3 mpa 400oc 40 m/s 2.5 mpa 300 m/s · pdf fileargon gas enters an adiabatic turbine at 900 kpa...

BaratuciHW #3

Problem : 5.35 - Adiabatic Steam Nozzle - 5 pts 21-Apr-11

a.) The temperature of the steam when it leaves the nozzle.b.) The ratio of the inlet cross-sectional area to the outlet cross-sectional area, A1 / A2.

Assume the process operates at steady-state.

Read :

Given : P1 3000 kPa P2 2500 kPa

T1 400oC v2 300 m/s

v1 40 m/s gc 1 kg-m/N-s2

Find : a.) T2 ???oC

b.) A1 / A2 ???

Diagram :

Assumptions : 1 -2 - Changes in potential energy are negligible.3 - The nozzle is adiabatic.4 - Flow work is the only form of work that crosses the system boundary.

Equations / Data / Solve :

part a.)

Eqn 1

Eqn 2

Eqn 3

If we knew H2, we could use the Steam Tables to determine T2.

So, let's solve Eqn 3 for H2 : Eqn 4

Because the nozzle is adiabatic and no shaft work crosses the system boundary and because we assumed that changes in potential energy are negligible, Eqn 1 can be simplified to :

Let's begin by applying the steady-state, SISO form of the 1st Law for open systems, with the nozzle as the system.

ENGR 224 - Thermodynamics

The keys to this problem are the 1st Law, the steam tables and the relationship between mass flow rate, volumetric flow rate, velocity, specific volume and cross sectional area for flow.

Steam at 3 MPa and 400oC enters an adiabatic nozzle at a velocity of 40 m/s and leaves at 2.5 MPa and 300 m/s. Determine…

The nozzle operates at steady-state, since no variable changes with respect to time.

The mass flow rate drops out and we can use the definition of the specific kinetic energy to obtain the following equation from Eqn 2.

ˆ ˆ ˆQ W m H E Esh kin pot

2 22 1

2 1C

v vˆ ˆH H 02g

kinˆ ˆ0 m H E

2 22 1

2 1C

v vˆ ˆH H2g

Nozzle

P1 = 3000 kPaT1 = 400 oCv1 = 40 m/s

P2 = 2500 kPaT2 = ? oCv2 = 300 m/s

Dr. Baratuci - ENGR 224 hw3-sp11.xlsm, 5.35 4/20/2011

Now, we can lookup H1 in the Steam Tables and plug values into Eqn 4 to evaluate H2 :

Ekin,1 0.80 kJ/kg H1 3231.7 kJ/kg

Ekin,2 45.0 kJ/kg H2 3187.5 kJ/kg

Now, because we know both P2 and H2, we can interpolate on the Steam Tables to determine T2.

At P2 = 2500 kPa : Hsat vap 2801.9 kJ/kg

At 2.5 MPa : T (oC) H (kJ/kg) V (m3/kg)300 2994.3 0.098937T2 3187.5 V2

400 3231.7 0.12012 T2 381.4oC

part b.)

Eqn 5

We can now solve Eqn 5 for the cross-sectional area ratio, A1 / A2 :

Eqn 6

We can evaluate the specific volume of the steam at both the nozzle inlet and outlet conditions :

V1 0.099379 m3/kg V2 0.11618 m3/kg

Now, we can plug values into Eqn 6 to complete this problem.

A1 / A2 6.416

Verify : None of the assumptions made in this problem solution can be verified.

Answers : a.) T2 381oC

b.) A1 / A2 6.42

Since H2 > Hsat vap, we can conclude that the steam leaving the nozzle is still a superheated vapor. So, we must interpolate in

the superheated steam table for 2.5 MPa.

The key to this part of the problem is the relationship between mass flow rate, volumetric flow rate, specific volume, cross-sectional area for flow and the average fluid velocity. At steady-state, the mass flow rate is constant, so m1 = m2.

1 1 1 2 2 21 2

1 1 2 2

V v A V v Am m

ˆ ˆ ˆ ˆV V V V

1 1 2

2 12

ˆA V vˆA vV

Dr. Baratuci - ENGR 224 hw3-sp11.xlsm, 5.35 4/20/2011

BaratuciHW #3

Problem : 5.54 - Adiabatic Gas Turbine - 5 pts 21-Apr-11

Read :

Given : P1 900 kPa P2 150 kPa

T1 450oC v2 150 m/s

723.15 K MW 39.95 g/molev1 80 m/s gc 1 kg-m/N-s2

A1 0.006 m2Wsh 250 kW

CoP 0.5203 kJ/kg-K R 8.314 J/mole-K

Find : T2 ???oC

Diagram :

Assumptions : 1 - The turbine operates at steady-state.2 - Change in potential energy is negligible.3 - The turbine is adiabatic.4 - Argon behaves as an ideal gas with constant specific heats.5 -


Apply the steady-state, SISO form of the 1st Law for open systems to the turbine.

Eqn 1

Eqn 2

We can express H in terms of the heat capacity and the temperature change using :

Eqn 3

Combining Eqns 2 & 3 yields : Eqn 4

We can solve Eqn 4 for T2, as follows : Eqn 5

Flow work and shaft work are the only forms of work that cross the system boundary.


Argon gas enters an adiabatic turbine at 900 kPa and 450oC with a velocity of 80 m/s and leaves at 150 kPa and a velocity of 150 m/s. The

inlet cross-sectional area is 60 cm2. If the power output of the turbine is 250 kW, determine the exit temperature of the argon. The process operates at steady-state and argon behaves as an ideal gas. The constant pressure specific heat of argon is 0.5203 kJ/kg-K.

The keys to this problem are the 1st Law and the Ideal gas EOS and the relationship between mass flow rate, volumetric flow rate, velocity, specific volume and cross sectional area for flow.

We can simplify Eqn 1 because the turbine is adiabatic and we have assumed that any change in potential energy is negligible.

sh kin potˆ ˆ ˆQ W m H E E

2 22 1

sh k 2 1c

v vˆ ˆ ˆ ˆW m H E m H H2g

2

1

T

2 1 P P 2 1T

ˆ ˆˆ ˆ ˆH H H C dT C T T

2 22 1

sh P 2 1c

v vˆW m C T T2g

2 2sh 2 1

2 1cP

v v1 WT T

ˆ 2gC m

2 22 1

sh P 2 1c

v vˆW m C T T2g

P1 = 900 kPaT1 = 450 oCv1 = 80 m/sA1 = 60 cm2

P2 = 150 kPaT2 = ? oCv2 = 150 m/sA1 = 60 cm2

Wsh = 250 kWQ = 0 kWCP = 0.5203 kJ/kg-K

Dr. Baratuci - ChemE 260 hw3-sp11.xlsm, 5.54 4/20/2011

Eqn 6

All we need to do now is determine the specific volume at the turbine inlet, V1.We can accomplish this using the ideal gas EOS, as follows.

