An ideal Diesel cycle has a compression ratio of 17 and a cutoff ratio of 1.3. Determine the maximum temperature of the air and the rate of heat addition to this cycle when it produces 140 kW of power and the state of the air at the beginning of the compression is 90 kPa and 578C. Use constant specific heats at room temperature.

Answers

Answer 1

Answer:

maximum temperature = 1322 k

rate of heat addition = 212 kw

Explanation:

compression ratio = 17

cut off ratio = 1.3

power produced = 140 Kw

state of air at the beginning of the compression = 90 kPa and 578 c

Determine the maximum temperature of air

attached below is the detailed solution

An Ideal Diesel Cycle Has A Compression Ratio Of 17 And A Cutoff Ratio Of 1.3. Determine The Maximum

Related Questions

The fins attached to a heat exchanger-surface are determined to have an effectiveness of 0.9. Do you think the rate of heat transfer from the surface has increased or decreased as a result of the addition of these fins?

Answers

Answer:

The rate of heat transfer has increased.

Explanation:

Heat transfer rate is the rate at which heat energy is dissipated to the ambient from a hot body. The rate of heat transfer is proportional to the available surface area for heat exchange. This means that the greater the exposed surface area for heat exchange, the greater the rate at which heat is lost to the ambient. In introducing the fins to the heat exchange system (fins have a large surface area to volume ratio for maximum exposure to the ambient), one maximizes the available surface area for heat exchange between the material and the ambient, increasing the rate of heat transfer.

Consider an ideal gas undergoing a constant pressure process from state 1 to state
2 in a closed system. The specific heat capacities for this material depend on temperature in
the following way, cv = aT^b , cp = cT^d , where the constants a, b, c and d are known. Calculate

the specific entropy change, (s2 − s1), from state 1 to state 2.

Answers

Answer:

[tex]s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})[/tex]

Explanation:

Hello,

In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy at constant pressure:

[tex]ds=c_p\frac{dT}{T}-Rg\ \frac{dP}{P}[/tex]

Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:

[tex]\int\limits^{s_2}_{s_1} {} \, ds=c\int\limits^{T_2}_{T_1} {T^{d-1}dT} \,-Rg\ \int\limits^{P_2}_{P_1} {\frac{dP}{P}} \,[/tex]

We obtain the expression to compute the specific entropy change:

[tex]s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})[/tex]

Best regards.

In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of line has resistance R. For a power line that supplies power to 10,000 households, we can conclude that:________

Answers

Answer:

IV > [tex]I^{2} R[/tex]

Explanation:

The current in the power line = I

The voltage in the power line = V

The resistance of the power line = R

Power supplied from the power house = P

power delivered to the households = [tex]p[/tex]

We know that the power supplied to a power line system is proportional to

P = IV    ....1

we also know that according to Ohm's law, the relationship between the voltage, resistance, and current through an electrical system is given as

V = IR    ....2

substituting equation 2 into equation 1, the power delivered to the households is proportional to the square of the current.

[tex]p[/tex] = [tex]I^{2} R[/tex]    ....3

The problem is that when power is delivered across a transmission line, some of the power is loss due to Joules heating effect of the power lines. This energy and power loss is proportional to [tex]I^{2}[/tex] therefore, the electrical power delivered to the households will be less than the electrical power supplied from the power station. This means that

P > [tex]p[/tex]

equating these two powers from equations 1 and equation 3, we have

IV > [tex]I^{2} R[/tex]

A charge is distributed uniformly along a long straight wire. The electric field 2 cm from the wire is 36 N/C. The electric field 4 cm from the wire is:

Answers

Answer:

New electric field = 18 N/C

Explanation:

Given:

Length (E1) = 2 cm

New length (E2) = 4 cm

Electric field =  36 N/C

Find:

New electric field

Computation:

New electric field = 36 [2 / 4]

New electric field = 36 [1/2]

