The boundary work is positive during an expansion process.
a.
False
b.
True
Answer:
True
Explanation:
During expansion process, the boundary work is positive while in case of contraction, the boundary work is negative. During expansion process, the work is done by the system while in case of compression process, work in done on the system
Hence, the given statement is true
Two consecutive, first order reactions (with reaction rate constant k1 and k2) take place in a perfectly mixed, isothermal continuous reactor (CSTR) A (k1) → B (k2) → C Volumetric flow rates (F) and densities are constant. The volume of the tank (V) is constant. The reactor operate at steady state and at constant temperature. The inlet stream to the reactor contains only A with CA,in = 10 kmol/m3. If k1 = 2 min-1, k2 = 3 min-1, and τ = V/F.= 0.5 min, determine the concentration of C in the stream leaving the reactor.
Answer:
3 kmol/m^3
Explanation:
Determine the concentration of C in the stream leaving the reactor
Given that the CTSR reaction ; A (k1) → B (k2) → C
K1 = 2 min^-1 , K2 = 3 min^-1 , time constant ; τ = V/F.= 0.5 min also n1 = n2
attached below is the detailed solution
concentration of C leaving the reactor= 3 kmol/mol^3
Given ; Ca = 5 kmol/m^3 , Cb = 2 kmol/m^3 ( from the attached calculations ) Cc = 3 kmol/m^3
Create a 6-bit full subtractor that uses the Borrow method to subtract two 6-bit binary numbers. You can use the proper basic sub-circuit.
Route Choice The cost of roadway improvements to the developer is a function of the amount of traffic being generated by the theater as well as the routes that these vehicles use in getting to/from the theater. Thus, you need to determine traffic demand on the available routes between the housing area (apartment complexes) and the proposed theater site. There are two potential routes from this housing area to the proposed theater site. The total flow on these two routes (origin-to-destination demand plus other traffic) is 5500 vehicles. These routes have the following speed and length characteristics:
Route Free Flow Speed(mi/h) Length(mi)
1 45 5.5
2 40 3.5
3 35 3.75
It is known that the individual route travel times increase (in units of minutes) according to the following functions (with x in units of 1000 vehicles per hour):
Route Travel time increase as a function of traffic volume
1 0.5x1
2 x2
3 0.25
Required:
Determine user equilibrium flows and travel times if the total flow on these routes (origin-to-destination demand plus other traffic) is 5500 veh/h.
3. A particle is projected to the right from the position S = 0, when an initial velocity of 8 m/s. If the acceleration of the particle is defined by the relation a = -0.5 v3/2, where a in m/s2 and v in m/s. Determine a) the distance the particle will have traveled when its velocity is 5 m/s b) the time when v = 1m/s c) the time require for the particle to travel 8m
Answer:
a) 3.5 m
b) 14 secs
c) 1.4 secs
Explanation:
a) Determine the distance the particle will travel
given velocity ( final velocity ) = 5 m/s
v^2 = u^2 + 2as
s = ( v^2 - u^2 ) / 2a
= ( 5^2 - 8^2 ) / 2 ( -0.5 * 5^3/2 )
= 3.5 m
b) Determine the time when v = 1m/s
V = u + at
1 = 8 + ( -0.5 * 1^3/2 ) * t
∴ t = 14 secs
c) Determine the time required for particle to travel 8 m
we will employ both equations above
V^2 = u^2 + 2as
s = 8 m , V = unknown , u = 8 m/s back to equation
V^2 = 8^2 + 2 ( - 1/2 * V^3/2 ) * 8
∴ V^2 + 8V^3/2 - 64 = 0
resolving the above equation
V = 3.478 m/s
now using the second equation
V = u + at
3.478 = 8 + ( - 1/2 * 3.478^3/2 ) * t
hence : t = 1.4 secs
A Class III two-lane highway is on level terrain, has a measured free-flow speed of 45 mi/h, and has 100% no-passing zones. During the peak hour, the analysis direction flow rate is 150 veh/h, the opposing direction flow rate is 100 veh/h, and the PHF-0.95. There are 5% large trucks and 10% recreational vehicles. Determine the level of service.
