Water flows steadily through a 3.28-in-diameter pipe and discharges through a 1.64-in-diameter nozzle to atmospheric pressure. The flow rate is 26.5 gpm. (a) Calculate the minimum static pressure required in the pipe to produce this flow rate. (b) Evaluate the axial force of the nozzle assembly on the pipe flange.

Answer:

The energy in pneumatic power systems is easily transmitted to gases, which absorb the energy of the system and lower its efficiency.

Explanation:

I did it on edge and got it right.

Answer:

The energy in pneumatic power systems is easily transmitted to gases, which absorb the energy of the system and lower its efficiency

Thanks for points!!! :D

Answer:

39.5 m/s

Explanation:

To solve this question, we would use Bernoulli's equation

P₁/ρ + V₁²/2 + gz₁ = P₂/ρ + V₂²/2 + gz₂,

where V₁ = 0 and z₁ = z₂, so that we have

P₁/ρ = P₂/ρ + V₂²/2

making V₂² subject of formula, we have that

V₂² = 2[P₁/ρ - P₂/ρ]

Then we finally go ahead and substitute the values such that

V₂² = 2 [(880000/1000) - (100000/1000)]

V₂² = 2 (880 - 100)

V₂² = 2(780)

V₂² = 1560

V₂ = √1560

V₂ = 39.5 m/s

Answer:

589.5 W

Explanation:

Given that

the thickness of the carbon steel and insulation is L(a) = L(c) = 0.00635 m

Width of the freezer, W = 3m

Thermal contact resistance, R(t.c) = 0.00025 m²

T(s.i) = -6°C = 267 K

T(s.o) = 22°C = 295 K

See attachment for calculation

Answer:

The correct answer will be "8640 kWh".

Explanation:

The given values are:

Rate (Ql)

= 5040 kJ/h

As we know,

⇒  [tex]COP = \frac{Cooling}{heat \ required}[/tex]

On putting the values, we get

⇒           [tex]=\frac{5040 \ kJ/h}{1 \ kW}[/tex]

⇒           [tex]=1.4[/tex]

Now,

⇒  [tex]Q = Ql + W[/tex]

        [tex]= 1 + 1.4[/tex]

        [tex]=2.4 \ kW[/tex]

        [tex]=2.4\times 3600[/tex]

        [tex]=8640 \ kWh[/tex]

Answer:

Number of necessary lanes = 3

Explanation:

Expected free flow speed = 55 mph

Volume of vehicles expected during peak-hour = 2700

PHF = 0.88

18% = 0.18  for heavy vehicles ( Pt )

Et ( for a rolling terrain ) = 3

Detailed solution attached below

First we calculate the Heavy-vehicle adjustment factor before determining the number of lanes

Answer:

0.9732

Explanation:

A signal splitter must be placed before a processor and a comparator be placed after a processor.

This contains a series combination of a splitter, processor and comparator. Let R1 represent the reliability of the first subsystem and R2 be the reliability of the added redundant system.

Given: R(processor) = 0.9, R(new components) = 0.95 (i.e. splitter, new processor, comparator)

R1 = series combination of splitter, processor and comparator

R1 = R(splitter) × R(processor) × R(comparator) = 0.95 × 0.9 × 0.95 = 0.81225

R2 = series combination of splitter, processor and comparator

R2 = R(splitter) × R(new processor) × R(comparator) = 0.95 × 0.95 × 0.95 = 0.8574

R2 is the redundant system which is connected in parallel with R1, hence:

R(system) = 1 - (1 - R1)(1 - R2) = 1 - (1 - 0.81225)(1 - 0.8574)

R(system) = 0.9732

Answer:

La diferencia media logarítimica de temperatura del intercambiador en paralelo es aproximadamente 75.466 ºC.

La diferencia media logarítmica de temperatura del intercambiador en contracorriente es aproximadamente 97.881 ºC.

Explanation:

De la teoría de Transferencia de Calor tenemos que un intercambiador de calor en paralelo presenta las siguientes dos características:

1) Tanto el fluido caliente como el fluido frío entran por el mismo lado.

2) Tanto el fluido caliente como el fluido frío salen por el mismo lado.

Mientras que el intercambiador de calor en contracorriente tiene que:

1) El fluido caliente y el fluido frío entran por lados opuestos.

2) El fluido caliente y el fluido frío salen por lados opuestos.

A continuación, anexamos los esquemas de perfil de cada intercambiador.

