Steam enters a turbine from a 2 inch diameter pipe, at 600 psia, 930 F, with a velocity of 620 ft/s. It leaves the turbine at 12 psia with a quality of 1.0, through an outlet duct 1 ft in diameter. Calculate the turbine power output


Answer 1


\dot W_(out) = 3374.289\,(BTU)/(s)


The model for the turbine is given by the First Law of Thermodynamics:

- \dot W_(out) + \dot m \cdot (h_(in) - h_(out)) = 0

The turbine power output is:

\dot W_(out) = \dot m\cdot (h_(in)-h_(out))

The volumetric flow is:

\dot V = (\pi)/(4) \cdot \left( (2)/(12)\,ft \right)^(2)\cdot (620\,(ft)/(s) )

\dot V \approx 13.526\,(ft^(3))/(s)

The specific volume of steam at inlet is:

State 1 (Superheated Steam)

\nu = 1.33490\,(ft^(3))/(lbm)

The mass flow is:

\dot m = (\dot V)/(\nu)

\dot m = (13.526\,(ft^(3))/(s) )/(1.33490\,(ft^(3))/(lbm) )

\dot m = 10.133\,(lbm)/(s)

Specific enthalpies at inlet and outlet are, respectively:

State 1 (Superheated Steam)

h = 1479.74\,(BTU)/(lbm)

State 2 (Saturated Vapor)

h = 1146.1\,(BTU)/(lbm)

The turbine power output is:

\dot W_(out) = (10.133\,(lbm)/(s) )\cdot (1479.1\,(BTU)/(lbm)-1146.1\,(BTU)/(lbm))

\dot W_(out) = 3374.289\,(BTU)/(s)

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In contrasting the read-evaluation loop and the notification-based paradigm for inter- active programs, construction of a pre-emptive dialog was discussed. How would a programmer describe a pre-emptive dialog by purely graphical means? (Hint: Refer to the discussion in Sec- tion 8.5 concerning the shift from external and independent dialog management to presentation control of the dialog)


The way a programmer describe a pre-emptive dialog by purely graphical means is; by producing a window that covers the entire screen to make it the currently selected window.

What is Pre - emptive Dialogue?

In a graphics - based interaction, it is supposed that the user can only interact with parts of the system that are visible. However, In a windowing system, the user can only direct input to a single window that was currently selected and the way to change that selected window is to indicate with some gesture within that window.

Finally, to create a pre-emptive dialog, the system would do so through the production of a window that covers the entire screen to make it the currently selected window. Thereafter, all user input would be directed to that window and the user would have no means of selecting any other window. Then the covering window will now pre-empt any other user action with the exception of that which it is defined to support.

Read more about dialogue at;


In an illustrations based communication, it is expected that the client can just associate with parts of the framework that are obvious. In a windowing framework, for instance, the client can just direct contribution to a solitary, at present chosen window, and the main methods for changing the chose window would be by demonstrating with some signal inside that window. To make a preemptive exchange, the framework can create a window that covers the whole screen and make it the right now chosen window. All client information would then be coordinated to that window and the client would have no methods for choosing another window. The 'covering' window in this way preempts some other client activity with the exception of that which it is characterized to help

Air enters the first compressor stage of a cold-air standard Brayton cycle with regeneration and intercooling at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The overall compressor pressure ratio is 10, and the pressure ratios are the same across each compressor stage. The temperature at the inlet to the second compressor stage is 300 K. The turbine inlet temperature is 1400 K. The compressor stages and turbine each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k = 1.4, calculate: a. the thermal efficiency of the cycle
b. the back work ratio
c. the net power developed, in kW
d. the rates of exergy destruction in each compressor stage and the turbine stage as well as the regenerator, in kW, for T 0 = 300 K.



a. \eta _(th) = 77.65%

b. bwr = 6.5%

c. 3538.986 kW

d. -163.169 kJ


a. The given property  are;

P₂/P₁ = 10, P₂ = 10 * 100 kPa = 1000 kPa

p₄/p₁ = 10

P₂/P₁ = p₄/p₃ = √10

p₂ = 100·√10

T_(2s) = T₁×(√10)^(0.4/1.4) = 300 × (√10)^(0.4/1.4) = 416.85 K

T₂ = T₁ + (T_(2s) - T₁)/\eta _c = 300 + (416.85 - 300)/0.8 = 446.0625 K

p₄ = 10×p₁ = 10×100 = 1000 kPa

p₄/p₃ = √10 =

p₃ = 100·√10

T₃ = 300 K

T₃/T_(4s) = (P₂/P₁)^((k - 1)/k) = (√10)^(0.4/1.4)

T_(4s) = T₃/((√10)^(0.4/1.4) ) = 300/((√10)^(0.4/1.4)) = 215.905 K

T₄ = T₃ + (T_(4s) - T₃)/\eta _c = 300 + (215.905- 300)/0.8 = 194.881 K

The efficiency = 1 - (T₄ - T₁)/(T₃ - T₂) = 1 - (194.881 -300)/(300 -446.0625 ) = 0.28

T₄ = 446.0625 K

T₆ = 1400 K

T_(7s)/T₆ = (1/√10)^(0.4/1.4)

T_(7s) = 1400×(1/√10)^(0.4/1.4)  = 1007.6 K

T₇ = T₆ - \eta _t(T₆ - T_(7s)) = 1400 - 0.8*(1400 - 1007.6) = 1086.08 K

T₈ = 1400 K

T₉ = 1086.08 K

T₅ = T₄ + \epsilon _(regen)(T₉ - T₄) = 446.0625 +0.8*(1086.08 - 446.0625) = 958.0765 K

