ME 201 Thermodynamics

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1 ME 0 Thermodynamics Solutions First Law Practice Problems. Consider a balloon that has been blown up inside a building and has been allowed to come to equilibrium with the inside temperature of 5 C and inside pressure of 00 kpa. The diameter of the balloon is measured and found to be 0.3 m. The balloon is then taken outside and allowed to come to equilibrium with the outside temperature of -5 C and outside pressure of 00 kpa. Determine the boundary work, heat transfer, and final balloon diameter. System Type: Closed System Substance Type: Ideal gas Process: Isobaric Final State: Fixed Q UNKNOWN W bnd UNKNOWN Conservation of Mass: m m st Law: m(u - u ) Q - W Boundary Work: P(v - v) State State T 5 C 98 K T -5 C 68 K P 00 kpa P 00 kpa D 0.3 m D 0.55 m V.50 x 0 m 3 V.034 x 0 m 3 m.345 x 0 kg m.345 x 0 kg u 0 kj/kg u -.6 kj/kg φ 6.7 kj/(kg K) φ kj/(kg K) v m 3 /kg v m 3 /kg Italicized values are from ideal gas relations or tables. Bold values are calculated. Approach: We begin by evaluating the properties at state and state by using the air tables and the ideal gas law. We can then use our boundary work equation to calculate the boundary work. Finally, we use the conservation of energy to determine the heat transfer. We start by determining our total volume at state. The volume of a sphere is given by V πr.50 x 0 m 3 Next we determine our mass. Using the ideal gas law we have P V (00)(.50 x 0 ) m.345 x 0 kg RT (0.87)(98) (.50 x 0 ) (.345 x 0 ) V v m m 3 /kg bnd

2 ME 0 Thermodynamics We can go to the air tables and find u 0 kj/kg φ 6.7 kj/(kg K) u -.6 kj/kg φ kj/(kg K) The specific volume at state must be given by the ideal gas law, so that RT (0.87)(68) 3 v m /kg P (00) The total final volume is then V m v (.345 x 0 )(0.769).034 x 0 m 3 The final diameter is /3 3V D 0.55 m 4 π The boundary work is calculated Wbnd P(V - V ) (00)(.034 x x 0 ) kj and the heat transfer is Q m(u - u ) + W kj bnd (.345 x 0 )(-.6-0) + (-0.06). Air at 800 K and 800 kpa enters an ideal turbine at.3 kg/s. The power output required of this turbine is 700 kw. Determine the exhaust temperature and pressure. System Type: Control Volume System Substance Type: Ideal gas Process: Isentropic Final State: Unknown Q 0 W sh 700 kw Conservation of Mass: m& m&.3 kg/s m& h - h - W& st Law: ( ) State State T 800 K T 55.7 K P 800 kpa P 44 kpa h 70.5 kj/kg h kj/kg φ kj/(kg K) φ kj/(kg K) Italicized values are from ideal gas relations or tables. Bold values are calculated. Approach: We begin by evaluating the properties at state by using the air tables and the ideal gas law. We can then use the conservation of energy to determine the enthalpy at state. The air tables will give T. Finally, the isentropic relation will give the exit pressure At state we can go to the air tables and find h 70.5 kj/kg φ kj/(kg K) From the first law we solve for h W& 700 h h - sh kj/kg m&.3 sh

3 ME 0 Thermodynamics The air tables can then give T 55.7 K φ kj/(kg K) The isentropic relation is P s 0 φ - φ R ln P Solving for P gives φ P P exp φ (800)exp R kpa 3. Refrigerant- as saturated liquid at 3 C enters a valve and exits at 0.5 MPa. What is the fluid phase at the exit? Solution System Type: Control Volume System Substance Type: phase change Process: Isenthalpic Final State: Unknown Q 0 st Law: h - h 0 State State T 3 C T C P MPa P 0.5 MPa h kj/kg h kj/kg Phase: sat.liq. Phase: phase, x 0.30 Italicized values are fromr4a tables. Bold values are calculated. Approach: We begin by evaluating the properties at state by using the refrigerant tables. We can then use the conservation of energy to determine the enthalpy at state. The refrigerant tables will give the fluid phase. At state we can go to the refrigerant tables and find h kj/kg P MPa From the first law we solve for h h h kj/kg At 0.5 MPa, we find from the tables h f 7.86 kj/kg h g kj/kg Since h is between these two values we have a two phase mixture at C and with quality h - h x f 0.30 h - h g f 3

