Instructor: Hacker Engineering 232 Sample Exam 3 Solutions Print your name neatly. If you forget to write your name, or if I can’t read your writing, you can lose up to 100 points. Answer all the questions that you can. Circle your answers. You must show your work. You will not receive credit for lucky guesses. Show your work as clearly as you can: if I can’t understand how you got an answer, I will not give you credit for it. Remember, I know how to solve the problem; and to make matters worse, I have a lot of training in following logical arguments! Warning: The definition of “little or no work” will be determined by the instructor, not the student. Grading rules. Each problem will be graded on a scale of 0 to 10 points. The algorithm is based on a series of steps that you should follow in solving the problem. You should try to learn these steps: in the future, you may find yourself facing difficult problems with no steps laid out for you to follow. On any problems, clarity is as important as the correct procedure and the correct answer. If you do not clearly label your steps, and your work within those steps, then I will grade your work as wrong. I will not waste time struggling to read an incoherent mess that is purported to be the solution; and after the exam is handed in and graded, I will not improve your grade based on your explanation of what you were trying to say. One purpose of these problems is to teach you how to lay out a logical argument that someone else with a technical background can follow. If in doubt, write it down! I refuse to look at it before you turn it in and tell you if there is enough detail! I also refuse to give you any hints, or tell you if you are on the right track. If you are unable to do this, then you should not and will not get credit for the problems. If you solve the wrong problem, you will receive 0 points, even if your solution to the wrong problem is correct. You are responsible for reading the problem correctly. If you think a problem is ambiguous, you should ask the instructor for clarification. For maximum partial credit be sure to write down: The basic problem set up (4 out of 10): (i) identify the control volume, (ii) what you are given, (iii) what you want, (iv) draw a picture, and (v) list any assumptions that you need to make to solve the problem: for example, “the pressure is constant”, “the system is a closed system”, etc. If there are multiple parts, be sure and label the answer for each part. If it is helpful draw a Tv and/or a Pv-diagram. The “Give up” option: If you have no idea how to solve the problem, you can write “Give up” and circle it instead of writing nonsense in hopes of getting partial credit. If I see “Give up”, I will ignore everything else that you’ve written and award you 3 out of 10 points for the problem. This is your reward for knowing what you don’t know, and for being honest about it. Thermodynamics 232, Exam 3 Copyright ©Wayne Hacker 2011. All rights reserved. 2 The three fundamental equations for a steady-state control volume with one inlet/one exit is m ˙ =m ˙i=m ˙e (Conservation of mass) (Equation 1) 1 ˙ cv − m∆ Q˙ cv = W ˙ h + V 2 + gz (Conservation of energy) 2 vm ˙ = AV (volumetric flow at each inlet and exit) (Equation 2) (Equation 3) The idea gas model for air is P v = Rair T, u = u(T ), h = h(T ), where P is pressure, v is specific volume, T is temperature (measured in kelvin), u is specific internal energy, h is specific enthalpy, and Rair kJ 8.314 R kmol · K ≈ 0.287 kJ , = = kg Mair kg · K 28.97 kmol where R is the universal gas constant, and Mair is the molecular mass. For moderate temperatures of T ≈ 200 K, to T ≈ 440 K, the specific heats cv and cp are approximately constant. Thus, ( ∆u ≈ cv ∆T , ∆h ≈ cp ∆T . Thermodynamics 232, Exam 3 Copyright ©Wayne Hacker 2011. All rights reserved. 