Water vapor at 10bar, 360°C enters a turbine operatingat steady state with a volumetric flow rate of 0.8m3/s and expandsadiabatically to an exit state of 1 bar, 160°C. Kinetic and potentialenergy effects are negligible. Determine for the turbine (a) the powerdeveloped, in kW, (b) the rate of entropy production, in kW/K, and (c)the isentropic turbine efficiency

Answer:

a.) 4.46 m/s

b.) 0.41 m/s

Explanation:

a) Given that the mass M of the bullet = 30g = 30/1000 = 0.03 kg

Velocity V = 450 m/s

From conservative of linear momentum,

Sum of momentum before impact = Sum of momentum after impact

0.03 × 450 = (0.03 + 3 ) × v₂

v₂ = 13.5/3.03 = 4.4554 m/s

Therefore the velocity of the bullet and B after the first impact = 4.46 m/s approximately

(b) To calculate the velocity of the carrier, you will consider the conservation of linear momentum again.

(m₁ + m₂)×v₂  = (m₁ + m₂ + m₃)×v₃

Where:

Mass of the carrier m₃ = 30 kg

Substitute all the parameters into the formula

3.03×4.4554 = (3.03 +30) × v₃

v₃ = 13.5 / 33.03 = 0.40872 m/s

Therefore the velocity of the carrier is 0.41 m/s approximately.

Answer:

Following are the conversion to this question:

Explanation:

In point (a):

[tex]\to \frac{36459080}{8} = 4557385 + \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{4557385}{8} = 569673 + \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{569673}{8} = 71209+ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{71209}{8}=8901+\ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{8901}{8}=1112+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{5}{8}\\\\\to \frac{1112}{8}=139+ \ \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{139}{8}=17+ \ \ \ \ \ \ \ \ \ \ \frac{3}{8}\\\\\to \frac{17}{8}=2+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\[/tex]

[tex]\to \frac{2}{8}=0+ \ \ \ \ \ \ \ \ \ \frac{2}{8}\\\\ \bold{(36459080)_{10}=(213051110)_8}[/tex]

In point (b):

[tex]\to \frac{20960032010}{16} = 13100020+ \ \ \ \ \ \ \ \ \ \frac{0}{16}\\\\\to \frac{13100020}{16} = 818751+ \ \ \ \ \ \ \ \ \ \frac{4}{16}\\\\\to \frac{818751}{16} = 51171+ \ \ \ \ \ \ \ \ \ \frac{15}{16}\\\\\to \frac{51171}{16}=3198+\ \ \ \ \ \ \ \ \ \ \ \frac{3}{16}\\\\\to \frac{3198}{16}=199+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{14}{1}\\\\\to \frac{199}{16}=12+ \ \ \ \ \ \ \ \ \ \ \frac{7}{16}\\\\\to \frac{12}{16}=0+ \ \ \ \ \ \ \ \ \ \ \frac{12}{16}\\\\ \bold{(20960032010)_{10}=(C7E3F40)_{16}}[/tex]

In point (c):

[tex]\to (2423233303003040)_s=(88757078520)_{10}\\\\\to \frac{88757078520}{25}= 3550283140+ \ \ \ \ \ \ \ \ \ \frac{20}{25}\\\\ \to \frac{3550283140}{25}= 142011325+ \ \ \ \ \ \ \ \ \ \frac{15}{25}\\\\\to \frac{142011325}{25}= 5680453+ \ \ \ \ \ \ \ \ \ \frac{0}{25}\\\\\to \frac{5680453}{25}= 227218+ \ \ \ \ \ \ \ \ \ \frac{3}{25}\\\\\to \frac{227218}{25}= 9088+ \ \ \ \ \ \ \ \ \ \frac{18}{25}\\\\\to \frac{9088}{25}= 363+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\[/tex]

[tex]\to \frac{363}{25}= 14+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\\to \frac{14}{25}= 0+ \ \ \ \ \ \ \ \ \ \frac{14}{25}\\\\\bold{(2423233303003040)_s=(EDDI30FK)_{25}}[/tex]

Symbols of Base 25 are as follows:

[tex]0, 1, 2, 3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I,J,K,L,M,N, \ and \ O[/tex]

Answer:

5.6 mm

Explanation:

Given that:

A cylindrical tank is required to contain a:

Gage Pressure P = 560 kPa

Allowable normal stress [tex]\sigma[/tex] = 150 MPa = 150000 Kpa.