4 ME 0 Thermodynamics 4. A tank contains Refrigerant- as saturated vapor at 00 kpa. There is heat transfer to the tank of 66 kj/kg. Determine the final temperature and pressure. System Type: Closed System Substance Type: phase change Process: Isotropic Final State: Unknown q 66 kj/kg W bnd 0 Conservation of Mass: m m st Law: u - u q State State T C T 80 C P 00 kpa P 50 kpa u kj/kg u 4.39 kj/kg v m 3 /kg v m 3 /kg phase: sat.vap. phase: sup.vap. Italicized values are from tables. Bold values are calculated. Approach: We begin by evaluating the properties at state by using the refrigerant tables. We can then use isotropic process to determine the specific volume at state. Next, we use the conservation of energy to determine the final internal energy. Finally, we go to the refrigerant tables and find the T and P. We can go to the refrigerant tables and find u kj/kg T C v m 3 /kg Since our process is isotropic, we have v v m 3 /kg The final internal energy is given by the st law or u q + u kj/kg So now we must go to the refrigerant tables and find the T and P that will correspond to these values of u and v. Since we had saturated vapor and we added heat, we will assume that we have gone to superheated vapor. Then scanning the tables we find that at 0.5 MPa and 80 C, we have u 7.67 kj/kg v m 3 /kg which is close enough. 5. Consider a 30 gallon hot water heater which is to provide water a 80 F and 0 psia. If a typical shower consumes 3 gallons/minute, last 5 minutes, and requires that the hot water should stay above 60 F, determine the heat transfer rate required. Water is supplied at 63 F and psia and gallons/minute. 4

5 ME 0 Thermodynamics System Type: Transient System Substance Type: Incompressible Process: Isotropic Final State: Fixed Inlet State: Fixed Exit State: Fixed Q UNKNOWN W bnd 0 m - m Conserv. of Mass: m& in - m& out t m u - mu st Law: m& h - m& h + & in in out out t State State In State Out State T 80 F 8. C T in 63 F 7. C T out 70 F 76.7 C T 60 F 7. C P 0 psia 37.9 kpa P in psia 5.7 kpa P out 0 psia 37.9 kpa P 0 psia 37.9 kpa u kj/kg h in 7.7 kj/kg h out kj/kg u kj/kg v m 3 /kg v in m 3 /kg v out m 3 /kg v m 3 /k m 0. kg m in 0.3 kg/s m out 0.89 kg/s m kg Note that we have used the average temperature at state and to fix our outlet temperature Italicized values are from steam tables. Bold values are calculated. Approach: We begin by evaluating the properties at all states by using the steam tables. We can then use the conservation of mass to determine the mass at state. Finally the st law is used to determine the heat transfer rate. Even though we will treat water as an incompressible substance in this case, we can use the steam tables to determine the specific volume and internal energy since for an incompressible substance they only depend on temperature, we can take the values for saturated liquid at the given temperature. For the enthalpy, we can use the enthalpy of saturated liquid and then bring it up to the appropriate pressure or h incompressible sub. h f + v f (P-P sat ) Our values are then entered on the table. To obtain the masses and mass flows we convert gallons to m 3 and divide by the specific volume. These values are entered on the table. The final state mass is given by conservation of mass or m t( m& in - m& out ) + m (5)(60)( ) kg Then the heat transfer rate becomes m u - m u Q& m& inhin + m& outhout t (50.80)(97.34) (0.)(343.88) (5)(60) 6.39 kw 90,046 Btu/hr - (0.3)(7.7) Q + (0.89)(30.95) 5

