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2 years ago

Which of the following is a variable expense for many adults?

Engineering

2 answers:

sertanlavr [38]2 years ago

7 0

But where are the options?

Allisa [31]2 years ago

5 0

Ans: Their monthly rent payment.

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A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °

SOVA2 [1]

Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30 (Btu/hr)/(ft ⋅ °F)= $\frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3} (Btu/s)/(ft.F)$

Thermal Conductivity is SI units:

$k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K$

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

$A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2$

Heat Q is:

$Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J$

In Btu:

$A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2$

Heat Q is:

$Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu$

PArt B:

At midpoint Length=L/2=0.1 m

$Q=\frac{k*A*(T_1-T_2)}{L}$

On rearranging:

$T_2=T_1-\frac{Q*L}{KA}$

$T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C$

4 0

2 years ago

Tensile testing provides engineers with the ability to verify and establish material properties related to a specific material.

Sedbober [7]

Answer:

True

Explanation:

Tensile testing which is also referred to as tension testing is a process which materials are subjected to so as to know how well it can be stretched before it reaches breaking point. Hence, the statement in the question is true

7 0

2 years ago

Refrigerant 134a enters an air conditioner compressor at 4 bar, 20 C, and is compressed at steady state to 12 bar, 80 C. The vol

sleet_krkn [62]

Answer:

$Q=15.7Kw$

Explanation:

From the question we are told that:

Initial Pressure $P_1=4bar$

Initial Temperature $T_1=20 C$

Final Pressure $P_2=12 bar$

Final Temperature $T_2=80C$

Work Output $W= 60 kJ/kg$

Generally Specific Energy from table is

At initial state

$P_1=4bar \& T_1=20 C$

$E_1=262.96KJ/Kg$

With

Specific Volume $V'=0.05397m^3/kg$

At Final state

$P_2=12 bar \& P_2=80C$

$E_1=310.24KJ/Kg$

Generally the equation for The Process is mathematically given by

$m_1E_1+w=m_2E_2+Q$

Assuming Mass to be Equal

$m_1=m_1$

Where

$m=\frac{V}{V'}$

$m=frac{0.06666}{V'=0.05397m^3/kg}$

$m=1.24$

Therefore

$1.24*262.96+60)=1.24*310.24+Q$

$Q=15.7Kw$

4 0

2 years ago

2. Determine the surface area of a primary settling tank sized to handle a maximum hourly flow of 0.570 m3/s at an overflow rate

Hitman42 [59]

Answer:

The surface area of the primary settling tank is 0.0095 m^2.

The effective theoretical detention time is 0.05 s.

Explanation:

The surface area of the tank is calculated by dividing the volumetric flow rate by the overflow rate.

Volumetric flow rate = 0.570 m^3/s

Overflow rate = 60 m/s

Surface area = 0.570 m^3/s ÷ 60 m/s = 0.0095 m^2

Detention time is calculated by dividing the volume of the tank by the its volumetric flow rate

Volume of the tank = surface area × depth = 0.0095 m^2 × 3 m = 0.0285 m^3

Detention time = 0.0285 m^3 ÷ 0.570 m^3/s = 0.05 s

7 0

2 years ago

Read 2 more answers

A circular bar will be subjected to an axial force (P) of 2000 lbf. The bar will be made of material that has a strength (S) of

schepotkina [342]

Answer:

$n = 2.36$

Explanation:

The stress experimented by the circular bar is:

$\sigma = \left[\frac{2000\, lbf}{\frac{\pi}{4}\cdot (0.5\,in)^{2}}\right]\cdot \left(\frac{1\,kpsi}{1000\,psi} \right)$

$\sigma = 10.186\,kpsi$

The safety factor is:

$n = \frac{24\,kpsi}{10.186\,kpsi}$

$n = 2.36$

5 0

2 years ago