¿Qué áreas del conocimiento me pueden<br> aportar a la ejecución del proyecto?

Even though the content of many alcohol blends doesn’t affect engine drive ability using gasoline with alcohol in warm weather m

Alchen [17]

Even though the content of many alcohol blends doesn't affect engine driveability, using gasoline with alcohol in warm weather may cause: decrease in fuel economy.

Mark brainliest
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3 0

3 years ago

In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle tempe

KIM [24]

Answer:

$\eta$ =0.60

Explanation:

Given :Take $\gamma$ =1.4 for air

$P_1=100 KPa ,T_1=300K$

$\frac{V_1}{V_2}$ =r ⇒ r=16

As we know that

$T_2=T_1(r^{\gamma-1})$

So $T_2=300\times 16^{\gamma-1}$

$T_2$ =909.42K

Now find the cut off ration $\rho$

$\rho=\frac{V_3}{V_2}$

$\frac{V_3}{V_2}=\frac{T_3}{T_2}$

$\rho=\frac{2031}{909.42}$

$\rho=2.23$

So efficiency of diesel engine

$\eta =1-\dfrac{\rho^\gamma-1}{\gamma\times r^{\gamma-1}(\rho-1)}$

Now by putting the all values

$\eta =1-\dfrac{2.23^{1.4}-1}{1.4\times 16^{1.4-1}(2.23-1)}$

So $\eta$ =0.60

So the efficiency of diesel engine=0.60

7 0

3 years ago

Air is saturated with water vapor at 35.0 oC and a total pressure of 1.50 atmospheres. If the molar flow rate of the dry air in

Marizza181 [45]

Answer:

11.541 mol/min

Explanation:

temperature = 35°C

Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa

note : partial pressure of water in mixture = saturation pressure of water at T = 35°c )

from steam table it is = 5.6291 Kpa

calculate the mole fraction of H $_{2}o$ ( YH $_{2}o$ )

= 5.6291 / 151.95

= 0.03704

calculate the mole fraction of air ( Yair )

= 1 - mole fraction of water

= 1 - 0.03704 = 0.9629

Now to determine the molar flow rate of water vapor in the stream

lets assume N = Total molar flow rate

NH $_{2}o$ = molar flow rate of water

Nair = molar flow rate of air = 300 moles /min

note : Yair * n = Nair

therefore n = 300 / 0.9629 = 311.541 moles /min

Molar flowrate of water

= n - Nair

= 311.541 - 300 = 11.541 mol/min

4 0

3 years ago

Steam enters an adiabatic turbine at 10MPa and 500 C and leaves at 10 kPa with a quality of 90%. Neglecting the changes in kinet

Amanda [17]

Answer:

flow ( m ) = 4.852 kg/s

Explanation:

Given:

- Inlet of Turbine

P_1 = 10 MPa

T_1 = 500 C

- Outlet of Turbine

P_2 = 10 KPa

x = 0.9

- Power output of Turbine W_out = 5 MW

Find:

Determine the mass ow rate required

Solution:

- Use steam Table A.4 to determine specific enthalpy for inlet conditions:

P_1 = 10 MPa

T_1 = 500 C ---------- > h_1 = 3375.1 KJ/kg

- Use steam Table A.6 to determine specific enthalpy for outlet conditions:

P_2 = 10 KPa -------------> h_f = 191.81 KJ/kg

x = 0.9 -------------> h_fg = 2392.1 KJ/kg

h_2 = h_f + x*h_fg

h_2 = 191.81 + 0.9*2392.1 = 2344.7 KJ/kg

- The work produced by the turbine W_out is given by first Law of thermodynamics:

W_out = flow(m) * ( h_1 - h_2 )

flow ( m ) = W_out / ( h_1 - h_2 )

- Plug in values:

flow ( m ) = 5*10^3 / ( 3375.1 - 2344.7 )

flow ( m ) = 4.852 kg/s

3 0

3 years ago

Pls paki answers naman Ito pls sa inyo

netineya [11]

:nanjan lang din yan sa binasa mo ne........

7 0

3 years ago

¿Qué áreas del conocimiento me pueden aportar a la ejecución del proyecto?