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3 years ago

I need your help to answer in the picture plss help me!

Engineering

1 answer:

Aloiza [94]3 years ago

3 0

this looks like its the different phases of a single cylinder 4 stroke engine what are you doing in the picture or assignment though matching the numbers to the descriptions on the side?

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Select the correct answer.

Ulleksa [173]

The answer is A. Immediately inform her colleague

4 0

3 years ago

Give two causes that can result in surface cracking on extruded products.

Andreas93 [3]

Answer:

1. High friction

2. High extrusion temperature

Explanation:

Surface cracking on extruded products are defects or breakage on the surface of the extruded parts. Such cracks are inter granular.

Surface cracking defects arises from very high work piece temperature that develops cracks on the surface of the work piece. Surface cracking appears when the extrusion speed is very high, that results in high strain rates and generates heat.

Other factors include very high friction that contributes to surface cracking an d chilling of the surface of high temperature billets.

6 0

3 years ago

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

Assoli18 [71]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

8 0

3 years ago

he circular stream of water from a faucet is observed to taper from a diameter of 21 mm to 12 mm in a distance of 52 cm. Determi

Gemiola [76]

Answer: $1.52$ × $10^{-3}$ $\frac{m^{3} }{s}$

Explanation:

Please kindly find the attached document for the answer.

5 0

3 years ago

At steady state, a reversible refrigeration cycle discharges energy at the rate QH to a hot reservoir at temperature TH, while r

ludmilkaskok [199]

Answer:

a) $COP_{R} = 25.014$ , b) $T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

Explanation:

a) The coefficient of performance of a reversible refrigeration cycle is:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)

$COP_{R} = \frac{275.15\,K}{286.15\,K-275.15\,K}$

$COP_{R} = 25.014$

b) The respective coefficient of performance is determined:

$COP_{R} = \frac{Q_{L}}{Q_{H}-Q_{L}}$

$COP_{R} = \frac{8.75\,kW}{10.5\,kW-8.75\,kW}$

$COP_{R} = 5$

But:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

The temperature at hot reservoir is found with some algebraic help:

$COP_{R} \cdot (T_{H}-T_{L})=T_{L}$

$T_{H}-T_{L} = \frac{T_{L}}{COP_{R}}$

$T_{H} = T_{L}\cdot \left(1+\frac{1}{COP_{R}} \right)$

$T_{H} = 273.15\,K \cdot \left(1+\frac{1}{5} \right)$

$T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

8 0

3 years ago

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I need your help to answer in the picture plss help me!​

I need your help to answer in the picture plss help me!