6 ME 0 Thermodynamics 6. A piston cylinder system contains gas at 300 K, 500 kpa, and 0.03 liters. The gas then undergoes a polytropic expansion with a polytropic exponent of.5 to 0.3 liters. Compare the work performed in kj for air as the gas versus hydrogen as the gas. We start this problem by working in air and then in hydrogen System Type: Closed System Substance Type: Ideal gas Process: Polytropic with n.5 Final State: UNKNOWN Q UNKNOWN (but not asked for) W bnd UNKNOWN Conservation of Mass: m m st Law: m(u - u ) Q - W Boundary Work: P v bnd - P v Wbnd (- n) State State T 300 K T NA P 500 kpa P 6.8 kpa V 3 x 0-5 m 3 V 3 x 0-4 m 3 Italicized values are from ideal gas relations. Bold values are calculated. Approach: At state we can determine the pressure from the polytropic relation. We can then use our boundary work equation to calculate the boundary work From the polytropic relationship we have P or solving for P n V P V V 0.03 P P (300) V 0.3 The boundary work is then given by n n kpa (6.8)(3 x 0 ) - (300)(3 x 0 ) Wbnd 0.34 kj (-.5) Since the calculation above never used the fact that air was our ideal gas, we will obtain the same boundary work for hydrogen as the gas. 7. In an open feedwater, subcooled liquid water is heated to saturated liquid by mixing directly with steam. In a given situation water at 0 MPa and 00 C enters an open feedwater heater at kg/s. Steam at 0 MPa and 350 C is available to be added to this water to produce saturated liquid at 0 MPa. How much steam in kg/s must be added? 6

7 ME 0 Thermodynamics System Type: Control Volume System Substance Type: Water (phase change substance) Process: Isobaric Inlet State: Fixed Exit State: Fixed Q 0 Conserv. of Mass: m& in - + m& in- - m& out 0 st Law: m& h + m& h - m& h 0 in - in- in- in- out out State In- State In- State Out T in- 00 C T in- 350 C T out 3 C P in- 0 MPa P in- 0 MPa P out 0 MPa h in kj/kg h in kj/kg h out kj/kg m& in- kg/s m& in kg/s m& out kg/s phase: sub.liq. phase: sup.vap. phase: sat.liq. Italicized values are from steam tables. Bold values are calculated. Approach: We begin by evaluating the properties at all states by using the steam tables. We can then use the conservation of mass to algebraically solve for the exit mass flow rate. Substituting this into the st law steam flow rate can be determined. The enthalpies are read from the steam tables and entered into our table above. Solving for m out from conservation of mass m& out m& in- + m& in- Substituting into the energy equation m& h + m& h - m& + m& h 0 ( ) in - in- in- in- in- in- out Solving for m in- m& ( h - h ) m& ( ) in- in- out in - hout - hin 4.37 kg 8. Most of the time during the winter Dr. Somerton turns down the thermostat to 50 F when he leaves in the morning. When he is in the house he likes to have the temperature at 68 F. The house may be considered to be composed of air, occupying a volume of 0,000 ft 3, and structural material (mostly wood with c P.76 kj/(kg K)) of,000 lb m. Determine the total heat transfer required to bring the house up to 68 F. What fraction of this total goes to heating up the air and what fraction goes to heating up the structural material? 7

