/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 Steam with a mass flux of \(600 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 28

Steam with a mass flux of \(600 \mathrm{lbm} / \mathrm{min}\) exits a turbine as saturated steam at 2 psia and passes through a condenser (a heat exchanger). What mass flux of cooling water is needed if the steam is to exit the condenser as saturated liquid and the cooling water is allowed a \(15^{\circ} \mathrm{F}\) temperature rise? SOLUTION The energy equations (4.75) are applicable to this situation. The heat transfer rate for the steam is, assuming no pressure drop through the condenser, $$ \dot{Q}_{s}=\dot{m}_{s}\left(h_{s 2}-h_{s 1}\right)=(600)(94.02-1116.1)=-613.200 \mathrm{Btu} / \mathrm{min} $$ This energy is gained by the water. Hence, $$ \dot{Q}_{w}=\dot{m}_{w}\left(h_{w 2}-h_{w 1}\right)=\dot{m}_{w} C_{p}\left(T_{w 2}-T_{w 1}\right) \quad \frac{613,200}{60}=\dot{m}_{w}(1.00)(15) \quad \dot{m}_{w}=681 \mathrm{lbm} / \mathrm{s} $$

Short Answer

Expert verified
681 lbm/s

Step by step solution

01

Calculate Heat Transfer Rate for Steam

Use the given mass flux of steam and the enthalpy values of saturated steam at 2 psia. The heat transfer rate is given by: \[ \dot{Q}_{s} = \dot{m}_{s} \left( h_{s2} - h_{s1} \right) \] where \( \dot{m}_{s} = 600 \, \mathrm{lbm} / \mathrm{min} \), \( h_{s2} = 94.02 \, \mathrm{Btu} / \mathrm{lbm} \) (saturated liquid at 2 psia), and \( h_{s1} = 1116.1 \, \mathrm{Btu} / \mathrm{lbm} \) (saturated steam at 2 psia).Calculating, \[ \dot{Q}_{s} = 600 \, \left( 94.02 - 1116.1 \right) = -613,200 \, \mathrm{Btu} / \mathrm{min} \]
02

Relate Heat Transfer Rate to Cooling Water

Heat lost by the steam is gained by the cooling water. Use this relationship to find the mass flux of cooling water. The formula is: \[ \dot{Q}_{w} = \dot{m}_{w} \left( h_{w2} - h_{w1} \right) = \dot{m}_{w} C_{p} \left( T_{w2} - T_{w1} \right) \]where \( C_{p} \) (specific heat of water) is approximately \(1.00 \, \mathrm{Btu}/\mathrm{lbm}/^{\circ}\mathrm{F} \), \( T_{w2} - T_{w1} = 15^{\circ} \mathrm{F} \), and \( \dot{Q}_{w} = \dot{Q}_{s} \).
03

Calculate Mass Flux of Cooling Water

Using the relationship established in Step 2, solve for the mass flux of cooling water: \[ \frac{613,200}{60} = \dot{m}_{w} (1.00) (15) \] since we need to convert \( \dot{Q}_{s} \) from \( \mathrm{min} \) to \( \mathrm{s} \):\[ \frac{613,200}{60} = 10,220 \, \mathrm{Btu} / \mathrm{s} \]Now, solving for \( \dot{m}_{w} \):\[ 10,220 = \dot{m}_{w} (15) \]\[ \dot{m}_{w} = 681.33 \, \mathrm{lbm} / \mathrm{s} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