The inner diameter of the tank = 3 m

In a closed cylinder  there exist both the circumferential stress and the longitudinal stress.

Circumferential stress [tex]\sigma = \dfrac{pd}{2t}[/tex]

Making thickness t the subject; we have

[tex]t = \dfrac{pd}{2* \sigma}[/tex]

[tex]t = \dfrac{560000*3}{2*150000000}[/tex]

t = 0.0056 m

t = 5.6 mm

For longitudinal stress.

[tex]\sigma = \dfrac{pd}{4t}[/tex]

[tex]t= \dfrac{pd}{4*\sigma }[/tex]

[tex]t = \dfrac{560000*3}{4*150000000}[/tex]

t = 0.0028  mm

t = 2.8 mm

From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value  with the maximum thickness = 5.6 mm

Answer:

the net work per cycle [tex]\mathbf{W_{net} = 0.777593696}[/tex]  Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume [tex]V_2 = 0.16 V_1[/tex]

the crankshaft N rotates at a speed of  2400 RPM.

At the beginning of the compression , temperature [tex]T_1[/tex] = 60 F = 519.67 R    

and;

Otto cycle with a pressure =  14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

[tex]V_1-V_2 = \dfrac{\pi}{4}D^2L[/tex]

[tex]V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)[/tex]

[tex]V_1-0.16V_1= 36.55714291[/tex]

[tex]0.84 V_1 =36.55714291[/tex]

[tex]V_1 =\dfrac{36.55714291}{0.84 }[/tex]

[tex]V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3[/tex]

[tex]V_1 = 0.02518 \ ft^3[/tex]

the mass in air ( lb) can be determined by using the formula:

[tex]m = \dfrac{P_1V_1}{RT}[/tex]

where;

R = 53.3533 ft.lbf/lb.R°

[tex]m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}[/tex]

m = 0.0018962 lb

From the tables  of ideal gas properties at Temperature 519.67 R

[tex]v_{r1} =158.58[/tex]

[tex]u_1 = 88.62 Btu/lb[/tex]

At state of volume 2; the relative volume can be determined as:

[tex]v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}[/tex]

[tex]v_{r2} = 158.58 \times 0.16[/tex]

[tex]v_{r2} = 25.3728[/tex]

The specific energy [tex]u_2[/tex] at [tex]v_{r2} = 25.3728[/tex] is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

[tex]v_{r3} = 0.1828[/tex]

[tex]u_3 = 1098 \ Btu/lb[/tex]

To determine the relative volume at state 4; we have:

[tex]v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}[/tex]

[tex]v_{r4} =0.1828 \times \dfrac{1}{0.16}[/tex]

[tex]v_{r4} =1.1425[/tex]

The specific energy [tex]u_4[/tex] at [tex]v_{r4} =1.1425[/tex] is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

[tex]W_{net} = Heat \ supplied - Heat \ rejected[/tex]

[tex]W_{net} = m(u_3-u_2)-m(u_4 - u_1)[/tex]

[tex]W_{net} = m(u_3-u_2- u_4 + u_1)[/tex]

[tex]W_{net} = m(1098-184.7- 591.84 + 88.62)[/tex]

[tex]W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)[/tex]

[tex]W_{net} = 0.0018962 \times (410.08)[/tex]

[tex]\mathbf{W_{net} = 0.777593696}[/tex]  Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the  four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

[tex]W = 4 \times N' \times W_{net[/tex]

where ;

[tex]N' = \dfrac{2400}{2}[/tex]

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec ×  0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

The net work per cycle and the power developed by this combustion engine are 0.7792 Btu and 88.20 hp.

Given the following data:

Note: The cylindrical volume is equal to [tex]0.16V_1[/tex]

First of all, we would determine the volume, mass and specific energy as follows:

[tex]V_1-V_2=\frac{\pi D^2L}{4} \\\\V_1-0.16V_1=\frac{3.142 \times 3.7^2 \times 3.4}{4}\\\\0.84V_1=36.56\\\\V_1=\frac{36.56}{0.84} \\\\V_1=43.52\;in^3 \;to \;ft^3 = 0.0252\;ft^3[/tex]

For the mass:

[tex]M=\frac{PV}{RT} \\\\M=\frac{2088 \times 0.0252}{53.3533 \times 519.67} \\\\M=\frac{52.6176}{27726.109411}[/tex]

M = 0.0019 lb.