Eqn 7

Solving Eqn 7 for the specific volume yields : Eqn 8

Plugging values into Eqns 5 & 6, in that order, yields :

V1 0.16722 m3/kg m 2.8705 kg/s

Finally, we can plug values back into Eqn 5 to complete this problem.

Ekin,1 3.20 kJ/kg

Ekin,2 11.3 kJ/kg

Ekin 8.05 kJ/kg T2 267.14 oC


Answers : T2 267.1 oC

So, we need to determine the mass flow rate. We can accomplish this because we are given the velocity and cross-sectional area for flow at the turbine inlet.

1 1 1

1 1

V v Am

ˆ ˆV V

1 1 1 1

mP V nRT RT

MW

1 11

1

V RTV

m P MW


BaratuciHW #3

Problem : 5.112 - Steam Flow in a HEX Tube - 5 pts 21-Apr-11

a.)

b.)

Read :

Given : State 1 State 2D 12 10 cmv 3 m/sP 2000 800 kPaT 300 250

oC

Find : a.) m ??? kg/s b.) Q ??? kW

Diagram :

Assumptions : 1 -2 - Changes in potential energy are negligible.3 - Flow work is the only form of work that crosses the system boundary.


Eqn 1

Eqn 2


This problem is an application of the steady-state form of the 1st Law for open systems.

Because both T & P are given at the inlet and outlet, the important properties specific enthalpy and specific volume can be directly determined from the steam tables, NIST Webbook or the TFT Plug-in.

The keys to the problem are the relationships between mass flow rate, volumetric flow rate, velocity, specific volume, cross-sectional area and pipe diameter.

Since we were asked to determine the mass flow rate and the heat transfer rate and both of these variables appear in the 1st Law for open systems (like this pipe!), let's begin by writing the 1st Law for open systems.

Since no moving parts were mentioned in the problem statement, it safe to assume that no shaft work crosses the system boundary. Since no elevation change was mentioned in the problem statement, it is safe to assume that changes in potential energy are negligible. As a result, Eqn 1 can be simplified to yield :

The mass flow rate of the steam

The rate of heat transfer

Steam enters a long, horizontal pipe with an inlet diameter of D1 = 12 cm at 2 MPa and 300oC with a velocity of 3 m/s. Farther downstream, the

conditions are 800 kPa and 250oC, and the diameter is D2 = 10 cm. Determine…

The pipe operates at steady-state, since no variable changes with respect to time.


2 22 1

kin 2 1C

v vˆ ˆ ˆ ˆQ m H E m H H2g

D2 = 10 cmP2 = 800 kPaT2 = 250

oC

D1 = 12 cmP1 = 2000 kPaT1 = 300

oC


Part a.)

Eqn 3

Eqn 4 A1 113.10 cm2

A1 0.01131 m2

We can also evaluate the inlet volumetric flow rate if we want : V1 0.03393 m3/s

Chemical : Water Units : SI_C V1 0.12547 m3/kg

m 0.2704 kg/s

Part b.)

H13023.0 kJ/kg H2

2949.5 kJ/kg

Last, we need to determine v2. We can do this by solving Eqn 3, as follows.

Eqn 5

V2 0.29314 m3/kg

Next, we can use Eqn 4 applied at the outlet of the pipe to determine A2.

Eqn 6 A2 78.54 cm2

A2 0.00785 m2

V2 0.07927 m3/s

v2 10.09 m/s

gC = 1 kg-m/N-s2 Q -19.87 kW


Answers : a.) m 0.270 kg/s b.) Q -19.9 kW

We can now complete this solution by plugging values into Eqn 2 to evaluate Q.

We can determine V2 using the steam tables, the NIST Webbook or the TFT Plug-in to evaluate H1 and H2. We can do this now

because T & P were given for the pipe effluent,state 2. I used the TFT Plug-in to obtain :

Plugging values into Eqn 4 yields :

Now, we plug values into Eqn 5 to get :

Now, we can plug values back into Eqn 3 to determine m and complete this part of the problem.

We can begin this part of the problem by using the steam tables, the NIST Webbook or the TFT Plug-in to evaluate H1 and H2.

We can do this now because T & P were given for both the inlet and outlet states. I used the TFT Plug-in to obtain :

Before we can use Eqn 4 to evaluate Q and complete this problem, we need to determine H1, H2 and v2.

Next, we need to determine the specific volume at state 1. We can use the steam tables, the NIST Webbook or the TFT Plug-in. We can do this because both P1 and T1 were given. I used the TFT Plug-

in to obtain :

We can determine the mass flow rate based on the information given at the pipe inlet using the following equation.

The velocity was given, so we need to determine the cross-sectional area for flow, A1. If we assume the cross-sectional area for

flow is circular, then we can determine A1 from :

Plugging values into Eqn 4 yields :

Eqn 2 shows that we need to determine ht e mass flow rate before we can use this form of the 1st Law to evaluate Q. So, it makes sense to complete part (a) now.

crossv AVm

ˆ ˆV V

21 1A D

4

22

2 2

ˆmV Vv

A A

22 2A D

4


BaratuciHW #3

Problem : 5.138 - Filling a Balloon with Helium - 10 pts 21-Apr-11

Read :

Given : V1 40 m3 Pin 125 kPa

P1 100 kPa Tin 25 oC

T1 17 oC 298.15 K

290.15 K P2 125 kPa

CV 3.1156 kJ/kg-K R 8.314 J/mol-K

CP 5.1926 kJ/kg-K MW 4.00 g/mole

Find : T2 ??? oC

Diagram :

Assumptions : 1-

2- The process is adiabatic.3-

4-

5-

A balloon initially contains 40 m3 of helium gas at atmospheric conditions of 100 kPa and 17oC. The balloon is connected by a valve to a large

reservoir that supplies helium gas at 125 kPa and 25oC. Now, the valve is opened and helium is alowed to enter the balloon until pressure equilibrium with the helium at the supply line is reached. The material of the balloon is such that its volume increases linearly with pressure. If no heat transfer takes place during this process, determine the final temperature of the helium in the balloon.

There are four keys to this problem. The first is that the He behaves as an ideal gas with constant heat capacities. This gives us a relatively simple relationship between P, V, T and moles or mass. It also simplifies the calculation of changes in U and H. The second key is to apply the transient form of the 1st Law to this adiabatic process. Although there is no shaft work inthis process, there is boundary work, and because the balloon expands during the process, Wb > 0. The third key is to make good use of the

properties of the balloon. Because P is linear with respect to volume, and we know the initial P and V and the final P, we can quickly determine the final volume. The final key to this problem is to recognize that you must choose a reference state so that you can evaluate U's and H's. Once you know how to use these four key aspects of this problem, the rest is just algebra !

