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ziro4ka [17]
3 years ago
11

stimate the maximum efficiency of an automobile engine that has a compression ratio of 5:1.0. Assume the engine operates accordi

ng to the Otto cycle and assume γ = 1.4.
Engineering
1 answer:
Fed [463]3 years ago
5 0

Answer:

Efficiency based on Otto cycle.

Effotto = 47.47%

Explanation:

Efficiency based on Otto cycle.

effotto = 1 – (V2 / V1)^γ-1

effotto = 1 – (1 / 5)^1.4 - 1

effotto = 47.47%

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Which option distinguishes the members of a software deployment process team most likely involved in the following scenario?
Alchen [17]

Answer:

A local bank, with several branches in three cities, requests changes to its mortgage calculation software.

5 0
2 years ago
P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diame
eimsori [14]

Answer:

a)  m=0.17kg/s

b)  Ma=0.89

Explanation:

From the question we are told that:

Pressure P=60kPa

Diameter d=3cm

Generally at sea level

T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4

Generally the Power series equation for Mach number is mathematically given by

\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}

\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}

Ma=0.89

Therefore

Mass flow rate

\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}

\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}

\rho=0.848kg/m^3

Generally the equation for Velocity at throat is mathematically given by

V=Ma(r*T_0\sqrt{T_e})

Where

T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}

T_e=248

Therefore

V=0.89(1.4*288\sqrt{248})\\\\V=284

Generally the equation for Mass flow rate is mathematically given by

m=\rho*A*V

m=0.84*\frac{\pi}{4}*3*10^{-2}*284

m=0.17kg/s

6 0
2 years ago
An ideal reheat Rankine cycle with water as the working fluid operates the boiler at 15,000 kPa, the reheater at 2000 kPa, and t
solniwko [45]

Answer:

See the explanation below.

Explanation:

First find the enthalpies h₁, h₂, h₃, h₄, h₅, and h₆.

Find h₁:

Using Saturated Water Table and given pressure p₁ = 100 kPa

h₁ = 417.5 kJ/kg

Find h₂:

In order to find h₂, add the w_{p} to h₁, where  w_{p}  is the work done by pump and h₁ is the enthalpy computed above h₁ = 417.5 kJ/kg.

But first we need to compute  w_{p} To computer  

Pressures:

p₁ = 100 kPa

p₂ = 15,000 kPa

and

Using saturated water pressure table, the volume of water v_{f} = 1.0432

Dividing 1.0432/1000 gives us:

Volume of water = v₁ =  0.001043 m³/kg

Compute the value of h₂:

h₂ = h₁ + v₁ (p₂ - p₁)

    = 417.5 kJ/kg + 0.001043 m³/kg ( 15,000 kPa - 100 kPa)

    =  417.5 + 0.001043 (14900)

    = 417.5 + 15.5407

    = 433.04 kJ/kg

Find h₃  

Using steam table:

At pressure p₃ = 15000 kPa

and Temperature = T₃ = 450°C

Then h₃ = 3159 kJ/kg

The entropy s₃ = 6.14 kJ/ kg K

Find h₄

Since entropy s₃ is equal to s₄ So

s₄ = 6.14 kJ/kgK

To compute h₄

s₄ = s_{f} + x_{4} s_{fg}

x_{4} = s_{4} -s_{f} /s_{fg}

x_{4} = 6.14 -  2.45 / 3.89

x_{4}   = 0.9497

The enthalpy h₄:

h₄ = h_{f} +x_{4} h_{fg}

    = 908.4 + 0.9497(1889.8)

    =  908.4 + 1794.7430

    = 2703 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₃ = 6.14 kJ/ kg K and pressure p₄ = 2000 kPa

Find h₅

Using steam table:

At pressure p₅ = 2000 kPa

and Temperature = T₅ = 450°C

Then h₅  = 3358 kJ/kg

Find h₆:

Since the entropy s₅ = 7.286 kJ/kgK is equal s₆ to  So

s₆ = 7.286 kJ/kgK = 7.29 kJ/kgK

To compute h₆

s₆ = s_{f} + x_{6} s_{fg}

x_{6} = s_{6} -s_{f} /s_{fg}

x_{6} = 7.29 - 1.3028 / 6.0562

x_{6}   = 0.988

The enthalpy h₆:

h₆ = h_{f} +x_{6} h_{fg}

    = 417.51 + 0.988 (2257.5)

    = 417.51 + 2230.41

  h₆ =  2648 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₅ = 7.286 kJ/kgK and pressure p₅ = 2000 kPa

Compute power used by pump:

P_{p} is found by using:

mass flow rate = m =  1.74 kg/s

Volume of water = v₁ =  0.001043 m³/kg

p₁ = 100 kPa

p₂ = 15,000 kPa

P_{p}  = ( m ) ( v₁ ) ( p₂ - p₁ )

     = (1.74 kg/s) (0.001043 m³/kg) (15,000 kPa - 100 kPa)

     = (1.74 kg/s) (0.001043 m³/kg) (14900)

     = 27.04

P_{p} = 27 kW

Compute heat added q_{a} and heat rejected q_{r}  from boiler using computed enthalpies:

q_{a} = ( h₃ - h₂ ) + ( h₅ - h₄ )

      = ( 3159 kJ/kg - 433.04 kJ/kg ) + ( 3358 kJ/kg - 2703 kJ/kg )

      = 2726 + 655

      = 3381  kJ/kg

q_{r} =  h₆ - h₁

  = 2648 kJ/kg - 417.5 kJ/kg

  = 2232 kJ/kg

Compute net work

W_{net} = q_{a} - q_{r}

       = 3381  kJ/kg - 2232 kJ/kg

       = 1150 kJ/kg

Compute power produced by the cycle

mass flow rate = m =  1.74 kg/s

W_{net} = 1150 kJ/kg

P = m * W_{net}

  = 1.74 kg/s * 1150 kJ/kg

  = 2001 kW

Compute rate of heat transfer in the reheater

Q = m * ( h₅ - h₄ )

   =  1.74 kg/s * 655

   =  1140 kW

Compute Thermal efficiency of this system

μ_{t} = 1 - q_{r} /  q_{a}

   = 1 - 2232 kJ/kg / 3381  kJ/kg

   = 1 - 0.6601

   = 0.34

   = 34%

7 0
3 years ago
What is the mass of a brass axle that has a volume of 0.318 cm? ​
NeX [460]

Answer:

2.7g

Explanation:

the mass of a brass axle that has a volume of 0.318 cm is 2.7g.

8 0
2 years ago
Write a method printShampooInstructions(), with int parameter numCycles, and void return type. If numCycles is less than 1, prin
kirill [66]

Answer:

// The method is defined with a void return type

// It takes a parameter of integer called numCycles

// It is declared static so that it can be called from a static method

public static void printShampooInstructions(int numCycles){

// if numCycles is less than 1, it display "Too few"

   if (numCycles < 1){

       System.out.println("Too few.");

   }

// else if numCycles is less than 1, it display "Too many"

    else if (numCycles > 4){

       System.out.println("Too many.");

   }

// else it uses for loop to print the number of times to display

// Lather and rinse

  else {

       for(int i = 1; i <= numCycles; i++){

           System.out.println(i + ": Lather and rinse.");

       }

       System.out.println("Done");

       

   }

}

Explanation:

The code snippet is written in Java. The method is declared static so that it can be called from another static method. It has a return type of void. It takes an integer as parameter.

It display "Too few" if the passed integer is less than 1. Or it display "Too much" if the passed integer is more than 4. Else it uses for loop to display "Lather and rinse" based on the passed integer.

8 0
3 years ago
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