mass flux
Mass flux represents the rate of mass flow per unit area through a given surface. In the context of our steam condenser problem, mass flux specifically refers to the amount of steam passing through the turbine and into the condenser per unit time. It is usually denoted by \( \dot{m}_{s} \), where 's' stands for steam. In this example, the mass flux of steam is given as \(600 \ \text{lbm}\text{/}\text{min} \). This value is essential as it is used to calculate the heat transfer rate. By understanding mass flux, you can determine how much energy is being transported within the steam and how it will affect the cooling process.
enthalpy
Enthalpy is a thermodynamic property representing the total heat content of a system. For a steam condenser, enthalpy changes are crucial in determining how much heat is transferred as the steam condenses. Enthalpy is typically denoted by 掳h掳. For the given exercise, the enthalpy of steam entering the condenser \( h_{s1} \) is \(1116.1 \ \text{Btu}\text{/}\text{lbm} \) and leaving it \( h_{s2} \) is \(94.02 \ \text{Btu}\text{/}\text{lbm} \) as a saturated liquid. When steam loses heat and condenses, its enthalpy changes. By calculating the difference in enthalpy, we can determine how much energy is being transferred.
heat transfer rate
The heat transfer rate, denoted as \( \dot{Q} \), represents the amount of thermal energy transferred per unit of time. It can be calculated by using the specific mass flux and enthalpy change. For steam in a condenser, the heat transfer rate can be calculated using the formula \( \dot{Q}_{s} = \dot{m}_{s}(h_{s2} - h_{s1}) \). In this case, the calculation is \( \dot{Q}_{s}=(600)(94.02-1116.1) = -613,200 \ \text{Btu}\text{/}\text{min} \). This negative sign indicates that energy is being transferred out of the steam. The same amount of energy is gained by the cooling water, ensuring energy conservation.
specific heat
Specific heat \(C_{p}\) is the amount of heat required to raise the temperature of one unit of mass of a substance by one degree Fahrenheit. For water in our steam condenser problem, the specific heat is about \(1.00 \ \text{Btu}\text{/}\text{lbm} \/ ^{\circ}F \). This value is vital when calculating the mass flux of cooling water that's needed to absorb the heat from the condensing steam. By knowing the specific heat and the temperature change of the water \( \Delta T_{w}=15^{\circ} \ \text{F} \), we can determine its heat capacity and use it to balance the energy equation for the system.
phase change
Phase change refers to the transition of a substance from one state of matter to another, such as from liquid to gas or from gas to liquid. In a steam condenser, the phase change occurs as steam condenses into liquid water. This transition releases a significant amount of latent heat, which is absorbed by the cooling water. The heat transfer during the phase change directly affects the enthalpy of the system. In the given problem, steam initially at \( h_{s1} = 1116.1 \ \text{Btu}\text{/}\text{lbm} \) condenses to \( h_{s2} = 94.02 \ \text{Btu}\text{/}\text{lbm} \), resulting in the calculated change in enthalpy. Grasping the concept of phase change is critical for understanding the energy dynamics within the condenser.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 6-V insulated battery delivers a 5 -A current over a period of \(20 \mathrm{~min}\). Calculate the heat transfer that must occur to return the battery to its initial temperature. The work done by the battery is \(W_{1-2}=V I \Delta t=(6)(5)[(20)(60)]=36 \mathrm{~kJ}\). According to the first law, this must equal \(-\left(U_{2}-U_{1}\right)\) since \(Q_{1-2}=0\) (the battery is insulated). To return the battery to its initial state, the first law, for this second process in which no work is done, gives $$ Q_{2-1}-W_{2-1}^{0}=\Delta U=U_{1}-U_{2} $$ Consequently, \(Q_{2-1}=+36 \mathrm{~kJ}\), where the positive sign indicates that heat must be transferred to the battery.

