PHYSICS:F/SCI.+ENGRS-W/WEBASSIGN
PHYSICS:F/SCI.+ENGRS-W/WEBASSIGN
10th Edition
ISBN: 9781337888479
Author: SERWAY
Publisher: CENGAGE L
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Chapter 21, Problem 15P

An electric generating station is designed to have an electric output power of 1.40 MW using a turbine with two-thirds the efficiency of a Carnot engine. The exhaust energy is transferred by heat into a cooling tower at 110°C. (a) Find the rate at which the station exhausts energy by heat as a function of the fuel combustion temperature Th. (b) If the firebox is modified to run hotter by using more advanced combustion technology, how does the amount of energy exhaust change? (c) Find the exhaust power for Th = 800°C. (d) Find the value of Th for which the exhaust power would be only half as large as in part (c). (e) Find the value of Th for which the exhaust power would be one-fourth as large as in part (c).

(a)

Expert Solution
Check Mark
To determine

The rate at which the station exhaust energy by heat as a function of the fuel combustion temperature TH.

Answer to Problem 15P

The rate at which the station exhaust energy by heat as a function of the fuel combustion temperature TH is 1.40(0.5TH+383TH383).

Explanation of Solution

The rate of work output of the engine is 1.40MW, thetemperature into the cooling tower is 110°C.

Write the formula to calculate the carnot efficiency of the engine.

    η=(THTCTH)=(1TCTH)

Here, η is the carnot efficiency of the engine, TC is the temperature into the cooling tower and TH is the fuel combustion temperature.

The actual efficiency of the engine is equal to two-thirds of the efficiency of the carnot engine.

    ηa=23η                                                                                (I)

Here, ηa is the actual efficiency of the engine.

Substitute (1TCTH) for η in equation (I) to find ηa,

    ηa=23(1TCTH)=23(THTCTH)

Write the formula to calculate the rate of heat input to the engine.

    ηa=WQQ=Wηa

Here, W is the rate of work output of the engine and Q is the rate of heat input to the engine.

Write the formula to calculate the rate at which the station exhaust energy by heat as a function of the fuel combustion temperature TH.

    QeΔt=QWΔt                                                                                       (II)

Here, Qe is the energy expelled by the engine, Δt is the time interval and QeΔt is the exhaust power.

Substitute Wηa for Q in equation (II) to find QeΔt,

    QeΔt=(WηaW)/Δt=WΔt(1ηa1)                                                                          (III)

Substitute 23(THTCTH) for ηa in equation (3) to find QeΔt,

    QeΔt=WΔt(123(THTCTH)1)=WΔt(32(THTHTC)1)=WΔt(TH+2TC2(THTC))=WΔt(0.5TH+TCTHTC)                                                                   (IV)

Conclusion:

Substitute 1.40MW for WΔt and 110°C for TC in equation (IV) to find QeΔt,

    QeΔt=1.40 MW(0.5TH+(110°C+273)KTH(110°C+273)K)=1.40(0.5TH+383TH383)

Thus, the rate at which the station exhaust energy by heat as a function of the fuel combustion temperature TH is 1.40(0.5TH+383TH383). The energy exhaust rate is in megawatts and the fuel temperature is in Kelvin.

(b)

Expert Solution
Check Mark
To determine

The effect on the amount of the energy if the firebox is modified to run hotter by using more advanced combustion technology.

Answer to Problem 15P

The  amount of the energy exhaust decreases as the fire box temperature increases.

Explanation of Solution

If the firebox is modified to run hotter by using more advanced combustion technology, the exhaust power increases by a factor of 0.5TH+383 and decreases by a factor of TH383. Thus, there is an overall decrease in power exhaust. Since power exhaust decreases the corresponding energy exhaust also should be decreasing.

Conclusion:

The  amount of the energy exhaust decreases as the fire box temperature increases.

(c)

Expert Solution
Check Mark
To determine

The exhaust power for TH=800°C.

