle tire inflated to a p = 800 kpa (gage), T = 37 o C Find: (a) Critical T (T*) and p (p*) (b) Speed of sound within tire Assumptions: (1) Air is ideal gas (1) Local isentropic stagnation conditions: () For ideal gas, speed of sound: c = (krt) 1/ (a) 1. Since there is no flow after the tire is pumped up, the given values are stagnation (b) conditions. T = T 0 = 37 o C + 73 = 310 K, and p = p 0 = 800 kpa (gage). For critical conditions, M = 1. Plug this and T 0 into nd of equations (1). 1 k T* = T0 1+ M 1 = 310K 1 (1) = 58. 3K + 3. For critical conditions, M = 1. Plug this and p 0 into 1 st of equations (1). p* = k /( k 1) k p0 1+ M 1 800kPa 1 (1) = = + 4. Again at critical conditions: /(0.4) c* = (krt*) 1/ = [(1.4)(87 J/kgK)(58.3 K)] 1/ = 3 m/s 43kPa

7 Problem 6 Given: Airplane traveling at: z = 3 km, M = 1.35, T = 303 K, U wind = 10 m/s Find: (a) Speed of the aircraft (b) Time between seeing the aircraft and hearing it Assumptions: (1) Air is ideal gas (1) Speed of sound: c = (krt) 1/ = V / M () Mach cone: (a) 1. Rearrange and solve equation (1) for V. (b) V = cm = (krt) 1/ M = (1.35)[(1.4)(87 J/kgK)(303 K)] 1/ = 471 m/s. Calculate the mach angle: α = 1 sin 1 = 1.35 o The distance the plane travels before we hear it = D = V earth t. Where V earth = V plane V wind = 471 m/s 10 m/s = 461 m/s. o And D = 3000 m / tan( α) = 3000m / tan(47.8 ) = 70m 4. Now rearrange above equation and solve for the time. t = D / V earth = 70 m / 461 m/s = 5.9 s

8 Problem 7 Given: Data on an adiabatic air compressor input: P 1 = 101kPa T 1 = (0+73)K output: P, gage = 650kPa T = (85+73)K Find: a) If this compressor is feasible b) Sketch the process on a Ts diagram Assumptions: (1) Air is an ideal gas here Ideal gasses: a) o 1. Look up c p and R b for air at 0 C in Table A.6.. Plug these and our givens into the equation for s s We know that entropy must be increasing for any real adiabatic process, so the compressor is feasible if s s 1 > 0. (And it is!) b)

9 Fox

10 Problem 8 Given: Steady flow of air through a turbine at m & = 0.5 kg / s Inlet: P 1 =.0 MPa T 1 = 1300 o C V 1 = 0 Outlet: P = 101 kpa T = 500 o C V = 00 m/s Find: a) Power produced by the turbine b) Label state points on a Ts diagram Assumptions: (1) Isentropic expansion through the turbine () Horizontal turbine (z 1 = z ) (3) Q & = 0 Isentropic (reversible adiabatic) process: Ideal gasses: Other equations: First law of thermodynamics: a) V V1 Q & W& = m& h + + gz h1 + + gz1 1. Use assumptions and 3 to simplify the first law of thermo.. Use the equation for an ideal gas with constant specific heats, h -h Look up c p in A Plug back into the eqn. for W & (will be positive because here defined as work out). b)

11 Fox (*note: difference in work definition)

12 Problem 9 Given: Balloon inflated isothermally from r = 5in to r = 7in. Flow is Q = 0.10 cfm of standard air (59 o F and 14.7psia). Balloon skin tension is σ = ka where k = 00 lbf/ft 3, and A = surface area of balloon. Find: a) Time required to inflate balloon 3 Assumptions: (1) Standard air ( ρ = lbm / ft ) () Ideal gas Ideal gasses: p = ρrt Other equations: Δm m& = ρq Δ t = m = ρv m& a) 1. Do a force balance on the balloon, where balloon skin tension balances pressure.. Calculate the pressure from this balance at the initial r = 5in. 3. Calculate the density from the first equation. 4. Find m for this first position. 5. Do 1-4 for balloon when r = 7in. 6. Calculate Δm = m 7 m Now find Δt using Δm. Fox

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14 Problem 10 Given: Gas storage reservoir with helium at T 0 = 000 K and p 0 = 6.0 MPa (gage). Find: a) T* for this stagnation condition b) P* for this stagnation condition c) V* for this stagnation condition Assumptions: (1) Ideal gas Critical conditions: (M = 1) a) 1. Look up the constant k for helium in table A.6.. Plug this and givens into the equation for T*. b) 1. Plug k and given into equation for p*. c) 1. Plug givens and k into eqn. for V*. Fox