3 Problem 1. (Diffuser) Air is flowing through a diffuser in a jet engine that is operating under steady-state conditions. At the inlet: Ti = 10◦ C, Pi = 80 kPa, Vi = 200 m/s, and the cross-sectional area of the inlet is Ai = 0.4 m2 . The air leaves the diffuser with a velocity that is very small compared with the inlet velocity (i.e., Ve Vi ). Assuming that we can model the air as an ideal gas (i.e., P v = Rair T , where Rair = R/Mair ), and that the velocity is normal to cross-sectional area at the inlet and exit, determine (a) the mass flow rate m ˙ through the diffuser and (b) the temperature of the air leaving the diffuser Te . Solution: We’ll start by writing down our control volume, what we’re given, what we want, and our assumptions. Control Volume: We’ll take our control volume around the diffuser. Recall a diffuser acts like and looks like a nozzle in reverse. Given: Want: Ti = 10◦ C (inlet temperature) P = 80 kPa (inlet pressure) i At the inlet: Vi = 200 m/s (inlet velocity) Ai = 4/10 m2 (inlet cross-sectional area) ( m ˙ e =? (mass-flow rate at exit) At the inlet: Te =? (exit temperature) Assumptions: 1. Steady-State Flow ⇒ d d mcv = Ecv = 0 dt dt 2. One inlet/One exit ⇒ m ˙i=m ˙e=m ˙ 3. Vi Ve ⇒ ∆KE ≈ 12 Vi2 4. Can ignore any effects in the change in gravitational P.E. 5. No moving mechanical parts in CV ⇒ ⇒ ∆PE = 0 ˙ cv = 0 W 6. Since the air is moving quickly, there is little time for significant heat transfer out of the device ⇒ Q˙ cv = 0 7. We’ll model the air as an ideal gas Notice that we are only given information at the inlet. Using fundamental equation 1 and 3 for a control volume with a single inlet/ single exit yields and equation for the mass flow rate m ˙ at the exit. m ˙e=m ˙i (fundamental equations 1) Ai Vi = (applying fundamental equation 3 at the inlet) vi Pi Ai Vi = (substituting for vi using the ideal gas law) Rair Ti (80 kPa)(4/10 m2 )(200 m/s) kg = ≈ 78.8 (0.287 kJ/(kg · K)((273 + 10) K) s Thermodynamics 232, Exam 3 Copyright ©Wayne Hacker 2011. All rights reserved. 4 ˙ e = 79kg/s . Thus, the answer to part (a) is m For part (b) we need to find Te . In general, this would require finding two independent properties of the system at the exit. However, since we have an ideal gas, part of the assumption of the ideal gas model is that the specific internal energy and specific enthalpy are independent of pressure. That is, they can be completely determined by temperature. In particular, he = h(Te ). Thus, we can use table A-22 to find Te , given he . This may require interpolation, since we’re essentially using the table of data to invert the function (i.e., Te = h−1 (he )). To find he we’ll need to apply fundamental equation 2, the conservation of energy for ˙ cv = ∆gz = 0. Under control volumes. From the assumptions, we have that Q˙ cv = W these conditions fundamental equation 2 becomes 1 2 0 = 0 − m∆ ˙ h+ V +0 2 1 ÷(−m) ˙ −−−−−−−→ (he − hi ) + (Ve2 − Vi2 ) = 0 2 1 2 Solve for he −−−−−−−−−−→ he = hi + (Vi − Ve2 ) 2 1 ≈ hi + Vi2 (By assumption: Ve Vi ) 2 1 = h(Ti ) + Vi2 (For an ideal gas: h = h(T )) . 2 Thus, 1 he ≡ h(Te ) = h(Ti ) + Vi2 . 2 We are given the value of Vi , and we can use table A-22 to find h(Ti ) = h(283K). From table A-22 we see that for a moderate temperature range of 200-440 K, the temperatures and the enthalpies are approximately the same value. That there is a linear relationship between h and T should be no surprise, since for an ideal gas over moderate temperature ranges we have ∆h = cp ∆T , where cp ≈ 1 kJ/(kg · K). 1 he = hi + Vi2 2 kJ m2 kg kJ 1 = 283 + (200)2 2 kg 2 s 3 m2 10 2 s kJ kJ = (283 + 20) = 303 . kg kg Since cp ≈ 1 over the range of temperatures considered, which is also verified by the tables, we find that he = cp Te ≈ Te (with units of enthalphy). Replacing he by Te (with units of he ) we find kJ 303 he kg Te = ≈ = 303 K = 30◦ C . kJ cp kg · K Thermodynamics 232, Exam 3 Copyright ©Wayne Hacker 2011. All rights reserved. 5 Problem 2. (Steam Turbine) ˙ out = 5000 kilowatts. The power output of an adiabatic steam turbine is known to be W Pi = 2 MPa (inlet pressure) ◦ Ti = 400 C (inlet temperature) The inlet conditions are: Vi = 50 m/s (inlet velocity) zi = 10 m (inlet height) m ˙ i = 5.73 kg/s (inlet mass flow) Pe = 15 kPa x = .9 e The exit conditions are: V e = 180 m/s ze = 6 m (exit pressure) (exit quality) (exit velocity) (exit height) Assumptions: The operating conditions for the turbine mechanism are 1. steady-state operating conditions 2. one inlet/one exit turbine 3. adiabatic turbine mechanism ⇒ No heat flow in or out of turbine Using the above information answer the following questions: (a) Sketch the location of state at the inlet and exit on a Pv-diagram. (b) Determine the work done per unit of mass of the steam flowing through the turbine. (c) In order to better understand the energy distribution throughout the turbine, compute the change in enthalpy ∆h, kinetic energy ∆KE, and potential energy ∆PE, for the turbine. Compare the magnitudes of these quantities and rank them from most important to least important effects. Thermodynamics 232, Exam 3 Copyright ©Wayne Hacker 2011. All rights reserved. 