Which process made making copies of technical drawings easier?

Engineering

1 answer:

kirill115 [55]2 years ago

4 0

Answer:

The Process of printing

Explanation:

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If it is desired to lay off a distance of 10,000' with a total error of no more than Â± 0.30 ft. If a 100' tape is used and the

Ira Lisetskai [31]

Answer:

± 0.003 ft

Explanation:

Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.

Let L' = our distance and L = our tape measure

So, L' = 100L

Now by error determination ΔL' = 100ΔL

Now ΔL' = ± 0.30 ft

ΔL = ΔL'/100

= ± 0.30 ft/100

= ± 0.003 ft

So, the maxim error per tape is ± 0.003 ft

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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

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