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2 years ago

Why is my lawn mower not starting?

Engineering

1 answer:

EleoNora [17]2 years ago

6 0

Answer:

Other possible causes include: Loose, Dirty or Disconnected Spark Plug in Your Lawn Mower: Check it out, clean off debris, re-connect and tighten. Dirty Air Filter: Clean or replace. Fuel Not Reaching the Engine: Tap the side of the carburetor to help the flow of gas.

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A steam power plant operates on an ideal Rankine cycle with two stages of reheat and has a net power output of 120 MW. Steam ent

uysha [10]

Answer:

a) 40.6%

b)72.19kg/s

Explanation:

The Rankine cycle with two reheat stages has 9 stages in total.

The maximum pressure will be at the first inlet stage of the HP turbine which is stage 3. The minimum pressure will be the exit stage of the condenser because the condenser operates under vacuum pressure which is stage 1.

The following assumptions can be made:

1 - Each component in the cycle is analyzed as an open system operating at steady-state.

2 - All of the processes are internally reversible.

3 - The turbine and pump operate adiabatically and are internally reversible, so they are also isentropic.

4 - Condensate exits the condenser as saturated liquid.

5 - The effluent from the HP turbine is a saturated vapor.

6 - No shaft work crosses the system boundary of the boiler or condenser.

7 - Changes in kinetic and potential energies are negligible

a) The thermal efficiency of the cycle is defined as the work of the cycle divided by the total heat input to the system. The stages that have heat input is stages 2-3, 4-5, 6-7.

For stage 2:

s₁=s₂ assuming isentropic

$s_1=0.4762 @ P_1=15MPa$

enthalpy will be a compressed liquid so after interpolation

$h_2=97.93+(0.4762-0.2932)((180.77-97.93)/(0.5666-0.2932))=153.38kJ/kg$

For stage 3:

Superheated steam @ T=500⁰C and P=15MPa

$h_3=3310.8kJ/kg$

Stage 4:

superheated vapor

P=5MPa

s₃=s₄=6.3480 kJ/kg, we must use interpolation to find h₄

$h_4=2925.7-(6.348-6.2111)((3069.3-2925.7)/(6.4516-6.2111))=3007.44kJ/kg$

Stage 5:

Superheated steam @ T=500⁰C and P₄=P₅=5 MPa

$h_5=3434.7kJ/kg$

Stage 6:

Superheated steam at P₆= 1MPa

s₅=s₆

$s_6=6.9781$

We find h₆ using interpolation from the steam tables:

$h_6=2943.1-(6.9781-6.9265)((3051.6-2943.1)/(7.1246-6.9265))=2970.67kJ/kg$

Stage 7:

P₇=P₆=1MPa

T=500⁰C superheated steam

$h_7=3479.1kJ/kg$

The heat into the cycle is:

$=(h_3-h_2)+(h_5-h_4)+(h_7-h_6)$

$=(3310.8-153.38)+(3434.7-3007.44)+(3479.1-2970.67)=4108.74kJ/kg$

We can determine the work out by the condenser from stage 9 to stage 1:

Stage 1:

saturated liquid P=5kPa

$h_1=137.75kJ/kg$

Stage 9:

We assume that its a saturated liquid with quality of 1 at 5kPa and

s₇=s₉ and after interpolation

$h_9=2568.53kj/kgK$

Qout = [/tex]2568.53-137.75=2430.79kJ/kg[/tex]

The thermal efficiency can be written in terms of qin and qout:

$n=1-(q_o/q_i)=1-2430.79/4093.11=0.4061$

Efficiency of 40.61%

The mass flow rate can be calculated from the Wnet:

$W_n=W_t-W_p$

Work of the turbines minus the work of the pumps:

$W_n=m((h_3-h_4)+(h_5-h_6)+(h_7-h_9)-(h_1-h_2)$

$120000=m(1662.33)$

$m=72.19$

mass flow rate of steam is 72.19 kg/s

7 0

2 years ago

the velocity of a particle is given by v=16t^2i+4t^3j+(5t+2k) m/s, where t is in seconds. if the particle is at the origin when

pshichka [43]

Answer:

80.16 m/s^2

at t=2 s

x=42.3 m

y=16 m

z=14 m

Explanation:

solution

The x,y,z components of the velocity are donated by the i,j,k vectors.

$v_{x}=16t^{2} \\v_{y}=4t^{3}\\v_{z}=5t+2$

acceleration is a derivative of velocity with respect to time.

$a_{x}=\frac{d}{dt} v_{x}=\frac{d}{dt}[16t^{2}]=32t\\a_{y}=\frac{d}{dt} v_{y}=\frac{d}{dt}[4t^{3}]=12t^{2} \\a_{z}=\frac{d}{dt} v_{z}=\frac{d}{dt}[5t+2]=5$

evaluate acceleration at 2 seconds

$a_{x} =32*2=64m/s^{2}\\ a_{y} =12*2^{2} =48m/s^{2}\\a_{z} =5m/s^{2}$

the magnitude of the acceleration is the square root of the sum of the square of each component of the acceleration.

$=\sqrt{a_{x}^2 +a_{y}^2+a_{z} ^2 } \\=\sqrt{64^2 +48^2+5 ^2 }\\=80.16m/s^2$

position is the integral of velocity with respect to time position at a time can be found by taking by taking the definite intergral of each component.

$x=\int\limits {v_{x} } \, dx=\int\limits^2_0 {{16t^2} \, dt=42.7m\\\\y=\int\limits {v_{y} } \, dx=\int\limits^2_0 {{4t^3} \, dt=16m\\\\\\\\\\z=\int\limits {v_{z} } \, dx=\int\limits^2_0 {{5t+2} \, dt=14m\\\\$