Answer:
LOS = A
Explanation:
Given all the parameters the level of service as seen from the attached graph
is LOS = A
To determine the LOS from the attached graph
calculate the trial value of Vp
Vp = V / PHF
= (100 + 150) / 0.95 = 263 pc/h
since the trial value of Vp = ( 0 to 600 ) pc/h . hence E.T = 1.7 , ER = 1
next we will calculate the flow rate
flow rate = 1 / [ ( 1 + PT(ET - 1 ) + PR ( ER - 1 ) ]
Fhr = 1 / 1.035 = 0.966 ≈ 1
next calculate the real value of Vp
Vp = V / ( PHF * N * Fhr * Fp )
= ( 100 + 150 ) / ( 0.95 * 2 * 1 * 1 )
Vp ≈ 126 pc/h/In
Next calculate the density
D = Vp / S = 126 / ( 45 * 1.61 ) = 1.74 pc/km/In
An FPC 4 m2 in area is tested during the night to measure the overall heat loss coefficient. Water at 60 C circulates through the collector at a flow rate of 0.06 l/s. The ambient temperature is 8 C and the exit temperature is 49 C. Determine the overall heat loss coefficient.
Answer:
- 14.943 W/m^2K ( negative sign indicates cooling )
Explanation:
Given data:
Area of FPC = 4 m^2
temp of water = 60°C
flow rate = 0.06 l/s
ambient temperature = 8°C
exit temperature = 49°C
Calculate the overall heat loss coefficient
Note : heat lost by water = heat loss through convection
m*Cp*dT = h*A * ( T - To )
∴ dT / T - To = h*A / m*Cp ( integrate the relation )
In ( [tex]\frac{49-8}{60-8}[/tex] ) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )
In ( 41 / 52 ) = 0.0159*h
hence h = - 0.2376 / 0.0159
= - 14.943 W/m^2K ( heat loss coefficient )
ihjpr2 ywjegnak'evsinawhe2'qwmasnh ngl,;snhy
Answer:
ummm ok?
Explanation:
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el protozoos es del reino protista?
Answer: Si (Yes)
Explanation:
Answer:
Protozoario o protozoo es un organismo unicelular y eucariota (con núcleo celular definido) perteneciente al Reino protista. Los protozoarios se encuentran junto con los protófitos o algas simples, generalmente acuáticas, dentro del Reino protista o también denominado Reino protoctista.
Explanation:
Identify the following formulas:
1. Slope Formula
2. Slope-Intercept Form
3. Standard Form
4. Point-Slope Formula
.) If the charges attracting each other in the problem above have equal magnitude, what is the magnitude of each charge?
Answer:
Not seeing any other information, the best answer I can give is 2m.
Explanation:
M = magnitude
You see, if they have an equal charge, and you add them, it'd be 2 * m, or 2m.
A source current of 10 mA is supplied to a parallel circuit consisting of the following resistors three resistors, a 2200 a 500 and a 1KO. What is the source voltage required to
supply the current
Can someone put each letter by the correct word for my automotive class !
Answer:
L = spindle
M = lower ball joint
part without the letter showing = steering knuckle
Explanation:
Al ejercer una fuerza de 50N sobre un resorte elastico esto se alarga desde los 15 cm hasta los 60cm¿cual es la constante elastica del resorte?
Answer:
Constante de resorte = 1.1 N/m
Explanation:
Dados los siguientes datos;
Fuerza = 50N
Extensión = 60cm - 15cm = 45cm
Para encontrar la constante del resorte;
Matemáticamente, la fuerza ejercida para estirar un resorte viene dada por la fórmula;
Fuerza = constante de resorte * extensión
Sustituyendo en la fórmula, tenemos;
50 = constante de resorte * 45
Constante de resorte = 50/45
Constante de resorte = 1.1 N/m
Consider a normal population distribution with the value of known.
