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Monica [59]
2 years ago
6

A 1020 Cold-Drawn steel shaft is to transmit 20 hp while rotating at 1750 rpm. Calculate the transmitted torque in lbs. in. Igno

re the effect of friction.If the shaft in Q2 was made of ASTM 30 cast iron, what would be the factor of safety
Engineering
1 answer:
velikii [3]2 years ago
4 0

Answer:

Question 1 A 1020 Cold-Drawn steel shaft is to transmit 20 hp while rotating at 1750 rpm. Calculate the transmitted torque in lbs. in. Ignore the effect of friction. Answer with three decimal points. 60.024 Question 2 Based on the maximum-shear-stress theory, determine the minimum diameter in inches for the shaft in Q1 to provide a safety factor of 3. Assume Sy = 57 Kpsi. Answer with three decimal points. 0.728 Question 3 If the shaft in Q2 was made of ASTM 30 cast iron, what would be the factor of safety? Assume Sut = 31 Kpsi, Suc = 109 Kpsi 0 2.1 O 2.0 O 2.5 0 2.4 2.3 O 2.2

Explanation:

hope it helps

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A heat engine that rejects waste heat to a sink at 520 R has a thermal efficiency of 35 percent and a second- law efficiency of
xeze [42]

Answer:

The source temperature is 1248 R.

Explanation:

Second law efficiency of the engine is the ratio of actual efficiency to the maximum possible efficiency that is reversible efficiency.

Given:  

Temperature of the heat sink is 520 R.

Second law efficiency is 60%.

Actual thermal efficiency is 35%.

Calculation:  

Step1

Reversible efficiency is calculated as follows:

\eta_{II}=\frac{\eta_{a}}{\eta_{rev}}

0.6=\frac{0.35}{\eta_{rev}}

\eta_{rev}=0.5834

Step2

Source temperature is calculated as follows:

\eta_{rev}=1-\frac{T_{L}}{T}

\eta_{rev}=1-\frac{520}{T}

0.5834=1-\frac{520}{T}

T = 1248 R.

The heat engine is shown below:

Thus, the source temperature is 1248 R.

6 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
Read 2 more answers
(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5
bezimeni [28]

Answer:

a.) 1.453MW/m2,  b.)  2,477,933.33 BTU/hr  c.) 22,733.33 BTU/hr  d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

                        50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, Q = \frac{109*0.5*100}{0.0075} = 726,666.667W

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, Q=\frac{1*0.5*100}{0.0075} = 6,666.67W

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, Q=\frac{109*0.5*100}{0.015} =363,333.33W

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0
3 years ago
The alternator produces direct current, which is changed to alternating current by the diodes.
exis [7]

Answer:

False

Explanation:

<u>Alternat</u>or produces <u>alternat</u>ing current  ....diodes convert this to direct current

8 0
2 years ago
An office building is served by an air-cooled chiller currently operating at 115 tons (404.5 kW). The measured chilled water sup
Andrei [34K]

Answer:

B.197 gpm and 12.4 L/s

Explanation:

Given that

Load Q = 404.5 KW

Water inlet temperature= 6.1 °C

Water outlet temperature= 13.9°C

We know that specific heat for water

C_p=4.187\ \frac{KJ}{kg.K}

Now from energy balance

Q=\dot{m}C_p\Delta T

by putting the values

Q=\dot{m}C_p\Delta T

404.5=\dot{m}\times 4.187(13.9-6.1)

\dot{m}=12.38\ \frac{kg}{s}     (1 Kg/s = 15.85 gal/min)

We can say that

\dot{m}=196.31\ \frac{gal}{min}

We know that

\dot{m}=\rho\times volume\ flow\ rate

12.38=1000 x volume flow rate

volume\ flow\ rate\ = 12.38\times 10^{-3}\ \frac{m^3}{s}

So

volume flow rate = 12.38 L/s

So the option B is correct.

8 0
3 years ago
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