New electric field = 18 N/C

/ Air enters a 20-cm-diameter 12-m-long underwater duct at 50°C and 1 atm at a

mean velocity of 7 m/s, and is cooled by the water outside. If the average heat

transfer coefficient is 85 W/m2

°C and the tube temperature is nearly equal to the

water temperature of 5°C, determine the exit temperature of air and the rate of heat

transfer.​

Answers

Answer:

A) EXIT TEMPERATURE = 14⁰C

b) rate of heat transfer of air = - 13475.78 = - 13.5 kw

Explanation:

Given data :

diameter of duct = 20-cm = 0.2 m

length of duct = 12-m

temperature of air at inlet= 50⁰c

pressure = 1 atm

mean velocity = 7 m/s

average heat transfer coefficient = 85 w/m^2⁰c

water temperature = 5⁰c

surface temperature ( Ts) = 5⁰c

properties of air at 50⁰c and at 1 atm

= 1.092 kg/m^3

Cp = 1007 j/kg⁰c

k = 0.02735 W/m⁰c

Pr = 0.7228

v  = 1.798 * 10^-5 m^2/s

determine the exit temperature of air and the rate of heat transfer

attached below is the detailed solution

Calculate the mass flow rate

= p*Ac*Vmean

= 1.092 * 0.0314 *  7 = 0.24 kg/s

"The transistor base-emitter voltage (VBE) a. increases with an increase in temperature. b. is not affected by temperature change. c. decreases with an increase in temperature. d. has no effect on collector current."

Answers

Answer:

C) Decreases with an increase in temperature

Explanation:

As the temperature of a transistor increases, the thermal runaway property of the transistor becomes more significant and the transistors, conducting more freely as a result of the rise in temperature, causes an increase in the collector current or leakage current. The transistor base-emitter voltage decreases as a result.

With increased heating due to heavy current flow, the transistor is damaged.

A rate of 0.42 minute per piece is set for a forging operation. The operator works on the job for a full eight-hour day and produces 1,500 pieces. Use a standard hour plan.

Required:
a. How many standard hours does the operator earn?
b. What is the operator's efficiency for the day?
c. If the base rate is 9.80 per hour, compute the earnings for the day.
d. What is the direct labor cost per piece at this efficiency?
e. What would be the proper piece rate (rate expressed in money) for this job, assuming that the above time standard is correct?

Answers

Answer:

b. What is the operator's efficiency for the day?

                                                      AND

e. What would be the proper piece rate (rate expressed in money) for this job, assuming that the above time standard is correct?

Explanation:

A four-cylinder four-stroke engine is modelled using the air standard Otto cycle (two engine revolutions per cycle). Given the conditions at state 1, total volume (V1) of each cylinder, compression ratio (r), rate of heat addition (Q), and engine speed in RPM, determine the efficiency and other values listed below. The gas constant for air is R =0.287 kJ/kg-K.

T1 = 300 K
P1 = 100 kPa
V1 = 500 cm^3
r = 10
Q = 60 kW
Speed = 5600 RPM

Required:
a. Determine the total mass (kg) of air in the engine.
b. Determine the specific internal energy (kJ/kg) at state 1.
c. Determine the specific volume (m^3/kg) at state 1.
d. Determine the relative specific volume at state 1.

Answers

Answer:

a) Mt = 0.0023229

b) = U1 = 214.07

c) = V₁  = 0.861 m³/kg

d) = Vr1 = 621.2

Explanation:

Given that

R = 0.287 KJ/kg.K, T1 = 300 K , P1 = 100 kPa , V1 = 500 cm³, r = 10 , Q = 60 kW , Speed N = 5600 RPM, Number of cylinders K = 4

specific heat at constant volume Cv = 0.7174 kJ/kg.K

Specific heat at constant pressure is 1.0045 Kj/kg.K

a)  To determine the total mass (kg) of air in the engine.

we say

P1V1 = mRT1

we the figures substitute

(100 x 10³) ( 500 x  10⁻⁶) = m ( 0.287 x  10³) ( 300 )

50 = m x 86100

m = 0.00005 / 86100 = 0.0005807 ( mass of one cylinder)

Total mass of 4 cylinder

Mt = m x k

Mt = 0.0005807 x 4

Mt = 0.0023229

b) To determine the specific internal energy (kJ/kg) at state 1

i.e at T1 = 300

we obtain the value of specific internal energy U1 at 300 K ( state 1) from the table ideal gas properties of air.

