Join With us. Today Updates. July July 9. Scientific Storytelling A science-writing initiative will empower Kenyon students to artfully articulate important scientific topics. Jul 17 Kenyon in Your Kitchen pm — pm Kenyon alumni working in food and drink industries are sharing their go-to happy hour pairings.

Just need a basic editor to be able… Jul 24 Kenyon in Your Kitchen pm — pm Kenyon alumni working in food and drink industries are sharing their go-to happy hour pairings. Take the Kenyon Virtual Tour Fully residential and one of the most beautiful anywhere, Kenyon's hilltop campus boasts buildings, a acre environmental center, hiking trails, and woods, all bordered by the Kokosing, one of Ohio's scenic rivers.

Start the virtual tour. For a polytropic exponent of 1. Using IT, plot the work and heat transfer per unit mass of air for polytropic exponents ranging from 1.

Investigate the error in the heat transfer introduced by assuming constant cv. FIND: Determine the heat transfer and work per unit mass of air for a constant specific heats and b variable specific heats. Plot the work and heat transfer per unit mass for a given range of polytropic exponents and investigate the error in heat transfer introduced by assuming constant cv. Note: The assumption of constant specific heat at K leads to a value that is 6.

Determine the heat transfer, in Btu per lb of steam, for a polytropic exponent of 1. Using IT, plot the heat transfer per unit mass of steam for polytropic exponents ranging from 1. Investigate the error in the heat transfer introduced by assuming ideal gas behavior. Note, the heat transfer value obtained using the ideal gas model is This is a significant error.

The channel has a rectangular cross-section of 0. Flow is one-dimensional. KNOWN: Mass flow rate, temperature, and quality of refrigerant a exiting a heat exchanger through 0. KNOWN: Pressure, temperature, and mass flow rate of air entering a combustion chamber through a 5 ft by 4 ft rectangular duct.

Combustion Chamber. Apply the ideal gas equation of state to solve for the specific volume of air. Substituting values to obtain specific volume yields. Air is modeled as an ideal gas. The tank is fitted with two inlet pipes, one delivering hot water at a mass flow rate of 0. Water exits through a single exit pipe at a mass flow rate of 2. Determine the mass of water, in lb, in the tank after 30 minutes. The control volume encloses the water in the tank and has two inlets and one exit.

The entering and exiting mass flow rates each remain constant. ANALYSIS: Apply a mass rate balance from the initial state when the tank contains lb of water until the final state after 30 minutes have elapsed and solve for the mass of water in the tank at the final state.

KNOWN: Air at specified initial temperature and pressure leaks from rigid tank until a final specified pressure is attained by the air remaining in the tank. The control volume is defined by the dashed line on the accompanying diagram.

Air can be modeled as an ideal gas. Pressure and volume are known at state 2. The mass in the tank at state 2, m2, equals the initial mass in the tank, m1, less the mass that leaks from the tank.

The gas constant, R, is the universal gas constant divided by the molecular weight of air. Substituting values and applying the appropriate conversion factor yield. Note the need to convert the final temperature from oR to oF to provide the answer in the requested units.

If the gate is raised for 24 hours, determine the volume of water, in gallons, provided for irrigation. Assume the density of river water is The control volume encloses the water in the field, which is initially empty. The control volume has one inlet through the gate and no exits. Water flows through the gate steadily and one-dimensionally. Density of river water is constant at ANALYSIS: Since mass of water provided for irrigation can be expressed in terms of water density and volume, determine the water volume based on the water mass and density.

Apply a mass rate balance from the initial state when the gate is initially closed mass of water in the field is zero until the final state after 24 hours have elapsed with the gate open and solve for the mass of water at the final state.

Both basins are initially empty. Water overflows from basin A into basin B. Dimensions of the basins are indicated on the figure. Determine the variation of water height in each basin as a function of time. Eventually, water overflows into basin B, which drains at a rate proportional to the water level. Basin B 5m. The control volume boundaries are shown on the accompanying diagram. Prior to basin A filling completely, there is no exiting mass flow.

