Answer:

**Answer:**

The power input, in kW is **-86.396 kW**

**Explanation:**

Given;

initial pressure, P₁ = 1.05 bar

final pressure, P₂ = 12 bar

initial temperature, T₁ = 300 K

final temperature, T₂ = 400 K

Heat transfer, Q = 6.5 kW

volumetric flow rate, V = 39 m³/min = 0.65 m³/s

mass of air, m = 28.97 kg/mol

gas constant, R = 8.314 kJ/mol.k

R' = R/m

R' = 8.314 /28.97 = 0.28699 kJ/kg.K

**Step 1:**

Determine the specific volume:

p₁v₁ = RT₁

**Step 2:**

determine the mass flow rate; m' = V / v₁

mass flow rate, m' = 0.65 / 0.81997

mass flow rate, m' = 0.7927 kg/s

**Step 3:**

using steam table, we determine enthalpy change;

h₁ at T₁ = 300.19 kJ /kg

h₂ at T₂ = 400.98 kJ/kg

Δh = h₂ - h₁

Δh = 400.98 - 300.19

Δh = 100.79 kJ/kg

**step 4:**

determine work input;

W = Q - mΔh

Where;

Q is heat transfer = - 6.5 kW, because heat is lost to surrounding

W = (-6.5) - (0.7927 x 100.79)

W = -6.5 -79.896

**W = -86.396 kW**

Therefore, the power input, in kW is **-86.396 kW**

An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formation B has a thickness of 2.0 m and a conductivity of 142 m/d. Formation C has a thickness of 34 m and a conductivity of 40 m/d. Assume that each formation is isotropic and homogeneous. Compute both the overall horizontal and vertical conductivities.

Given that the debouncing circuit is somewhat expensive in terms of hardware (2 NAND gates, 2 resistors, and a double-pole, single throw switch), describe applications where you would require switch-debouncing circuits as well as applications where you would not need to include the additional hardware for switch debouncing (in other words, applications where you can tolerate switch bouncing). Note, you cannot use the clock and clear inputs of our lab as example applications; instead you need to think of other examples.

A horizontal curve on a single-lane highway has its PC at station 1+346.200 and its PI at station 1+568.70. The curve has a superelevation of 6.0% and is designed for 120 km/h. The limiting value for coefficient of side friction at 120 km/h is 0.09. What is the station of the PT? Remember that 1 metric station = 1000 m.

what is an example of an innovative solution to an engineering problem? Explain briefly why you chose this answer.

write an interface downloadable that has a method "geturl" that returns the url of a downloadable object

Given that the debouncing circuit is somewhat expensive in terms of hardware (2 NAND gates, 2 resistors, and a double-pole, single throw switch), describe applications where you would require switch-debouncing circuits as well as applications where you would not need to include the additional hardware for switch debouncing (in other words, applications where you can tolerate switch bouncing). Note, you cannot use the clock and clear inputs of our lab as example applications; instead you need to think of other examples.

A horizontal curve on a single-lane highway has its PC at station 1+346.200 and its PI at station 1+568.70. The curve has a superelevation of 6.0% and is designed for 120 km/h. The limiting value for coefficient of side friction at 120 km/h is 0.09. What is the station of the PT? Remember that 1 metric station = 1000 m.

what is an example of an innovative solution to an engineering problem? Explain briefly why you chose this answer.

write an interface downloadable that has a method "geturl" that returns the url of a downloadable object

**Answer:**

The pressure rise in the tire when the air temperature in the tire rises to 50°C is 337.43 KPa.

The amount of air that must be bled off to restore pressure 0.007 Kg

**Explanation:**

Knowing

T1 = 25°C = 298 K

T2 = 50°C = 323 K

volume of the tire = V = 0.025

P = 210 kPa (gage)

Pabs = 210 + 101 = 311 KPa

Before the temperature rise

P1 V1 = m1 R1 T1

m1 =

After the temperature rise

P2 =

after bleeding the pressure and the volume returns to its first value

P1 = P2 and V1 = V2

then

m2 =

m2 =

mbleed = m1 - m2 --> mbleed = 0.91 - 0.84 = 0.007 Kg

P2 = 337.43 KPa

mbleed = 0.007 Kg

**Answer:**

answer is 0 ok answer is 0

**Explanation:**

The way a programmer describe a **pre-emptive dialog** by purely graphical means is; by producing a window that covers the entire screen to make it the currently selected window.

