Describe the make-up of an internal combustion engine.<br> Pls answer quickly.

One good way to improve your gas Milage is to ___.

VashaNatasha [74]

Answer: B

Explanation:

One good way to improve your gas mileage is to accelerate smoothly and directly to a safe speed.

Hope this helps!

5 0

3 years ago

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What unit of measurment would be used to measure current?

Alex_Xolod [135]

Answer:

The S. I unit of current is Amphere

5 0

3 years ago

Viefleur [7K]

Answer:

Both come from the sun

Both are reusable sources

and both don't cause pollution

Explanation:

3 0

2 years ago

A rigid 14-L vessel initially contains a mixture of liquid water and vapor at 100°C with 12.3 percent quality. The mixture is th

tigry1 [53]

Answer:

Q = 65.388 KJ

Explanation:

To calculate the heat required for the given process Q, we recall the energy balance equation.

Therefore, : Q = Δ U = m (u₂ - u₁) ..................equation (1)

We should note that there are no kinetic or potential energy change so the heat input in the system is converted only to internal energy.

Therefore, we will start the equation with the mass of the water (m) using given the initial percentage quality as x₁ = 0.123 and initial temperature t₁ = 100⁰c , we can them determine the initial specific volume v₁ of the mixture. For the calculation, we will also need the specific volume of liquid vₙ = 0.001043m³/kg and water vapour (vₐ) = 1.6720m³/kg

Therefore, u₁ = vₙ + x₁ . ( vₐ - vₙ)

u₁ = 0.001043m³/kg + 0.123 . ( 1.6720m³/kg - 0.001043m³/kg)

u₁ = 0.2066m³/kg

Moving forward, the mass of the vapor can then be calculated using the given volume of tank V = 14 L but before the calculation, we need to convert the volume to from liters to m³.

Therefore, V = 14L . 1m² / 1000L = 0.014 m³

Hence, m = V / u₁

0.014m³ / 0.2066 m³/kg

m = 0. 0677 kg

Also, the initial specific internal energy u₁ can be calculated using the given the initial given quality of x₁ , the specific internal energy of liquid water vₐ = 419.06 kj / kg and the specific internal energy of evaporation vₐₙ = 2087.0 kj/kg.

Therefore, u₁ = vₐ + x₁ . vₐₙ

u₁ = 419.06 kj / kg + 0.123 . 2087.0 kj/kg

u₁ = 675.76 kj/kg

For the final specific internal energy u₂, we first need to calculate the final quality of the mixture x₂ . The tank is rigid meaning the volume does not change and it is also closed meaning the mass does not change.from this, we can conclude the the specific volume also does not change during the process u₁ = u₂. This allows us to use the given final temperature T₂ = 180⁰c to determine the final quality x₂ of the mixture. for the calculation, we will also need the specific volume of liquid vₙ=0.001091m³/kg and vapor vₐ = 0.39248m³/kg

Hence, x₂ = u₂ - vₙ / uₐ

x₂ = 0.2066 m³/kg - 0.001091m³/kg / 0.39248m³/kg

x₂ = 0.524

Moving forward to calculate the final internal energy u₂, we have :

u₂ = vₙ + x₂ . vₙₐ

u₂ = 631.66 kj/kg + 0.524 . 1927.4 kj/kg

u₂ = 1641.62 kj/kg

We now return to equation (1) to plug in the values generated thus far

Q = m (u₂ - u₁)

0. 0677 kg ( 1641.62 kj/kg - 675.76 kj/kg)

Q = 65.388KJ

7 0

3 years ago

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1 A long, uninsulated steam line with a diameter of 100 mm and a surface emissivity of 0.8 transports steam at 150°C and is expo

Alexeev081 [22]

Answer:

a) q' = 351.22 W/m

b) q'_total = 1845.56 W / m

c) q'_loss = 254.12 W/m

Explanation:

Given:-

- The diameter of the steam line, d = 100 mm

- The surface emissivity of steam line, ε = 0.8

- The temperature of the steam, Th = 150°C

- The ambient air temperature, T∞ = 20°C

Find:-

(a) Calculate the rate of heat loss per unit length for a calm day.

Solution:-

- Assuming a calm day the heat loss per unit length from the steam line ( q ' ) is only due to the net radiation of the heat from the steam line to the surroundings.

- We will assume that the thickness "t" of the pipe is significantly small and temperature gradients in the wall thickness are negligible. Hence, the temperature of the outside surface Ts = Th = 150°C.

