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SOLUTION MANUAL

CHAPTER 7

Borgnakke Sonntag

Fundamentals of

Thermodynamics

Borgnakke Sonntag

8e

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CONTENT CHAPTER 7

SUBSECTION PROB NO.

In-Text Concept Questions a - g Concept Problems 1- Steady State Reversible Processes Single Flow 13- Multiple Devices and Cycles 36- Transient Processes 49- Reversible Shaft Work, Bernoulli Equation 57- Steady State Irreversible Processes 82- Transient irreversible Processes 115- Device efficiency 128- Review Problems 158- Problems re-solved with Pr and vr from Table A.7.2: 17, 31, 34, 55, 80, 174

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7.a A reversible adiabatic flow of liquid water in a pump has increasing P. How about T? Solution:

Steady state single flow: s (^) e = s (^) i + ⌡

i

e dq T + s^ gen^ = s^ i^ + 0 + 0 Adiabatic (dq = 0) means integral vanishes and reversible means s (^) gen = 0, so s is constant. Properties for liquid (incompressible) gives Eq.6.

ds =

C

T dT then constant s gives constant T.

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7.b A reversible adiabatic flow of air in a compressor has increasing P. How about T?

Solution:

Steady state single flow: s (^) e = s (^) i + ⌡

i

e dq T + s^ gen^ = s^ i^ + 0 + 0 so s is constant. Properties for an ideal gas gives Eq.6.15 and for constant specific heat we get Eq.6.16. A higher P means a higher T, which is also the case for a variable specific heat, recall Eq.6.19 using the standard entropy.

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7.d A flow of water at some velocity out of a nozzle is used to wash a car. The water then falls to the ground. What happens to the water state in terms of V , T and s?

let us follow the water flow. It starts out with kinetic and potential energy of some magnitude at a compressed liquid state P, T. As the water splashes onto the car it looses its kinetic energy (it turns in to internal energy so T goes up by a very small amount). As it drops to the ground it then looses all the potential energy which goes into internal energy. Both of theses processes are irreversible so s goes up. If the water has a temperature different from the ambient then there will also be some heat transfer to or from the water which will affect both T and s.

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7.e In a steady state single flow s is either constant or it increases. Is that true? Solution:

No.

Steady state single flow: s (^) e = s (^) i + ⌡

i

e (^) dq T + s^ gen Entropy can only go up or stay constant due to s (^) gen , but it can go up or down due to the heat transfer which can be positive or negative. So if the heat transfer is large enough it can overpower any entropy generation and drive s up or down.

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7.g A polytropic flow process with n = 0 might be which device?

As the polytropic process is Pvn^ = C, then n = 0 is a constant pressure process. This can be a pipe flow, a heat exchanger flow (heater or cooler) or a boiler.

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Concept Problems

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Which process will make the statement in concept question e) on page 330 true?

Solution:

If the process is said to be adiabatic then: Steady state adiabatic single flow: s (^) e = s (^) i + s (^) gen ≥ si

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A reversible process in a steady flow with negligible kinetic and potential energy changes is shown in the diagrams. Indicate the change h (^) e - h (^) i and transfers w and q as positive, zero or negative

i

P

v

T

s

e

i

e

dw = – v dP > 0 P drops so work is positive out. dq = T ds = 0 s is constant, and process reversible so adiabatic. he - hi = q – w = 0 – w < 0 so enthalpy drops

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A reversible steady isobaric flow has 1 kW of heat added with negligible changes in KE and PE, what is the work transfer?

P = C: Shaft work Eq. 7.14:

dw = – v dP + ΔKE + ΔPE – T ds (^) gen = 0 + 0 + 0 – 0 = 0

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An air compressor has a significant heat transfer out. See example 9.4 for how high T becomes if there is no heat transfer. Is that good, or should it be insulated?

That depends on the use of the compressed air. If there is no need for the high T, say it is used for compressed air tools, then the heat transfer will lower T and result in lower specific volume reducing the work. For those applications the compressor may have fins mounted on its surface to promote the heat transfer. In very high pressure compression it is done in stages between which is a heat exchanger called an intercooler.

This is a small compressor driven by an electric motor. Used to charge air into car tires.

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To increase the work out of a turbine for a given inlet and exit pressure how should the inlet state be changed?

w = – ∫ v dP + ….. Eq.7.

For a given change in pressure boosting v will result in larger work term. So for larger inlet T we get a larger v and thus larger work. That is why we increase T by combustion in a gasturbine before the turbine section.

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An irreversible adiabatic flow of liquid water in a pump has higher P. How about T? Solution:

Steady state single flow: s (^) e = s (^) i + ⌡

i

e dq T + s^ gen^ = s^ i^ + 0 + s^ gen so s is increasing. Properties for liquid (incompressible) gives Eq.6. where an increase in s gives an increase in T.