At a temperature of 519.67 R, the relative volume and specific energy are:

[tex]v_{r1}=158.58\\\\u_1 = 88.62\;Btu/lb[/tex]

For the relative volume at the second state, we have:

[tex]v_{r2}=v_{r1}\times \frac{V_2}{V_1} \\\\v_{r2}=158.58\times 0.16\\\\v_{r2}=25.3728[/tex]

Note: At 25.3728, specific energy ([tex]u_2[/tex]) is 184.7 Btu/lb.

At a maximum temperature of 519.67 R, the relative volume and specific energy are:

[tex]v_{r3}=0.1828\\\\u_3 = 1098\;Btu/lb[/tex]

For the relative volume at state 4, we have:

[tex]v_{r4}=v_{r3}\times \frac{V_1}{V_3} \\\\v_{r4}=0.1828\times \frac{1}{0.16}\\\\v_{r4}=1.1425[/tex]

Note: At 1.1425, specific energy ([tex]u_4[/tex]) is 591.84 Btu/lb.

Now, we can calculate the net work per cycle by using this following formula:

[tex]W=Heat\;supplied -Heat\rejected\\\\W=m(u_3-u_2)-m(u_4-u_1)\\\\W=0.0019(1098-184.7)-0.0019(591.84-88.62)\\\\W=1.73527-0.956118[/tex]

W = 0.7792 Btu.

In a four-cylinder, four-stroke internal combustion engine, power is given by this formula:

[tex]W=4N'W_{net}[/tex]

But;

[tex]N'=\frac{N}{2 \times 60} \\\\N'=\frac{2400}{120} \\\\N'=20\;cycle/sec[/tex]

Substituting the given parameters into the formula, we have;

[tex]W=4 \times 20 \times 0.7792[/tex]

W = 62.336 Btu/sec.

In horsepower:

W = 88.20 hp.

Read more on net work here: https://brainly.com/question/10119215

Answer:

No, the velocity profile does not change in the flow direction.

Explanation:

In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.

Answer:

[tex]W_s =[/tex] 283.181 hp

Explanation:

Given that:

Air enters a compressor operating at steady state at a pressure [tex]P_1[/tex] =  176.4 lbf/in.^2  and Temperature [tex]T_1[/tex] at 260°F

Volumetric flow rate V = 424 ft^3/min

Air exits at a pressure [tex]P_2[/tex]  = 15.4 lbf/in.^2 and Temperature [tex]T_2[/tex] at 80°F.

Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings; since heat is released to the surrounding; then:

[tex]Q_{cv}[/tex] = -6800 Btu/h  = - 1.9924 kW

Using the steady  state  energy in the process;

[tex]h_2 - h_1 + g(z_2-z_1)+ \dfrac{1}{2}(v^2_2-v_1^2) = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}[/tex]

where;

[tex]g(z_2-z_1) =0[/tex]  and  [tex]\dfrac{1}{2}(v^2_2-v_1^2) = 0[/tex]

Then; we have :

[tex]h_2 - h_1 = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}[/tex]

[tex]h_2 - h_1 = \dfrac{Q_{cv} - W_s}{m}[/tex]

[tex]{m}(h_2 - h_1) ={Q_{cv} - W_s}[/tex]

[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex] ----- (1)

Using the relation of Ideal gas equation;

P₁V₁ = mRT₁

Pressure [tex]P_1[/tex] =  176.4 lbf/in.^2   = ( 176.4 ×  6894.76 ) N/m² = 1216235.664 N/m²

Volumetric flow rate V = 424 ft^3/min = (424 ×  0.0004719) m³  /sec

= 0.2000856 m³  /sec

Temperature = 260°F = (260°F − 32) × 5/9 + 273.15 = 399.817 K

Gas constant R=287 J/kg K

Then;

1216235.664 N/m² × 0.2000856 m³  /sec = m × 287 J/kg K × 399.817 K

[tex]m = \dfrac { 1216235.664 N/m^2 \times 0.2000856 m^3 /sec } {287 J/kg K \times 399.817 K }[/tex]

m = 2.121 kg/sec

The change in enthalpy:

[tex]m(h_1-h_2) = m * C_p * \Delta T= m* C_p * ( T_1 -T_2)[/tex]

[tex]= 2.121* 1.005* ( 399.817 -299.817)[/tex]

= 213.1605 kW

From (1)

[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex]

[tex]W_s =[/tex]  - 1.9924 kW + 213.1605 kW

[tex]W_s =[/tex] 211.1681  kW

[tex]W_s =[/tex] 283.181 hp

The power input is [tex]W_s =[/tex] 283.181 hp

Answer:

dx/dt = kx(9000-x) where k > 0

Explanation:

Number of students in the campus, n = 9000

Number of students who have contracted the flu = x(t) = x

Number of students who have bot yet contracted the flu = 9000 - x

Number of Interactions between those that have contracted the flu and those that are yet to contract it = x(9000 - x)

The rate of spread of the disease = dx/dt

Note: the rate at which the disease spread is proportional to the number of interactions between those that have contracted the flu and those that have not contracted it.