Ahora, la Diferencia Media Logarítimica de Temperatura ([tex]\Delta T_{lm}[/tex]), medida en grados Celsius, queda definida como sigue:

[tex]\Delta T_{lm} = \frac{\Delta T_{1}-\Delta T_{2}}{\ln \frac{\Delta T_{1}}{\Delta T_{2}} }[/tex] (Eq. 1)

Donde [tex]\Delta T_{1}[/tex] y [tex]\Delta T_{2}[/tex] son las diferencias de temperatura de los fluidos en cada extremo del intercambiador, medido en grados Celsius.

Procedemos a determinar esas diferencias y la Diferencia Media Logarítimica de Temperatura para cada configuración:

Intercambiador en paralelo

[tex]\Delta T_{1} = 180\,^{\circ}C-3\,^{\circ}C[/tex]

[tex]\Delta T_{1} = 177\,^{\circ}C[/tex]

[tex]\Delta T_{2} = 78\,^{\circ}C - 55\,^{\circ}C[/tex]

[tex]\Delta T_{2} = 23\,^{\circ}C[/tex]

[tex]\Delta T_{lm} = \frac{177\,^{\circ}C-23\,^{\circ}C}{\ln \frac{177\,^{\circ}C}{23\,^{\circ}C} }[/tex]

[tex]\Delta T_{lm} \approx 75.466\,^{\circ}C[/tex]

La diferencia media logarítimica de temperatura del intercambiador en paralelo es aproximadamente 75.466 ºC.

Intercambiador en contracorriente

[tex]\Delta T_{1} = 180\,^{\circ}C-55\,^{\circ}C[/tex]

[tex]\Delta T_{1} = 125\,^{\circ}C[/tex]

[tex]\Delta T_{2} = 78\,^{\circ}C-3\,^{\circ}C[/tex]

[tex]\Delta T_{2} = 75\,^{\circ}C[/tex]

[tex]\Delta T_{lm} = \frac{125\,^{\circ}C-75\,^{\circ}C}{\ln \frac{125\,^{\circ}C}{75\,^{\circ}C} }[/tex]

[tex]\Delta T_{lm} \approx 97.881\,^{\circ}C[/tex]

La diferencia media logarítmica de temperatura del intercambiador en contracorriente es aproximadamente 97.881 ºC.

Answer:

Follows are the solution to this question:

Explanation:

Its difference in power supply will be smaller whenever the pressure transducer is being performed. Its transducer sensitivity(s) is often used to calculate its number of changes occurring at voltage output of ([tex]v_o[/tex]) when the supply tension ([tex]v_s[/tex]) changes with the measured pressure ([tex]p_m[/tex]) and the rated tension of transducers ([tex]p_s[/tex]).

[tex]v_o= (s) V_s (\frac{P_m}{P_s})[/tex]

Answer:

1819.4 ft

Explanation:

Let us assume  the coefficient of side friction is 0.10

The radius of curve is given by the formula:

[tex]R=\frac{V^2}{g(f_s+\frac{e}{100} )}[/tex]

Where R is the minimum radius of a horizontal curve, V is the speed, f is coefficient of side friction and e/100 = super elevation rate, g =acceleration due to gravity

Given that:

V = 70 mph = (70 * 1.467) ft/s = 102.69 ft/s, g =32.2 ft/s², e/100 = 0.08, f= 0.1

[tex]R=\frac{V^2}{g(f_s+\frac{e}{100} )}=\frac{102.69^2}{32.2(0.1+0.08)}=1819.4\ ft[/tex]

the minimum radius of a horizontal curve is 1819.4 ft

Answer:200%

Explanation:

so it can melt to use it later?

Answer:

24 cans

Explanation:

The number of glycol solution needed to prevent water from freezing is 42 galloons. The glycol solution comes in cans of 1.75 gallons.

Therefore 1 can of glycol = 1.75 gallon.

Number of cans of glycol needed = Total gallons of glycol needed / number of gallon present in each can

Number of cans of glycol needed = 42 gallons / (1.75 gallon per can)

Number of cans of glycol needed = 24 cans

Therefore a total of 24 cans are needed.

