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sleet_krkn [62]

3 years ago

9

A triangular roadside channel is poorly lined with riprap. The channel has side slopes of 2:1 (H:V) and longitudinal slope of 2.

5 percent. Determine the flow rate (cubic meters per second) in the channel if flow is uniform and the top width of the flowing channel is 5 meters.

1 answer:

Oliga [24]3 years ago

8 0

Answer:

Q = 14.578 m³/s

Explanation:

Given

We use the Manning Equation as follows

Q = (1/n)*A*(∛R²)*(√S)

where

Q = volumetric water flow rate passing through the stretch of channel (m³/s for S.I.)

A = cross-sectional area of flow perpendicular to the flow direction, (m² for S.I.)

S = bottom slope of channel, m/m (dimensionless) = 2.5% = 0.025

n = Manning roughness coefficient (empirical constant), dimensionless = 0.023

R = hydraulic radius = A/P (m for S.I.) where :

A = cross-sectional area of flow as defined above,

P = wetted perimeter of cross-sectional flow area (m for S.I.)

we get A as follows

A = (B*h)/2

where

B = 5 m (the top width of the flowing channel)

h = (B/2)*(m) = (5 m/2)*(1/2) = 1.25 m (the deep)

A = (5 m*1.25 m/2) = 3.125 m²

then we find P

P = 2*√((B/2)²+h²) ⇒ P = 2*√((2.5 m)²+(1.25 m)²) = 5.59 m

⇒ R = A/P ⇒ R = 3.125 m²/5.59 m = 0.559 m

Substituting values into the Manning equation gives:

Q = (1/0.023)*(3.125 m²)*(∛(0.559 m)²)*(√0.025)

⇒ Q = 14.578 m³/s

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Consider a Carnot heat pump cycle executed in a steady-flow system in the saturated mixture region using R-134a flowing at a rat

attashe74 [19]

Answer:

7.15

Explanation:

Firstly, the COP of such heat pump must be measured that is,

$COP_{HP}=\frac{T_H}{T_H-T_L}$

Therefore, the temperature relationship, $T_H=1.15\;T_L$

Then, we should apply the values in the COP.

$=\frac{1.15\;T_L}{1.15-1}$

$=7.67$

The number of heat rejected by the heat pump must then be calculated.

$Q_H=COP_{HP}\times W_{nst}$

$=7.67\times5=38.35$

We must then calculate the refrigerant mass flow rate.

$m=0.264\;kg/s$

$q_H=\frac{Q_H}{m}$

$=\frac{38.35}{0.264}=145.27$

The $h_g$ value is 145.27 and therefore the hot reservoir temperature is 64° C.

The pressure at 64 ° C is thus 1849.36 kPa by interpolation.

And, the lowest reservoir temperature must be calculated.

$T_L=\frac{T_H}{1.15}$

$=\frac{64+273}{1.15}=293.04$

$=19.89\°C$

the lowest reservoir temperature = 258.703 kpa

So, the pressure ratio should be = 7.15

8 0

3 years ago

A 20-cm-long rod with a diameter of 0.250 cm is loaded with a 5500 N weight. If the diameter decreases to 0.210 cm, determine th

ss7ja [257]

Answer:

1561.84 MPa

Explanation:

L=20 cm

d1=0.21 cm

d2=0.25 cm

F=5500 N

a) σ= F/A1= 5000/(π/4×(0.0025)^2)= 1018.5916 MPa

lateral strain= Δd/d1= (0.0021-0.0025)/0.0025= -0.16

longitudinal strain (ε_l)= -lateral strain/ν = -(-0.16)/0.3

(assuming a poisson's ration of 0.3)

ε_l =0.16/0.3 = 0.5333

b) σ_true= σ(1+ ε_l)= 1018.5916( 1+0.5333)

σ_true = 1561.84 MPa

ε_true = ln( 1+ε_l)= ln(1+0.5333)

ε_true= 0.4274222

The engineering stress on the rod when it is loaded with a 5500 N weight is 1561.84 MPa.

7 0

3 years ago

Air in a large tank at 300C and 400kPa, flows through a converging diverging nozzle with throat diameter 2cm. It exits smoothly

-Dominant- [34]

Answer:

The answer is " $3.74 \ cm\ \ and \ \ 0.186 \frac{kg}{s}$ "

Explanation:

Given data:

Initial temperature of tank $T_1 = 300^{\circ}\ C= 573 K$

Initial pressure of tank $P_1= 400 \ kPa$

Diameter of throat $d* = 2 \ cm$

Mach number at exit $M = 2.8$

In point a:

calculating the throat area:

$A*=\frac{\pi}{4} \times d^2$

$=\frac{\pi}{4} \times 2^2\\\\=\frac{\pi}{4} \times 4\\\\=3.14 \ cm^2$

Since, the Mach number at throat is approximately half the Mach number at exit.