In a **graphics - based** interaction, it is supposed that the user can only interact with parts of the system that are visible. However, In a **windowing system**, the user can only direct input to a single window that was currently selected and the way to change that selected window is to indicate with some gesture within that window.

Finally, to create a **pre-emptive dialog**, the system would do so through the production of a window that covers the entire screen to make it the **currently selected window**. Thereafter, all user input would be directed to that **window** and the user would have no means of selecting any other window. Then the covering window will now pre-empt any other user action with the exception of that which it is defined to **support.**

Read more about **dialogue** at; brainly.com/question/5962406

**Answer:**

In an illustrations based communication, it is expected that the client can just associate with parts of the framework that are obvious. In a windowing framework, for instance, the client can just direct contribution to a solitary, at present chosen window, and the main methods for changing the chose window would be by demonstrating with some signal inside that window. To make a preemptive exchange, the framework can create a window that covers the whole screen and make it the right now chosen window. All client information would then be coordinated to that window and the client would have no methods for choosing another window. The 'covering' window in this way preempts some other client activity with the exception of that which it is characterized to help

Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

b. The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.

Answer:

1)ΔL = 0.616 mm

2)Δd = 0.00194 mm

Explanation:

We are given;

Force; F = 52900 N

Initial length; L_o = 207 mm = 0.207 m

Diameter; d_o = 19.2 mm = 0.0192 m

Elastic modulus; E = 61.4 GPa = 61.4 × 10^(9) N/m²

Now, from Hooke's law;

E = σ/ε

Where; σ is stress = force/area = F/A

A = πd²/4 = π × 0.0192²/4

A = 0.00009216π

σ = 52900/0.00009216π

ε = ΔL/L_o

ε = ΔL/0.207

Thus,from E = σ/ε, we have;

61.4 × 10^(9) = (52900/0.00009216π) ÷ (ΔL/0.207)

Making ΔL the subject, we have;

ΔL = (52900 × 0.207)/(61.4 × 10^(9) × 0.00009216π)

ΔL = 0.616 × 10^(-3) m

ΔL = 0.616 mm

B) Poisson's ratio is given as;

υ = ε_x/ε_z

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

υ = (Δd/d_o) ÷ (ΔL/L_o)

Making Δd the subject gives;

Δd = (υ × d_o × ΔL)/L_o

We are given Poisson's ratio to be 0.34.

Thus;

Δd = (0.34 × 19.2 × 0.616)/207

Δd = 0.00194 mm

**Answer:**

**tensile stress at yield = 254 MPa**

**ultimate stress = 477 MPa**

**average stress = 892 MPa**

**Explanation:**

**Given data in question**

bar yields load = 80 kN

load maximum = 150 kN

load fail = 70 kN

diameter of steel (D) = 20 mm i.e. = 0.020 m

diameter of breaking point (d) = 10 mm i.e. 0.010 m

**to find out **

tensile stress at the yield point , ultimate tensile stress and average stress at the breaking point

**solution**

in 1st part we calculate tensile stress at the yield point by this formula

tensile stress at yield = yield load / area

tensile stress at yield = 80 ×10³ / /4 × D²

tensile stress at yield = 80 ×10³ / /4 × 0.020²

**tensile stress at yield = 254 MPa**

in 2nd part we calculate ultimate stress by given formula

ultimate stress = maximum load / area

ultimate stress = 150 ×10³ / /4 × 0.020²

**ultimate stress = 477 MPa**

In 3rd part we calculate average stress at breaking point by given formula

average stress = load fail / area

average stress = 70 ×10³ / /4 × d²

average stress = 70 ×10³ / /4 × 0.010²

**average stress = 892 MPa**