U1 = 214.07

c) To determine the specific volume (m³/kg) at state 1.

we say

V₁ = V1/m

V₁ = (500 x  10⁻⁶) / 0.0005807

V₁  = 0.861 m³/kg

d) To determine the relative specific volume at state 1.

To obtain the value of relative specific volume at 300 K ( i.e state 1) from the table ideal gas properties of air.

At T1 = 300 k

Vr1 = 621.2

Q1: You have to select an idea developing an application like web/mobile or industrial, it should be based on innovative idea, not just a simple CRUD application. After selecting the idea do the following: 1) How your project will be helpful and what problem this project addresses. (10-Marks) 2) Write down the requirements. (10Marks) 3) List the functional and non-functional requirements of your project. (10marks) 4) Which process model you will follow for this project and why? (10marks) 5) Draw the Level 0, and level 1 DFD of your application. (20marks)

Answers

Answer:

Creating an app is both an expression of our self and a reflection of what we see is missing in the world. We find ourselves digging deep into who we are, what we would enjoy working on, and what needs still need to be fulfilled. Generating an app idea for the first time can be extremely daunting. Especially with an endless amount of possibilities such as building a church app.

The uncertainty has always spawned a certain fear inside creators. The fear of creating something no one will enjoy. Spending hundreds of dollars and hours building something which might not bring back any real tangible results. The fear of losing our investment to a poor concept is daunting but not random. But simple app ideas are actually pretty easy to come by.

Great app idea generation is not a gift given to a selected few, instead, it is a process by which any of us are able to carefully explore step by step methods to find our own solution to any problem. Whether you are a seasoned creator or a novice, we have provided a few recommendations to challenge and aid you as you create your next masterpiece.

if I am right then make me brainliest

Consider atmospheric air at 25 C and a velocity of 25 m/s flowing over both surfaces of a 1-m-long flat plate that is maintained at 125 C. Determine the rate of heat transfer per unit width from the plate for values of the critical Reynolds number corresponding to 105 , 5 105 , and 106 .

Answers

Answer:

Explanation:

Temperature of atmospheric air To = 25°C = 298 K

Free  stream velocity of air Vo = 25 m/s

Length and width of plate = 1m

Temperature of plate Tp = 125°C = 398 K

We know for air, Prandtl number Pr = 1

And for air, thermal conductivity K = 24.1×10?³ W/mK

Here, charectorestic dimension D = 1m

 

Given value of Reynolds number Re = 105

For laminar boundary layer flow over flat plate

= 3.402

Therefore, hx = 0.08199 W/m²K

So, heat transfer rate q = hx×A×(Tp – To)

                                          = 0.08199×1×(398 – 298)

Two small balls A and B with masses 2m and m respectively are released from rest at a height h above the ground. Neglecting air resistance, which of the following statements are true when the two balls hit the ground?
(a) The kinetic energy of A is the same as the kinetic energy of B
(b) The kinetic energy of A is half the kinetic energy of B.
(c) The kinetic energy of A is twice the kinetic energy of B.
(d) The kinetic energy of A is four times the kinetic energy of B Explain your answer why.
What is the definition of General Plan Motion? What would be the effective methodology or approach to solve a rigid body kinematics problem?

Answers

Answer:

The kinetic energy of A is twice the kinetic energy of B

Explanation:

The true statement when the two balls hit the ground is, the kinetic energy of A is twice the kinetic energy of B. The correct option is (c).

What is kinetic energy?

Kinetic energy is the energy of motion, which can be seen as an item or subatomic particle moving. Kinetic energy exists in every moving object and particle.

Kinetic energy is demonstrated by a person walking, a soaring baseball, a crumb falling from a table, and a charged particle in an electric field.