Answer:
it is alba kk
Explanation:
rbeacuse wrong
In a device to produce drinking water, humid air at 320C, 90% relative humidity and 1 atm is cooled to 50C at constant pressure. If the duty on the unit is 2,200 kW of heat is removed from the humid air, how much water is produced and what is the volumetric flow rate of air entering the unit
Answer: hello your question lacks some data below is the missing data
Air at 32C has H = 0.204 kJ/mol and at 50C has H = -0.576 kJ/mol
H of steam can be found on the steam tables – vapor at 32C and 1 atm; vapor at 5C and liquid at 5C. Assume the volume of the humid air follows the ideal gas law.
H of water liquid at 5C = 21 kJ/kg; vapor at 5C = 2510.8 kJ/kg; H of water vapor at 32C = 2560.0 kJ/kg
Answer :
a) 34.98 lit/min
b) 1432.53 m^3/min
Explanation:
a) Calculate how much water is produced
density of water = 1 kg/liter
First we will determine the mass of condensed water using the relation below
inlet mass - outlet vapor mass = 0.0339508 * n * 18/1000 ----- ( 1 )
where : n = 57241.57
hence equation 1 = 34.98 Kg/min
∴ volume of water produced = mass of condensed water / density of water
= 34.98 Kg/min / 1 kg/liter
= 34.98 lit/min
b) calculate the Volumetric flow rate of air entering the unit
applying the relation below
Pv = nRT
101325 *V = 57241.57 * 8.314 * 305
∴ V = 1432.53 m^3/min
) A flow is divided into two branches, with the pipe diameter and length the same for each branch. A 1/4-open gate valve is installed in line A, and a 1/3-closed ball valve is installed in line B. The head loss due to friction in each branch is negligible compared with the head loss across the valves. Find the ratio of the velocity in line A to that in line B (include elbow losses for threaded pipe fittings).
Answer:
Va / Vb = 0.5934
Explanation:
First step is to determine total head losses at each pipe
at Pipe A
For 1/4 open gate valve head loss = 17 *Va^2 / 2g
elbow loss = 0.75 Va^2 / 2g
at Pipe B
For 1/3 closed ball valve head loss = 5.5 *Vb^2 / 2g
elbow loss = 0.75 * Vb^2 / 2g
Given that both pipes are parallel
17 *Va^2/2g + 0.75*Va^2 / 2g = 5.5 *Vb^2 / 2g + 0.75 * Vb^2 / 2g
∴ Va / Vb = 0.5934
Which of the following identifies the beginning phase of the engineering design process?
structural analysis
visual analysis
recognizing specifications and limitations
brainstorming possible designs
Answer:
<❤)structural analysi(❤>
Explanation:
(♨️)BRAINLEIST PLEASE(♨️)
Answer:
Structural analysis
Explanation:
please help with my economics problem
Answer:
You first get a new job, and make a new company and then by amazon to traumatize Jeff Bezos after his divorce
Explanation:
Cast irons containing graphite are formed under a metastable eutectic reaction. 2) ___ High carbon steels are easily welded. 3) ___ Hardenability refers to the ease with which a steel can be quenched to form pearlite. 4) ___ In the AISI system for designating steels, the first two numbers refer to the major alloying elements of the steel. 5) ___ Quenching hardens most steels while tempering increases the toughness.
Answer:
Stating which of the above is TRUE or FALSE;
1) True
2) False
3) False
4) True
5) True
Explanation:
1)True ___ Cast irons containing graphite are formed under a metastable eutectic reaction.
2) False___ High carbon steels are easily welded.
3) False ___ Hardenability refers to the ease with which a steel can be quenched to form pearlite.
4) True ___ In the AISI system for designating steels, the first two numbers refer to the major alloying elements of the steel.
5) True ___ Quenching hardens most steels while tempering increases the toughness.
In a shear box test on sand a shearing force of 800 psf was applied with normal stress of 1750 psf. Find the major and minor principal stresses.
Answer:
The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.