Helium behaves as an ideal gas with constant CP and CV in this process.

The volume of He contained in the balloon increases linearly as the pressure within the balloon increases.

Uniform state: The state of the He in the balloon is uniform at any time during this process.

Uniform flow: The properties and flow rate of the air entering the balloon are constant over the cross-sectional area of the tube entering the balloon and the properties are also constant with respect to time. The flow rate into the balloon does not need to be constant in order to solve this problem.


V1 = 40 m3

P1 = 100 kPaT1 = 17

oC

V2 = ?? m3

P2 = 125 kPaT2 = ???

oC

Pin = 125 kPaT1 = 25

oC

Dr. Baratuci - ChemE 260 hw3-sp11.xlsm , 5.138 4/20/2011


Eqn 1

Eqn 2

Eqn 3

Eqn 3 can be simplified because no mass leaves the balloon.

Eqn 4

Next, we can apply the Ideal Gas EOS to the He in the balloon in both the initial and final states.

Eqn 5

Eqn 6 Eqn 7

m1 6.633 kg

Next, let's evaluate the boundary work done during this process from its definition.

Eqn 8

Eqn 9

Eqn 10

Solving Eqn 10 for V2 yields : Eqn 11

V2 50 m3

Now, we can plug V2 back into Eqn 9 to evaluate Wb : Wb 1125 kJ

This equation applies because the state within the system is indeed uniform and because the properties of the mass crossing the system boundary are constant. It is relevant because we know that U2 depends on T2 and T2 is what we are trying to determine.

Let's begin by writing the transient form of the 1st Law for uniform state, uniform flow processes:

We can simplify Eqn 1 because the process is adiabatic, no mass leaves the system (the balloon) and there is no shaft work in this process (only boundary work).

Next, we can write the integral form of the transient mass balance or conservation of mass equation:

We can immediately evaluate m1, but we cannot evaluate m2

because T2 is the unknown we are trying to determine.

Because the volume of the balloon is linear with respect to pressure, we get the following equation for the boundary work.

In order to evaluate Wb, we need to determine V2. We can do this by making use of the fact that volume is proportional pressure.

2 2 1 1 b in inˆ ˆ ˆm U m U W m H

2 1 inm m m

mP V nRT RT

MW

1 11

1

P Vm MW

RT 2 2

22

P Vm MW

RT

2 1 in outm m m m

2

1

V

bV

W PdV

2 1b 2 1

P PW V V

2

2 2

1 1

P V

P V

22 1

1

PV V

P

2 2 1 1 non flow in in out outˆ ˆ ˆ ˆm U m U Q W m H m H


Now, we need to write equation for the U's and H's in Eqn 2 in terms of known and unknow T's.

We can do this using the ideal gas heat capacities, as follows.

Eqn 12

Eqn 13

Eqn 14

Tref 290.15 K

Pref 100 kPa Uref 0 kJ/kg

Next, we need to determine Href from the definition of enthalpy.

Eqn 15

Eqn 16

But, Uref = 0 and the He behaves as an ideal gas, therefore:

Eqn 17 Href 603.1 kJ/kg

Now, because the heat capacities are constants, we can evaluate the integrals in Eqns 12 - 14.

Eqn 18 Eqn 19

Eqn 20

Hin 644.6 kJ/kg

Now, we can substitute Eqns 4, 6, 7, & 18 - 20 back into Eqn 2 to get :

Eqn 21

Distribute T2 :

Eqn 22

We cannot evaluate U2 immediately because

we do not know T2, but we can evaluate Hin.

The good news is that the only unknown left in Eqn 21 is T2. The bad news is that solving for T2 involves a bit of algebra. Here is

one way to solve Eqn 21 for T2.

Next, we must choose a reference state. It is often helpful to choose a state that actually exists in a problem to be the reference state. So, I will choose U = 0 for the initial state of He gas in the balloon.

2

ref

To

2 ref VT

ˆˆ Û U C dT

1

ref

To

1 ref VT

ˆˆ Û U C dT

2

ref

To

2 ref PT

ˆˆ ˆH H C dT

ref ref ref refˆ ˆ ˆH U P V

refref ref

R TˆP VMW

refref

R TH

MW

o2 V 2 ref

Û C T T

oin P in ref

ˆH C T T

1U 0 kJ /kg

o2 2 2 2V 2 ref b 1 in

2 2

P V P Vˆ ˆMW C T T W MW m HRT RT

o ref2 2 2 2 inV b 1 in

2 2

ˆTP V P V Hˆ ˆMW C 1 W MW m HR T R T


Collect the T2 terms on the right side of the equation :

Eqn 23

Solve for T2 : Eqn 24

Plug in values : 3006.98 kg-K

T2 315.30 K m2 9.54 kg

T2 42.15 oC U2 78.35 kJ/kg

Verify :

V1 24.12 L/mole

V2 20.97 L/mole Vin 19.83 L/mole


The only assumption that we can verify is the ideal gas assumption. We can verify this by determining the molar volume of He in the initial, final and inlet states and comparing the results to 5 L/mole (He is a noble gas).

Since the molar volume in all three states is significantly greater than 5 L/mole, the ideal gas assumption is valid.

o o2 2 2 2V b 1 in V ref in

2

P V P V1ˆ ˆˆ ˆMW C W m H MW C T HR T R

o2 2V ref in

2o2 2V b 1 in

P V ˆ ˆMW C T HR

TP V ˆ ˆMW C W m H

R

2 2P VMW

R


BaratuciHW #3

Problem : 5.142 - Charging a Cylinder with a Spring-Loaded Piston - 8 pts 21-Apr-11

Read :

Given : V1 0 m3 P (kPa) V (m3)Tin 200

oC Spring 300 0Pin = P2 1500 kPa Properties 3000 5

Find : T2 ???oC

Diagram :

Assumptions : 1 -

2 -

3 -4 -


An adiabatic piston-and-cylinder device equipped with a spring maintains the pressure inside at 300 kPa when the volume is zero and 3000

kPa when the volume is 5 m3. The device is connected to a steam line maintained at 1500 kPa and 200oC and initially the volume is zero. Determine the final temperature (and quality if appropriate) when the valve is opened and steam in the line is allowed to enter the cylinder until the pressure inside matches the pressure in the line. Also determine the total work produced during this adiabatic filling process.

This problem is a transient mass and energy balance on a uniform-state, uniform-flow process.

