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marshall27 [118]
3 years ago
14

An Otto cycle with air as the working fluid has a compression ratio of 8.2. Under cold air standard conditions, what is the ther

mal efficiency of this cycle?
Engineering
1 answer:
Tresset [83]3 years ago
5 0

Answer:

Under cold air standard conditions, the thermal efficiency of this cycle is 56.9 percent.

Explanation:

From Thermodynamics we remember that thermal efficiency of the ideal Otto cycle (\eta_{th}), dimensionless, is defined by the following formula:

\eta_{th} = 1-\frac{1}{r^{\gamma-1}} (Eq. 1)

Where:

r - Compression ratio, dimensionless.

\gamma - Specific heat ratio, dimensionless.

Please notice that specific heat ratio under cold air standard conditions is \gamma = 1.4.

If we know that r = 8.2 and \gamma = 1.4, then thermal efficiency of the ideal Otto cycle is:

\eta_{th} = 1-\frac{1}{8.2^{1.4-1}}

\eta_{th} = 0.569

Under cold air standard conditions, the thermal efficiency of this cycle is 56.9 percent.

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Steam enters a turbine at 120 bar, 508oC. At the exit, the pressure and quality are 50 kPa and 0.912, respectively. Determine th
Archy [21]

Answer:

The turbine produces <u>955.53 KW</u> power.

Explanation:

Taking the turbine as a system. Applying Law of Conservation of Energy:

m(h₁ - h₂) - Heat Loss = P

where,

m = mass flow rate of steam = 1.31 kg/s

h₁ = enthalpy at state 1 (120 bar and 508°C)

h₂ = enthalpy at state 2 (50 KPa and x = 0.912)

Heat Loss = 225 KW

P = Power generated by turbine = ?

First, we find h₁ from super steam tables:

At,

T = 508°C

P = 120 bar = 12000 KPa = 12 MPa

we find that steam is in super-heated state with enthalpy:

Due to unavailibility of values in chart we approximate the state to 500° C and 12.5 MPa:

h₁ = 3343.6 KJ/kg

Now, for state 2, we have:

P = 50 KPa and x = 0.912

From saturated steam table:

h₂ = hf₂ + x(hfg₂) = 340.54 KJ/kg + (0.912)(2304.7 KJ/kg)

h₂ = 2442.4 KJ/kg

Now, using values in the conservation equation:

(1.31 kg/s)(3343.6 KJ/kg - 2442.4 KJ/kg) - 225 KW = P

<u>P = 955.53 KW</u>

5 0
3 years ago
The biggest advantage of sketches is that
steposvetlana [31]

Answer:

They communicate ideas very quickly.

Explanation:

8 0
3 years ago
A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy a
Ratling [72]

Answer:

Explanation:

From  the given question:

Using the distortion energy theory to determine the  factors of safety  FOS can be expressed  by the relation:

\dfrac{Syt}{FOS}= \sqrt{ \sigma x^2+\sigma  y^2-\sigma x \sigma y+3 \tau_{xy^2}}

where; syt = strength in tension and compression = 350 MPa

The maximum shear stress theory  can be expressed as:

\tau_{max} = \dfrac{Syt}{2FOS}

where;

\tau_{max} =\sqrt{ (\dfrac{\sigma x-\sigma  y}{2})^2+ \tau _{xy^2

a. Using distortion - energy theory formula:

\dfrac{350}{FOS}= \sqrt{94^2+0^2-94*0+3 (-75)^2}}

\dfrac{350}{FOS}=160.35

{FOS}=\dfrac{350}{160.35}

FOS = 2.183

USing the maximum-shear stress theory;

\dfrac{350}{2 FOS}  =\sqrt{ (\dfrac{94-0}{2})^2+ (-75)^2

\dfrac{350}{2 FOS}  =88.51

\dfrac{350}{ FOS}  =2 \times 88.51

{ FOS}  =\dfrac{350}{2 \times 88.51}

FOS = 1.977

b. σx = 110 MPa, σy = 100 MPa

Using distortion - energy theory formula:

\dfrac{350}{FOS}= \sqrt{ 110^2+100^2-110*100+3(0)^2}

\dfrac{350}{FOS}= \sqrt{ 12100+10000-11000

\dfrac{350}{FOS}=105.3565

FOS=\dfrac{350}{105.3565}

FOS =3.322

USing the maximum-shear stress theory;

\dfrac{350}{2 FOS}  =\sqrt{ (\dfrac{110-100}{2})^2+ (0)^2

\dfrac{350}{2 FOS}  ={ (\dfrac{110-100}{2})^2

\dfrac{350}{2 FOS}  =25

FOS = 350/2×25

FOS = 350/50

FOS = 70

c. σx = 90 MPa, σy = 20 MPa, τxy =−20 MPa

Using distortion- energy theory formula:

\dfrac{350}{FOS}= \sqrt{ 90^2+20^2-90*20+3(-20)^2}

\dfrac{350}{FOS}= \sqrt{ 8100+400-1800+1200}

\dfrac{350}{FOS}= 88.88

FOS = 350/88.88

FOS = 3.939

USing the maximum-shear stress theory;

\dfrac{350}{2 FOS}  =\sqrt{ (\dfrac{90-20}{2})^2+ (-20)^2

\dfrac{350}{2 FOS}  =\sqrt{ (35)^2+ (-20)^2

\dfrac{350}{2 FOS}  =\sqrt{ 1225+ 400

\dfrac{350}{2 FOS}  =40.31

FOS}  =\dfrac{350}{2*40.31}

FOS = 4.341

7 0
3 years ago
Technician A says that hoods are designed with reinforcements to prevent folding during a collision. Technician B says that some
-BARSIC- [3]

Technician A is wrong.

  • Usually, hoods have what is called "Crush Zones" underneath the panels. The function of the Crush Zone is to prevent the hoods, during a collision, from entering into the passenger space.

  • The crush zones allow the hoods to fold instead.

Technician B is right.

  • Automobile producers now make use of a hybrid form of hood that consists of fiberglass reinforced with plastic.

  • They are mostly used for trucks that have a low volume of production.

  • The hood is built using a process called Resin Transfer Model (RTM).

See the link below for more about automobile engineering:

brainly.com/question/4822721

6 0
2 years ago
A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric c
Paraphin [41]

Answer:

10.8\ \text{lb/ft^2}

101.96\ \text{lb/ft}^2

Explanation:

v_1 = Velocity of car = 65 mph = 65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}

\rho = Density of air = 0.00237\ \text{slug/ft}^3

v_2=0

P_1=0

h_1=h_2

From Bernoulli's law we have

P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}

The maximum pressure on the girl's hand is 10.8\ \text{lb/ft^2}

Now v_1 = 200 mph = 200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}

P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2

The maximum pressure on the girl's hand is 101.96\ \text{lb/ft}^2

5 0
2 years ago
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