Answer:

[tex]0.00273\ \text{Pa s}[/tex]

3.944

Explanation:

[tex]\tau[/tex] = Shearing stress = [tex]0.54\ \text{N/m}^2[/tex]

[tex]\dot{\gamma}[/tex] = Rate of shearing strain = [tex]198\ \text{s}^{-1}[/tex]

(a) Viscosity is given by

[tex]\mu_b=\dfrac{\tau}{\dot{\gamma}}\\\Rightarrow \mu_b=\dfrac{0.54}{198}\\\Rightarrow \mu_b=0.00273\ \text{Pa s}[/tex]

The apparent viscosity of the blood is [tex]0.00273\ \text{Pa s}[/tex]

(b) Viscosity of water at 37 ˚C = 0.0006922 Pa s = [tex]\mu_w[/tex]

The ratio is

[tex]\dfrac{\mu_b}{\mu_w}=\dfrac{0.00273}{0.0006922}\\\Rightarrow \dfrac{\mu_b}{\mu_w}=3.944[/tex]

The required ratio is 3.944

Answer:

The specimen would not fracture having a flexural strength ( 300 MPa ) > ( 286.5 MPa )

Explanation:

The expression to represent  the stress on a cylindrical specimen  by a three-point transverse bending test

б =  [tex]\frac{F_{f}L }{\pi R^3}[/tex]  ---------------- ( 1 )

where : R = radius, Ff = flexural load,

L = distance between loads ( two loads )

Given data:

7500 N =  flexural load

(15 * 10^-3) m =  distance between loads

(5 * 10^-3) m  =  radius

input the given values into equation 1

б = 286.5 * 10^6 N/m^2   =  286.5 MPa

Note : the flexural strength of the cylindrical specimen is 300MPa

Therefore the specimen would not fracture having a flexural strength ( 300 MPa ) > ( 286.5 MPa )

Answer:

1417.05 kw

Explanation:

From the question,

The Mass flow rate = 3kg

The Saturated temperature = 275⁰c

The Thermal efficiency = 30%

The enthalpy of vaporization is 1574.5kj/kg

Rate of heat transfer = 3x1574.5 = 4723.5 kW

Thermal efficiency of engine =

0.30 x 4723.5

= 1417.05 kW

This answer is the net power output of this engine, in kW.

Thank you!

Answer:

The specific volume of R-134a at the exit of the expansion valve is 0.014655 cubic meters per kilogram.

Explanation:

From Thermodynamics, we know that expansion valves are devices that work at steady state, in which heat and work interactions are usually non-existent and changes in gravitational potential and translational kinetic energy. From First Law of Thermodynamics and the Principle of Mass Conservation we get that expansion valve is represented by the following model:

[tex]\dot H_{in}-\dot H_{out} = 0[/tex] (Eq. 1)

[tex]\dot m_{in}-\dot m_{out} = 0[/tex] (Eq. 2)

Where:

[tex]\dot H_{in}[/tex], [tex]\dot H_{out}[/tex] - Rates of change of inlet and output enthalpies, measured in kilowatts.

[tex]\dot m_{in}[/tex], [tex]\dot m_{out}[/tex] - Mass flow rates at inlet and outlet, measured in kilograms per second.

From definition of enthalpy and (Eq. 2), we simplify (Eq. 1) as follows:

[tex]u_{in} + P_{in}\cdot \nu_{in} = u_{out} + P_{out}\cdot \nu_{out}[/tex] (Eq. 3)

Where:

[tex]u_{1}[/tex], [tex]u_{2}[/tex] - Specific internal energies at inlet and outlet, measured in kilojoules per kilogram.

[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Inlet and outlet pressures, measured in kilopascals.

[tex]\nu_{1}[/tex], [tex]\nu_{2}[/tex] - Inlet and outlet specific volumes, measured in cubic meters per kilogram.

Now we get the following information from property charts:

Inlet (Saturated Liquid)

[tex]T = 40\,^{\circ}C[/tex]

[tex]P = 1017.1\,kPa[/tex]

[tex]\nu = 0.0008720\,\frac{m^{3}}{kg}[/tex]

[tex]u = 107.39\,\frac{kJ}{kg}[/tex]

[tex]h = 108.28\,\frac{kJ}{kg}[/tex]

[tex]x = 0[/tex]

Outlet (Liquid-Vapor Mix)

[tex]T = 6\,^{\circ}C[/tex]

[tex]P = 362.23\,kPa[/tex]

[tex]\nu = 0.014655\,\frac{m^{3}}{kg}[/tex]

[tex]u = 102.968\,\frac{kJ}{kg}[/tex]

[tex]h = 108.28\,\frac{kJ}{kg}[/tex]

The specific volume of R-134a at the exit of the expansion valve is 0.014655 cubic meters per kilogram.