[tex]\frac{dx}{dt} \alpha [x(9000 -x)]\\[/tex]

Introducing a constant of proportionality, k:

dx/dt = kx(9000-x) where k > 0

Answer:

HVACR Industry Trade Groups

Explanation:

Answer:

- 1.19 lb/ft^3

Explanation:

You are given the following information;

Radius r = 20 ft

Speed V = 100 ft/s

You should use Bernoulli equation pertaining to streamline. That is, normal to streamline.

The pressure gradient = dp/dn

Where air density rho = 0.00238 slugs per cubic foot.

Please find the attached files for the solution and diagram.

Answer:

2.47  ft

Explanation:

Given that:

The initial temperature of air = 80°F

Diameter of the pipe = 72 ft

average velocity [tex]v_{air}[/tex] of the air flow through the pipe =  34 ft/s

The objective is to determine the diameter of the  pipe to  be used to move water at:

At a temperature = 60°F   &

An average velocity [tex]v_{water}[/tex] of 71 ft/s

Assuming Reynolds number similarity is enforced;

where :

kinematic viscosity (V_air) of air at 80 °F  (V_air)  = 1.69 × 10⁻⁴ ft²/s

kinematic viscosity of water  at 60 °F (V_water) = 1.21 × 10⁻⁵ ft²/s

The diameter of the pipe can be calculated by using the expression:

[tex]D_{water} = \dfrac{V_{water}}{V_{air}}*\dfrac{v_{air}}{v_{water}}* D_{air}[/tex]

[tex]D_{water} = \dfrac{1.21*10^{-5} \ ft^2/s}{1.69*10^{-4} \ ft^2/s}*\dfrac{34 \ ft/s}{71 \ ft/s}* 72 \ ft[/tex]

[tex]D_{water} =[/tex] 2.4686  ft

[tex]D_{water} =[/tex] 2.47 ft   ( to two decimal places)

Thus; diameter pipe to be use to move water at the given temperature and average velocity is 2.47 ft

Answer:

2.47  ft

Explanation:

Given that:

The initial temperature of air = 80°F

Diameter of the pipe = 72 ft

average velocity  of the air flow through the pipe =  34 ft/s

The objective is to determine the diameter of the  pipe to  be used to move water at:

At a temperature = 60°F   &

An average velocity  of 71 ft/s

Assuming Reynolds number similarity is enforced;

where :

kinematic viscosity (V_air) of air at 80 °F  (V_air)  = 1.69 × 10⁻⁴ ft²/s

kinematic viscosity of water  at 60 °F (V_water) = 1.21 × 10⁻⁵ ft²/s

The diameter of the pipe can be calculated by using the expression:

2.4686  ft

2.47 ft   ( to two decimal places)

Thus; diameter pipe to be use to move water at the given temperature and average velocity is 2.47 ft

Answer:

(A)0.122 kW (B) Actual cost = 1.056 $/day, Theoretical cost =  0.234 $/day

Explanation:

Solution

Given that:

The coefficient of performance is =3

Heat transfer = 6000kJ/hr

Temperature = 20°C

Cost of electricity = 8 cents per kW-hr

Now

The next step is to find the power input to the refrigerator and compare with the power input considered by a reversed refrigeration cycle operating between reservoirs at the two temperatures.

