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Chapter 21, Problem 15P

(a)

To determine

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

(a)

Expert Solution
Check Mark

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

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

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.

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 (1)

Here,

ηa is the actual efficiency of the engine.

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

ηa=23(1TCTH)=23(THTCTH)

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.

Q is the rate of heat input to the engine.

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

QeΔt=QW (2)

Here,

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

Substitute Wηa for Q in equation (2) to find QeΔt ,

QeΔt=WηaW=W(1ηa1) (3)

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

QeΔt=W(123(THTCTH)1)=W(32(THTHTC)1)=W(TH+2TC2(THTC))=W(0.5TH+TCTHTC) (4)

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

QeΔt=W(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) .

Conclusion:

Therefore, 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) .

(b)

To determine

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

(b)

Expert Solution
Check Mark

Answer to Problem 15P

The  amount of the energy exhaust change if the firebox is modified to run hotter by using more advanced combustion technology because the exhaust power 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 amount of the energy exhaust change because the exhaust power is inversely proportional to the fire box temperature. So, the exhaust power decreases as the fire box temperature increases.

Conclusion:

The amount of the energy exhaust change if the firebox is modified to run hotter by using more advanced combustion technology because the exhaust power decreases as the fire box temperature increases.

(c)

To determine

The exhaust power for TH=800°C .

(c)

Expert Solution
Check Mark

Answer to Problem 15P

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

Explanation of Solution

Given information: 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 (4), the formula to calculate the exhaust power for TH=800°C .

QeΔt=W(0.5TH+TCTHTC)

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

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

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

Conclusion:

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

(d)

To determine

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

(d)

Expert Solution
Check Mark

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

Given information: 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) (5)

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 (4), 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) (6)

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 .

Conclusion:

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

(e)

To determine

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

(e)

Expert Solution
Check Mark

Answer to Problem 15P

The value of TH not exists because the exhaust energy can not be that small.

Explanation of Solution

Given information: 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) (7)

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.

From equation (4), the formula to calculate the value of TH for which the exhaust power would be one-fourth 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) (8)

Substitute 1.40MW for W , 0.466MW for (QeΔt) , 110°C for TC in equation (8) to find TH ,

TH=(110°C+273)K×(1+(0.466MW×11.40MW))((0.466MW×11.40MW)0.5)=(383)K×(1+(0.333MW))((0.333MW)0.5)=3057.119K3.05×103K

Thus, the value of TH for which the exhaust power would be one-fourth as large as in part (c) is 3.05×103K . In this the value of TH not exists because the exhaust energy can not be that small.

Conclusion:

Therefore, the value of TH not exists because the exhaust energy can not be that small.

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Chapter 21 Solutions

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term

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