All categories

3 years ago

If a weld is laying into a joint with a concave weld contour with undercutting issues, what might be the cause?

Engineering

1 answer:

Katarina [22]3 years ago

4 0

Answer:bbbb

Explanation:

You might be interested in

Steam enters a radiator at 16 psia and 0.97 quality. The steam flows through the radiator, is con- densed, and leaves as liquid

AnnZ [28]

Answer:

5.328Ibm/hr

Explanation:

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature.

for this case we can define the following equation for mass flow using the first law of thermodynamics

$m=\frac{Q}{h1-h2}$

where

Q=capacity of the radiator =5000btu/hr

m = mass flow

then using thermodynamic tables we found entalpy in state 1 and 2

h1(x=0.97, p=16psia)=1123btu/lbm

h2(x=0, p=16psia)=184.5btu/lbm

solving

$Q=\frac{5000}{1123-184.5} =5.328Ibm/hr$

3 0

3 years ago

A geothermal pump is used to pump brine whose density is 1050 kg/m3 at a rate of 0.3 m3/s from a depth of 200 m. For a pump effi

nasty-shy [4]

Answer:

Input power of the geothermal power will be 686000 J

Explanation:

We have given density of brine $\rho =1050kg/m^3$

Rate at which brine is pumped $V=0.3m^3/sec$

So mass of the pumped per second

Mass = volume × density = $1050\times 0.3=315$ kg/sec

Acceleration due to gravity $g=9.8m/sec^2$

Depth h = 200 m

So work done $W=mgh=315\times 9.8\times 200=617400J$

Efficiency is given $\eta =0.9$

We have to fond the input power

So input power $=\frac{617400}{0.9}=686000J$

So input power of the geothermal power will be 686000 J

5 0

3 years ago

water flows in a horizontal constant-area pipe; the pipe diameter is 75 mm and the average flow speed is 5 m/s. At the pipe inle

Veronika [31]

Answer:

Head loss = 28.03 m

Explanation:

According to Bernoulli's theorem for fluids we have

$\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=Constant$

Applying this between the 2 given points we have

$\frac{P_{1}}{\gamma _{w}}+\frac{V_{1}^{2}}{2g}+z_{1}=\frac{P_{2}}{\gamma _{w}}+\frac{V_{2}^{2}}{2g}+z_{2}+h_{l}$

Here $h_{l}$ is the head loss that occurs

$\therefore h_{l}=\frac{P_{1}}{\gamma _{w}}+\frac{V_{1}^{2}}{2g}+z_{1}-\frac{P_{2}}{\gamma _{w}}-\frac{V_{2}^{2}}{2g}-z_{2}$

Since the pipe is horizantal we have $z_{1}-z_{2}=0$

Applying contunity equation between the 2 sections we get

$A_{1}V_{1}=A_{2}V_{2}\\\\\therefore V_{1}=V_{2}(\because A_{1}=A_{2})$

Since the cross sectional area of the both the sections is same thus the speed

is also same

Using this information in the above equation of head loss we obtain

$h_{l}=\frac{1}{\gamma _{w}}(P_{1}-P_{2})$

Applying values we get

$h_{l}=\frac{1}{9810}\times (275\times 10^{3})m\\\\\therefore h_{l}=28.03m$

3 0

4 years ago

A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.

Nataliya [291]

Answer:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

Explanation:

We can assume that the general formula for the drag force is given by:

$D= C_D \frac{\rho}{2}V^2 A$

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

$D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)$

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

$D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})$

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

$\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D1} \frac{\rho}{2}V^2 (wh) V$