\eta _(th) =(((T₆ - T₇) + (T₈ - T₉)) -((T₂ - T₁) + (T₄ - T₃)))/((T₆ - T₅) + (T₈ - T₇))

(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300)))/((1400 -958.0765 ) + (1400 -1086.08 )) = 0.7765

\eta _(th) = 77.65%

b. Back work ratio, bwr = bwr = (w_(c,in))/(w_(t,out))

((446.0625 - 300)+(194.881 - 300))/((1400 - 1086.08) + (1400 -1086.08 ))

40.9435/627.84 = 6.5%

c. w_(net, out) = c_p[(T_6 -T_7) + (T_8 - T_9)] - [(T_2 - T_1) + (T_4 -T_3)]

Power developed is given by the relation;

\dot m \cdot w_(net, out)

\dot m \cdot w_(net, out)= 6*1.005*(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300))) = 3538.986 kW

d. Exergy destruction = 6*(1.005*(300-446.0625 ) - 300*1.005*(-0.3966766)

-163.169 kJ

1. The presentation of data is becoming more and more important in today's Internet. Some people argue that the TCP/IP protocol suite needs to add a new layer to take care of the presentation of data. If this new layer is added in the future, where should its position be in the suite?



It is important to add presentation layer after the physical layer, so that the data along with it's headers can be translated, when the receiver machine is applying a set of different characters.

Data compression is also required to reduce the space that is occupied by data during transmission, now once the presentation is added to the physical layer, data from the physical layer can be compressed at the presentation layer and sent by improving the throughput.



The presentation of data involves the following as shown below:

Presentation of data comprises of the task like translating between receiver and sender devices so that machines with different capabilities sets can communicate with one another.

It involves encoding  and decoding of data to provide data security that is been transmitted by different machines.

Data sometimes needs to compressed for efficiency improvement  for transmission.

The physical layer of the TCP/IP protocol suite is responsible or refers to the transmission of physical data over a physical medium

It is good or important to add presentation layer after the physical layer, so that the data along with it's headers can be translated, when the receiver machine is applying a set of different characters.

Data encryption at this stage is good for security instead of encrypting the data at upper/higher layers.

Hence, it is advisable to add presentation layer after the physical layer in the TCP/IP suite.



The layer ought to be embedded between Layer 2 and 3.


Applications often communicate with each other. This cannot be successful if they don't see data the same way. The Presentation Layer in the Open Systems Interconnection defines how data is presented and is often processed in the TCP/IP applications.

While the Presentation Layer does not exist as a different layer in the TCP/IP protocol order of arrangement, it is important to note that the Network Layer is also known referred to as the TCP/IP’s Network Layer.

Therefore, if the presentation of the data layer will be separated, it should be between layer 2 and 3.  


At full load, a commercially available 100hp, three phase induction motor operates at an efficiency of 97% and a power factor of 0.88 lag. The motor is supplied from a three-phase outlet with a line voltage rating of 208V.a. What is the magnitude of the line current drawn from the 208 V outlet? (1 hp = 746 W.) b. Calculate the reactive power supplied to the motor.



I = Line Current = 242.58 A

Q = Reactive Power = 41.5 kVAr


Firstly, converting 100 hp to kW.

Since, 1 hp = 0.746 kW,

100 hp = 0.746 kW x 100

100 hp = 74.6 kW

The power of a three phase induction motor can be given as:

P_(in)  = √(3) VI Cos\alpha\n


P in = Input Power required by the motor

V = Line Voltage

I = Line Current

Cosα = Power Factor

Now, calculating Pin:

efficiency = \frac{{P_(out)} }{P_(in) }\n0.97 = (74.6)/(P_(in) ) \nP_(in) = (74.6)/(0.97)\n  P_(in) = 76.9 kW

a) Calculating the line current:

P_(in) = √(3)VICos\alpha   \n76.9 * 1000= √(3)*208*I*0.88\nI = (76.9*1000)/(√(3)*208*0.88 )\nI =   242.58 A

b) Calculating Reactive Power:

The reactive power can be calculated as:

Q = P tanα


Q = Reactive power

P = Active Power

α = power factor angle


Cos\alpha =0.88\n\alpha =Cos^(-1)(0.88)\n\alpha=28.36


Q = 76.9 * tan (28.36)\nQ = 76.9 * (0.5397)\nQ = 41. 5 kVAr

Which of these factors would help the environment?a. Betty drinks only bottled water instead of soda.
b. Greg changes all of the light bulbs in the office to energy efficient bulbs.
c. John keeps his off topic conversations down to 10 minutes a day.
d. Doug turns the air conditioning down to cool the office during the hot summer months.
Please select the best alswer from the choices provided



B. Greg changes all of the light bulbs in the office to energy efficient bulbs.


Energy efficient bulbs help you to reduce the carbon footprint of your office / house and last up to 12 times as long as traditional bulbs, using less electricity to emit the same amount of light as a traditional bulb.




B. Suppose R1 is a fuse which burns out due to a sudden surge of current, thus, it essentially becomes an open switch. How do the currents change after this?



The currents becomes 0


when the fuse burns out due to a sudden surge of current and becomes an open switch (with a resistance of Infinity ∞) this automatically reduces the currents through it to zero