4.29 Refrigerant R134a enters the compressor of the simple refrigeration cycle of Fig. \(4.19\) as a saturated vapor at \(120 \mathrm{kPa}\), enters the condenser at \(800 \mathrm{kPa}\) and \(50^{\circ} \mathrm{C}\), exits the condenser as a saturated liquid, and exits the evaporator as a saturated vapor. Calculate \((a)\) the lowest cycle temperature, \((b)\) the heat transferred from the condenser, \((c)\) the cooling load \(\dot{Q}_{E},(d)\) the necessary compressor horsepower, and \((e)\) the coefficient of performance for this refrigerator. The refrigerant mass flow rate is \(0.2 \mathrm{~kg} / \mathrm{s}\). The enthalpies are of primary interest into and out of the four control volumes. The IRC Fluid Property Calculator will be used to obtain the enthalpies, but, if, for some reason, it is not available, Appendix D can be used (the IRC Fluid Property Calculator gives slightly different values than does Table D). They are listed in the following: State 1 (saturated vapor): \(\quad P_{1}=P_{4}=120 \mathrm{kPa}, x=1 . \therefore h_{1}=237 \mathrm{~kJ} / \mathrm{kg}\) State 2 (superheated vapor): \(\quad P_{2}=800 \mathrm{kPa}, T_{2}=50^{\circ} \mathrm{C} . \therefore h_{2}=287 \mathrm{~kJ} / \mathrm{kg}\) State 3 (saturated liquid): \(\quad P_{3}=P_{2}=800 \mathrm{kPa}, x_{3}=0 . \therefore h_{3}=95.5 \mathrm{~kJ} / \mathrm{kg}\) State 4 (quality region): \(\quad P_{4}=P_{1}=120 \mathrm{kPa}, h_{4}=h_{3}=95.5 \mathrm{~kJ} / \mathrm{kg}\) Using the above enthalpies, the quantities of interest can be determined. (a) The lowest cycle temperature is \(T_{4}\) (refer to Fig. 4.19). The IRC Fluid Property Calculator, with \(P_{4}=120 \mathrm{kPa}\) and \(h_{4}=95.5 \mathrm{~kJ} / \mathrm{kg}\), provides \(T_{4}=-22.3^{\circ} \mathrm{C}\). (b) The heat transferred from the condenser (a heat exchanger) is $$ \dot{Q}_{C}=\dot{m}\left(h_{2}-h_{3}\right)=0.2 \mathrm{~kg} / \mathrm{s} \times(287-95.5) \text { But } / \mathrm{lbm}=38.3 \mathrm{~kJ} / \mathrm{s} $$ (c) The cooling load provided by the evaporator (a heat exchanger) is $$ \dot{Q}_{E}=\dot{m}\left(h_{1}-h_{4}\right)=0.2(237-95.5)=28.3 \mathrm{~kJ} / \mathrm{s} $$ (d) The compressor work requirement is $$ \dot{W}_{\text {Comp }}=\dot{m}\left(h_{2}-h_{1}\right)=0.2 \times(287-237)=10 \mathrm{~kW} \quad \text { or } \quad 13.4 \mathrm{hp} $$ (e) The coefficient of performance for this ideal cycle is $$ \mathrm{COP}=\frac{\text { desired effect }}{\text { required input }}=\frac{\dot{Q}_{E}}{\dot{W}_{\text {Comp }}}=\frac{28.3}{10}=2.83 $$ If the desired effect was to heat a space, \(\dot{Q}_{C}\) would be used as the numerator in COP.