Answer: 0.0000084

Explanation:

Given data:

Cross sectional area = 9.18mm*13.22mm

Force N = 3310

Elastic modulus = 79 GPa

Solution:

Area = 9.18mm * 13.22mm

= 121.36mm^2

= π(d)^2

= π(0.121)^2

= 0.05m^2

Stress = F/A

= 3310N/0.05m^2

= 66200Nm^2

Strain = Stress/ modulus of elastic

= 66200/7.9*10^10

= 0.0000084

Answer:

mechanical engineer

Explanation:

Answer:

hello your question is incomplete attached below is the complete question

Given you have a . 24 bit CPU The contents of memory are Address 412 413 Value0x120x34 Ox56 0x78 0x9A OxBC OxDEOxF1 0x23 414 419 418 420 415 416417 What is the value of the word in HEX beginning at address 414 if the machine is Little Endian? If it is Big Endian? Write your answer in HEX . Hex digits that are alphabetical should be written in uppercase For example A and not a Make sure that Ox appears at the beginning of your answer Do not put whitespace in your answer

Answer:  Little Endian: 0x9A7856.

                Big Endian: 0x56789A.

Explanation:

Given data :

24 bit CPU  ( size of each word = 24 bit )

next we have to determine how many bytes are contained in each 24 bit word

= 24 / 8 = 3 bytes

with this information a word in Hex beginning at 414 will end at 416

when we use the Little Endian representation the last byte of the binary representation which is 416 will be stored first followed by 415 and then 414

hence the word beginning at 414 using little Endian = 0x9A7856.

For Big Endian the process is vice versa

Hence the word beginning at 414 using Big Endian  = 0x56789A.

Answer:

The correct answer is True

Explanation:

Most business organizations especially those that are into sales, call centers, even individuals that are into marketing use CRM (Customer Relationship Management) software to drive their business forward by enabling them to interact more with customers and strive to provide better customer service. This is achievable because The software (CRM) which is a business tool stores customers' information ranging from names, contact info, sales, and other important information based on the need of the salespeople (organization) to track, meet and respond to customers needs. CRM benefits range from getting to know more about prospective or existing customers to building a better relationship with customers so as to retain them and allow better communication between customers in a bid to meet their demands.

Answer:

8.66142

Explanation:

Answer:

8.66

Explanation:

hope it helps

Answer:

C

Explanation:

Answer:

Construction engineers often focus on a specific type of construction project. Some of these specialties are: Building - commercial housing or business building. Electrical - electrical systems. Mechanical - plumbing, heating or air conditioning. Highway or Heavy - bridges, airports, highways or water waste.

Explanation:

Answer:

Drain current = 4050 μA

Explanation:

Determine the drain current in an NMOS transistor

Given data :

Kn = 400 μA/V^2

Vt = 0.5 v

λ = 0.02 V^−1

Vgs = 5 V

Vds = 6 v

Id = ?

The drain current can be calculated using the formula below

[tex]I_{D} = \frac{1}{2} U_{n} C_{OX} \frac{W}{L} ( V_{GS} - V_{t} ) ^2[/tex]

     = [tex]\frac{1}{2} Kn ( Vgs - Vt)^2[/tex]

     = 1/2 ( 400 μA/V^2) ( 5 - 0.5 )^2

     = (200 μA/V^2) * (20.25 v^2)

     = 4050 μA

Answer:

The  average reactive power is [tex]I_{rms}V_{rms} cos \phi[/tex]

Explanation:

For a single-phase AC circuit;

the peak current in the circuit is given as  [tex]I_o[/tex]

the root-mean-square current is given by;

[tex]I_{rms} = \frac{I_o}{\sqrt{2}}[/tex]

the peak voltage in the circuit is given as [tex]V_o[/tex]

the root-mean-square voltage is given by;

[tex]V_{rms} = \frac{V_o}{\sqrt{2} }[/tex]

The average reactive power is given by

[tex]P_{avg} = I_{rms}V_{rms} cos \phi[/tex]

where;

[tex]cos \phi \ is \ the \ power \ factor[/tex]

Therefore, the  average reactive power is [tex]I_{rms}V_{rms} cos \phi[/tex]

Answer:

hello your question is incomplete attached below is the complete question

Explanation:

attached below is the detailed solution

Taking into account that EI is constant and the Beam is made out of a single uniform material which is prismatic