The process is a quasi-equilibrium process. This allows us to calculate the boundary work done using the integral of P dV.

The spring is a linear spring. The process path is a straight line on a PV Diagram.Uniform state: The state of the steam in the cylinder is uniform at any time during this process.

Uniform flow: The properties and flow rate of the steam entering the cylinder are constant over the cross-sectional area of the pipe entering the cylinder and the properties are also constant with respect to time. The flow rate into the cylinder does not need to be constant in order to solve this problem.

The twist is that boundary work is done against the spring as the cylinder fills with steam and the pressure rises.The main complication is that in the final state, the temperature is unknown (T2) and both the final specific volume and the final

internal energy depend on T2. This is a significant complication. It makes the soution of this problem iterative. This iterative

solution is easily accomplished using the TFT Plug-in for thermodynamic properties along with the Solver add-in for excel. If either or both of these tools is unavailable, the solution becomes one of trial and error. The NIST Webbook can make trial and error a bit less painful compared to suing paper steam tables.

V1 = 0 m3

P1 = 300 kPaV2 = ??? m

3

P2 = 1500 kPa



The filling of the cylinder is a transient process because the mass of water inside the cylinder increases as a function of time.

Eqn 1

In this problem the only form of non-flow work is boundary work done against the spring pushing on the back of the piston.

Eqn 2

The transient mass balance or conservation of mass equation for this process is:

Eqn 3

Eqn 4

We can use Eqn 4 to eliminate min from Eqn 2 : Eqn 5

We need to solve Eqn 5 for U2 because if we know the value of U2, we can use thermodynamic data to determine T2.

Eqn 6

Because the spring is linear, the process path is linear on a PV Diagram and Eqn 6 can be simplified to obtain :

Eqn 7

P (kPa) V (m3)300 0

1500 V2

3000 5 V2 2.222 m3

Now, we can plug values into Eqn 7 to evaluate Wb : Wb 2000 kJ

Chemical : Water Hin 2796.3 kJ/kgUnits : SI_C

Now, let's try to determine m2. We know that mass and volume are related by : Eqn 8

Eqn 9

The characteristics of the spring given in the problem statement lead us to conclude that P1 = 300 kPa when the cylinder

begins to fill with steam and the pressure in the cylinder increases linearly with pressure until it reaches P2 = 1500 kPa.

The linear characteristics of the spring also allow us to conclude that V1 = 0 m3 and that V2 can be determined by linear

interpolation, as follows.

The next step in resolving Eqn 5 is to evaluate Hin. We can do this using the steam tables, the NIST Webbook or the TFT

Plug-in. I used the TFT plug-in to obtain :

This introduces a new level of complexity into Eqn 5 because the specific volume in state 2 depends on T2 !

So, next, let's try to evaluate the boundary work. Because we have assumed the process to be quasi-equilibrium, we can evaluate Wb using :

Eqn 3 can be simplified because there is initially no mass inside the cylinder (m1 = 0) and no mass flows out of the cylinder

(mout = 0).

So, let's begin by writing the transient form of the 1st Law as it applies to systems undergoing uniform flow and uniform state processes.

Eqn 1 can be further simplified because the cylinder is initially empty (m1 = 0), the process is adiabatic (Q = 0) and there is no

flow out of the balloon (mout = 0). The resulting form of the 1st Law is :

2 2 1 1 non flow in in out outˆ ˆ ˆ ˆm U m U Q W m H m H

2 2 b in inˆ ˆm U W m H

2 1 in outm m m m

2 inm m

2 2 b 2 inˆ ˆm U W m H

2

1

V

bV

W PdV

2 1b 2 1

P PW V V

2

22

2

Vm

V=

2 22 b in

2 2

V Vˆ ˆU W Hˆ ˆV V


Let's rearrange Eqn 9, as follows : Eqn 10

T2 (oC) V2 (m

3/kg) U2 (kJ/kg) m2 (kg)210 0.13660 2618.4 16.268

Now, use Solver to minimize the /error/ by changing T2. The results are shown below.

233.86 0.14594 2665.0 15.227

T2 (oC) V2 (m

3/kg) U2 (kJ/kg) m2 (kg)210 0.13660 2618.4 16.268220 0.14059 2638.3 15.806230 0.14447 2657.7 15.382240 0.14825 2676.5 14.990235 0.14637 2667.1 15.183231 0.14485 2659.6 15.342232 0.14523 2661.5 15.302233 0.14561 2663.3 15.262234 0.14599 2665.2 15.222

233.5 0.14580 2664.3 15.242233.75 0.14589 2664.8 15.232233.86 0.14594 2665.0 15.227

The result is : T2 233.9 oC



2012.2 12.22003.7 3.72000.0 0.0

2063.5 63.52029.2 29.21995.2 4.8

1796.5 203.51961.5 38.52098.1 98.1

2895.0 895.02497.2 497.22132.9 132.9

2000.0 0.0

If you do not use the TFT Plug-in or yu do not use Solver, you can create a table by guessing T2 repeatedly until Eqn 10 is

satisfied. Such a table is shown below.

This problem can be solved very quickly using the TFT Plug-in and the "Solver" add-in for Excel, as follows.The problem is that because both specific U2 and specific V2 depend on T2, Eqn 10 must be solved iteratively !

2895.0 895.0

b 2 in 2ˆ ˆW m H U

b 2 in 2ˆ ˆerror W m H U 2 in 2

ˆ ˆm H U

b 2 in 2ˆ ˆerror W m H U 2 in 2

ˆ ˆm H U


BaratuciHW #3

Problem : WB-1 - Effluent Pressure in a Non-Adiabatic Steam Diffuser - 5 pts 21-Apr-11

Read :

Given : P1 14.7 psia x2 1 kg vap/kg

T1 300oF v2 0 ft/s

v1 500 ft/s Q/m -19.59 Btu.lbm

gc 32.174 lbm-ft/lbf-s2 Epot 0 kJ/kg

1 Btu 778.17 ft-lbf

Find : P2 ??? psia

Diagram :

Assumptions : 1 - The diffuser operates at steady-state.2 - Changes in potential energy are negligible.3 - Flow work is the only form of work that crosses the system boundary.


Apply the steady-state, SISO form of the 1st Law for open systems to the diffuser.

Eqn 1

Eqn 2

Where : Eqn 3

Eqn 4


Steam enters a diffuser at a pressure of 14.7 psia, a temperature of 300oF and a velocity of 500 ft/s. Steam exits the diffuser as a saturated vapor with negligible kinetic energy. Heat transfer occurs from the steam to the surroundings at a rate of 19.59 Btu/lbm of flowing steam.

Neglecting potential energy effects, determine the exit pressure in psia. Assume the diffuser operates at steady-state.