Answer to Problem 15P

The exhaust power for TH=800°C is 1.87MW.

Explanation of Solution

 The rate of work output of the engine is 1.40MW, fuel combustion temperature is 800°C, the temperature into the cooling tower is 110°C.

From equation (IV), write the formula to calculate the exhaust power for TH=800°C.

    QeΔt=W(0.5TH+TCTHTC)

Conclusion:

Substitute 1.40MW for W, 800°C for TH, 110°C for TC in equation (4) to find QeΔt,

    QeΔt=1.40MW(0.5(800°C+273)K+(110°C+273)K(800°C+273)K(110°C+273)K)=1.40MW(0.5(1073)K+(383)K(1073)K(383)K)=1.8656MW1.87MW

Thus, the exhaust power for TH=800°C is 1.87MW.

(d)

Expert Solution
Check Mark
To determine

The value of TH for which the exhaust power would be only half as large as in part (c).

Answer to Problem 15P

The value of TH for which the exhaust power would be only half as large as in part (c) is 3.84×103K.

Explanation of Solution

 The rate of work output of the engine is 1.40MW, the temperature into the cooling tower is 110°C.

Write the expression for the exhaust power whuch would be only half as large as in part (c).

    (QeΔt)=12(QeΔt)                                                                             (V)

Here, (QeΔt) is the exhaust power whuch would be only half as large as in part (c).

Substitute 1.86MW for (QeΔt) in equation (5) to find (QeΔt),

    (QeΔt)=12(1.86MW)=0.933MW

Thus, the exhaust power whuch would be only half as large as in part (c) is 0.933MW.

From equation (IV), Write the formula to calculate the value of TH for which the exhaust power would be only half as large as in part (c).

    (QeΔt)=W(0.5TH+TCTHTC)(THTC)(QeΔt)×1W=0.5TH+TCTH(((QeΔt)×1W)0.5)=TC(1+(QeΔt)×1W)TH=TC(1+(QeΔt)×1W)(((QeΔt)×1W)0.5)                                  (VI)

Conclusion:

Substitute 1.40MW for W, 0.933MW for (QeΔt), 110°C for TC in equation (6) to find TH,

    TH=(110°C+273)K×(1+(0.933MW×11.40MW))((0.933MW×11.40MW)0.5)=(383)K×(1+(0.666MW))((0.666MW)0.5)=3844.832K3.84×103K

Thus, the value of TH for which the exhaust power would be only half as large as in part (c) is 3.84×103K.

(e)

Expert Solution
Check Mark
To determine

The value of TH for which the exhaust power would be one-fourth as large as in part (c).

Answer to Problem 15P

No temperature value will provide an exhaust power of one-fourth of the value in part (c).

Explanation of Solution

The rate of work output of the engine is 1.40MW, the temperature into the cooling tower is 110°C.

Write the expression for the exhaust power whuch would be one-fourth as large as in part (c).

    (QeΔt)=14(QeΔt)                                                                  (VII)

Here, (QeΔt) is the exhaust power whuch would be one-fourth as large as in part (c).

Substitute 1.86MW for (QeΔt) in equation (7) to find (QeΔt),

    (QeΔt)=14(1.86MW)=0.466MW

Thus, the exhaust power whuch would be one-fourth as large as in part (c) is 0.466MW which is too small.

Conclusion:

From equation (IV), Write the formula to calculate the value of TH for which the exhaust power would be one-fourth as large as in part (c).

    limTH(QeΔt)=limTH1.40MW(0.5+383THTHTH383TH)=1.40MW(0.51)=0.700MW          (VIII)

Thus, the given exhaust power is lesser than the minimum possible exhaust power 0.700MW. Since it is impossible to obtain a lesser exhaust power, no anwer exists.

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Chapter 21 Solutions

PHYSICS:F/SCI.+ENGRS-W/WEBASSIGN

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