5. For liquid water, 𝜐<𝜐f(T) (Eq. 3.11) and Eq. 3.13 is used to evaluate specific enthalpy.
6. g 5 9.81 m/s2.
10 m Pump
+–
1
2
D2 = 3 cm
T1 = 20°C p1 = 1 atm D1 = 12 cm (AV)1 = 0.83 m3/min
Fig. E4.6 Analysis:
(a) A mass rate balance reduces at steady state to read m#
25m#
1. The common mass flow rate at the inlet and exit, m#
, can be evaluated using Eq. 4.4b together with 𝜐<𝜐f(20°C)51.001831023 m3/kg from Table A-2. That is, m# 5 AV
𝜐 5a 0.83 m3/min
1.001831023 m3/kgb `1 min 60 s ` 513.8kg
s
Pump A.21 – Tabs a, b, & c
A A
Compressor A.20 – Tabs a, b, & c
c c c c EXAMPLE 4.6 c
166 Chapter 4 Control Volume Analysis Using Energy
Thus, the inlet and exit velocities are, respectively, V15 m#𝜐
A1 5 (13.8 kg/s)(1.001831023 m3/kg)
𝜋(0.12 m)2/4 51.22 m/s V25 m#𝜐
A2
5 (13.8 kg/s)(1.001831023 m3/kg)
𝜋(0.03 m)2/4 519.56 m/s
(b) To calculate the power input, begin with the one-inlet, one-exit form of the energy rate balance for a control volume at steady state, Eq. 4.20a. That is
05Q#
cv2W#
cv1m# c(h12h2)1aV122V22
2 b1g(z12z2)d
➋ Introducing Q#
cv5(0.05)W#
cv, and solving for W#
cv, W#
cv5 m#
0.95c(h12h2)1aV212V22
2 b1g(z12z2)d (a)
Using Eq. 3.13, the enthalpy term is expressed as
h12h25[hf(T1)1𝜐f(T1)[p12psat(T1)]]
2 [hf(T2)1𝜐f(T2)[p22psat(T2)]] (b) Since there is no significant change in temperature, Eq. (b) reduces to
h12h25𝜐f(T)(p12p2)
As there is also no significant change in pressure, the enthalpy term drops out of the present analysis. Next, evaluating the kinetic energy term
V122V22
2 5
[(1.22)22(19.56)2]am sb2
2 ` 1 N
1 kg?m/s2` ` 1 kJ
103 N?m` 5 20.191 kJ/kg Finally, the potential energy term is
g(z12z2)5(9.81 m/s2)(0210)m` 1 N
1 kg?m/s2` ` 1 kJ
103 N?m` 5 20.098 kJ/kg Inserting values into Eq. (a)
W#
cv5a13.8 kg/s
0.95 b[020.19120.098]akJ
kgb `1 kW 1 kJ/s` 5 24.2 kW
where the minus sign indicates that power is provided to the pump.
➊ Alternatively, V1 can be evaluated from the volumetric flow rate at 1. This is left as an exercise.
➋ Since power is required to operate the pump, W#
cv is negative in accord with our sign convention. The energy transfer by heat is from the control volume to the surroundings, and thus Q#
cv is negative as well. Using the value of W#
cv
found in part (b), Q#
cv5(0.05)W#
cv5 20.21 kW.
Ability to…
❑ apply the steady-state energy rate balance to a control volume.
❑apply the mass flow rate expression, Eq. 4.4b.
❑develop an engineering model.
❑ retrieve properties of liquid water.
✓Skills Developed
➊
If the nozzle were removed and water exited directly from the hose, whose diameter is 5 cm, determine the velocity at the exit, in m/s, and the power required, in kW, keeping all other data unchanged.
Ans. 7.04 m/s, 1.77 kW.
4.8 Compressors and Pumps 167
4.8.3 Pumped-Hydro and Compressed-Air Energy Storage
Owing to the dictates of supply and demand and other economic factors, the value of electricity varies with time. Both the cost to generate electricity and increasingly the price paid by consumers for electricity depend on whether the demand for it is on-peak or off-peak. The on-peak period is typically weekdays—for example from 8 a.m. to 8 p.m., while off-peak includes nighttime hours, weekends, and major holidays.
Consumers can expect to pay more for on-peak electricity. Energy storage methods benefiting from variable electricity rates include thermal storage (see box on p. 98) and pumped-hydro and compressed-air storage introduced in the following box.
Economic Aspects of Pumped-Hydro and Compressed-Air Energy Storage
Despite the significant costs of owning and operating utility-scale energy storagy systems, various economic strategies, including taking advantage of differing on- and off-peak electricity rates, can make pumped-hydro and compressed-air storage good choices for power generators. In this discussion, we focus on the role of variable electricity rates.
In pumped-hydro storage, water is pumped from a lower reservoir to an upper reservoir, thereby storing energy in the form of gravitational potential energy. (For simplicity, think of the hydropower plant of Fig. 4.10 operating in the reverse direction.) Off-peak electricity is used to drive the pumps that deliver water to the upper reservoir. Later, during an on-peak period, stored water is released from the upper reservoir to generate electricity as the water flows through turbines to the lower reservoir. For instance, in the summer water is released from the upper reservoir to generate power to meet a high daytime demand for air conditioning;
while at night, when demand is low, water is pumped back to the upper reservoir for use the next day. Owing to friction and other nonidealities, an overall input-to-output loss of electricity occurs with pumped-hydro storage and this adds to operating costs. Still, differ- ing daytime and nighttime electricity rates help make this technology viable.
In compressed-air energy storage, compressors powered with off-peak electricity fill an underground salt cavern, hard-rock mine, or aquifer with pressurized air drawn from the atmosphere. See Fig. 4.12. When electricity demand peaks, high-pressure com- pressed air is released to the surface, heated by natural gas in combustors, and expanded through a turbine generator, generating electricity for distribution at on-peak rates.