This is a straightforward application of the 1st Law to a non-adiabatic diffuser operating at steady-state with negligible changes in potential energy.

The keys to this probelm are the 1st Law and the relationship between mass flow rate, volumetric flow rate, velocity, specific volume and cross sectional area for flow.

Because no shaft work crosses the system boundary and because we assumed that changes in potential energy are negligible, Eqn 1 can be simplified to :

If we knew the value of H2, we could determine P2 because we know that the effluent is saturated vapor, x2 = 1.0. Toward that

end, we can solve Eqn 1 for the unknown H2.


kin 2 1 kinˆ ˆ ˆ ˆ ˆQ m H E m H H E

2 22 1

kinC

v vE

2g

2 1 kin

Qˆ ˆ ˆH H Em

Diffuser

P2 = ? psiax2 = 1 lbm vap/lbmv2 = 0 ft/s

P1 = 14.7 psiav1 = 500 ft/sT1 = 300 oF

Dr. Baratuci - ChemE 260 hw3-sp11.xlsm, WB-1 4/20/2011

Because we know both T1 and P1, we can lookup H1 in the Steam Tables.

T1 300oF

Tsat 211.95oF

H1 1192.7 Btu/lbm

Ekin,1 3885 ft-lbf/lbm Ekin,2 0.0 kJ/kg4.993 Btu/lbm 0.0 Btu/lbm

Ekin -4.993 Btu/lbm

Now, we can use Eqn 4 to evaluate H2. H2 1178.1 Btu/lbm

P (psia) H (Btu/lbm)

60 1177.8P2 1178.1

65 1179.4 P2 60.95 psia


Answers : P2 60.9 psia

Next, we can plug values into Eqn 3 because we know v1 and we have assumed that the effluent kinetic energy is negligible.

Now, because we know both x2 = 1 and the value of H2, we can interpolate on the saturated vapor column of the saturation

pressure table to determine P2.

Since T1 > Tsat, we can conclude that the feed is a superheated vapor.

We can determine the phase of the feed stream by comparing T1 to Tsat at P1 = 14.7 psia.

In the superheated steam table for 14.7 psia, we find the value of H1.


BaratuciHW #3

Problem : WB-2 - Steady-State, Polytropic Air Compressor - 5 pts 21-Apr-11

Read :

Given : P1 14.7 psia P2 120 psia

T1 70 oF v2 700 ft/s529.67 oR 1 ft2 = 144 in2

V1 500 ft3/s R 1545.37 ft-lbf/lbmole-oR 1.34 MW 28.97 lbm/lbmole

Find : T2 ???oF D2 ??? in

Diagram :

Assumptions : 1 - The compressor operates at steady-state.2 - Changes in potential energy are negligible.3 - Shaft work and flow work are the only forms of work that cross the system boundary.4 - The cross-sectional area for flow in the exit pipe is circular.5 - Air behaves as an ideal gas throughout the process. This assumption will be verified.


Eqn 1

Eqn 2

Eqn 3

Plugging values into Eqns 3 & 1 yields : V1 13.348 ft3/lbm

m 0.6243 lbm/s

Now that we know the mass flow rate, we can use it to determine the diameter of the exit pipe, as follows.

Begin by solving Eqn 1 for the volumetric flow rate of stream 2. Eqn 4


Air enters a compressor at steady-state with a pressure of 14.7 psia, a temperature of 70°F, and a volumetric flow rate of 500 ft3/min. The air velocity in the exit pipe is 700 ft/s and the exit pressure is 120 psia. If each unit mass of air passing through the compressor undergoes a

process described by PV1.34 = constant, determine the exit temperature, in oF, and the diameter of the exit pipe in inches.

Th e key to this problem is that the mass flow rate is the same at the outlet as it is at the inlet. So, let's begin by seeing if we have enough information about the compressor feed to evaluate the mass flow rate.

The volumetric flow rate of the feed is given, so we need to determine the specific volume of the feed. We can accomplish this by applying the Ideal Gas EOS.

Solving Eqn 2 for the specific volume yields :

P1 = 14.7 psiaT1 = 70

oFV1 = 500 ft

3/min

P2 = 120 psiaT2 = ???

oFv2 = 700 ft/sD2 = ??? in

Compressor

1

1

Vm

V=

1 1 1 1

mP V nRT RT

MW

1 11

1

V RTV

m P MW

2 2ˆV m V=

The key to determining the specific volume of the effluent is the polytropic process path: P V1.34 = constant.

Both the feed and the inlet states lie on this process path. Therefore:

Eqn 5

We can now solve Eqn 5 for V2 : Eqn 6

Next, we can use the volumetric flow rate and the given velocity to determine the cross-sectional area of the exit pipe, A2 :

Eqn 6 Eqn 7

Assuming the exit pipe is circular in cross-section : Eqn 8

Solving Eqn 8 for D2 yields : Eqn 9

Now, we can plug values into Eqns 6, 4, 7 & 9 to complete this part of the problem.

V2 2.786 ft3/lbm A2 0.002484 ft2

V2 1.739 ft3/s D2 0.0562 ft

D2 0.675 in

Eqn 10 Eqn 11

Plugging values into Eqn 11 yields : T2 902.35 oR

T2 442.68 oF

Verify : Eqn 12

Plugging values into Eqn 12 yields : V1 386.7 ft3/lbmole

V2 80.70 ft3/lbmole

Answers : T2 443 oF D2 0.675 in

We can now apply the Ideal gas EOS to determine T2 because we know P2 and V2. Apply Eqn 3 to the compressor effluent and

solve for T2, as follows :

We can test the Ideal Gas assumption by determining the molar volume of the feed and the effluent from :

The rule of thumb is that a gas be considered to be an ideal gas, to an accuracy of 2 significant figures, if the molar volume is

greater than 350 ft3/lbmole. So, it is very reasonable to treat the compressor feed as an ideal gas. What about the compressor

effluent ? For diatomic and noble gases the molar volume must only be greater than or equal to 70 ft3/lbmole. Air can sometimes be treated as a diatomic gas since its main consitituents are O2 and N2. To the extent that this rule can be applied to air, the ideal gas assumption is valid for the compressor effluent.

2 2 2V v A 2

22

VA

v

22 2A D

4

2 2

4D A

1.34 1.341 1 2 2

ˆ ˆP V constant P V= =

1/1.34

12 1

2

Pˆ ˆV VP

æ öç ÷= ç ÷ç ÷ç ÷è ø

22

2

RTV

P MW 2 2

2

ˆP VT

R / MW

ˆV V MW

BaratuciHW #3

Problem : WB-3 - Analysis of a Two-Stage, Adiabatic Turbine - 6 pts 21-Apr-11

a.)

b.)

