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3 years ago

- You have a bin wrench turning a 1/2 13 UNC bolt. You overcome 1200 lbs of resistance when you

Engineering

1 answer:

andrew11 [14]3 years ago

6 0

is my guess but yeah

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Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.

nikitadnepr [17]

Answer:

a) $m_e= 3.05 Kg$

b) $\rho=1072.3kg/m^3$

c) $m_e= 3.05 Kg$

Explanation:

From the question we are told that:

Beaker Mass $m_b=1.20$

Liquid Mass $m_l=1.85$

Balance D:

Mass $m_d=3.10$

Balance E:

Mass $m_e=7.50$

Volume $v=4.15*10^{-3}m^3$

Generally the equation for Liquid's density is mathematically given by

$m_e=m_b+m_l+(\rho*v)$

$\rho=\frac{7.50-(1.2+1.85)}{4.15*10^{-3}}$

$\rho=1072.3kg/m^3$

Generally the equation for D's Reading at A pulled is mathematically given by

m_d = mass of block - mass of liquid displaced

$m_d=m- (\rho *v )$

$m=3.10+ (1072.30 *4.15*10^{-3}m^3 )$

$m=18.10kg$

Generally the equation for E's Reading at A pulled is mathematically given by

$m_e=m_b+m_l$

$m_e = 1.20 + 1.85$

$m_e= 3.05 Kg$

6 0

3 years ago

A stainless steel (AISI 304) tube used to transport a chilled pharmaceutical has an inner diameter of 36 mm and a wall thickness

Bingel [31]

Answer:

a) heat gain per unit tube length = $\frac{23-6}{1.35} = 12.6W/m$

b) heat gain per unit tube length = $\frac{23-6}{2.20} = 7.7W/m$

Explanation:

Assumptions:

Constant properties
Steady state conditions
Negligible effect of radiation
Negligible constant resistance between tube and insulation
one dimensional radial conduction

a) What is the heat gain per unit tube length

$R_{conv,i}'=\frac{1}{2\pi r_1h_i}$

$d_1=36mm$ Therefore $r_1=\frac{d_1}{2} =36/2=18mm=18*10^{-3}$

$r_2=2mm=2*10^{-3}m$

$k_{st}=14.2W/m.k$

$h_o=6W/m^2$

$h_i=400W/m^2$

$R_{conv,i}'=\frac{1}{2\pi * 1.8*10^{-3}*400}= 0.221m.K/W$

$R_{cond,st}'=\frac{ln(r_2/r_1)}{2\pi k_{st}} =\frac{ln(20/18)}{2\pi *14.2} =1.18*10^{-3}m.K/W$

$R_{conv,o}'=\frac{1}{2\pi r_2h_0}=\frac{1}{2\pi *2*10^{-3}*6}=1.33m.K/W$

$R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.33=1.35m.K/W$

heat gain per unit tube length = $\frac{23-6}{1.35} = 12.6W/m$

b) What is the heat gain per unit length if a 10-mm-thick layer of calcium silicate insulation (k_ins = 0.050 W/m.K) is applied to the tube

$r_3=r_1+r_2+10mm=30mm=0.03m$

$R_{conv,i}'$ and $R_{cond,st}'$ are the same, but $R_{conv,o}'$ changes.

Therefore:

$R_{conv,o}'=\frac{1}{2\pi r_3h_0} = \frac{1}{2\pi *0.03*6}=0.88m.K/W$

$R_{conv,ins}'=\frac{ln(r_3/r_)}{2\pi k_{ins}} =\frac{ln(30/20)}{2\pi *0.05} =1.29m.K/W$

The total resistance $R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,ins}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.29+0.88=2.20m.K/W$

heat gain per unit tube length = $\frac{23-6}{2.20} = 7.7W/m$

8 0

3 years ago

Read 2 more answers

Design a calculator circuit which can perform either addition or subtraction on a 4-digit binary input and generate a 5-digit bi

uysha [10]

Answer: See attachment below

Explanation:

7 0

3 years ago

The power supply converts the wall outlet AC power into DC power. T or F

Alika [10]

Answer:

True

Explanation:

All computer parts require DC power to operate, and wall outlets provide AC Power.

7 0

3 years ago

Water flows in a constant diameter pipe with the following conditions measured:

Burka [1]

Answer:

a) $h_L=-3.331ft$

b) The flow would be going from section (b) to section (a)

Explanation:

1) Notation

$p_a =31.1psi=4478.4\frac{lb}{ft^2}$

$p_b =27.3psi=3931.2\frac{lb}{ft^2}$

For above conversions we use the conversion factor $1psi=144\frac{lb}{ft^2}$

$z_a =56.7ft$

$z_a =68.8ft$

$h_L =?$ head loss from section

2) Formulas and definitions

For this case we can apply the Bernoulli equation between the sections given (a) and (b). Is important to remember that this equation allows en energy balance since represent the sum of all the energies in a fluid, and this sum need to be constant at any point selected.

The formula is given by:

$\frac{p_a}{\gamma}+\frac{V_a^2}{2g}+z_a =\frac{p_b}{\gamma}+\frac{V_b^2}{2g}+z_b +h_L$

Since we have a constant section on the piple we have the same area and flow, then the velocities at point (a) and (b) would be the same, and we have just this expression:

$\frac{p_a}{\gamma}+z_a =\frac{p_b}{\gamma}+z_b +h_L$

3)Part a

And on this case we have all the values in order to replace and solve for $h_L$

$\frac{4478.4\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+56.7ft=\frac{3931.2\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+68.8ft +h_L$

$h_L=(71.769+56.7-63-68.8)ft=-3.331ft$

4)Part b

Analyzing the value obtained for $\h_L$ is a negative value, so on this case this means that the flow would be going from section (b) to section (a).

5 0

3 years ago