ENGINEERING COLLEGE

Determine the degrees of superheat of steam at 101.325 kPa and 170°C. a. 50°C b. 70°C c. 60°C d. 80°C

Answers

Answer 1

Answer:

Answer:

b) 70°C

Explanation:

Given that super heat temperature at 101.325 KPa= 170°C.

We know that saturation temperature at 101.325 KPa is 100°C.

Degree of super heat =Super heat temperature - saturation temperature ( at constant pressure)

Now by putting the values

Degree of super heat=170-100

Degree of super heat=70°C.

So our option b is right.

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(d) Arches NP is known for its spectacular arches that develop in the jointed areas of the park. Placemark Problem 2d flies you to Landscape Arch, the arch with the largest span in Arches NP. If the stresses that stretched the rock to form the joints were oriented perpendicular to the joint surfaces and the rock fins that contain the arches, what was the direction that the rocks were stretched? ☐ N-S
☐ E-W
☐ NW-SE
☐ NE-SW

Answers

Answer:

☐ NE-SW

Explanation:

Based on the description, the rock direction is North East - South West (NE-SW). Rocks generally can expand or compress depending on the type and magnitude of stress applied on the rocks. However, if the applied stress is sufficiently high, cracks and fractures will be created on the rock and it can ultimately lead to the formation of particles.

An automobile engine consumes fuel at a rate of 27.4 L/h and delivers 55 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.73 g/cm3, deter- mine the efficiency of this engine in percentage(

Answers

Answer:

The efficiency of the engine is 22.5%.

Explanation:

Efficiency = power output ÷ power input

power output = 55 kW

power input = specific energy×volumetric flow rate×density

specific energy = 44,000 kJ/kg

volumetric flow rate = 27.4 L/h = 27.4 L/h × 1000 cm^3/1 L × 1 h/3600 s = 7.61 cm^3/s

density = 0.73 g/cm^3 = 0.73 g/cm^3 × 1 kg/1000g = 7.3×10^-4 kg/cm^3

power input = 44,000 kJ/kg × 7.61 cm^3/s × 7.3×10^-4 kg/cm^3 = 244.4332 kJ/s = 244.4332 kW

Efficiency = 55 ÷ 244.4332 = 0.225 × 100 = 22.5%

Considering only (110), (1 1 0), (101), and (10 1 ) as the possible slip planes, calculate the stress at which a BCC single crystal will yield if the critical resolved shear stress is 50 MPa and the load is applied in the [100] direction.

Answers

Solution :

i. Slip plane (1 1 0)

Slip direction -- [1 1 1]

Applied stress direction = ( 1 0 0 ]

τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)

τ = σ cos Φ cos λ

$\cos \phi = ((1,0,0) \cdot (1,1,0))/(1 * \sqrt2)$

$=(1)/(\sqrt2 )$

$\cos \lambda = ((1,0,0) \cdot (1,-1,1))/(1 * \sqrt3)$

$=(1)/(\sqrt3 )$

τ = σ cos Φ cos λ

∴ $50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3)$

σ = 122.47 MPa

ii. Slip plane --- (1 1 0)

Slip direction -- [1 1 1]

$\cos \phi = ((1, 0, 0) \cdot (1, -1, 0))/(1 * \sqrt2) =(1)/(\sqrt2)$

$\cos \lambda = ((1, 0, 0) \cdot (1, 1, -1))/(1 * \sqrt3) =(1)/(\sqrt3)$

τ = σ cos Φ cos λ

∴ $50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3)$

σ = 122.47 MPa

iii. Slip plane --- (1 0 1)

Slip direction --- [1 1 1]

$\cos \phi = ((1, 0, 0) \cdot (1, 0, 1))/(1 * \sqrt2) =(1)/(\sqrt2)$

$\cos \lambda = ((1, 0, 0) \cdot (1, 1, -1))/(1 * \sqrt3) =(1)/(\sqrt3)$

τ = σ cos Φ cos λ

∴ $50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3)$

σ = 122.47 MPa

iv. Slip plane -- (1 0 1)

Slip direction ---- [1 1 1]

$\cos \phi = ((1, 0, 0) \cdot (1, 0, -1))/(1 * \sqrt2)=(1)/(\sqrt2)$

$\cos \lambda = ((1, 0, 0) \cdot (1, -1, 1))/(1 * \sqrt3) =(1)/(\sqrt3)$

τ = σ cos Φ cos λ

∴ $50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3)$

σ = 122.47 MPa

∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0

One piece of evidence that supports the Theory of Plate Tectonics is the discovery of what in both South America and Africa? The ancient atmosphere in both places was identical. The rates of weathering of rock are similar. Fossil remains of the same land-dwelling animal. Plants on both continents have similar flowers.

Answers

Answer: Fossil remains of the same land-dwelling animal.

Explanation: Fossil remains which were found to belong to same land dwelling animals, in South America and Africa was used as evidence to help support the theory of Tectonics plates, what this theory simply means is that the whole continents of earth were once fused together until a tectonic plate caused it’s division. Since same remains were found in Africa and South America this shows that both continents were once fused together.

Answer:

Fossil remains of the same land-dwelling animal

Explanation:

Fossil remains tell us where the animals once lived and how by the movement of plate spearated their remaind that was burried thousands of years ago.

Which of the following expressions causes an implicit conversion between types? Assume variable x is an integer, t is a float, and name is a string.Group of answer choices7.5 + (x / 2)x + 2 * x"Hello, " + str(name)print(str(t))

Answers

x + 2 * x is the correct option. The above-selected option demonstrates implicit conversion, which is an automated type of conversion. Thus, option B is correct.

The series of conversions are necessary to change the type of a function call's argument to that of the parameter with the same name in the function declaration is known as an implicit conversion sequence. For each parameter, the compiler tries to identify an implicit conversion sequence.

If both user-defined conversion sequences A and B contain the same user-defined conversion function or constructor, and if the second standard conversion sequence of A is superior to the second standard conversion sequence of B, then user-defined conversion sequence A is preferable to user-defined conversion sequence B.

Learn more about implicit conversion here:

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A cube with 1 m on a side is located in the positive x-y-z octant in a Cartesian coordinate system, with one of its points located at the origin. Find the total charge contained in the cube if the charge is given by p_v = x^2 ye^-z mC/m^3

Answers

Answer:

4.61 mC

Explanation:

The cube has 1 m side in the positive x-y-z octant in a Cartesian coordinate system, with one of its points located at the origin. The charge density is given as:

$\rho_v=x^2ye^(-z) \ mC/m^3$

Charge density is the charge per unit length or area or volume. It is the amount of charge in a particular region.

The charge Q is given as:

$Q=\int\limits_v {\rho_v} \, dv \nQ=\int\limits_v {\rho_v} \, dv=\int\limits^2_(x=0)\int\limits^2_(y=0)\int\limits^2_(z=0) {x^2ye^(-z)} \, dxdydz\n$

$Q=\int\limits^2_(x=0) {x^2} \, dx \int\limits^2_(y=0) {y} \, dy \int\limits^2_(z=0) {e^(-z)} \, dz \n\nQ=((1)/(3) [x^3]^2_0)((1)/(2) [y^2]^2_0)(-1 [e^(-z)]^2_0)\n\nQ=(-1)/(6) ([x^3]^2_0)( [y^2]^2_0)( [e^(-z)]^2_0)\n\nQ=(-1)/(6)[2^3-0^3][2^2-0^2][e^(-2)-e^0]\n\nQ=(-1)/(6)(8)(4)(0.1353-1)=(-1)/(6)(8)(4)(-0.8647)\n\nQ=4.61\ mC$

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