5-72C Yes, if the mixing chamber is losing heat to the surrounding medium.5-73C Under the conditions of no heat and work interactions between the mixing chamber and the surrounding medium.5-74C Under the conditions of no heat and work interactions between the heat exchanger and the surrounding medium.5-75 A hot water stream is mixed with a cold water stream. For a specified mixture temperature, the mass flow rate of cold water is to be determined.Assumptions 1 Steady operating conditions exist. 2 The mixing chamber is well-insulated so that heat loss to the surroundings is negligible. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5 There are no work interactions.Properties Noting that T < Tsat @ 250 kPa = 127.41°C, the water in all three streams exists as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus,
5-50 EES Problem 5-49 is reconsidered. The effect of the turbine exit pressure on the power output of the turbine as the exit pressure varies from 10 kPa to 200 kPa is to be investigated. The power output is to be plotted against the exit pressure.Analysis The problem is solved using EES, and the solution is given below."Knowns "T[1] = 450 [C]P[1] = 10000 [kPa]Vel[1] = 80 [m/s]P[2] = 10 [kPa]X_2=0.92Vel[2] = 50 [m/s]m_dot[1]=12 [kg/s]Fluid$=`Steam_IAPWS`*************0405060708*************30P[2] [kPa]T[2][C]"Property Data"h[1]=enthalpy(Fluid$,T=T[1],P=P[1])h[2]=enthalpy(Fluid$,P=P[2],x=x_2)T[2]=temperature(Fluid$,P=P[2],x=x_2)v[1]=volume(Fluid$,T=T[1],p=P[1])v[2]=volume(Fluid$,P=P[2],x=x_2)"Conservation of mass: "m_dot[1]= m_dot[2]"Mass flow rate"m_dot[1]=A[1]*Vel[1]/v[1]m_dot[2]= A[2]*Vel[2]/v[2]"Conservation of Energy - Steady Flow energy balance"m_dot[1]*(h[1]+Vel[1]^2/2*Convert(m^2/s^2, kJ/kg)) = m_dot[2]*(h[2]+Vel[2]^2/2*Convert(m^2/s^2, kJ/kg))+W_dot_turb*convert(MW,kJ/s)DELTAke=Vel[2]^2/2*Convert(m^2/s^2, kJ/kg)-Vel[1]^2/2*Convert(m^2/s^2, kJ/kg)
Conservation of Mass5-1C Mass, energy, momentum, and electric charge are conserved, and volume and entropy are not conserved during a process.5-2C Mass flow rate is the amount of mass flowing through a cross-section per unit time whereas the volume flow rate is the amount of volume flowing through a cross-section per unit time.5-3C The amount of mass or energy entering a control volume does not have to be equal to the amount of mass or energy leaving during an unsteady-flow process.5-4C Flow through a control volume is steady when it involves no changes with time at any specified position.5-5C No, a flow with the same volume flow rate at the inlet and the exit is not necessarily steady (unless the density is constant). To be steady, the mass flow rate through the device must remain constant.5-6E A garden hose is used to fill a water bucket. The volume and mass flow rates of water, the filling time, and the discharge velocity are to be determined.Assumptions 1 Water is an incompressible substance. 2 Flow through the hose is steady. 3 There is no waste of water by splashing.Properties We take the density of water to be 62.4 lbm/ft3 (Table A-3E).Analysis (a) The volume and mass flow rates of water are
4-110 Heat is transferred to a piston-cylinder device containing air. The expansion work is to be determined.Assumptions 1 There is no friction between piston and cylinder. 2 Air is an ideal gas.QAir0.5 kgΔT=5°CProperties The gas contant for air is 0.287 kJ/kg.K (Table A-2a).Analysis Noting that the gas constant represents the boundary work for a unit mass and a unit temperature change, the expansion work is simply determined fromkJ4-111 Solar energy is to be stored as sensible heat using phase-change materials, granite rocks, and water. The amount of heat that can be stored in a 5-m3 = 5000 L space using these materials as the storage medium is to be determined.Assumptions 1 The materials have constant properties at the specified values. 2 No allowance is made for voids, and thus the values calculated are the upper limits.Analysis The amount of energy stored in a medium is simply equal to the increase in its internal energy, which, for incompressible substances, can be determined from )(12TTmcU−=Δ.(a) The latent heat of glaubers salts is given to be 329 kJ/L. Disregarding the sensible heat storage in this case, the amount of energy stored is becomesΔUsalt = mhif = (5000 L)(329 kJ/L) = 1,645,000 kJThis value would be even larger if the sensible heat storage due to temperature rise is considered.
4-74 A number of brass balls are to be quenched in a water bath at a specified rate. The rate at which heat needs to be removed from the water in order to keep its temperature constant is to be determined.Assumptions 1 The thermal properties of the balls are constant. 2 The balls are at a uniform temperature before and after quenching. 3 The changes in kinetic and potential energies are negligible.Properties The density and specific heat of the brass balls are given to be ρ = 8522 kg/m3 and cp = 0.385 kJ/kg.°C.Analysis We take a single ball as the system. The energy balance for this closed system can be expressed as)()(21out12balloutenergies etc. potential, kinetic, internal,in Changesystemmass and work,heat,by nsferenergy traNet outinTTmCQuumUQEEE−=−=Δ=−Δ=−4342143421The total amount of heat transfer from a ball iskJ/ball 88.9C)74120(C)kJ/kg. 385.0)(kg 558.0()(kg 558.06m) 05.0()kg/m 8522(621out333=°−°=−=====TTmcQDmππρρVWaterbath,50°CBrassballs,120°CThen the rate of heat transfer from the balls to the water becomeskJ/min 988=×==)kJ/ball 88.9(balls/min) 100(ballballtotalQnQ&&Therefore, heat must be removed from the water at a rate of 988 kJ/min in order to keep its temperature constant at 50°C since energy input must be equal to energy output for a system whose energy level remains constant. That is, .