The mass rate balance reduces to. Water Height LA [m] 0. At the exit, the pressure is 3 bar, the temperature is The air behaves as an ideal gas. KNOWN: Air flows through a one-inlet, one-exit control volume with known pressure, temperature, and velocity at the inlet and exit.

The control volume shown on the accompanying figure is at steady state. For the inlet, state 1, the mass flow rate can be determined from given data and the ideal gas equation of state. At the exit, the refrigerant is a saturated vapor at a temperature of —4oC. The evaporator flow channel has constant diameter. If the mass flow rate of the entering refrigerant is 0. KNOWN: Refrigerant a flows through a constant-diameter evaporator entering as a saturated mixture at given temperature, quality, and velocity and exiting as a saturated vapor at a given temperature.

Apply the quality relation to determine the specific volume at state 1. Substituting to determine specific volume. Since the diameter is constant throughout the channel, the inlet and exit areas are the same. The mass flow rate is the same at the inlet and the exit based on the mass rate balance for one- inlet, one-exit, steady flow.

The diameter is the same at the inlet and exit since the diameter is constant through the evaporator. Substituting values and applying the appropriate conversion factor. KNOWN: An air compressor operates at steady state with specified inlet pressure and volumetric flow rate and exit velocity and pressure. The air undergoes a polytropic process, pv1. The ratio of the specific volumes can be determined from the polytropic relationship, pv1. The inlet cross-sectional area is 6 cm2.

The control volume shown with the schematic is at steady state. The refrigerant enters the tube with a quality of 0. The refrigerant exits the tube at 9 bar as a saturated liquid.

Specify whether the heat transfer is to or from the air. FIND: a the area at the inlet, in m2, and b the heat transfer between the nozzle and its surroundings, in kW. To achieve an exit velocity of Model nitrogen as an ideal gas. Since enthalpy values for nitrogen are provided on a molar basis in Table A, the energy rate balance is expressed in terms of molar enthalpy.

For negligible potential energy effects, determine the exit temperature, in K. No stray heat transfer occurs between the air and its surroundings. Model the air as an ideal gas. For an ideal gas, specific enthalpy is a function of only temperature. Substituting values and solving for exit specific enthalpy. Heat transfer occurs at a rate of 20 kW from the air to cooling water circulating in a water jacket enclosing the compressor.

Model air as an ideal gas. Substituting values and solving for mass flow rate give. The temperature change of the blood is negligible as it flows through the pump. The pump requires 20 W of power input. Modeling the blood as an incompressible substance with negligible kinetic and potential energy effects, determine the pressure change, in kPa, of the blood as it flows through the pump.

No stray heat transfer occurs between the blood and its surroundings. The temperature of blood does not change as it flows through the pump. Model the blood as an incompressible substance. Substituting into the energy rate balance. Thus, volumetric flow rate, AV , is constant.

Substituting m. Water enters the well- insulated pump operating at steady state at a rate of 0. The inlet pressure and temperature are Water can be modeled as an incompressible substance with constant density of Neglecting kinetic and potential energy effects, determine the temperature change, in oR, as the water flows through the pump.

Comment on this change. No stray heat transfer occurs between the water and its surroundings. The change in specific enthalpy of an incompressible substance with constant specific heat is given by Eq. Solving for mass flow rate gives. A separate stream of steam enters inlet 2 at 10 bar and oC.

Saturated liquid at 10 bar exits the feedwater heater at exit 3. KNOWN: Liquid water at given pressure and temperature and steam at given pressure and temperature enter a feedwater heater.

Saturated liquid exits the feedwater heater at given pressure. Heat transfer and kinetic and potential energy effects can be neglected.

Neglecting heat transfer and kinetic and potential energy effects and recognizing no work is associated with a feedwater heater, the energy balance simplifies to. At inlet 1, the water is compressed liquid. At inlet 2, the steam is superheated. At exit 3, the water is saturated liquid.

Steam enters inlet 1 with flow rate of 0. Steam enters inlet 3 with flow rate of 1. Steam exits the tank at 1 bar.

The rate of heat transfer from the collection tank is 40 kW. Since steam at inlet 2 is superheated vapor, specific enthalpy, h2, is determined from Table A Substituting values into the energy rate balance and solving for the exit specific enthalpy yield.

