Chapter 4: Energy Equation for a Control Volume BITS Pilani Hyderabad Campus Problem 4.12 An empty bath tub has its drain closed and is being filled with water from the faucet at a rate of 10 kg/min. After 10 min, the tub is filled up, the drain is opened and 4 kg/min flows out; at the same time, the inlet flow is reduced to 2 kg/min. Determine the time from the very beginning when the tub will be empty. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-2 Solution During the first 10 min, we have ππππ£ ππ‘ = πππ = 10 ππ/πππ, and Δπ = πππ Δπ‘1 = 10 × 10 = 100 ππ Thus, we end up with 100 ππ after 10 min. For the remaining time period, we have, ππππ£ = πππ − πππ’π‘ = 2 − 4 = −2 ππ/πππ ππ‘ The overall mass flowing out is given as: Δπ2 −100 Δπ2 = ππππ‘ Δπ‘2 ⇒ Δπ‘2 = = = 50 πππ ππππ‘ −2 Thus, the total time required is given as: Δπ‘π‘ππ‘ = Δπ‘1 + Δπ‘2 = 10 + 50 = 60 πππ BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-3 Problem 4.13 Air at 25°πΆ and 101 πππ flows in a 100 × 150 ππ rectangular duct in a heating system. The mass flow rate is 0.015 ππ/π . What is the velocity of the air flowing in the duct, and what is the volumetric flow rate? BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-4 Solution Assume a constant velocity across the cross sectional area of the duct. The cross sectional area of the duct is given as π΄ = 100 × 150 × 10−6 π2 = 0.015 π2 As air behaves as an ideal gas, π π 0.287 × 298.15 π£= = = 0.8472 π3 /ππ π 101 The volumetric flow rate is given as π = ππ£ = 0.015 × 0.8472 = 0.01271 π3 /π ππ This volumetric flow rate is given as the product of the cross-sectional area and the velocity of flow, which gives, π½= π 0.01271 = = 0.84733 π/π π΄ 0.015 Thus, the air flowing through the duct is 0.84733 m/s, and the volumetric flow rate is 0.01271 m3/sec. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-5 Problem 4.15 Nitrogen gas flowing in a 50 mm diameter pipe at 300 K and 200 kPa, at the rate of 0.05 kg/s, encounters a partially closed valve. If there is a pressure drop of 100 kPa across the valve and joints and essentially no temperature change, what are the velocities upstream and downstream of the valve? BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-6 Solution The area of cross section of the pipe is given as π π × π 2 = × 0.052 = 0.0019635 π2 4 4 From table B.6.2., the specific volume of Nitrogen at the inlet of the valve is given as π΄= π£π = 0.44503 π3 /ππ Similarly, the specific volume of nitrogen at the outlet of the valve is given as π£π = 0.89023 π3 /ππ Thus, the velocity at the inlet of the valve is given as ππ£π 0.05 × 0.44503 ππ = = = 11.33257 π/π π΄ 0.0019635 And the velocity at the exit of the valve is given as ππ = ππ£π 0.05 × 0.89023 = = 22.66947 π/π π΄ 0.0019635 Thus, the nitrogen flows at 11.33 m/s at the inlet of the valve and 22.67 m/s at the exit of the valve. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-7 Problem 4.17 Nitrogen gas flows into a convergent nozzle at 200 πππ, 400 πΎ and a very low velocity. It flows out of the nozzle at 100 πππ and 300 πΎ. If the nozzle is insulated, find the exit velocity. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-8 Solution As the nozzle is insulated, the process is an adiabatic process. Thus, from the continuity eqn at steady state, πππ = πππ’π‘ = π Thus, from the energy equation at steady state, 1 1 β1 + 2 π½12 = β2 + 2 π½22 due to adiabatic process, π = 0, and π = 0 As the inlet velocity is negligible, π½1 = 0. Thus, 1 2 β1 = β2 + π½2 2 Thus, π½22 = 2 β1 − β2 = 2 × 1000 × (415.31 − 311.16) ⇒ π½22 = 208300 ⇒ π½2 = 456.399 π/π Thus, the exit velocity of Nitrogen is 456.399 m/s. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-9 Problem 4.19 The front of a jet engine acts as a diffuser, receiving air at 900 km/h, -5 °C, and 50 kPa, bringing it to 80 m/s relative to the engine before entering the compressor (see Fig.). If the flow area is reduced to 80% of the inlet area, find the temperature and pressure in the compressor inlet. Assume constant specific heats and ideal gas behavior. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-10 Solution Here, the control volume is the diffuser, at steady state, with 1 inlet and 1 exit. Also, as the diffuser is insulated, there is no heat flow or work done. From the continuity equation, we have, ππ = ππ = π΄π½ π£ The energy equation thus gives, ππ 1 2 1 2 βπ + π½π = ππ βπ + π½π 2 2 Thus, we have, 1 2 1 2 1 900 × 1000 βπ − βπ = πΆπ ππ − ππ = ππ − ππ ⇒ πΆπ Δπ = 2 2 2 3600 ⇒ πΆπ Δπ = 28050 2 1 − 80 2 2 π½ ππ½ 28.050 = 28.050 ⇒ Δπ = = 27.938 °πΆ ππ ππ 1.004 ⇒ ππ = −5 + 27.938 = 22.938°πΆ BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-11 Now, using the continuity equation, we have, π΄π π½π π΄π π½π π΄π π½π = ⇒ π£π = π£π π£π π£π π΄π π½π 0.8 × 80 π£π = π£π × = π£π × 0.256 1 × 250 Using the ideal gas law, we have, ππ£ = π π. This gives, π£π = ⇒ ππ = ππ π ππ π ππ = × 0.256 ππ ππ ππ 1 296.12 1 × = 50 × × = 215.685 πππ ππ 0.256 268.15 0.256 Thus, the air exits the diffuser at an exit pressure of 215.685 kPa. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-12 Problem 4.21 The wind is blowing horizontally at 30 m/s in a storm at Po, 20°C toward a wall, where it comes to a stop (stagnation condition) and leaves with negligible velocity similar to a diffuser with a very large exit area. Find the stagnation temperature from the energy equation. Assume constant specific heats. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-13 Solution From the continuity equation, we have, ππ = ππ = π΄π½ π£ From the energy equation, we have at constant mass, 1 2 β1 + π½1 = β2 2 From the equation of enthalpy, we have, 1 πΆπ π1 + π½12 = πΆπ π2 2 1 π½12 ⇒ π2 = π1 + 2 πΆπ 1 302 ⇒ π2 = 293 + × = 293.45 πΎ 2 1.004 × 1000 Thus, the stagnation temperature of the air is 293.45 K. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-14 Problem 4.23 Saturated Liquid R-410a at 25°C is throttled to 400 kPa in a refrigerator. What is the exit temperature? Find the percentage increase in the volume flow rate. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-15 Solution Given that the flow is a steady throttle flow, and as the nozzle is insulated, there is no heat transfer. Also, changes in kinetic and potential energy is zero. Thus, we have, βπ = βπ Given that the R-410a enters in the saturated liquid state, we have, βπ = βπ25°πΆ = 97.59 ππ½/ππ This enthalpy remains the same at the exit state. From table B.4.2., at 400 kPa, we see that the exit R-410a remains in the saturated state. Thus, we have ππ = ππ ππ‘ 400 πππ = −19.98°πΆ ≈ − 20°πΆ, βπ = 97.59 = βππ + π₯π βπππ ⇒ π₯π = βπ − βππ 97.59 − 28.24 = = 0.28463 βπππ 243.65 The specific volume at the outlet is given as π£π = π£π + π₯π π£ππ = 0.000803 + 0.28463 × 0.06400 = 0.01902 π3 /ππ The specific volume at the inlet is given as π£π = π£π25°πΆ = 0.000944 π3 /ππ As π = ππ£, we get the ratio of the volume flow rates as, ππ ππ = ππ£π π£π 0.01902 = = = 20.148 ππ£π π£π 0.000944 BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-16 Problem 4.26 Helium is throttled from 1.2 MPa, 30 °C to a pressure of 100 kPa. The diameter of the exit pipe is so much larger than that of the inlet pipe that the inlet and exit velocities are equal. Find the exit temperature of the helium and the ratio of the pipe diameters. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-17 Solution Given that the throttling process occurs at the steady state, heat and work done is zero. Also, as the inlet and exit velocities are equal, ΔπΎπΈ = ΔππΈ = 0 From the basic equation of throttling process, we have, βπ = βπ ⇒ πΆπ ππ = πΆπ ππ ⇒ ππ = ππ = 30°πΆ The mass flow rate is given as π= π΄π½ π΄π½ = π π π£ π As there is no change in the mass flow rate, we have, π, π½, π as constant, which gives, π·π ππ = π·π ππ 1 2 = 1.2 0.1 1 2 = 3.464 Thus, the ratio of diameters of the exit to the inlet is 3.464. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-18 Problem 4.28 An adiabatic steam turbine has an inlet of 2 kg/s water at 2000 kPa and 400 °C with a velocity of 15 m/s. The exit as saturated vapor at 10 kPa, and velocity is very low. Find the specific work and the power produced. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-19 Solution It is given that the turbine is adiabatic. Thus, the heat transfer is zero, and the potential energy also remains constant. Thus, we have, π π12 = 0, π½π = 15 , π½π = 0 π If we take a unit mass flowing through the control volume, the energy eqn becomes, π12 π22 π12 + β1 + = π€12 + β2 + 2 2 From table B.1.3., the inlet enthalpy of the water is β1 = 3247.60 ππ½. ππ−1 . From table B.1.2., the exit enthalpy of the water is β2 = 2584.63 ππ½. ππ−1 . Thus, the energy eqn becomes, π€12 π12 152 = β1 − β2 + = 3247.60 − 2584.63 + = 663.0825 ππ½/ππ 2 2000 The power output thus becomes ππ‘ = ππ€12 = 2 × 663.0825 = 1326.165 ππ Thus, the turbine has a power output of 1326.165 kW. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-20 Problem 4.32 A wind turbine with a rotor diameter of 20 m takes 40 % of the kinetic energy out as shaft work on a day with a temperature of 20 °C and a wind speed of 35 km/h. Determine the power produced. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-21 Solution From the continuity eqn, we have ππ = ππ = π At a constant mass flow rate, we have the energy equation as 1 2 1 2 π βπ + π½π + πππ = π βπ + π½π + πππ + π12 2 2 Given that the wind turbine converts 40% of the kinetic energy as work, giving, π12 1 2 = 0.4 × π π½π 2 The mass flow rate is given as π = ππ΄π½ = π΄ π½π π£π π Where the area of cross-section is π΄ = 4 π·2 = 0.25 × π × 202 = 314.159 π2 The specific volume is given by the ideal gas law, i.e., π£π = π ππ 0.287 × 293.15 = = 0.83034 π3 /ππ ππ 101.325 BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-22 The wind speed is π½π = 35 ππ βπ = 9.72222 π/π Thus, the mass flow rate is given by π = ππ΄π½ = π΄ π½π 314.159 × 9.72222 = = 3678.4 ππ/π π£π 0.83034 The kinetic energy of this mass flow of air is given as 1 π 1 π½ = 9.72222 2 π 2 2 = 47.261 π½/ππ Thus, the power produced by the windmill is given as π12 = 0.4 × π 1 2 π½ π½π = 0.4 × 3678.4 × 47.261 = 69537.945 = 69.538 ππ 2 ππ. π Thus, the wind turbine produces a power output of 69.538 kW. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-23 Problem 4.35 An adiabatic compressor in a commercial refrigerator receives R-410a at -10 °C and x = 1. The exit is at 800 kPa and 60 °C. Neglect kinetic energies and find the specific work. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-24 Solution Given that the compressor is adiabatic, and that there is no change in the kinetic energy of the fluid, we have, π12 = 0, ΔπΎπΈ = 0, ΔππΈ = 0 Assuming a unit mass flow rate, we have the enthalpies of the inlet and outlet flow rates as βπ = 275.78 ππ½. ππ−1 and βπ = 338.44 ππ½. ππ−1 The energy equation is given as ππ2 ππ2 π12 + βπ + + πππ = π€12 + βπ + + πππ 2 2 Which when substituted with the given data gives, βπ = π€12 + βπ ⇒ π€12 = βπ − βπ Thus, the work done is, π€12 = 275.78 − 338.44 = −62.66 ππ½. ππ−1 The negative sign symbolizes the work input to the system. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-25 Problem 4.38 An adiabatic compressor receives R-410a as saturated vapor at 400 kPa and brings it to 2000 kPa, 60 °C. Then a cooler brings it to a state of saturated liquid at 2000 kPa (see Fig.). Find the specific compressor work and the specific heat transfer in the cooler. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-26 Solution Taking the control volume as the compressor, assuming adiabatic conditions, and assuming no change in kinetic and potential energies, we have the energy equation as π€12 = β2 − β1 At the initial state of entry, from table B.4.2., we have β1 = 271.90 ππ½. ππ−1 At the exit of compressor, from table B.4.2., we have β2 = 320.62 ππ½. ππ−1 At the exit of the cooler, from table B.4.1., we have, by interpolation, β3 = 106.14 + 114.95 − 106.14 × 2000 − 1885.1 2140.2 − 1885.1 = 110.11 ππ½. ππ−1 Substituting the values in the energy equation, we have, π€12 = β1 − β2 = −48.72ππ½. ππ−1 For the cooler, we have π23 = β3 − β2 = 110.11 − 320.62 = −210.51 ππ½ ππ The negative symbol indicate the loss of heat. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-27 Problem 4.43 A boiler section boils 3 kg/s saturated liquid water at 4000 kPa to saturated vapor. Find the heat transfer in the process. Represent the process on P-v and T-v diagram. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-28 Solution Taking the boiler as the control volume, we have a constant pressure system. As there is no velocity or elevation change, we have zero work done. Thus, π12 = 0 The continuity eqn is π1 = π2 = π The enthalpies at the initial and final states of water, from table B.1.2., we have, β1 = 1087.29 ππ½. ππ−1 and β2 = 2801.38 ππ½. ππ−1 Thus, the energy equation gives, π12 + πβ1 = π12 + πβ2 ⇒ π12 = π β2 − β1 ⇒ π12 = 3 × 2801.38 − 1087.29 = 5142.27 ππ The Positive sign indicates that heat is added to the water in this process. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-29 Problem 4.48 (HW) A chiller cools liquid water for air-conditioning purposes. Assume that 2.5 kg/s water at 20 °C 100 kPa is cooled to 5 °C in a chiller. How much heat transfer (kW) is needed? How much will be the heat transfer (kW) if constant specific heat is considered for liquid water? BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-30 Solution The control volume here is the Chiller, at a steady state flow with heat transfer, and the kinetic energy and potential energy changes are neglected. As there is no work done in the chiller, we have the energy equation as πππ’π‘ = βπ − βπ From table B.1.1., the inlet and exit enthalpies of the water in this system is βπ = 83.94 ππ½. ππ−1 and βπ = 20.98 ππ½. ππ−1 This gives the heat transfer as πππ’π‘ = 83.94 − 20.98 = 62.96 ππ½. ππ−1 Thus, the rate of heat transfer in the chiller is equal to πππ’π‘ = ππππ’π‘ = 2.5 × 62.96 = 157.4 ππ½. π −1 BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-31 Problem 4.53 The main water line into a tall building has a pressure of 600 kPa at 5 m below ground level, as shown in Fig. A pump brings the pressure up so that the water can be delivered at 200 kPa at the top floor 150 m above ground level. Assume a flow rate of 10 kg/s liquid water at 10 β¦C and neglect any difference in kinetic energy and internal energy u. Find the pump work. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-32 Solution The control volume is the pipe from inlet at −5π, all the way to the exit at +150π, 200 πππ. Here, work is done on the water, but there is no heat transfer to the water. From the energy equation, we have 1 1 βπ + π½2π + πππ = βπ + π½2π + πππ + π€ 2 2 As the temperature of the water is not changing, the internal energy remains constant, and the changes in enthalpy is the difference in the ππ£ terms. Thus, the work done can be written as 1 2 π€ = βπ − βπ + π½π − π½2π + π ππ − ππ 2 As the flows are at very low velocities, ⇒ π€ = ππ π£π − ππ π£π + π(ππ − ππ ) −5 − 150 1000 ⇒ π€ = 0.4 + −1.519 = −1.119 ππ½/ππ ⇒ π€ = 600 × 0.001 − 200 × 0.001 + 9.806 × Thus, overall rate of work done by the pump for a flow of 10 ππ/π is, π = ππ€ = 10 × −1.119 = −11.19 ππ BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-33 Problem 4.55 (HW) A steam pipe for a 300 m tall building receives superheated steam at 200 kPa at ground level. At the top floor the pressure is 125 kPa and the heat loss in the pipe is 110 kJ/kg. What should the inlet temperature be so that no water will condense inside the pipe? BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-34 Solution The control volume is the pipe length from the ground level to 300 m. There is no change in kinetic energy change, with single inlet and exit flows. Thus, the energy equation becomes, π + βπ = βπ + πππ For a condition of no condensation, we need to have a minimum state of saturated vapor for the water. Thus, the exit enthalpy is βπ = βπ 125 πππ = 2685.35 ππ½. ππ−1 Thus, the inlet enthalpy is given as βπ = βπ + πππ − π = 2685.35 + 9.807 × 300 − −110 = 2798.29 ππ½. ππ−1 1000 Interpolating at 200 kPa for the entry temperature at the calculated enthalpy is ππ = 164.504°πΆ Thus, the superheated steam needs to be at a temperature of 164.504°C at 200 kPa. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-35 Problem 4.57 Cogeneration is often used where a steam supply is needed for industrial process energy. Assume that a supply of 5 kg/s steam at 0.5 MPa is needed. Rather than generating this from a pump and boiler, the setup in Fig. is used to extract the supply from the high-pressure turbine. Find the power the turbine now cogenerates in this process. Assume an adiabatic turbine system. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-36 Solution The control volume here is the turbine with 1 inlet and 2 outlet flows. As the turbine is adiabatic, πππ£ = 0. The continuity and energy eqns thus become, π1 = π2 + π3 ππΆπ + π1 β1 = π2 β2 + π3 β3 + ππ At the initial state, we have, π1 = 20 ππ/π , β1 = 3373.63 ππ½/ππ At the intermediate tapping, we have, π2 = 5 ππ/π , β2 = 2748.67 + 2855.37 − 2748.67 × 155 − 151.86 200 − 151.86 = 2755.63 ππ½/ππ At the exit state, we have, π3 = 15 ππ/π , β3 = 251.38 + 0.9 × 2358.33 = 2373.88 ππ½/ππ Thus, the power output of the turbine is calculated as ππ = 20 × 3373.63 − 5 × 2755.63 − 15 × 2373.88 = 18086.25 ππ ≈ 18.086 ππ Thus, the turbine system outputs a power of 18.086 MW. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-37 Problem 4.60 A compressor receives 0.05 kg/s R-410a at 300 kPa, -20 °C and 0.1 kg/s R410a at 500 kPa, 0 °C. The exit flow is at 1000 kPa, 60 °C, as shown in Fig. Assume it is adiabatic, neglect kinetic energies, and find the required power input. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-38 Solution The control volume is the entire compressor, with two inlets and one exit. Given that its adiabatic, π13 = 0, and kinetic and potential energy being neglected, we have Continuity equation: π1 + π2 = π3 ⇒ π3 = 0.05 + 0.1 = 0.15 ππ/π The energy equation thus becomes, π13 + π1 β1 + π2 β2 = ππ + π3 β3 From table B.4.2., we have the values of the enthalpies as β1 = 275.46 ππ½/ππ β2 = 287.84 ππ½/ππ β3 = 335.75 ππ½/ππ Thus, the work done is given as ππ = π1 β1 + π2 β2 − π3 β3 ⇒ ππ = 0.05 × 275.46 + 0.1 × 287.84 − 0.15 × 335.75 ⇒ ππ = −7.8055 ππ Thus, the compressor requires a power input of 7.8055 kW. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-39 Problem 4.63 (HW) A condenser (heat exchanger) brings 1 kg/s water flow at 10 kPa from 300β¦C to saturated liquid at 10 kPa, as shown in Fig. The cooling is done by lake water at 20β¦C that returns to the lake at 30β¦C. For an insulated condenser, find the flow rate of cooling water. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-40 Solution The control volume here is the heat exchanger. As the system is at steady state, the mass flow rates are constant for inflow and outflow. The energy eqn gives, πππππ β20 + πβππ‘ β300 = πππππ β30 + πβππ‘ βπ10 πππ From table B.1.1., the enthalpy at the inflow of lake water is β20 = 83.96 ππ½/ππ, and the enthalpy at the outflow of lake water is β30 = 125.79 ππ½/ππ For the process water, we use the table B.1.2., and B.1.3., For the inflow at 10kPa and 300°πΆ, we have the superheated vapour, and from table B.1.3., we have, β300 = 3076.5 ππ½/ππ For the outflow at 10 kPa and saturated liquid, we have βπ10 πππ = 191.83 ππ½/ππ The mass flow rate of the lake water is given as πππππ β300 − βπ10 πππ 3076.5 − 191.83 = πβππ‘ =1× = 69 ππ/π β30 − β20 125.79 − 83.96 BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-41 Problem 4.67 An automotive radiator has glycerin at 95β¦C enter and return at 55β¦C as shown in Fig. Air flows in at 20β¦C and leaves at 25β¦C. If the radiator should transfer 25 kW, what is the mass flow rate of the glycerin and what is the volume flow rate of air in at 100 kPa? BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-42 Solution If we take a control volume around the whole radiator then there is no external heat transfer – it is all between the glycerin and the air. Thus, we take a control volume around each flow separately. Thus, for glycerin, from table A.4., πβπ + −π = πβπ ππππ¦ = −π −π −25 = = = 0.258 ππ/π βπ − βπ πΆπππ¦ ππ − ππ 2.42 55 − 95 For air, from table A.5., πβπ + π = πβπ ππππ = π π 25 = = = 4.98 ππ/π βπ − βπ πΆπππ ππ − ππ 1.004 25 − 20 The volumetric flow rate of air is given as π£π = ππππ π ππ 0.287 × 293 = = 0.8409 π3 /ππ ππ 100 = ππ£π = 4.98 × 0.8409 = 4.19 π3 /π BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-43 Problem 4.71 A cooler in an air conditioner brings 0.5 kg/s air at 35°C to 5°C, both at 101 kPa and it then mix the output with a flow of 0.25 kg/s air at 20°C, 101 kPa sending the combined flow into a duct. Find the total heat transfer in the cooler and the temperature in the duct flow. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-44 Solution Cooler section (no πΎ) Energy equation: πβ1 = πβ2 + πππππ ππΆπππ = π β1 − β2 = ππΆπ π1 − π2 = 0.5 ∗ 1.004 ∗ 35 − 5 = ππ. ππ ππΎ Thus, the total heat transfer in the cooler section is 15.06 kW. Mixing section (no πΎ, πΈ) Continuity Equation: π2 + π3 = π4 --------- 1 Energy Equation: π2 β2 + π3 β3 = π4 β4 ----------2 π4 = π2 + π3 = 0.5 + 0.25 = 0.75 ππ/π From Eq 1 and 2, π4 β4 = π2 + π3 β4 = π2 β2 + π3 β3 π2 β4 − β2 + π3 β4 − β3 = 0 π2 πΆπ π4 − π2 + π3 πΆπ π4 − π3 = 0 ⇒ π4 = π2 π2 +π3 π3 π2 +π3 π π 0.5 0.25 =π2 π2 + π3 π3 = 0.75 ⋅ 5 + 0.75 ⋅ 20 = ππ°C 4 4 Thus, the temperature in the duct flow is equal to 10°C. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-45 Problem 4.76 The following data are for a simple steam power plant shown below. State 6 has ππ=0.92 and velocity of 200 m/s. The rate of steam flow is 25 kg/s, with 300 kW of power input to the pump. Piping diameters are 200 mm from the steam generator to the turbine and 75mm from the condenser to the economizer and steam generator. Determine the velocity at state 5 and the power output of the turbine. Stat e 1 2 3 4 5 P, kPa 6200 6100 5900 5700 5500 10 T, ΛC 45 175 500 490 h, kJ/kg 194 744 3426 3404 6 7 9 40 168 BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-46 Solution Turbine inlet area π΄5 = π 4 × 0.2 2 = 0.03142 π2 ; π£5 = 0.06163 π3 ππ ππ£5 25 ππ π × 0.06163 π3 ππ π5 = = = 49 π π π΄5 0.03142 π2 β6 = 191.83 + 0.92 × 2392.8 = 2393.2 ππ½ ππ 1 π€π = β5 − β6 + (π52 −π62 ) 2 492 − 2002 = 3404 − 2393.2 + = 992 ππ½/ππ 2 × 1000 ππ = ππ€π = 25 ππ π × 992 ππ½ ππ = 24800 ππ The power output of the turbine is 24.8 MW. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-47 Problem 4.82 A modern jet engine has a temperature after combustion of about 1500 K at 3200 kPa as it enters the turbine section (state 3 in fig.). The compressor inlet is at 80 kPa, 260 K (state 1) and the outlet (state 2) is at 3300 kPa, 780 K; the turbine outlet (state 4) into the nozzle is at 400kPa, 900 K and the nozzle exit (state 5) is at 80 kPa, 640 K. Neglect any heat transfer and neglect kinetic energy except out of the nozzle. Find the compressor and turbine specific work terms and the nozzle exit velocity. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-48 Solution From the properties of air chart, β1 = 260.32 ; β2 = 800.28; β3 = 1635.80; β4 = 933.15 ; β5 = 649.53 kJ/kg all values in Energy equation for compressor : π€πΆ ππ = β2 − β1 = 800.28 − 260.32 = 539.36 ππ½/ππ Energy equation for turbine : π€π = β3 − β4 = 1635.80 − 933.15 = 702.65 ππ½/ππ Energy equation for nozzle : 1 2 β4 = β5 + π5 2 1 2 π = β4 − β5 2 5 π5 = (2 × 283.62 × 1000)1/2 = 753 π/π Thus, the specific work input of the compressor is 539.36 kJ/kg, the work output of the turbine is 702.65 kJ/kg, and the exit velocity of the nozzle is 753 m/s. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-49 Problem 4.86 A tank contains 2 π3 of air 100 πππ and 300 πΎ. A pipe flowing air at 1000 πππ, 300 πΎ is connected to the tank is connected to the tank and the tank is filled slowly till it reaches 1000 kPa. Find the heat transfer to reach a final temperature of 300 πΎ. Assume ideal gas behaviour throughout. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-50 Solution The control volume here is the tank itself, and it is a transient problem as it is being filled up. From the continuity equation, π2 − π1 = πππ The energy equation yields, π2 π’2 − π1 π’1 = π12 − π12 + πππ βππ As the tank is at constant volume, π12 = 0 In the tank, the temperature is constant, thus, the internal energy doesn’t change between the two states. Thus, π’2 = π’1 = π’ππ = π’300 βππ = π’ππ + ππ ππ = π’ππ + π πππ The masses of the air at the two states inside the tank is, π1 = π2 = π1 π1 100 × 2 = = 2.32288 ππ π π1 0.287 × 300 π2 π2 1000 × 2 = = 23.22880 ππ π π2 0.287 × 300 Thus, the energy eqn gives π12 = π2 π’2 − π1 π’1 − πππ βππ = π1 + πππ π’1 − π1 π’1 − πππ π’ππ − πππ π πππ ⇒ π12 = π1 π’1 − π1 π’1 + πππ π’1 − πππ π’ππ − πππ π πππ = −πππ π πππ ⇒ π12 = − 23.22880 − 2.32288 × 0.287 × 300 = −1799.999 ππ½ BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-51 Problem 4.89 A 1 π3 tank contains ammonia at 150 πππ and 25°πΆ. The tank is attached to a line in which ammonia flows at 1200 πππ and 60°πΆ. The valve is opened and mass flows in till the tank is half full of liquid, at 25°πΆ. Calculate the heat transferred from the tank in this process. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-52 Solution The control volume here is the tank. The process is a transient process as the ammonia flows in. From continuity eqn, πππ = π2 − π1 At state 1(inside the tank), ammonia exists as a superheated vapor. Thus, interpolating between 20°πΆ and 30°πΆ in table B.2.2., we have π£1 = 0.9552 π3 /ππ π π’1 = 1380.6 ππ½/ππ, 1 π1 = π£ = 0.9552 = 1.047 ππ 1 Given that at state 2, the 1 π3 tank is half full with liquid. Thus, it is in saturated state in the tank at 25°πΆ. Thus, from table B.2.1., we have π£π = 0.001658 π3 /ππ π£π = 0.12813 π3 /ππ The masses are, ππππ = 0.5 0.001658 = 301.568 ππ, ππ£ππ = π₯= 0.5 0.12813 = 3.902 ππ, ππ‘ππ‘ = 305.47 ππ ππ£ππ = 0.01277 ππ‘ππ‘ BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-53 From the continuity equation, πππ = π2 − π1 = 305.47 − 1.047 = 304.423 ππ From table B.2.1., π’2 = 296.59 + 0.01277 × 1038.4 = 309.85 ππ½/ππ At the inlet of the tank, from table B.2.2., βππ = 1553.3 ππ½/ππ Thus, from the energy equation, we have πππ£ + πππ βππ = π2 π’2 − π1 π’1 ∴ πππ£ = 305.47 × 309.85 − 1.047 × 1380.6 − 304.423 × 1553.3 πππ£ = −379 655.855 ππ½ Thus, the heat transferred from the tank is 379 655.855 kJ. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-54 Problem 4.95 In a glass factory a 2 m wide sheet of glass at 1500 K comes out of the final rollers, which fix the thickness at 5 mm with a speed of 0.5 m/s (see Fig. P4.95). Cooling air in the amount of 20 kg/s comes in at 290 K from a slot 2 m wide and flows parallel with the glass. Suppose this setup is very long, so that the glass and air come to nearly the same temperature (a co-flowing heat exchanger); what is the exit temperature? Assume constant specific heat. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-55 Solution From the energy equation, we have, πππππ π βππππ π 1 + ππππ βπππ2 = πππππ π βππππ π 3 + ππππ βπππ4 The mass flow rate of the glass in the sheet is given as πππππ π = ππ = ππ΄π½ = 2500 × 2 × 0.005 × 0.5 = 12.5 ππ/π Thus, the energy equation becomes, πππππ π πΆππππ π π3 − π1 + ππππ πΆππππ π4 − π2 = 0 At the exit, both the glass and air come to the same temperature, giving us π4 = π3 As πΆππππ π = 0.80 ππ½. ππ−1 . πΎ −1 , and πΆππππ = 1.004 ππ½. ππ−1 . πΎ −1 , the exit temperature is πππππ π πΆππππ π π1 + ππππ πΆππππ π2 12.5 × 0.80 × 1500 + 20 × 1.004 × 290 π4 = π3 = = πππππ π πΆππππ π + ππππ πΆππππ 12.5 × 0.80 + 20 × 1.004 ⇒ π4 = π3 = 692.26 πΎ Thus, the temperature at the exit of the glass sheet is equal to 692.26 K. BITS F111 Themodynamics, BITS, Pilani-Hyderabad Campus P-56 BITS Pilani Hyderabad Campus Thank You