Calculate the Mach number at throat.

$M*=\frac{M}{2}\\\\=\frac{2.8}{2}\\\\=1.4$

Calculate the exit area using isentropic flow equation.

$\frac{A}{A*}= (\frac{\gamma -1}{2})^{\frac{\gamma +1}{2(\gamma -1)}} (\frac{1+\frac{\gamma -1}{2} M*^2}{M*})^{\frac{\gamma +1}{2(\gamma -1)}}$

Here: $\gamma$ is the specific heat ratio. Substitute the values in above equation.

$\frac{A}{3.14}= (\frac{1.4-1}{2})^{-\frac{1.4+1}{2(1.4 -1)}} (\frac{1+\frac{1.4-1}{2} (1.4)^2}{1.4})^{\frac{1.4+1}{2(1.4-1)}} \\\\A=\frac{\pi}{4}d^2 \\\\10.99=\frac{\pi}{4}d^2 \\\\d = 3.74 \ cm$

exit diameter is 3.74 cm

In point b:

Calculate the temperature at throat.

$\frac{T*}{T}=(1+\frac{\Gamma-1}{2} M*^2)^{-1}\\\\\frac{T*}{573}=(1+\frac{1.4-1}{2} (1.4)^2)^{-1}\\\\T*=411.41 \ K$

Calculate the velocity at exit.

$V*=M*\sqrt{ \gamma R T*}$

Here: R is the gas constant.

$V*=1.4 \times \sqrt{1.4 \times 287 \times 411.41}\\\\=569.21 \ \frac{m}{s}$

Calculate the density of air at inlet

$\rho_1 =\frac{P_1}{RT_1}\\\\=\frac{400}{ 0.287 \times 573}\\\\=2.43\ \frac{kg}{m^3}$

Calculate the density of air at throat using isentropic flow equation.

$\frac{\rho}{\rho_1}=(1+\frac{\Gamma -1}{2} M*^2)^{-\frac{1}{\Gamma -1}} \\\\\frac{\rho *}{2.43}=(1+\frac{1.4-1}{2} (1.4)*^2)^{-\frac{1}{1.4-1}} \\\\\rho*= 1.045 \ \frac{kg}{m^3}$

Calculate the mass flow rate.

$m= \rho* \times A* \times V*\\\\= 1.045 \times 3.14 times 10^{-4} \times 569.21\\\\= 0.186 \frac{kg}{s}$

5 0

2 years ago

How much energy does it take to boil water for pasta? For a one-pound box of pasta

ArbitrLikvidat [17]

Answer:

a. 164 °F b. 91.11 °C c. 1439.54 kJ

Explanation:

a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?

Since the starting temperature is 48°F and the final temperature which water boils is 212°F, the number of degrees Fahrenheit we would need to raise the temperature is the difference between the final temperature and the initial temperature.

So, Δ°F = 212 °F - 48 °F = 164 °F

b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?

To find the degree change in Celsius, we convert the initial and final temperature to Celsius.

°C = 5(°F - 32)/9

So, 48 °F in Celsius is

°C₁ = 5(48 - 32)/9

°C₁ = 5(16)/9

°C₁ = 80/9

°C₁ = 8.89 °C

Also, 212 °F in Celsius is

°C₂ = 5(212 - 32)/9

°C₂ = 5(180)/9

°C₂ = 5(20)

°C₂ = 100 °C

So, the number of degrees in Celsius you must raise the temperature is the temperature difference between the final and initial temperatures in Celsius.

So, Δ°C = °C₂ - °C₁ = 100 °C - 8.89 °C = 91.11 °C

c. [2 pts] How much energy is required to heat the four quarts of water from

48°F to 212°F (boiling)?

Since we require 15.8 kJ for every degree Celsius of temperature increase of the four quarts of water, that is 15.8 kJ/°C and it rises by 91.11 °C, then the amount of energy Q required is Q = amount of heat per temperature rise × temperature rise = 15.8 kJ/°C × 91.11 °C = 1439.54 kJ

8 0

3 years ago

To do a research, a researcher needs to be clear about what he/she is doing, why he/she is doing it and the associated implicati

Mashutka [201]

Answer:

True

Explanation:

To carry out a research in order to that it should yield the desirable result, a researcher must be very clear about his objective for the research.

The researcher must be very much clear about all the do's and don'ts and all the where, why, what sort of thing, i.e., the researcher must be clear about his/her intentions of carrying out the research, and what research is being carried out by him/her.

The one doing the research must be clear about the implications of the research.

Unless all these things are not clear at the researcher's end the research will not give the desired results.

4 0

3 years ago