Air travels through the \(4 \times 2 \mathrm{~m}\) test section of a wind tunnel at \(20 \mathrm{~m} / \mathrm{s}\). The gage pressure in the test section is measured to be \(-20 \mathrm{kPa}\) and the temperature \(20^{\circ} \mathrm{C}\). After the test section, a diffuser leads to a 6-m-diameter exit pipe. Estimate the velocity and temperature in the exit pipe. SOLUTION The energy equation (4.72) for air takes the form $$ V_{2}^{2}=V_{1}^{2}+2 C_{p}\left(T_{1}-T_{2}\right)=20^{2}+(2)(1.00)\left(293-T_{2}\right) $$ The continuity equation, \(\rho_{1} A_{1} V_{1}=\rho_{2} A_{2} V_{2}\), yields $$ \frac{P_{1}}{R T_{1}} A_{1} V_{1}=\rho_{2} A_{2} V_{2} \quad \therefore \rho_{2} V_{2}=\left[\frac{80}{(0.287)(293)}\right]\left[\frac{8}{\pi(3)^{2}}\right](20)=5.384 \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s} $$ The best approximation to the actual process is the adiabatic quasi- equilibrium process. Using (4.49), letting \(\rho=1 / v\), we have $$ \frac{T_{2}}{T_{1}}=\left(\frac{\rho_{2}}{\rho_{1}}\right)^{k-1} \quad \text { or } \quad \frac{T_{2}}{\rho_{2}^{0.4}}=\frac{293}{[80 /(0.287)(293)]^{0.4}}=298.9 $$ The above three equations include the three unknowns \(T_{2}, V_{2}\), and \(\rho_{2}\). Substitute for \(T_{2}\) and \(V_{2}\) back into the energy equation and find $$ \frac{5.384^{2}}{\rho_{2}^{2}}=20^{2}+(2)(1.00)\left[293-(298.9)\left(\rho_{2}^{0.4}\right)\right] $$ This can be solved by trial and error to yield \(\rho_{2}=3.475 \mathrm{~kg} / \mathrm{m}^{3}\). The velocity and temperature are then $$ V_{2}=\frac{5.384}{\rho_{2}}=\frac{5.384}{3.475}=1.55 \mathrm{~m} / \mathrm{s} \quad T_{2}=(298.9)\left(\rho_{2}^{0.4}\right)=(298.9)(3.475)^{0.4}=492 \mathrm{~K} \quad \text { or } \quad 219^{\circ} \mathrm{C} $$

The pressure of \(200 \mathrm{~kg} / \mathrm{s}\) of water is to be increased by \(4 \mathrm{MPa}\). The water enters through a 20 -cm-diameter pipe and exits through a \(12-\mathrm{cm}\)-diameter pipe. Calculate the minimum horsepower required to operate the pump. SOLUTION The energy equation (4.68) provides us with $$ -\dot{W}_{p}=\dot{m}\left(\frac{\Delta P}{\rho}+\frac{V_{2}^{2}-V_{1}^{2}}{2}\right) $$ The inlet and exit velocities are calculated as follows: $$ V_{1}=\frac{\dot{m}}{\rho A_{1}}=\frac{200}{(1000)(\pi)(0.1)^{2}}=6.366 \mathrm{~m} / \mathrm{s} \quad V_{2}=\frac{\dot{m}}{\rho A_{2}}=\frac{200}{(1000)(\pi)(0.06)^{2}}=17.68 \mathrm{~m} / \mathrm{s} $$ The energy equation then gives $$ \dot{W}_{P}=-200\left[\frac{4000000}{1000}+\frac{(17.68)^{2}-(6.366)^{2}}{2}\right]=-827200 \mathrm{~W} \quad \text { or } \quad 1109 \mathrm{hp} $$ Note: The above power calculation provides a minimum since we have neglected any internal energy increase. Also, the kinetic energy change represents only a 3 percent effect on \(\dot{W}_{P}\) and could be neglected.

\(\mathrm{R} 134 \mathrm{a}\) enters a valve at \(800 \mathrm{kPa}\) and \(30^{\circ} \mathrm{C}\). The pressure downstream of the valve is measured to be \(60 \mathrm{kPa}\). Calculate the internal energy downstream. SOLUTION The energy equation across the valve, recognizing that heat transfer and work are zero, is \(h_{1}=h_{2}\). The enthalpy before the valve is that of compressed liquid. The enthalpy of a compressed liquid is essentially equal to that of a saturated liquid at the same temperature. Hence, at \(30^{\circ} \mathrm{C}\) in Table D.1, \(h_{1}=91.49 \mathrm{~kJ} / \mathrm{kg}\). Using Table D. 2 at \(60 \mathrm{kPa}\) we find $$ h_{2}=91.49=h_{f}+x_{2} h_{f g}=3.46+221.27 x_{2} \quad \therefore x_{2}=0.398 $$ The internal energy is then $$ u_{2}=u_{f}+x_{2}\left(u_{g}-u_{f}\right)=3.14+0.398[(206.12-3.14)]=83.9 \mathrm{~kJ} / \mathrm{kg} $$
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Astrophysics

Read Explanation

Electrostatics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Classical Mechanics

Read Explanation

Physical Quantities And Units

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.