From Table A-3,. Water exits the tank at Saturated vapor at 1. Neglecting kinetic and potential energy effects, determine the rate of heat transfer, in kW, and its associated direction. At exit 2 steam is saturated vapor.

At exit 3 steam is saturated liquid. Since the rate of heat transfer has a negative sign, heat transfer is from the steam separator to the surroundings. Substituting values and solving for quality give Btu Btu Since the inlet and exit diameters are the same, the areas cancel as well as the gas constant. Solving for exit pressure yields. As water flows through the valve, the change in its temperature, stray heat transfer with the surroundings, and potential energy effects are each negligible.

Operation is at steady state. The temperature of water does not change as it flows through the valve. The kinetic energy of the water increases slightly as it flows from the inlet to the exit of the valve. Saturated liquid Refrigerant a enters the valve at a pressure of 9 bar and is throttled to a pressure of 2 bar. The refrigerant then enters the heat exchanger, exiting at a temperature of 10oC with no significant decrease in pressure.

Stray heat transfer and kinetic and potential energy effects can be ignored. Determine a the temperature, in oC, of the refrigerant at the exit of the valve. KNOWN: Refrigerant a flows through a throttling valve in series with a heat exchanger while liquid water flows through the same heat exchanger in a separate line. Each component operates at steady state. No stray heat transfer occurs between the components and their surroundings. For both components, kinetic and potential energy effects can be ignored.

From Table A,. Substituting values gives. Refrigerant a flows through a throttling valve and a heat exchanger while air flows through a fan and the same heat exchanger. Data for steady-state operation are given on the figure. There is no significant heat transfer between any of the components and the surroundings.

KNOWN: An air-conditioning system section operates with refrigerant a flowing through a throttling value and heat exchanger and air flowing through a fan and the same heat exchanger. All components operate at steady state. No stray heat transfer occurs between any of the components and their surroundings. For all components, kinetic and potential energy effects can be ignored.

Define control volume A to encompass the fan and the heat exchanger. Since the air and refrigerant do not mix in control volume A, the steady state mass balance reduces to. The enthalpy at state 4 can be determined by analyzing the throttling valve with the energy rate balance. Steady-state operating data for the turbine and pump are labeled on the figure.

Heat transfer from the water to its surroundings occurs at a rate of 2 kW. For the turbine, heat transfer with the surroundings and potential energy effects are negligible. Kinetic energy effects at all numbered states can be ignored. Determine a The power required by the pump, in kW, to supply water to the inlet of the mixing chamber. KNOWN: A steam turbine drives a pump through which water flows to a mixing chamber located 25 m higher than the pump.

Define control volume A to encompass the pump and the line to the mixing chamber. Define control volume B to encompass the turbine. For both control volumes A and B, kinetic energy effects can be ignored.

No stray heat transfer occurs between control volume B and its surroundings. For control volume B, potential energy effects can be ignored. Substituting values and solving yield. Substituting values and solving yield kJ 1 Steady-state operating data are provided on the figure. Heat transfer with the surroundings can be neglected, as can all kinetic and potential energy effects. The air is modeled as an ideal gas. Determine a the total power required by both compressors, in kW.

Control volumes at steady state enclose the compressors and heat exchanger. For each control volume, heat transfer with the surroundings is negligible and kinetic and potential effects can be ignored.

Steam flows into the tank through a valve until 2. Conditions within the steam line remain constant. V There is no power, and no mass exits the control volume. Noting hi remains constant, combining mass and energy rate balances results in. Since the sign associated with heat transfer is positive, heat transfer is into the control volume.

An electric resistor transfers energy to the air in the tank at a constant rate for 5 minutes, after which time the pressure in the tank is 1 bar and the temperature is oC. Modeling air as an ideal gas, determine the power input to the tank, in kW. The condition of the air entering the tank remains constant. There is no heat transfer, and no mass exits the control volume.

The tank is initially evacuated. Steam is allowed to flow into the tank until the pressure inside is p. The condition of the steam entering the tank remains constant. Control Volume Temperature vs. Control Volume Pressure 22 Water in the tank of the heater initially has a temperature of oF.

