All categories

2 years ago

How are sedimentary - gravity flows different from fluid - gravity flows

Engineering

1 answer:

Nimfa-mama [501]2 years ago

3 0

Answer:

Sediment gravity flows are mixtures of water and sediment particles where the gravity acting on the sediment particles moves the fluid, in contrast to rivers, where the fluid moves the particles.

Explanation:

You might be interested in

Gtjffs

grandymaker [24]

the required documents is 3000

4 0

2 years ago

Common car loan duration

chubhunter [2.5K]

Answer:

In 2019, the average term length was 69 months for new cars and 65 months for used vehicles. Most car loans are available in 12 month increments, lasting between two and eight years. The most common loan terms are 24, 36, 48, 60, 72, and 84 months, according to Autotrader

Explanation:

4 0

2 years ago

Engineers will redesign their products when they find flaws. TRUE O False

nataly862011 [7]

Answer:

true

Explanation:

6 0

2 years ago

Consider a 1.5-m-high and 2.4-m-wide glass window whose thickness is 6 mm and thermal conductivity is k = 0.78 W/m⋅K. Determine

Bess [88]

Answer:

The steady rate of heat transfer through the glass window is 707.317 watts.

Explanation:

A figure describing the problem is included below as attachment. From First Law of Thermodynamics we get that steady rate of heat transfer through the glass window is the sum of thermal conductive and convective heat rates, all measured in watts:

$\dot Q_{total} = \dot Q_{cond} + \dot Q_{conv, in} + \dot Q_{conv, out}$ (Eq. 1)

Given that window is represented as a flat element, we can expand (Eq. 1) as follows:

$\dot Q_{total} = \frac{T_{i}-T_{o}}{R}$ (Eq. 2)

Where:

$T_{i}$ , $T_{o}$ - Indoor and outdoor temperatures, measured in Celsius.

$R$ - Overall thermal resistance, measured in Celsius per watt.

Now, we know that glass window is configurated in series and overall thermal resistance is:

$R = R_{cond} + R_{conv, in}+R_{conv, out}$ (Eq. 3)

Where:

$R_{cond}$ - Conductive thermal resistance, measured in Celsius per watt.

$R_{conv, in}$ , $R_{conv, out}$ - Indoor and outdoor convective thermal resistances, measured in Celsius per watt.

And we expand the expression as follows:

$R = \frac{l}{k\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d}$

$R = \frac{1}{w\cdot d}\cdot \left(\frac{l}{k}+\frac{1}{h_{i}}+\frac{1}{h_{o}} \right)$ (Eq. 4)

Where:

$w$ - Width of the glass window, measured in meters.

$d$ - Length of the glass window, measured in meters.

$l$ - Thickness of the glass window, measured in meters.

$k$ - Thermal conductivity, measured in watts per meter-Celsius.

$h_{i}$ , $h_{o}$ - Indoor and outdoor convection coefficients, measured in watts per square meter-Celsius.

If we know that $w = 2.4\,m$ , $d = 1.5\,m$ , $l = 0.006\,m$ , $k = 0.78\,\frac{W}{m\cdot ^{\circ}C}$ , $h_{i} = 10\,\frac{W}{m^{2}\cdot ^{\circ}C}$ and $h_{o} = 25\,\frac{W}{m^{2}\cdot ^{\circ}C}$ , the overall thermal resistance is:

$R = \left[\frac{1}{(2.4\,m)\cdot (1.5\,m)}\right] \cdot \left(\frac{0.006\,m}{0.78\,\frac{W}{m\cdot ^{\circ}C} }+\frac{1}{10\,\frac{W}{m^{2}\cdot ^{\circ}C} }+\frac{1}{25\,\frac{W}{m^{2}\cdot ^{\circ}C} } \right)$

$R = 0.041\,\frac{^{\circ}C}{W}$

Now, we obtain the steady rate of heat transfer from (Eq. 2): ( $R = 0.041\,\frac{^{\circ}C}{W}$ , $T_{i} = -5\,^{\circ}C$ , $T_{o} = 24\,^{\circ}C$ )

$\dot Q_{total} = \frac{24\,^{\circ}C-(-5\,^{\circ}C)}{0.041\,\frac{^{\circ}C}{W} }$

$\dot Q_{total} = 707.317\,W$

The steady rate of heat transfer through the glass window is 707.317 watts.

6 0

2 years ago

You are to design the control for an automatic candy vending machine. The candy bars inside the machine cost 25 cents, and the m

labwork [276]

Answer:

I believe that it is E. Write Verilog HDL models for the machine based on the state diagram in (a) and the D flip-flop sequential circuit that you implement in (b).

6 0

2 years ago