Read :

Given : P1 3000 kPa P2 = P3 = P4 500 kPaT1 400

oC T2 = T3 = T4 180oC

V1 85 m3/min P5 6 kPav4 20 m/s x5 0.9 kg vap/kg

Ws,tot 11400 kW

Find : a.) m4 ??? kg/sm5 ??? kg/s

b.) D4 ??? m

Diagram : See above.

Assumptions : 1 - The entire process and each device operate at steady-state.2 - Changes in potential energy are negligible.3 - Flow work is the only form of work in the splitter.4 - Shaft work and flow work are the only forms of work in the turbines.5 - Each device in the process is adiabatic.


Part a.)

Eqn 1

Eqn 2

Eqn 1 can be simplified because we have assumed that the turbines are adiabatic and that changes in kinetic and potential energies are negligible.


A well-insulated two-stage turbine operating at steady-state is shown in the diagram. Steam enters at 3 MPa and 400oC with a volumetric

flow rate of 85 m3/min. Some steam is extracted from the turbine at a pressure of 0.5 MPa and a temperature of 180oC. The rest expands to a pressure of 6 kPa and a quality of 90%. The total power developed by the turbine is 11,400 kW. Changes in kinetic and potential energies are negligible. Determine:

Let's begin by applying the steady-state, SISO form of the 1st Law for open systems to each turbine.

The key to this problem is the splitter. It is just like a tee in a pipe. So, the properties of streams 2, 3 & 4 are the same.

Apply the 1st Law to the HP Turbine and determine Wsh,HP. The remaining work is Wsh,LP. Apply the 1st Law to the LP

Turbine to determine m5. Use a mass balance on the entire system or just the splitter to determine m4.

A faster solution can be done by applying the 1st Law to the entire system and solving the equation simultaneously with a mass balance on the entire system.

The diameter in meters of the duct through which the 0.5 MPa steam is extracted if the velocity there is 20 m/s.

The mass flow rate of the steam at each of the two exits.


P1 = 3000 kPaT1 = 400 oCV1 = 85 m3/min

P4 = 500 kPaT4 = 180 oC

P2 = 500 kPaT2 = 180 oC

P3 = 500 kPaT3 = 180 oC P5 = 6 kPa

x5 = 0.9 kg vap/kg


shˆW m H


Applying Eqn 2 to each turbine yields :

Eqn 3

Eqn 4

Eqn 5

We can solve Eqn 5 for the unknown m5 :

Eqn 6

Start by comparing the stream temperatures to the Tsat value at each stream pressure.

Stream

P(kPa)

Tsat(P)

(oC)

T

(oC)

1 3000 233.85 4002, 3, 4 500 151.83 180

H1 3231.7 kJ/kg

At 500 kPa : T (oC) H (kJ/kg)

151.83 2748.1180 H2

200 2855.8 H2,3,4 2811.1 kJ/kg

Psat

(kPa)Hsat vap

(kJ/kg)Hsat vap

(kJ/kg)

5 137.75 2560.7 Hsat liq 150.2 kJ/kg6 Hsat liq Hsat vap Hsat vap 2566.0 kJ/kg

7.5 168.8 2574.0 H5 2324.4 kJ/kg

We can determine m1 from the given volumetric flow rate, using:

Eqn 7

V1 0.099379 m3/kg m1 14.255 kg/s

Finally plugging values into Eqn 6 yields : m5 11.105 kg/s

The total shaft work produced by this system is the sum of the shaft work produced by the turbines because the stream splitter does not produce or consume any shaft work.

We still need to evaluate m1 and all of the H's. Let's begin by evaluating all the enthalpies we need in Eqn 6.

Since both T1 and T2,3,4 are greater that the associated Tsat value, we conclude that streams 1, 2, 3 & 4 are superheated

vapors.

No interpolation is required to determine H1. We can read it right out of the superheated steam table.

We must interpolate between Tsat = 151.83oC and 200oC on the superheated steam table for 500 kPa in order to determine

H2,3,4.

We must also interpolate on the saturation pressure table to determine H5 because there is not a row for Psat = 6 kPa.

We can read the specific volume at state 1 directly from the superheated steam tables and plug values into Eqn 7.

1sh,HP 1 2ˆ ˆW m H H

5sh,LP 3 5ˆ ˆW m H H

1 5sh,tot sh,HP sh,LP 1 2 3 5ˆ ˆ ˆ ˆW W W m H H m H H

11

1

Vm

V

1sh,tot 1 25

3 5

ˆ ˆW m H Hm

ˆ ˆH H


Eqn 8

Solving Eqn 8 for m4 yields : Eqn 9

Plugging values into Eqn 8 yields : m4 3.151 kg/s

Part b.)

Eqn 10

We can solve Eqn 10 for the cross-sectional area for flow for stream 4.

Eqn 11

First we need to evaluate V4 by interpolation on the superheated steam table for 500 kPa.

At 500 kPa : T (oC) V (m3/kg)

151.83 0.37481180 V4

200 0.42503 V2,3,4 0.40418 kJ/kg

Plugging values into Eqn 11 yields : A4 0.06367 m2

Assuming the cross-sectional area for flow is circular : Eqn 12

Solving Eqn 12 for D4 yields : Eqn 13

Plugging values into Eqn 12 yields : D4 0.2847 m

D4 28.47 cm


Answers : a.) m5 11.1 kg/s m4 3.15 kg/s

b.) D4 28.5 cm

In order to determine m4, we need to write a mass balance equation. You can think of the mass balance as being written on the entire process or just on the stream splitter. In either case, the equation is :

The key to this part of the problem is the relationship between mass flow rate, volumetric flow rate, specific volume, cross-sectional area for flow and the average fluid velocity. At steady-state, the mass flow rate is constant, so m1 = m2.

1 4 5m m m

4 1 5m m m

4 4 44

4 4

V v Am

ˆ ˆV V

4 44

4

ˆm VA

v

24 4A D

4

44

4AD


BaratuciHW #3

Problem : WB-4 - Analysis of an Adiabatic Steam De-Superheater - 8 pts 21-Apr-11

a.)

b.)

Read : This is a straightforward application of the steady-state, MIMO form of the 1st Law.

Given : P1 3000 kPa T1 320oC

P3 2500 kPa T3 200oC

P5 2000 kPa m1 15.0 kg/sx5 1.0 kg vap/kg

Find : a.) m3 ??? kg/s

b.) Plot m3 as a function of T3 for T3 = 20 to 220oC.

Diagram : See above.