Explanation:
The major and minor principal stresses are given as follows:
[tex]\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}[/tex]
[tex]\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}[/tex]
Here
[tex]\sigma_x[/tex] is the normal stress which is 1750 psf[tex]\sigma_y[/tex] is 0[tex]\tau_{xy}[/tex] is the shear stress which is 800 psfSo the formula becomes
[tex]\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}[/tex]
Similarly, the minimum normal stress is given as
[tex]\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}[/tex]
The maximum shear stress is given as
[tex]\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}[/tex]
Determine the slopes and deflections at points B and C for the beam shown below by the moment-area method. E=constant=70Gpa I=500 (10^6)mm^4
Answer:
hello your question is incomplete attached below is the complete question
answer :
Slopes : B = 180 mm , C = 373 mm
Deflection: B = 0.0514 rad , C = 0.077 rad
Explanation:
Given data :
I = 500(10^6) mm^4
E = 70 GPa
The M / EI diagram is attached below
Deflection angle at B
∅B = ∅BA = [ 150 (6) + 1/2 (300)*6 ] / EI
= 1800 / ( 500 * 70 ) = 0.0514 rad
slope at B
ΔB = ΔBA = [ 150(6)*3 + 1/2 (300)*6*4 ] / EI
= 6300 / ( 500 * 70 ) = 0.18 m = 180 mm
Deflection angle at C
∅C = ∅CA = [ 1800 + 300*3 ] / EI
= 2700 / ( 500 * 70 )
= 2700 / 35000 = 0.077 rad
Slope at C
ΔC = [ 150 * 6 * 6 + 1/2 (800)*6*7 + 300(3) *1.5 ]
= 13050 / 35000 = 373 mm
Design a septic system with gravity flow subsurface soil adsorption for a three-bedroom house with two baths and a basement that is currently occupied by a family of four. The building site is depicted in topographic map A and in the cross sectional transects A3-C3, A2-C2, A1-C1. Based on an initial site investigation, soil percolation tests were conducted in areas C1 and A3 to assess their suitability as locations for the absorption bed. The pertinent data from these tests is given below.
Area Cl Water Level Drop in Inches per 30 Minute Interval Percolation Rate (min/in Hole 1 13.56.57.361.24.221.221.211.2 65 .92.75.71 2.2 1.781.1.78 63 65 3 2.441.46 1.02 7 4 51.85 1.5 .92.87.8 1.76.781.6 1.441.36.28 1.26 1.26 .78 .78.78 Average: Area A3 Water Level Drop in Inches per 30 Minute Interval Percolation Rate (min/in Hole 1 3 5 1.340.940.900.88 0.840.78 0.76 0.75 2 1.45 1.18 0.98 0.92 0.90.88 0.880.86 3 0.980.88 0.840.81 0.820.800.790.77 1.140.960.90.870.850.84 0.82 0.83 1.211.02 0.920.880.85 0.830.82 0.81 4 Average:
Your solution to this exercise should include the following and should be in accord with the State of Virginia "Sewage Handling and Disposal Regulations" (A summary of the appropriate regulations is available within the course notes):________.
a) The design of an appropriately sized septic tank. This design should include a fully labeled sketch that notes all appurtenances
b) The design for a complete gravity, subsurface trench drainfield system. Depict your system on the layout sketch for the site and note all setback distances. Discuss the reasons behind your choice of a location for the drainfield,
c) If you were to utilize a bed-based design, how would the surficial area of your drainfield change?
An acid-base indicator is usually a weak acid with a characteristic color in the protonated (acid) and deprotonated (conjugate base) forms. In this assignment, you will monitor the color of an acetic acid solution containing Bromocresol Green as an indicator, as the pH is changed and then you will use your data to calculate the ionization constant, Ka, for the bromocresol green indicator and compare to an accepted value.
1. Start Virtual Chemlab, select Acid-Base Chemistry, and then select lonization Constants of Weak Acids from the list of assignments. The lab will open in the Titration laboratory. Bottles of 0.1104 M NaOH and 0.1031 M HAC (acetic acid) will be on the lab bench. The buret will be filled with the NaOH solution and a beaker containing 10.00 mL of the HAc solution will be on the stir plate. The stir plate will be on, Bromocresol Green indicator will have been added to the beaker, and a calibrated pH probe will also be in the beaker so the pH of the solution can be monitored.
