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dalvyx [7]
3 years ago
7

An engine indicator is used to determine the following: (a) speed (d) m.e.p, and IHP (e) BHP (b) temperature (c) volume of cylin

der
Engineering
1 answer:
Greeley [361]3 years ago
8 0

Answer:

Option (d) MEP and IHP

Explanation:

MEP stands for Mean Effective Pressure and IHP stands for Indicated Horse Power

In engines (Internal Combustion), engine indicator is generally to indicate the indicate the changes in pressure inside the cylinder of an Internal Combustion Engine or IC engines. Once, Mean Effective Pressure of the engine is calculated it further helps to calculate the Horse power and both these quantities, i.e., MEP and IHP are displayed on the engine indicator.

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

w_{c}=4 \cdot 013 \mathrm{kw}

(b)

Refrigerator capacity

Q_{i n}=\dot{m}\left(h_{1}-h_{4}\right)=\left(\frac{g \cdot s}{60} k_{0} / s\right) \times(234 \cdot 45-71 \cdot 33) \frac{k J}{k_{8}}

Q_{i n}=23 \cdot 108 \mathrm{kW}\\1 ton of retregiration =3.51 k \omega

\ Q_{in} =6 \cdot 583 \text { tons }

(c)

Cop:

\beta=\frac{\left(h_{1}-h_{4}\right)}{\left(h_{2}-h_{1}\right)}=\frac{Q_{i n}}{\omega_{c}}=\frac{23 \cdot 108}{4 \cdot 013}

\beta=5 \cdot 758

3 0
3 years ago
As project manager, you approve a team member’s request to change the order of their tasks because they think it will be more ef
Oduvanchick [21]

The project is going to scope if the situation happens. Option A is correct.

<h3 /><h3>What is the function of a project manager?</h3>

Project managers are in charge of organizing, planning, and guiding the execution of certain projects for an organization .

As the project manager, you grant a team member's request to rearrange their work in a way they believe will increase productivity.

However, this modification interferes with another team member's workflow since they now have to complete two more activities that are unrelated to the project's objective. The project will be within its scope.

Hence option A is correct.

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6 0
2 years ago
A water pump delivers 3 hp of shaft power when operating. If the pressure differential between the outlet and the inlet of the p
Natali [406]

Answer:

Mechanical Efficiency =  83.51%

Explanation:

Given Data:

Pressure difference = ΔP=1.2 Psi

Flow rate = V=8ft^3/s\\

Power of Pump = 3 hp

Required:

Mechanical Efficiency

Solution:

We will first bring the change the units of given data into SI units.

P=1.2*6.895 = 8.274KPa\\V=8*0.00283=0.226 m^3/s\\P=3*0.746=2.238KW

Now we will find the change in energy.

Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.

Thus change in energy is

=(Mass * change in P)/density\\= \frac{M*P}{p}\\\\

As we know that Mass = Volume x density

substituting the value

Energy = Volume * density x ΔP / density

Change in energy = Volumetric flow x ΔP

Change in energy = 0.226 x 8.274 = 1.869 KW

Now mechanical efficiency = change in energy / work done by shaft

Efficiency = 1.869 / 2.238

Efficiency = 0.8351 = 83.51%

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