Thus

(A)The coefficient of performance is given below:

COP = Heat transfer from freezer/Power input

3 =6000/P

P =6000/3

P= 2000

P =  2000 kJ/hr = 2000/(60*60) kW

= 2000 (3600)kW

= 0.55 kW

Thus

The ideal coefficient of performance = T_low/(T_high - T_low)

= (0+273)/(20-0)

= 13.65

So,

P ideal = 6000/13.65 = 439.6 kJ/hr

= 439.6/(60*60) kW

= 0.122 kW

(B)For the actual cost we have the following:

Actual cost = 0.55 kW* 0.08 $/kW-hr = $ 0.044 per hour

= 0.044*24 $/day

= 1.056 $/day

For the theoretical cost we have the following:

Theoretical cost = 0.122 kW* 0.08 $/kW-hr = $ 0.00976 per hour

= 0.00976*24 $/day

= 0.234 $/day

Answer:

hello the answer options are missing here are the options

A)The thickness of the heated region near the plate is increasing

B)The velocities near the plates are increasing

C)The fluid temperature near the plate are increasing

ANSWER : all of the above

Explanation:

Laminar flow  is the flow of a type of fluid across the surface of an object following regular paths and it is unlike a turbulent flow which flows in irregular paths (encountering fluctuations)

For laminar flow over a hot flat plate, the local heat transfer coefficient decreases with distance because :

Answer:

i) Truth Table:

A      |     B     |     C     |     O

0      |     0     |     0     |      0

0      |     0     |     1      |      0

0      |     1      |     0     |      0

0      |     1      |     1      |      0    (condition 2 not satisfied)

1       |     0     |     0     |      0

1       |     0     |     1      |      1    (both conditions satisfied)

1       |     1      |     0     |      1    (both conditions satisfied)

1       |     1      |     1      |      1    (both conditions satisfied)

ii) The minimized sum of products (SOP) expression is

O = AC + AB

Explanation:

We have three inputs A, B and C

Let O is the output.

We are given two conditions to open the vault door:

1. At  least two people must insert their keys into the assigned locks at the same  time.

2. The trainee bank teller (C) can only open the vault when the bank  manager (A) is present in the opening.

i) Construct the Truth Table

A      |     B     |     C     |     O

0      |     0     |     0     |      0

0      |     0     |     1      |      0

0      |     1      |     0     |      0

0      |     1      |     1      |      0    (condition 2 not satisfied)

1       |     0     |     0     |      0

1       |     0     |     1      |      1    (both conditions satisfied)

1       |     1      |     0     |      1    (both conditions satisfied)

1       |     1      |     1      |      1    (both conditions satisfied)

ii) SOP Expression using Karnaugh-Map:

A 3 variable Karnaugh-map is attached.

The minimized sum of products (SOP) expression is

O = AC + AB

The orange pair corresponds to "AC" and the purple pair corresponds to "AB"

Bonus:

The above expression may be realized by using two AND gates and one OR gate.  

Please refer to the attached logic circuit diagram.

Answer:

A ball bearing has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify.... ... has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify appropriate minimum and maximum shaft diameters to provide a locational interference fit.

Explanation:

Answer:

A) V_peak ≈ 165 V

B) V_rms ≈ 24 V

C) V_peak ≈ 311 V

D) V_peak ≈ 311 KV

Explanation:

Formula for RMS value is given as;

V_rms = V_peak/√2

Formula for peak value is given as;

V_peak = V_rms x √2

A) At RMS value of 117 V, peak value would be;

V_peak = 117 x √2

V_peak = 165.46 V

V_peak ≈ 165 V

B) At peak value of 33.9 V, RMS value would be;

V_rms = 33.9/√2

V_rms = 23.97 V

V_rms ≈ 24 V

C) At RMS value of 220 V, peak value is;

V_peak = 220 × √2

V_peak = 311.13 V

V_peak ≈ 311 V

D) At RMS value of 220 KV, peak value is;

V_peak = 220 × √2

V_peak = 311.13 KV

V_peak ≈ 311 KV

Answer:

the answer is

Explanation:

Answer:

The answer is 2.25 kΩ

Explanation:

Solution

Given that:

The resistance reading on a DMM'S meter face = 22.5 ohms

The range selector switch = R * 100 range,

We now have to find the actual measure resistance of the circuit which is given below:

The actual measured resistance of the circuit is=R * 100

= 22.5 * 100

=2.25 kΩ

Hence the measured resistance of the circuit is 2.25 kΩ

Answer:

The answer is B. Avoid making a large wake.

Explanation:

When passing a fishing boat it is important to maintain a minimal wake due to the dangers a large wake could pose to the fishing boat you are passing, it is part of maintaining safety on the water.

You can not pass on the sides with the fishing lines also, and you are supposed to communicate to the fishing boat before taking the appropriate action.

Answer:

A 3-phase, 50 Hz, 110 KV overhead line has conductors

Explanation:

hope it will helps you

Answer:

100000000000000000000000