The definition of a General Plan of Motion is every point on the body has a different path. As a result, we must relate the forces to the acceleration of the body's center of mass, as well as the moments to the angular accelerations.

Therefore, the correct option is (c), The kinetic energy of A is twice the kinetic energy of B.

To learn more about kinetic energy, refer to the link:

https://brainly.com/question/26472013

#SPJ2

When replacing a timing belt, many experts and vehicle manufacturers recommend that all of the following should be replaced except the

A. water pump
B. camshaft oil seal(s).
C. camshalt sprocket
D. tensioner assembly

Answers

Answer:

Correct Answer:

A. water pump

Explanation:

Timing belt in a vehicle helps to ensure that crankshaft, pistons and valves operate together in proper sequence. Timing belts are lighter, quieter and more efficient than chains that was previously used in vehicles.

Most car manufacturers recommended that, when replacing timing belt, tension assembly, water pump, camshaft oil seal should also be replaced with it at same time.

The structure of a house is such that it loses heat at a rate of 5400 kJ/h per degree Cdifference between the indoors and outdoors. A heat pump that requires a power input of 6 kW isused to maintain this house at 21 C. Determine the lowest outdoor temperature for which the heatpump can meet the heating requirements of this house

Answers

Answer: Tl = - 13.3°C

the lowest outdoor temperature is - 13.3°C

Explanation:

Given that;

Temperature of Th = 21°C = 21 + 273 = 294 K

the rate at which heat lost is Qh = 5400 kJ/h°C

the power input to heat pump Wnet = 6 kw

The COP of a reversible heat pump depends on the temperature limits in the cycle only, and is determined by;

COPhp = Th/(Th - Tl)

COPhp = Qh/Wnet

Qh/Wnet = Th/(Th -Tl)

the amount of heat loss is expressed as

Qh = 5400/3600(294 - Tl)

the temperature of sink

( 5400/3600(294 - Tl)) / 6 = 294 / ( 294 - Tl)

now solving the equation

Tl = 259.7 - 273

Tl = - 13.3°C

so the lowest outdoor temperature is - 13.3°C

An air-conditioner which uses R-134a operates on the ideal vapor compression refrigeration cycle with a given compressor efficiency.
--Given Values--
Evaporator Temperature: T1 (C) = 9
Condenser Temperature: T3 (C) = 39
Mass flow rate of refrigerant: mdot (kg/s) = 0.027
Compressor Efficiency: nc (%) = 90

a) Determine the specific enthalpy (kJ/kg) at the compressor inlet.
Your Answer =
b) Determine the specific entropy (kJ/kg-K) at the compressor inlet
Your Answer =
c) Determine the specific enthalpy (kJ/kg) at the compressor exit
Your Answer =
d) Determine the specific enthalpy (kJ/kg) at the condenser exit.
Your Answer =
e) Determine the specific enthalpy (kJ/kg) at the evaporator inlet.
Your Answer =
f) Determine the coefficient of performance for the system.
Your Answer =
g) Determine the cooling capacity (kW) of the system.
Your Answer =
h) Determine the power input (kW)to the compressor.
Your Answer =

Answers

Answer:

A) 251.8 kj/kg

B) 0.9150 kj/kg-k

C) 155.4 kj/kg

F) 1.50

G) 3.95 kw

H) 2.6 kw

Explanation:

Given conditions :

air conditioner : R -134a

compressor efficiency (nc) = 90%.