Assumptions :1 - The desuperheater is adiabatic. Q = 0.2 - Changes in kinetic and potential energies are negligible.3 - The desuperheater operates at steady-state.4 -

Interpolation between saturated liquid and the lowest pressure subcooled liquid table is required in order to determine the enthalpy of the subcooled liquid entering the de-superheater.


As shown in the diagram, 15 kg/s of steam enters a de-superheater operating at steady-state at 30 bar and 320oC where it is mixed with liquid water at 25 bar and temperature T3 to produce saturated vapor at 20 bar. Heat transfer between the device and its surroundings and

changes in kinetic and potential energies are negligible.

If T3 = 200oC, determine the mass flow

rate of stream 3.

Plot the mass flow rate of stream 3 in kg/s as a function of T3 as T3 ranges from

20 to 220oC.

The de-superheater behaves much like a mixer except the outlet pressure is not necessarily the same as the inlet stream pressures.

No shaft work or boundary crosses the boundary of the system because the desuperheater has a constant volume and has no moving parts.

P1 = 3000 kPaT1 = 320 oCm1 = 15 kg/sSuperheatedVapor

P3 = 2500 kPaT3 = 200 oCm3 = ??? Kg/sSubcooledLiquid

P5 = 2000 kPaT5 = Tsatx5 = 1.0 kg vap/kgSaturated Vapor



Part a.)

Eqn 1

Eqn 2

Eqn 3

Now, we can use Eqn 3 to eliminate m5 from Eqn 2 :

Eqn 4

Now, we can solve for m3 because that is the variable we need to determine.

Eqn 5

At 3 MPa : T (oC) H (kJ/kg)

300 2994.3320 H1

400 3231.7 H1 3041.8 kJ/kg

We can verify that stream 3 is a subcooled liquid by comparing T3 to Tsat at P3 = 2.5 MPa.

T3 200oC Tsat(2.5 MPa) = 233.95

oC

Because T3 < Tsat, we conclude that stream 3 is a subcooled liquid.

At 200oC : P (MPa) H (kJ/kg)

1.5549 852.272.5 H1

5.0 853.68 H3 852.66 kJ/kg

H5 2801.9 kJ/kg

Now, we can plug values back into Eqn 5 to complete the problem.

m3 1.846 kg/s

Finally, we need to evaluate H5. Because it is a saturated vapor, this value comes straight out of the saturated pressure table in

the steam tables with no calculations required.

There are two equivalent approaches to solving this problem. We could assume that the valves are isenthalpic so that H2 = H1

and H4 = H3 and then appply the 1st Law to the desuperheater alone. Or we could choose the desuperheater AND the two

valves as our system and apply the 1st Law. The result is the same either way.

So, let's begin by writing the 1st Law applied to the larger system that includes the de-superheater and the two valves. This is a steady-state MIMO process, so the appropriate form of the 1st Law is :

We can simplify Eqn 1 by assuming that the desuperheater and the valves are adiabatic, have no shaft work crossing the system boundary and that changes in kinetic and potential energies are negligible.

Next we can apply a mass balance to the system consisting of the de-superheater and the two valves :

Now, all we need to do is to determine the specific enthalpies of all three streams and plug these values into Eqn 5 to complete the problem.

Stream 1 is a superheated vapor and although we have a table for 3 MPa, we must interpolate because there is not a row for

320oC.

There is no subcooled liquid table for 2.5 MPa. Therefore, in order to determine H3, we must interpolate between the enthalpy

of saturated liquid at T3 = 200oC at Psat and subcooled liquid at T3 = 200oC and 5 MPa.

1 3 51 3 5ˆ ˆ ˆm H m H m H

1 3 5m m m

1 3 1 31 3 5ˆ ˆ ˆm H m H m m H

1 53 1

5 3

ˆ ˆH Hm m

ˆ ˆH H

#outlets #inlets

out, j in,iS out kin,out pot ,out in kin,in pot ,injj 1 i 1 i

ˆ ˆ ˆ ˆ ˆ ˆQ W m H E E m H E E


Part b.)

T3 H3 m3 Chemical = WateroC kJ/kg kg/s Units = SI_C

20 85.824 1.32530 127.58 1.34540 169.28 1.36750 210.99 1.38960 252.72 1.41270 294.51 1.43580 336.38 1.45990 378.32 1.485100 420.36 1.511110 462.53 1.538120 504.85 1.566130 547.34 1.596140 590.04 1.627150 632.97 1.659160 676.17 1.693170 719.67 1.728180 763.50 1.765190 807.71 1.804200 852.35 1.846210 897.49 1.889220 943.20 1.936


Answers: a.) m3 1.85 kg/s b.) See plot, above.

In order to construct a plot of cooling water mass flow rate as a function of the temperature of the cooling water, we must repeat the procedure from part (a) for each data point we want to plot. The only values that change are H3 and m3 (determined using

Eqn 5). The interpolation is just too painful. I decided to use the TFT Plug-in for thermodynamic properties of the subcooled liquid.The table of results is shown below.

1.00

1.10

1.20

1.30

1.40

1.50

1.60

1.70

1.80

1.90

2.00

20 70 120 170 220

Req

uir

ed C

oo

lin

g W

ater

Flo

w R

ate

(k

g/s

)

Cooling Water Temperature (oC)

De-Superheater Performance Analysis


BaratuciHW #3

Problem : WB-5 - Waste Heat Steam Generator - 8 pts 21-Apr-11

CP air = 7.05 Btu/lbmole-oF.

Read :

Given : P1 42 psia V4 3000 ft3/min

T1 220oF R 0.7302 atm-ft3/lbmol-oR

P2 40 psia T5 280oF

T2 320oF CP,air 7.05 Btu/lbmole-oF

P3 1 psia 0.2431 Btu/lbm-oFx3 0.9 kg vap/kg MWair 28.97 lbm/lbmoleP4 1 atm 1 hP 2545 Btu/hT4 360

oF

819.67oR

Find : Wsh ??? hP

Assumptions : 1 - The HEX and turbine are operating at steady-state.2 - Changes in potential energy are negligible.3 - The HEX and the turbine are both adiabatic.4 - Flow work is the only form of work that crosses the HEX boundary.5 -


At steady-state, water enters the waste heat recovery steam generator shown in the diagram at 42 psia and 220oF and exits at 40 psia and

320oF. The steam is then fed into a turbine from which it exits at 1 psia and a quality of 90%.

Air from an oven exhaust enters the steam generator at

360oF and 1 atm with a volumetric flow rate of 3000

ft3/min and exits at 280oF and 1 atm. Ignore all heat exchange with the surroundings and any changes in potential and kinetic energies. Determine the power developed by the turbine in horsepower.

The key here is to aply the 1st Law to the HEX to determine the mass flow rate of the water in streams 1, 2 & 3. Then, apply the 1st Law to the turbine to determine its power output in hP.