TAKE NOTE...
Cost refers to the amount paid to produce a good or service. Price refers to what consumers pay to acquire that good or service.
t
+ Compressor –
Atmospheric air Air and combustion products
Turbine +
– Off-peak electricity in
On-peak electricity out
Compressed air out
Cavern
Air in Air out
Compressed air in
Fuel to combustor (not shown)
Generator
Fig. 4.12 Compressed-air storage.
168 Chapter 4 Control Volume Analysis Using Energy
4.9 Heat Exchangers
Heat exchangers have innumerable domestic and industrial applications, including use in home heating and cooling systems, automotive systems, electrical power gen- eration, and chemical processing. Indeed, nearly every area of application listed in Table 1.1 (p. 3) involves heat exchangers.
One common type of heat exchanger is a mixing chamber in which hot and cold streams are mixed directly as shown in Fig. 4.13a. The open feedwater heater, which is a component of the vapor power systems considered in Chap. 8, is an example of this type of device.
Another common type of exchanger is one in which a gas or liquid is separated from another gas or liquid by a wall through which energy is conducted. These heat exchang- ers, known as recuperators, take many different forms. Counterflow and parallel tube- within-a-tube configurations are shown in Figs. 4.13b and 4.13c, respectively. Other con- figurations include cross-flow, as in automobile radiators, and multiple-pass shell-and-tube condensers and evaporators. Figure 4.13d illustrates a cross-flow heat exchanger.
heat exchanger
(a) (b)
(c) (d)
Fig. 4.13 Common heat exchanger types. (a) Direct contact heat exchanger.
(b) Tube-within-a-tube counterflow heat exchanger. (c) Tube-within-a-tube parallel flow heat exchanger. (d) Cross-flow heat exchanger.
BIOCONNECTIONS Inflatable blankets such as shown in Fig. 4.14 are used to prevent subnormal body temperatures (hypothermia) during and after surgery. Typically, a heater and blower direct a stream of warm air into the blanket. Air exits the blanket through perforations in its surface. Such thermal blankets have been used safely and without incident in millions of surgical procedures. Still, there are obvious risks to patients if temperature controls fail and overheating occurs.
Such risks can be anticipated and minimized with good engineering practices.
Warming patients is not always the issue at hospitals; sometimes it is cooling, as in cases involving cardiac arrest, stroke, heart attack, and overheating of the body (hyper- thermia). Cardiac arrest, for example, deprives the heart muscle of oxygen and blood, causing part of it to die. This often induces brain damage among survivors, including irreversible cognitive disability. Studies show when the core body temperature of cardiac
4.9 Heat Exchangers 169
patients is reduced to about 338C, damage is limited because vital organs function more slowly and require less oxygen. To achieve good outcomes, medical specialists say cooling should be done in about 20 minutes or less. A system approved for cooling cardiac arrest victims includes a disposable plastic body suit, pump, and chiller. The pump provides rapidly flowing cold water around the body, in direct contact with the skin of the patient wearing the suit, then recycles coolant to the chiller and back to the patient.
These biomedical applications provide examples of how engineers well versed in thermodynamics principles can bring into the design process their knowledge of heat exchangers, temperature sensing and control, and safety and reliability requirements.
Heater- blower unit
Fig. 4.14 Inflatable thermal blanket.
4.9.1 Heat Exchanger Modeling Considerations
As shown by Fig. 4.13, heat exchangers can involve multiple inlets and exits. For a control volume enclosing a heat exchanger, the only work is flow work at the places where matter enters and exits, so the term W#
cv drops out of the energy rate balance.
In addition, the kinetic and potential energies of the flowing streams usually can be ignored at the inlets and exits. Thus, the underlined terms of Eq. 4.18 (repeated below) drop out, leaving the enthalpy and heat transfer terms, as shown by Eq. (a). That is,
05Q#
cv2W#
cv1ai m#
iahi1 V2i
2 1gzib2ae m#
eahe1 Ve2 2 1gzeb 05Q#
cv1 ai m#
ihi2ae m#
ehe (a)
Although high rates of energy transfer within the heat exchanger occur, heat transfer with the surroundings is often small enough to be neglected. Thus, the Q#
cv term of Eq. (a) would drop out, leaving just the enthalpy terms. The final form of the energy rate balance must be solved together with an appropriate expression of the mass rate bal- ance, recognizing both the number and type of inlets and exits for the case at hand.
4.9.2 Applications to a Power Plant Condenser and Computer Cooling
The next example illustrates how the mass and energy rate balances are applied to a condenser at steady state. Condensers are commonly found in power plants and refrigeration systems.
A A
Heat_Exchanger A.22 – Tabs a, b & c
170 Chapter 4 Control Volume Analysis Using Energy
Evaluating Performance of a Power Plant Condenser
Steam enters the condenser of a vapor power plant at 0.1 bar with a quality of 0.95 and condensate exits at 0.1 bar and 458C. Cooling water enters the condenser in a separate stream as a liquid at 208C and exits as a liquid at 358C with no change in pressure. Heat transfer from the outside of the condenser and changes in the kinetic and potential energies of the flowing streams can be ignored. For steady-state operation, determine (a) the ratio of the mass flow rate of the cooling water to the mass flow rate of the condensing steam.
(b) the rate of energy transfer from the condensing steam to the cooling water, in kJ per kg of steam passing through the condenser.
SOLUTION
Known: Steam is condensed at steady state by interacting with a separate liquid water stream.
Find: Determine the ratio of the mass flow rate of the cooling water to the mass flow rate of the steam and the rate of energy transfer from the steam to the cooling water.
Schematic and Given Data:
Engineering Model: