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2 years ago

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Oil, with density of 900 kg/m3 and kinematic viscosity of 0.00001 m2/s, flows at 0.2 m3/s through 500 m of 200-mm-diameter cast-

iron pipe. determine (a) the head loss and (b) the pressure drop if the pipe slows down at 10o in the flow direction.

1 answer:

Vsevolod [243]2 years ago

5 0

Just trying to personally understand this is making me shed a few tears. Good luck with finding someone to answer your question!

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(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th

arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation ( $\dot Q$ ), measured in BTU per hour, is represented by this formula:

$\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4})$ (1)

Where:

$\epsilon$ - Emissivity, dimensionless.

$A$ - Surface area of the student, measured in square feet.

$\sigma$ - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

$T_{s}$ - Temperature of the student, measured in Rankine.

$T_{b}$ - Temperature of the bus, measured in Rankine.

If we know that $\epsilon = 0.90$ , $A = 16.188\,ft^{2}$ , $\sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}$ , $T_{s} = 554.07\,R$ and $T_{b} = 527.67\,R$ , then the heat transfer rate due to electromagnetic radiation is:

$\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]$

$\dot Q = 417.492\,\frac{BTU}{h}$

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

$Q = \dot Q \cdot \Delta t$ (2)

Where $\Delta t$ is the heat transfer time, measured in hours.

If we know that $\dot Q = 417.492\,\frac{BTU}{h}$ and $\Delta t = \frac{1}{3}\,h$ , then the net energy transfer is:

$Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)$

$Q = 139.164\,BTU$

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

7 0

2 years ago

Gas is kept in a 0.1 m diameter cylinder under the weight of a 100 kg piston that is held down by a spring with a stiffness k =

Artyom0805 [142]

Answer:

The spring is compressed by 0.275 meters.

Explanation:

For equilibrium of the gas and the piston the pressure exerted by the gas on the piston should be equal to the sum of weight of the piston and the force the spring exerts on the piston

Mathematically we can write

$Force_{pressure}=Force_{spring}+Weight_{piston}$

we know that

$Force_{pressure}=Pressure\times Area=300\times 10^{3}\times \frac{\pi \times 0.1^2}{4}=750\pi Newtons$

$Weight_{piston}=mass\times g=100\times 9.81=981Newtons$

Now the force exerted by an spring compressed by a distance 'x' is given by $Force_{spring}=k\cdot x=5\times 10^{3}\times x$

Using the above quatities in the above relation we get

$5\times 10^{3}\times x+981=750\pi \\\\\therefore x=\frac{750\pi -981}{5\times 10^{3}}=0.275meters$

5 0

2 years ago

An escalator handles a steady load of 26 people per minute in elevating them from the first to the second floor through a vertic

photoshop1234 [79]

Answer:

$\eta = 70.711\,\%$

Explanation:

The power needed to make the escalator working is obtained by means of the Work-Energy Theorem:

$\dot W = \dot U_{g}$

$\dot W = \dot n \cdot m_{p}\cdot g \cdot \Delta y$

$\dot W = \left(26\,\frac{persons}{min}\right)\cdot (124\,lbm)\cdot \left(32.174\,\frac{ft}{s^{2}}\right)\cdot \left(\frac{1\,lbf}{32.174\,\frac{lbm\cdot ft}{s^{2}} } \right)\cdot (27.5\,ft)$

$\dot W = 88660\,\frac{lbf\cdot ft}{min}\,\left(2.687\,hp\right)$

The mechanical efficiency of the escalator is:

$\eta = \frac{2.687\,hp}{3.8\,hp}\times 100\,\%$

$\eta = 70.711\,\%$

3 0

2 years ago

A three-point bending test is performed on a glass specimen having a rectangular cross section of height d 5 mm (0.2 in.) and wi

Anon25 [30]

Answer:

The flexural strength of a specimen is = 78.3 M pa

Explanation:

Given data

Height = depth = 5 mm

Width = 10 mm

Length L = 45 mm

Load = 290 N

The flexural strength of a specimen is given by

$\sigma = \frac{3 F L}{2 bd^{2} }$

$\sigma = \frac{3(290)(45)}{2 (10)(5)^{2} }$

$\sigma =$ 78.3 M pa

Therefore the flexural strength of a specimen is = 78.3 M pa

4 0

2 years ago

Suppose there are 93 packets entering a queue at the same time. Each packet is of size 4 MiB. The link transmission rate is 1.4

Ghella [55]

Answer:

0.19s

Explanation:

Queueing delay is the time a job waits in a queue before it can be executed. it is the difference in time betwen when the packet data reaches it destination and the time when it was executed.

Queueing delay =(N-1) L /2R

where N = no of packet =93

L = size of packet = 4MB

R = bandwidth = 1.4Gbps = 1×10⁹ bps

4 MB = 4194304 Bytes

(93 - 1)4194304 / 2× 10⁹

queueing delay =192937984 ×10⁻⁹

=0.19s

5 0

2 years ago