Which of the following is NOT one of the 3 technology bets we have made?

A vacuum pump is used to drain a basement of 20 °C water (with a density of 998 kg/m3 ). The vapor pressure of water at this tem

lord [1]

Answer:

The maximum theoretical height that the pump can be placed above liquid level is $\Delta h=9.975\,m$

Explanation:

To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature. As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:

$\frac{\Delta P}{\rho}+g\, \Delta h +\frac{1}{2} \Delta v^2 =0$

( $\rho$ stands here for density, $h$ for height)

Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:

$\frac{\Delta P}{\rho}+g\, \Delta h =0$

$\Delta P= -g\, \rho\, \Delta h$

This means that pressure drop is proportional to the suction lift's height.

We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.

That means:

$\Delta P = 2.34\,kPa- 100 \,kPa = -97.66 \, kPa\\$

We insert that into our last equation and get:

$\frac{ \Delta P}{ -g\, \rho\,}= \Delta h\\\Delta h=\frac{97.66 \, kPa}{998 kg/m^3 \, \, 9.81 m/s^2} \\\Delta h=9.975\,m$

And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.

8 0

3 years ago

An aircraft is in a steady level turn at a flight speed of 200 ft/s and a turn rate about the local vertical of 5 deg/s. Thrust

notka56 [123]

Answer:

L= 50000 lb

D = 5000 lb

Explanation:

To maintain a level flight the lift must equal the weight in magnitude.

We know the weight is of 50000 lb, so the lift must be the same.

L = W = 50000 lb

The L/D ratio is 10 so

10 = L/D

D = L/10

D = 50000/10 = 5000 lb

To maintain steady speed the thrust must equal the drag, so

T = D = 5000 lb

5 0

3 years ago

Consider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 k

In-s [12.5K]

Consider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 km, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.

Required:

a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?

b. Find the average rate of CO emissions during this two-hour period.

c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pmConsider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 km, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.

Required:

a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?

b. Find the average rate of CO emissions during this two-hour period.

c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pmConsider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 km, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.

Required:

a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?

b. Find the average rate of CO emissions during this two-hour period.

c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pm

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3 0

3 years ago

Someone has suggested that the air-standard Otto cycle is more accurate if the two polytropic processes are replaced with isentr

omeli [17]

Answer:

q_net,in = 585.8 KJ/kg

q_net,out = 304 KJ/kg

n = 0.481

Explanation:

Given:

- The compression ratio r = 8

- The pressure at state 1, P_1 = 95 KPa

- The minimum temperature at state 1, T_L = 15 C

- The maximum temperature T_H = 900 C

- Poly tropic index n = 1.3

Find:

a) Determine the heat transferred to and rejected from this cycle

b) cycle’s thermal efficiency

Solution:

- For process 1-2, heat is rejected to sink throughout. The Amount of heat rejected q_1,2, can be computed by performing a Energy balance as follows:

W_out - Q_out = Δ u_1,2

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

c_v*(T_2 - T_L) = R*(T_2 - T_L)/n-1 - q_1,2

- Using polytropic relation we will convert T_2 = T_L*r^(n-1):

c_v*(T_L*r^(n-1) - T_L) = R*(T_1*r^(n-1) - T_L)/n-1 - q_1,2

- Hence, we have:

q_1,2 = T_L *(r^(n-1) - 1)* ( (R/n-1) - c_v)

- Plug in the values:

q_1,2 = 288 *(8^(1.3-1) - 1)* ( (0.287/1.3-1) - 0.718)

q_1,2= 60 KJ/kg

- For process 2-3, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

Q_in = Δ u_2,3

q_2,3 = u_3 - u_2

q_2,3 = c_v*(T_H - T_2)

- Again, using polytropic relation we will convert T_2 = T_L*r^(n-1):

q_2,3 = c_v*(T_H - T_L*r^(n-1) )

q_2,3 = 0.718*(1173-288*8(1.3-1) )

q_2,3 = 456 KJ/kg

- For process 3-4, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

q_3,4 - w_in = Δ u_3,4

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

c_v*(T_4 - T_H) = - R*(T_4 - T_H)/1-n + q_3,4

- Using polytropic relation we will convert T_4 = T_H*r^(1-n):

c_v*(T_H*r^(1-n) - T_H) = -R*(T_H*r^(1-n) - T_H)/n-1 + q_3,4

- Hence, we have:

q_3,4 = T_H *(r^(1-n) - 1)* ( (R/1-n) + c_v)

- Plug in the values:

q_3,4 = 1173 *(8^(1-1.3) - 1)* ( (0.287/1-1.3) - 0.718)

q_3,4= 129.8 KJ/kg

- For process 4-1, heat is lost from the system. The Amount of heat rejected q_4,1, can be computed by performing a Energy balance as follows:

Q_out = Δ u_4,1

q_4,1 = u_4 - u_1

q_4,1 = c_v*(T_4 - T_L)

- Again, using polytropic relation we will convert T_4 = T_H*r^(1-n):

q_4,1 = c_v*(T_H*r^(1-n) - T_L )

q_4,1 = 0.718*(1173*8^(1-1.3) - 288 )

q_4,1 = 244 KJ/kg

- The net gain in heat can be determined from process q_3,4 & q_2,3:

q_net,in = q_3,4+q_2,3

q_net,in = 129.8+456

q_net,in = 585.8 KJ/kg

- The net loss of heat can be determined from process q_1,2 & q_4,1:

q_net,out = q_4,1+q_1,2

q_net,out = 244+60

q_net,out = 304 KJ/kg

- The thermal Efficiency of a Otto Cycle can be calculated:

n = 1 - q_net,out / q_net,in

n = 1 - 304/585.8

n = 0.481

6 0

3 years ago

A plate of an alloy steel has a plane-strain fracture toughness of 50 MPa√m. If it is known that the largest surface crack is 0.

Ivahew [28]

Answer:

option B is correct. Fracture will definitely not occur

Explanation:

The formula for fracture toughness is given by;

K_ic = σY√πa

Where,

σ is the applied stress

Y is the dimensionless parameter

a is the crack length.

Let's make σ the subject

So,

σ = [K_ic/Y√πa]

Plugging in the relevant values;

σ = [50/(1.1√π*(0.5 x 10^(-3))]

σ = 1147 MPa

Thus, the material can withstand a stress of 1147 MPa

So, if tensile stress of 1000 MPa is applied, fracture will not occur because the material can withstand a higher stress of 1147 MPa before it fractures. So option B is correct.

8 0

3 years ago