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2 years ago

Vocabulary review for shop

Engineering

1 answer:

kaheart [24]2 years ago

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Hazard is a possible source of danger

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A slender rod AB, of weight W, is attached to blocks A and B, which move freely in the guides shown. The blocks are connected by

gregori [183]

Answer:

(a) T = W/2(1-tanθ) (b) 39.81°

Explanation:

(a) The equation for tension (T) can be derived by considering the summation of moment in the clockwise direction. Thus:

Summation of moment in clockwise direction is equivalent to zero. Therefore,

T*l*(sinθ) + W*(l/2)*cosθ - T*l*cosθ = 0

T*l*(cosθ - sinθ) = W*(l/2)*cosθ

T = W*cosθ/2(cosθ - sinθ)

Dividing both the numerator and denominator by cosθ, we have:

T = [W*cosθ/cosθ]/2[(cosθ - sinθ)/cosθ] = W/2(1-tanθ)

(b) If T = 3W, then:

3W = W/2(1-tanθ),

Further simplification and rearrangement lead to:

1 - tanθ = 1/6

tanθ = 1 - (1/6) = 5/6

θ = tan^(-1) 5/6 = 39.81°

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2 years ago

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MAXImum [283]

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3 years ago

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For which of 'water' flow velocities at 200C can we assume that the flow is incompressible ? a.1000 km per hour b. 500 km per ho

ad-work [718]

Answer:d

Explanation:

Given

Temperature $=200^{\circ}\approc 473 K$

Also $\gamma for air=1.4$

R=287 J/kg

Flow will be In-compressible when Mach no.<0.32

Mach no. $=\frac{V}{\sqrt{\gamma RT}}$

(a) $1000 km/h\approx 277.78 m/s$

Mach no. $=\frac{277.78}{\sqrt{1.4\times 287\times 473}}$

Mach no.=0.63

(b) $500 km/h\approx 138.89 m/s$

Mach no. $=\frac{138.89}{\sqrt{1.4\times 287\times 473}}$

Mach no.=0.31

Mach no. $=\frac{555.55}{\sqrt{1.4\times 287\times 473}}$

Mach no.=1.27

(d) $200 km/h\approx 55.55 m/s$

Mach no. $=\frac{55.55}{\sqrt{1.4\times 287\times 473}}$

Mach no.=0.127

From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.

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3 years ago

An ideal vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated va

Ksju [112]

Explanation:

Note: Refer the diagram below

Obtaining data from property tables

State 1:

$\left.\begin{array}{l}P_{1}=1.25 \text { bar } \\\text { Sat - vapour }\end{array}\right\} \begin{array}{l}h_{1}=234.45 \mathrm{kJ} / \mathrm{kg} \\S_{1}=0.9346 \mathrm{kJ} / \mathrm{kgk}\end{array}$

State 2:

$\left.\begin{array}{l}P_{2}=5 \text { bor } \\S_{2}=S_{1}\end{array}\right\} \quad h_{2}=262.78 \mathrm{kJ} / \mathrm{kg}$

State 3:

$\left.\begin{array}{l}P_{3}=5 \text { bar } \\\text { Sat }-4 q\end{array}\right\} h_{3}=71-33 \mathrm{kJ} / \mathrm{kg}$

State 4:

Throttling process $h_{4}=h_{3}=71.33 \mathrm{kJ} / \mathrm{kg}$

(a)

Magnitude of compressor power input

$\dot{w}_{c}=\dot{m}\left(h_{2}-h_{1}\right)=\left(8 \cdot 5 \frac{\mathrm{kg}}{\min } \times \frac{1 \mathrm{min}}{\csc }\right)(262.78-234 \cdot 45)\frac{kj}{kg}$