When someone turns on the shower faucet, water flows from the tank at a rate of 0. If the water within the tank is well mixed, the temperature at any time can be taken as uniform throughout.

The tank is well insulated so stray heat transfer with the surroundings is negligible. Neglecting kinetic and potential energy effects, assuming negligible change in pressure from inlet to exit of the tank, and modeling water as an incompressible substance with density of Pressure change from inlet to exit is negligible.

Water is modeled as an incompressible substance with density of The state of the water entering the tank remains constant. Water in the tank is well-mixed. Stray heat transfer with the surroundings is negligible.

For an incompressible substance with negligible pressure change, Eqs. Using the given water density which can be obtained from Table AE , the mass of water in the control volume is. Entering in IT the given values and the equation to determine control volume temperature T as a function of time and varying the range for time from 0 s to s 20 minutes yield the plot below.

Time Note that continuous use of hot water results in lower water temperature and eventually cold showers! As the water is heated at constant volume the pressure rises to 2 bar and the quality becomes With further heating a pressure-regulating valve keeps the pressure constant in the cooker at 2 bar by allowing saturated vapor at 2 bar to escape.

Neglecting kinetic and potential energy effects a determine the quality of the H2O at the initial onset of vapor escape state 2 and the amount of heat transfer, in kJ, to reach this state.

FIND: a Determine the quality of the H2O when at the initial onset of vapor escape state 2 and the amount of heat transfer, in kJ, to reach this state, b determine the final mass in the cooker, in kg, and the additional amount of heat transfer, in kJ, if heating continues from state 2 until the final quality is 1. For part a the closed system during process is defined by the dashed line on the accompanying figure.

For part b the control volume during process is defined by the dashed line on the accompanying figure. For part b , saturated vapor exits the control volume at 2 bar. Since the mass is constant during process and with assumption 2 in the engineering model, the energy rate balance for process reduces to. The mass of water can be determined from the volume of the cooker and the specific volume of the water at state 1.

Since state 2 is a two-phase liquid- vapor mixture, quality is determined by. There is no work, and no mass enters the control volume. Noting he remains constant, combining mass and energy rate balances results in.

Solving for the mass gives V 0. Now using the Explore button, sweep x3 from 0. The following plots are constructed from the data. Final Mass vs. Quality Heat Transfer vs. Quality 0. We see from the plots that as the quality of the mixture within the tank increases the mass in the tank drops rapidly while heat transfer to the tank rapidly increases.

An electrical resistor inside the tank maintains the temperature at oF. The control volume is defined by the dashed line on the accompanying figure. Carbon dioxide is modeled as an ideal gas. The mass of the resistor is small enough to be ignored. ANALYSIS: a The mass of the carbon dioxide withdrawn over a time interval equals the difference between the initial amount of mass in the tank and the mass in the tank at a later time:.

One can either hand-sketch a plot based on sample data or use IT to generate the plots. Results using IT are:. Final Pressure Work vs.

Final Pressure 1. Work Btu 0. Steam is withdrawn slowly from the tank until the pressure drops to p. Heat transfer to the tank contents maintains the temperature at oC.

At each instant, pressure is uniform throughout the steam. Specific enthalpy he is nearly linear with the mass in the tank. For each mass in the interval 7. These values allow the plot of he vs. Now using the Explore button, sweep m from 7. The following plot is constructed from the data. Exit Enthalpy vs. As this variation is very 1 nearly linear, the average value of he is appropriate to evaluate this term.

Now using the Explore button, sweep p from kPa to kPa in steps of Heat Transfer vs. Pressure 11, 10, 9, Heat Transfer kJ. The pressure within the tank decreases as mass is withdrawn while the temperature remains constant.

Once the burner is turned on the water is gradually heated at a rate of 0. After a period of time, the water starts boiling and continues to do so until all of the water has evaporated. Determine a the time required for the onset of evaporation, in s. FIND: a The time required for the onset of evaporation, in s, and b the time required for all of the water to evaporate, in s, once evaporation starts.

Process State 1 State 2. For part a the liquid water inside the pot defines the system during process as shown by the dashed line on the accompanying figure. Add to Cart.

Daisie D. Shapiro Michael J.