2. What is the color and pH of the solution?
3. On the buret, the horizontal position of the orange handle is off for the stopcock. Open the stopcock by pulling down on the orange handle. The vertical position delivers solution the fastest with three intermediate rates in between (slow dropwise, fast dropwise, and slow stream). Turn the stopcock to the second position or fast drop-wise addition. Observe the color of the solution and close the stopcock when you observe the change in color by double clicking on the center of the stopcock.
4. What is the color and pH of the solution now?
5. Continue to add NaOH as before or at a faster rate. What is the final color and pH of the solution after all of the NaOH is added?
6. An acid-base indicator is usually a weak acid with a characteristic color in the protonated and deprotonated forms. Because bromocresol green is an acid, it is convenient to represent its rather complex formula as HBOG. HBOG ionizes in water according to the following equation:
HBOG + H2O = BCG + H3O+
(yellow) (blue)
The K. (the equilibrium constant for the acid) expression is:
Ka = [BCG-][H3O+]/[HBCG)
When [BCG-[ = [HBCG), then Ka = [H3O+). If you know the pH of the solution, then the [H3O+] and Ka can be determined.
What would be the color of the solution if there were equal concentrations of HBCG and BCG-?
What is the pH at the first appearance of this color?
What is an estimate for the Ka for bromocresol green?
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 75 MPa (68.25 ksi). If the plate is exposed to a tensile stress of 361 MPa (52360 psi) during use, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.03 for Y.
Answer:
Explanation:
From the given information:
Strain fracture toughness [tex]K_k[/tex]= 75 MPa[tex]\sqrt{m}[/tex]
Tensile stress [tex]\sigma[/tex] = 361 MPa
Value of Y = 1.03
Thus, the minimum length of the critical interior surface crack which will result to fracture can be determined by using the formula:
[tex]a_c = \dfrac{1}{\pi} ( \dfrac{k_k}{\sigma Y})^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ \dfrac{75 \times \sqrt{10^3}}{361 \times 1.03 } \Big]^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ 6.378474693\Big]^2 \\ \\ \mathbf{ a_c = 12.95 \ mm}[/tex]
calculate force and moment reactions at bolted base O of overhead traffic signal assembly. each traffic signal has a mass 36kg, while the masses of member OC and AC are 50Kg and 55kg, respectively. The mass center of mmber AC at G.
Answer:
The free body diagram of the system is, 558 368 368 508 O ?? O, Consider the equilibrium of horizontal forces. F
Explanation:
I hope this helps you but I think and hope this is the right answer sorry if it’s wrong.
Methane (CH4) at 298 K, 1 atm enters a furnace operating at steady state and burns completely with 140% of theoretical air entering at 400 K, 1 atm. The products of combustion exit at 500 K, 1 atm. The flow rate of the methane is 1.4 kg/min. Kinetic and potential energy effects are negligible and air can be modeled as 21% O2 and 79% N2 on a molar basis.
Required:
Determine the dew point temperature of the products, in K.
14. The top plate of the bearing partition
I
a. laps the plate of the exterior wall.
b. is a single member.
c. butts the top plate of the exterior wall.
d. is applied after the ceiling joists are
installed.
Answer:
d. is applied after the ceiling joists are
installed.
hmmmmmmmm i already put the photo as attachment its
Answer:
letse see
Explanation:
Air initially at 120 psia and 500o F is expanded by an adiabatic turbine to 15 psia and 200o F. Assuming air can be treated as an ideal gas and has variable specific heat. a) Determine the specific work output of the actual turbine (Btu/lbm). b) Determine the amount of specific entropy generation during the irreversible process (Btu/lbm R). c) Determine the isentropic efficiency of this turbine (%).