6 Problem 3. (Compressor) Air enters a rotating one inlet/one exit compressor operating at a steady-state. The compressor loses heat at a rate of Qcv = −1 KJ/s. The technicians have given the following information at the inlet and exit of the compressor: Pi = 1 bar (inlet pressure) T = 300 K (inlet temperature) i The inlet conditions are: Vi = 10 m/s (inlet velocity) Ai = 0.1 m2 (inlet cross-sectional area) Pe = 10 bar The exit conditions are: Te = 400 K Ve = 1 m/s (exit pressure) (exit temperature) (exit velocity) ˙ cv that must be supplied to Using the ideal-gas model, determine the amount of power W the compressor. Thermodynamics 232, Exam 3 Copyright ©Wayne Hacker 2011. All rights reserved. 7 Problem 4. (The Shell-and Tube Heat Exchanger) An isolated hotel in the wilds of Wyoming uses a geothermal heat exchanger to provide hot water for the hotels needs. In addition to using the geothermal water to heat the tap water, the hotel would like to use the leftover geothermal water to heat other devices. The hotel has hired you to figure out the temperature of the geothermal water at the exit of the inner tube T1,e . The inner cylinder contains the hot geothermal water (the heating source) that is moving at a steady mass flow rate of m ˙ 1 = 0.3kg/s through the inner cylinder. The initial temperature at the inlet of the geothermal water is known to be T1,i = 140◦ C (assuming the it is initially at the underground temperature), and the specific heat of the geothermal water is known to be c1p = 4.31kJ/(kg·◦ C). The outer cylinder contains the cold tap water that is to be heated. The cold water in the outer cylinder is also moving at a steady mass flow rate of m ˙ 2 = 0.2kg/s. The temperature at the inlet and exit of outer cylindrical pipe is measure at T2,i = 25◦ C and T2,e = 60◦ C, respectively, and the specific heat of water is c2p = 4.18kJ/(kg·◦ C). The outer boundary of the outer cylinder is well-insulated so that only a negligible amount of energy is lost to the surroundings. There are no moving mechanical parts in the device and it is assumed to operate under steady-state conditions. Each pipe can be treated as a separate system having one inlet and one exit. We shall also ignore any change in kinetic, or potential energies in the system as the fluid passes through each pipe. Under these conditions and assumptions derive a formula the exit temperature of the geothermal water. Solution: given: ˙ in = 0.3kg/s m Inner cylinder: Tin,i = 140◦ C cp,in = 4.31kJ/(kg·◦ C) m ˙ out = 0.2kg/s T ◦ out,i = 25 C Outer cylinder: Tout,e = 60◦ C cp,out = 4.18kJ/(kg·◦ C) want: T1,e =? Note: As we will soon see, in order to find the temperature at the exit we’ll need to solve for the rate of heat transfer in the heat exchanger, which will require taking two separate control volumes: one around the inner tube and one around the outer tube. assumptions: • We will assume steady-state operating conditions ⇒ dEcv = 0. dt ˙ cv = 0. • No moving parts ⇒ W • Insulated unit ⇒ Q˙ cv,ext = 0. • Ignore any changes in K.E. and P.E. ⇒ ∆PE = ∆KE = 0. • The fluid is incompressible ⇒ u = u(T ), ρ = constant, and cp ≈ constant. Thermodynamics 232, Exam 3 Copyright ©Wayne Hacker 2011. All rights reserved. 8 The steady-state one-inlet/one-exit assumption gives: m ˙ 1,i = m ˙ 1,e = m ˙1 m ˙ 2,i = m ˙ 2,e = m ˙2 (inner tube condition) , (outer tube condition) . (0.1a) (0.1b) For each tube, the first law of thermodynamics (in the energy rate form) reduces to Q˙ cv = m ˙ ∆h = m ˙ cp ∆T Q˙ cv = m ˙ cp (Te − Ti ) . (0.2) Equation (0.2) will be our governing equation for modeling the heat flow in each pipe separately. That is, if we take our control volumes to be the inner and outer tubes, respectively, then we have Q˙ 1,cv = m ˙ 1 c1p (T1,e − T1,i ) (inner tube governing equation) , Q˙ 2,cv = m ˙ 2 c2p (T2,e − T2,i ) (outer tube governing equation) . (0.3a) (0.3b) Since the device is insulated, no heat escapes through the outer walls of the outer tube (i.e., Q˙ cv,ext = 0). Thus, the energy transfer occurs between the inner wall of the outer tube and the outer wall of the inner tube. That is, Q˙ 1,cv = −Q˙ 2,cv . Substituting this result into equation (0.3a), and equating this resulting equation with (0.3b) yields m ˙ 2 c2p (T2,e − T2,i ) = Q˙ 2,cv = −m ˙ 1 c1p (T1,e − T1,i ) m ˙ 1 c1p (T1,e − T1,i ) ⇒ T2,e = T2,i − m ˙ 2 c2p Substituting in the above given values into equation (0.4) gives T2,e ≈ 117.4◦ C. (0.4)

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