- The net heat loss per unit length due to radiation is given by:

q' = ε*σ*( π*d )* [ Ts^4 - T∞^4 ]

Where,

σ: the stefan boltzmann constant = 5.6703 10-8 (W/m2K4)

Ts: The absolute pipe surface temperature = 150 + 273 = 423 K

T∞:The absolute ambient air temperature = 20 + 273 = 293 K

Therefore,

q' = 0.8*(5.6703 10-8)*( π*0.1 )* [ 423^4 - 293^4 ]

q' = (1.4251*10^-8)* [ 24645536240 ]

q' = 351.22 W / m ... Answer

Find:-

(b) Calculate the rate of heat loss on a breezy day when the wind speed is 8 m/s.

Solution:-

- We have an added heat loss due to the convection current of air with free stream velocity of U∞ = 8 m/s.

- We will first evaluate the following properties of air at T∞ = 20°C = 293 K

Kinematic viscosity ( v ) = 1.5111*10^-5 m^2/s

Thermal conductivity ( k ) = 0.025596

Prandtl number ( Pr ) = 0.71559

- Determine the flow conditions by evaluating the Reynold's number:

Re = U∞*d / v

= ( 8 ) * ( 0.1 ) / ( 1.5111*10^-5 )

= 52941.56574 ... ( Turbulent conditions )

- We will use Churchill - Bernstein equation to determine the surface averaged Nusselt number ( Nu_D ):

$Nu_D = 0.3 + \frac{0.62*Re_D^\frac{1}{2}*Pr^\frac{1}{3} }{[ 1 + (\frac{0.4}{Pr})^\frac{2}{3} ]^\frac{1}{4} }*[ 1 + (\frac{Re_D}{282,000})^\frac{5}{8} ]^\frac{4}{5} \\\\Nu_D = 0.3 + \frac{0.62*(52941.56574)^\frac{1}{2}*(0.71559)^\frac{1}{3} }{[ 1 + (\frac{0.4}{0.71559})^\frac{2}{3} ]^\frac{1}{4} }*[ 1 + (\frac{52941.56574}{282,000})^\frac{5}{8} ]^\frac{4}{5} \\\\$

$Nu_D = 0.3 + \frac{127.59828 }{ 1.13824 }*1.27251 = 142.95013$

- The averaged heat transfer coefficient ( h ) for the flow of air would be:

$h = Nu_D*\frac{k}{d} \\\\h = 143*\frac{0.025596}{0.1} \\\\h = 36.58951 W/m^2K$

- The heat loss per unit length due to convection heat transfer is given by:

q'_convec = h*( π*d )* [ Ts - T∞ ]

q'_convec = 36.58951*( π*0.1 )* [ 150 - 20 ]

q'_convec = 11.49493* 130

q'_convec = 1494.3409 W / m

- The total heat loss per unit length ( q'_total ) owes to both radiation heat loss calculated in part a and convection heat loss ( q_convec ):

q'_total = q_a + q_convec

q'_total = 351.22 + 1494.34009

q'_total = 1845.56 W / m ... Answer

Find:-

For the conditions of part (a), calculate the rate of heat loss with a 20-mm-thick layer of insulation (k = 0.08 W/m ⋅ K)

Solution:-

- To reduce the heat loss from steam line an insulation is wrapped around the line which contains a proportion of lost heat within.

- A material with thermal conductivity ( km = 0.08 W/m.K of thickness t = 20 mm ) was wrapped along the steam line.

- The heat loss through the lamination would be due to conduction " q'_t " and radiation " q_rad":

$q'_t = 2*\pi*k \frac{T_h - T_o}{Ln ( \frac{r_2}{r_1} )}$

q' = ε*σ*( π*( d + 2t) )* [ Ts^4 - T∞^4 ]

Where,

T_o = T∞ = 20°C

T_s = Film temperature = ( Th + T∞ ) / 2 = ( 150 + 20 ) / 2 = 85°C

r_2 = d/2 + t = 0.1 / 2 + 0.02 = 0.07 m

r_1 = d/2 = 0.1 / 2 = 0.05 m

- The heat loss per unit length would be:

q'_loss = q'_rad - q'_cond

- Compute the individual heat losses:

$q'_t = 2*\pi*0.08 \frac{150 - 85}{Ln ( \frac{0.07}{0.05} )}\\\\q'_t = 0.50265* \frac{65}{0.33647}\\\\q'_t = 97.10 W/m$

Therefore,

q'_loss = 351.22 - 97.10

q'_loss = 254.12 W / m .... Answer

- If the wind speed is appreciable the heat loss ( q'_loss ) would increase and the insulation would become ineffective.

6 0

3 years ago

Describe the make-up of an internal combustion engine. Pls answer quickly.