Answer:
a) specific work output of the actual turbine is 73.14 Btu/lbm
b) the amount of specific entropy generation during the irreversible process is 0.050416 Btu/lbm°R
c) Isentropic efficiency of the turbine is 70.76%
Explanation:
Given the data in the question;
For an adiabatic turbine; heat loss Q = 0
For Initial State;
p₁ = 120 psia
T₁ = 500°F = 959.67°R
from table; { Gas Properties of Air }
At T₁ = 959.67°R
[tex]s_1^0[/tex] = 0.74102 Btu/lbm°R
[tex]h_1[/tex] = 230.98 Btu/lbm
For Finial state;
p₂ = 15 psia
T₂ = 200°F = 659.67°R
[tex]s^0_{2a[/tex] = 0.64889 Btu/lbm°R
[tex]h_{2a[/tex] = 157.84 Btu/lbm
we know that R for air is 0.06855 Btu/lbm.R
a)
The specific work output of the actual turbine Wₐ is;
W[tex]_a[/tex] = [tex]h_1[/tex] - [tex]h_{2a[/tex]
we substitute
W[tex]_a[/tex] = 230.98 - 157.84
W[tex]_a[/tex] = 73.14 Btu/lbm
Therefore, specific work output of the actual turbine is 73.14 Btu/lbm
b)
amount of specific entropy generation during the irreversible process.
To determine the entropy generation [tex]S_{gen[/tex];
[tex]S_{gen[/tex] = ΔS = [tex]s_{2a[/tex] - [tex]s_1[/tex] = [tex]s^0_{2a[/tex] - [tex]s_1^0[/tex] - R ln([tex]\frac{p_2}{p_1}[/tex])
we substitute in our values
[tex]S_{gen[/tex] = 0.64889 - 0.74102 - 0.06855 ln([tex]\frac{15}{120}[/tex])
[tex]S_{gen[/tex] = 0.64889 - 0.74102 + 0.1425457
[tex]S_{gen[/tex] = 0.050416 Btu/lbm°R
Therefore, the amount of specific entropy generation during the irreversible process is 0.050416 Btu/lbm°R
c)
Isentropic efficiency of turbine η[tex]_{is[/tex]
η[tex]_{is[/tex] = {actual work output] / [ ideal work output ] = ([tex]h_1[/tex] - [tex]h_{2a[/tex] ) / ( [tex]h_1[/tex] - [tex]h_{2s[/tex] )
Now, for an ideal turbine;
ΔS = 0 = [tex]s_{2s[/tex] - [tex]s_1[/tex]
so, [tex]s_{2s[/tex] - s₁ = [tex]s^0_{2s[/tex] - [tex]s_1^0[/tex] - R ln([tex]\frac{p_2}{p_1}[/tex])
0 = [tex]s^0_{2s[/tex] - [tex]s_1^0[/tex] - R ln([tex]\frac{p_2}{p_1}[/tex])
[tex]s^0_{2s[/tex] = [tex]s_1^0[/tex] + R ln([tex]\frac{p_2}{p_1}[/tex])
we substitute
[tex]s^0_{2s[/tex] = 0.74102 + 0.06855 ln([tex]\frac{15}{120}[/tex])
[tex]s^0_{2s[/tex] = 0.74102 - 0.1425457
[tex]s^0_{2s[/tex] = 0.59847 Btu/lbm°R
Now, from table; { Gas Properties of Air }
At [tex]s^0_{2s[/tex] = 0.59847 Btu/lbm°R; [tex]h_{2s[/tex] = 127.614 Btu/lbm
η[tex]_{is[/tex] = [( [tex]h_1[/tex] - [tex]h_{2a[/tex] ) / ( [tex]h_1[/tex] - [tex]h_{2s[/tex] )] × 100%
we substitute
η[tex]_{is[/tex] = [( 230.98 - 157.84 ) / ( 230.98 - 127.614 )] × 100%
η[tex]_{is[/tex] = [ 73.14 / 103.366] × 100%
η[tex]_{is[/tex] = 0.70758 × 100%
η[tex]_{is[/tex] = 70.76%
Therefore, Isentropic efficiency of the turbine is 70.76%