Flow work and shaft work are the only forms of work that cross the turbine boundary.

The analysis of the turbine is straightforward. The analysis of the HEX can be approached in at least two different ways that yield the same result. The key is that the HEX does not exchange heat with the surroundings. Therefore, all of the heat that leaves the air enters the water.

Assume each device is adiabatic, changes in kinetic and potential energies are negligible and the air behaves as an ideal gas with a constant CP.

P3 = 1 psiax3 = 0.9 lbm vap/lbmSaturated Mixture

P2 = 40 psiaT2 = 320 oFSuperheated Vapor

P1 = 42 psiaT1 = 220 oFSubcooled Liquid

P4 = 1 atmT4 = 360 oFV4 = 3000 ft3/minIdeal gas

P5 = 1 atmT5 = 280 oFIdeal gas



Part a.)

Eqn 1

Eqn 2

We can obtain H3 right away because we know the pressure and the quality.

Eqn 3

Plugging values into Eqn 3 at 1 psia yields : Hsat liq 69.73 Btu/lbm

Hsat vap 1105.4 Btu/lbm

H3 1001.8 Btu/lbm

Next, let's determine H2. We can do this because we know both T2 and P2.

At P2 = 40 psia : Tsat 267.22oF

T2 320oF

At 40 psia : T (oF) H (Btu/lbm)

300 1186.9320 H2

350 1212.0 H2 1196.9 Btu/lbm

Now, all we need is the value of m1 and we can plug values into Eqn 2 to complete this solution.

We can approach the HEX in either of two ways.Method 1: Use the MIMO form of the 1st Law with Q = 0 and solve for m1.Method 2:

Method 1 :

Eqn 4

Eqn 5

Eqn 6


Now, we can group the air and water terms, as follows, and make use of the fact that m1 = m2 and m4 = m5.

We can determine m1 by applying the 1st Law to the HEX because all of the heat leaving the air enters the water to produce the

saturated steam in stream 2.

Use the SISO form of the 1st Law applied to the air only to determine the rate of heat thransfer from the air into the

water. Then, apply the SISO form of the 1st Law applied to the water to evaluate m1.

We can simplify Eqn 4 by assuming that the HEX is adiabatic (all the heat leaving the air enters the water), has no shaft work crosses the system boundary and that changes in kinetic and potential energies are negligible.

Since T2 > Tsat, we can conclude that the steam entering the turbine is a superheated vapor. So, we

must interpolate in the superheated steam table for 40 psia.

Eqn 1 can be simplified because we have assumed that the turbine is adiabatic and that changes in kinetic and potential energies are negligible.

Let's begin by applying the steady-state, SISO form of the 1st Law for open systems to the turbine.

1sh kin potˆ ˆ ˆQ W m H E E

1 1sh 2 3ˆ ˆ ˆW m H m H H

sat sat satmix vap liq

ˆ ˆ ˆH x H 1 x H

#outlets #inlets

out, j in,iS out kin,out pot ,out in kin,in pot ,injj 1 i 1 i

ˆ ˆ ˆ ˆ ˆ ˆQ W m H E E m H E E

1 4 2 51 4 2 5ˆ ˆ ˆ ˆm H m H m H m H

1 42 1 4 5ˆ ˆ ˆ ˆm H H m H H

4 51 4

2 1

ˆ ˆH Hm m

ˆ ˆH H


Method 2 :

Eqn 8

Simplifying Eqn 8 based on these assumptions yields :

Eqn 9

When we apply the 1st Law to the water flowing through the HEX, we get the following equations.

Eqn 10

Eqn 11

Eqn 12

Combining Eqns 9, 11 & 12 yields :

Eqn 13

Now, all we need to do is evaluate all the terms on the right-hand side of Eqn 7.

Eqn 8


Plugging values into Eqn 8 yields : m4 145.21 lbm/min

Eqn 10

We can determine the specific heat using : Eqn 11

Plugging values into Eqn 11 and then Eqn 10 yields : CP,air 0.24336 Btu/lbm-oFH4 - H5 19.468 Btu/lbm

Apply the 1st Law to the air stream assuming no shaft work and negligible changes in kinetic and potentail energies.

The connection between the air energy balance and the water energy balance is that all of the energy that leaves the air enters the water. Because of our sign convention for heat transfer, Qair < 0 and Qwater > 0. Expressing these ideas in terms of an

equation yields :

Eqn 13 is identical to Eqn 6 ! So it doesn't matter which method you use, you still end up using Eqn 7 to evaluate m1.

Let's begin by evaluating m4. We can do this because we know the volumetric flow rate of the air and we have assumed it is an

ideal gas.

Next, we need to evaluate the change in enthalpy of the air using the constant CP value given in the problem statement.

4

44

mP V n R T R T

MW

44

P Vm MW

R T

4

5

T

4 5 P P 4 5T

ˆ ˆˆ ˆH H C dT C T T

PP

CC

MW

4shair kin potˆ ˆ ˆQ W m H E E

4air 5 4ˆ ˆQ m H H

1shwater kin potˆ ˆ ˆQ W m H E E

1water 2 1ˆ ˆQ m H H

water airQ Q

1 42 1 4 5ˆ ˆ ˆ ˆm H H m H H


Now, we need to determine H1 so we can finally use Eqn 7.Since we know T1 and P1, we can use the steam tables to determine H1.

P (psia) Tsat (oF)

40 267.2242 Tsat T1 220 oF45 274.42 Tsat 270.10 oF

Begin by determining the specific enthalpy of subcooled liquid water at 500 psia and 220oF.

P (psia) Tsat (oF) H (Btu/lbm)

500 250 219.6500 220500 200 169.2 H(500 psia, 220oF) = 189.37 Btu/lbm

Tsat (oF) P (psia) H (Btu/lbm)

220 17.201 188.25220 42500 220 189.4 H1 188.39 Btu/lbm

Finally plugging all the values back into Eqn 7 yields : m1 2.803 lbm/min

Note : Qwater = -Qair = 2827.0 Btu/min

Part b.) We can now go all the way back to Eqn 2 and plug in numbers to determine Wsh for the turbine.

Ws 546.88 Btu/min Ws 12.89 hP


Answers : Ws 12.9 hP

Use this result and the specific enthalpy of saturated liquid water at 220oF to interpolate and

determine H1 at 42 psia and 220oF.

Since T1 < Tsat, we can conclude that the water entering the HEX is a subcooled liquid. So, we must interpolate between the

saturated liquid and the subcooled liquid table for 500 psia.


3 mpa 400oc 40 m/s 2.5 mpa 300 m/s · pdf fileargon gas enters an adiabatic turbine at 900 kpa...

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