T1 = 9⁰c,  T3 = 39⁰c, mass flow rate = 0.027 kg/s

A) Specific enthalpy at the compressor inlet

at T = 9⁰c the saturated vapor (x) = 1

from the R-134a property table

h1 = 251.8 kj/kg

B ) specific entropy ( kj/kg-k) at the compressor inlet

at T = 9⁰c the saturated vapor (x) = 1

s = 0.9150 kj/kg-k ( from the R-134a property table )

C) specific enthalpy at the compressor exit

at T3 = 39⁰c , s2 = s1

has = 165.12 kj/kg

h2 = 155.4 kj/kg

attached below is the remaining solution to some of the problems

Input resistance of a FET is very high due to A) forward-biased junctions have high impedance B) gate-source junction is reverse-biased C) drain-source junction is reverse-biased D) none of the above

Answers

Answer:

B) gate-source junction is reverse-biased

Explanation:

FET is described as an electric field that controls the specific current and is being applied to a "third electrode" which is generally known as "gate". However, only the electric field is responsible for controlling the "current flow"   in a specific channel and then the particular device is being "voltage operated" that consists of high "input impedance".

In FET, the different "charge carriers" tend to enter a particular channel via "source" and exits through "drain".

After a capacitor is fully chargerd, a small amount of current will flow though it. what is this current called?

Answers

Answer:

  leakage

Explanation:

That current is "leakage current."

You have accumulated several parking tickets while at school, but you are graduating later in the year and plan to return to your home in another jurisdiction. A friend tells you that the authorities in your home jurisdiction will never find out about the tickets when you re-register your car and apply for a new license. What should you do?

Answers

Answer:

pay off the parking tickets

Explanation:

In the scenario being described, the best thing to do would be to pay off the parking tickets. The parking tickets stay under your name, and if they are not paid in time can cause problems down the road. For starters, if they are not paid in time the amount will increase largely which will be harder to pay. If that increased amount is also not paid, then the government will suspend your licence indefinitely which can later lead to higher insurance rates.

Summary of Possible Weather and Associated Aviation Impacts for Geographic/Topographic Categories Common in the Western United States.
Geographic/Topographic Descriptive Summary of Potential Aviation Impacts
Category of a Possible Weather That Could Impact Based on Weather
of Airport Location Aviation Operations
Along the US West coast,
with steep mountains to the east
(An example of this category is
Santa Barbara Airport, located
on the Southern California Coast,
at an elevation of 10 feet).
Within a valley in elevated terrain
surrounded by high mountains
(An example of this category is
Friedman Memorial Airport, located
in Central Idaho, at an elevation of 5300 feet).
In elevated terrain on the leeside of
high mountains
(An example of this category is Northern Colorado
Regional Airport, located in northern Colorado,
at an elevation of 5000 feet, on the leeside
of the Rocky mountains).

Answers

Answer: answer provided in the explanation section.

Explanation:

Weather phenomenons that would impart Aviation Operations in Santa Barbara -

1. Although winters are cold, wet, and partly cloudy here. It is in general favorable for flying. But sometimes strong winds damage this pleasant weather.

2.  The Sundowner winds cause rapid warming and a decrease in relative humidity. The wind speed is very high surrounding this area for this type of wind.  

3. Cloud is an important factor that affects aviation operations. Starting from April, here the sky is clouded up to November. The sky is overcast (80 to 100 percent cloud cover) or mostly cloudy (60 to 80 percent) 44% on a yearly basis. Thus extra cloud cover can trouble aviation operations.

4. The average hourly wind speed can also be a factor. This also experiences seasonal variations, these variations are studied carefully in the aviation industry. The windier part of the year starts in January and ends in June. In April, the wind speed can reach 9.5 miles per hour.

This and more are some factors to look into when considering wheather conditions that would affect aviation operations.

I hope this was a bit helpful. cheers

After clamping a buret to a ring stand, you notice that the set-up is tippy and unstable. What should you do to stabilize the set-up

Answers

Answer:

Move the buret clamp to a ring stand with a larger base.

Explanation:

A right stand is used for titration experiments in the laboratory. It holds the burette firmly during experiments so that accurate readings can be taken.

The right stand is made up of support base, vertical stainless steel, clamp with adjustable screw that holds on to the vertical rod.

The clamp is used to hold the burette in place.

If after clamping a buret to a ring stand, you notice that the set-up is tippy and unstable, the best action will be to move the buret clamp to a ring stand with a larger base.

The larger base provides a better center of gravity and stabilises the setup

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