8 ME 0 Thermodynamics System Type: Closed System Substance Type: Ideal gas (air) and Incompressible (wood) Process: Isobaric Final State: Fixed Q UNKNOWN W bnd 0 Conservation of Mass: m m st Law: U - U Q State State T 50 F 0 C T 68 F 0 C V,a 0,000 ft m 3 V,a 0,000 ft m 3 m,a kg m,a kg m,w,000 lb m kg m,w,000 lb m kg Italicized values are from ideal gas relations. Bold values are calculated. Approach: We begin by determining the mass of air in the house. Then we can use the st law to calculate the heat transfer required. We determine our mass of air assuming a pressure of 4.7 psia. Using the ideal gas law we have P V,a (00)(83.) m, a kg RT (0.87)(83) We can calculate the heat transfer from the energy equation U - U m c (T - T ) + m c (T - T ) Q a v,a w P,w (348.7)(0.79)(0-0) + (4989.5)(.76)(0-0) 89,794 kj 85,65 Btu of which 98% goes into the structural material and % into the air. 9. Engine oil enters an ideal pump at 00 F and psia and leaves at 7 psia. The oil flow rate is 0.04 lb m /s. The pump inlet has a diameter of 5 inches and the pump outlet has a diameter of inches. What pumping power is required? 8

9 ME 0 Thermodynamics System Type: Control Volume System Substance Type: Incompressible Process: Isentropic, can't neglect KE Final State: Unknown Q 0 W sh UNKNOWN Conservation of Mass: m& m& 0.04 lbm/s r r v v st Law: m& h + - h - - W& sh State State T 00 F 37.8 C T 37.8 C 00 F P psia 8.7 kpa P 7 psia 7. kpa v r ft/s v r 0.03 ft/s Bold values are calculated. Approach: We begin by using our process description to fix state. Then from the diameter and mass flow information the velocities are obtained. Finally, the st law is used to calculate the power. Since the process is isentropic and we have an incompressible substance, we write T s - s cp,avg ln T which yields T T 37.8 C The velocities can be calculated from one of our continuity relations r m& m& v ρ A c πd ρ 4 Using a density for engine oil of 04 kg/m 3, we find v r 0.08 m/s v r 0.08 m/s The enthalpy change for an incompressible substance is given by - h c (T T ) + v (P P ) h P,avg avg Then the shaft work is calculated (using the density rather than the specific volume) r r P - P v - v W& - m sh & + ρ (0.08) - (0.08) kw.93 Btu/hr - (0.08) 0 3 9

10 ME 0 Thermodynamics 0. The exhaust process for an internal combustion engine may be modeled as transient system undergoing an isobaric process with boundary work. Just before the exhaust valve opens the cylinder of 0.5 liters contains air at 00 kpa and 500 K. At the end of exhaust the volume is liters. Assume that the process is adiabatic. Determine the final temperature and mass and the boundary work. System Type: Transient System Substance Type: Ideal gas Process: Isobaric Outlet State: UNKNOWN Final State: UNKNOWN Q 0 W bnd P(V -V ) Conserv. of Mass: m -m - m out st Law: m u -m u - m out h out - W bnd State State Out State T 500 K T out T P 00 kpa P out 00 kpa P 00 kpa V 5 x 0-4 m 3 V out NA V 8.33 x 0-5 m 3 Bold values are calculated. Approach: We begin by calculating the boundary work. We then use the ideal gas law and constant specific heat to determine the enthalpy and internal energies. The boundary work is then W bnd P(V -V ) (00)[8.33 x x 0-4 ] kj Now we want to solve for the final temperature, but we note that in the first law four different things depend on T : u, m, h out, and m out. We can eliminate m out with the help of our conservation of mass so that m u -m u - (m -m )h out - W bnd Since the exhaust air temperature is changing during the process, we will use our linear average approach to determine h out, or T + T hout hair at The mass at the final state will be given by PV m RT So substituting RT PV T + T u mu - m hair at - Wbnd PV RT When we consider that the air tables provide the relationships among u and T and h and T, we see that we do have a well posed problem, but we can't do the algebra, so we will have to solve the problem by iteration. Our process will be Guess T Evaluate h air and u from the air tables Calculate u from the above equation 0

11 ME 0 Thermodynamics Compare the u at the guessed value of T to the calculated value of u Re-guess T and repeat the process until the difference between the values of u at the guessed T and the value of the calculated u is negligible I set this up in an Excel spreadsheet and the results are shown below T,guess u,guess T out h out m u u,guess -u,calc E E E So it would appear that our final temperature is about 65 K.

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