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Aliun [14]
3 years ago
13

Test question for my account

Engineering
1 answer:
Ierofanga [76]3 years ago
5 0

Answer:

Well, I hope your account is working good now :)

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The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature
Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

P_{1} = 5 Mpa

T_{1} = 1623°C

                       = 1896 K

V_{1} = 0.05 m^{3}

Also given \frac{V_{2}}{V_{1}} = 20

Therefore, V_{2} = 1  m^{3}

R = 0.27 kJ / kg-K

C_{V} = 0.8 kJ / kg-K

Also given : P_{1}V_{1}^{1.25}=C

   Therefore, P_{1}V_{1}^{1.25} = P_{2}V_{2}^{1.25}

                     5\times 0.05^{1.25}=P_{2}\times 1^{1.25}

                     P_{2} = 0.1182 MPa

a). Work transfer, δW = \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

                                  \left [\frac{5\times 0.05-0.1182\times 1}{1.25-1}  \right ]\times 10^{6}

                              = 527200 J

                             = 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

   = \frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

  =\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W

  =\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200

  = 197.7 kJ

6 0
3 years ago
Write a C program that asks the user to enter two numbers, obtains the two numbers from the user and prints the sum, product, di
Bogdan [553]

Answer:

View Image

Explanation:

Initialize your variable as a float or double since you're going to be using fractions in your answer.

User scanf() to get user input.

Print out the sum, product, quotient, and difference between the two numbers.

8 0
3 years ago
An insulated piston-cylinder device initially contains 0.16 m2 of CO2 at 150 kPa and 41 °C. Electric resistance heater supplied
lyudmila [28]

Answer:

I=0.3636

Explanation:

See the attached picture for explanation.

4 0
3 years ago
A 600 MW power plant has an efficiency of 36 percent with 15
ololo11 [35]

Answer:

401.3 kg/s

Explanation:

The power plant has an efficiency of 36%. This means 64% of the heat form the source (q1) will become waste heat. Of the waste heat, 85% will be taken away by water (qw).

qw = 0.85 * q2

q2 = 0.64 * q1

p = 0.36 * q1

q1 = p /0.36

q2 = 0.64/0.36 * p

qw = 0.85 *0.64/0.36 * p

qw = 0.85 *0.64/0.36 * 600 = 907 MW

In evaporation water becomes vapor absorbing heat without going to the boiling point (similar to how sweating takes heat from the human body)

The latent heat for the vaporization of water is:

SLH = 2.26 MJ/kg

So, to dissipate 907 MW

G = qw * SLH = 907 / 2.26 = 401.3 kg/s

8 0
3 years ago
Read 2 more answers
We have considered heat and work to be path-dependent. However, if all heat transfer with surroundings is performed using a reve
NeX [460]

Answer:

path-dependent. However, if all heat transfer with surroundings is performed using a reversible heat transfer device (some type of reversible Carnot-type device), work can be performed by the heat transfer device during heat transfer to the surroundings. The net heat transferred to the surroundings and the net work done will be independent of the path. Demonstrate this by calculating the work and heat interactions for the system, the heat transfer device, and the sum for each of the following paths where the surroundings are at Tsurr = 273 K. The state change is from state 1, P1 = 0.1 MPa, T1 = 298 K and state 2, P2 = 0.5 MPa and T2 which will be found in part (a). CP = 7R/2. a. Consider a state change for an ideal gas in a piston/cylinder. Find T2 by an adiabatic reversible path. Find the heat and work such that no entropy is generated in the universe. This is path a. Sketch path a qualitatively on a P-V diagram. b. Now consider a path

3 0
3 years ago
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