So far in your lifetime, about how much garbage have you contributed

A boiler is used to heat steam at a brewery to be used in various applications such as heating water to brew the beer and saniti

Natalija [7]

Answer:

net boiler heat = 301.94 kW

Explanation:

given data

saturated steam = 6.0 bars

temperature = 18°C

flow rate = 115 m³/h = 0.03194 m³/s

heat use by boiler = 90 %

to find out

rate of heat does the boiler output

solution

we can say saturated steam is produce at 6 bar from liquid water 18°C

we know at 6 bar from steam table

hg = 2756 kJ/kg

and

enthalpy of water at 18°C

hf = 75.64 kJ/kg

so heat required for 1 kg is

=hg - hf

= 2680.36 kJ/kg

and

from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg

so here mass flow rate is

mass flow rate = $\frac{0.03194}{0.315}$

mass flow rate m = 0.10139 kg/s

so heat required is

H = h × m

here h is heat required and m is mass flow rate

H = 2680.36 × 0.10139

H = 271.75 kJ/s = 271.75 kW

now 90 % of boiler heat is used for generate saturated stream

so net boiler heat = $\frac{H}{0.90}$

net boiler heat = $\frac{271.75}{0.90}$

net boiler heat = 301.94 kW

5 0

3 years ago

Steep safety ramps are built beside mountain highway to enable vehichles with fedective brakes to stop safely. a truck enters a

Veronika [31]

Answer:

a. 6 seconds

b. 180 feet

Explanation:

Images attached to show working.

a. You have the position of the truck so you integrate twice. Use the formula and plug in the time t = 7 sec. Check out uniform acceleration. The time at which the truck's velocity is zero is when it stops.

b. Determine the initial speed. Plug in the time calculated in the previous step. From this we can observe that the truck comes to a stop before the end of the ramp.

7 0

3 years ago

Pls help and solve this problem. I don't have an idea to solve this.

lions [1.4K]

Answer:

MIS HIEVOSTES bbbbbbbbbbbb MIS HUEVOTES

7 0

2 years ago

Write a naive implementation (i.e. non-vectorized) of matrix multiplication, and then write an efficient implementation that uti

erik [133]

Answer:

import numpy as np

import time

def matrixMul(m1,m2):

if m1.shape[1] == m2.shape[0]:

t1 = time.time()

r1 = np.zeros((m1.shape[0],m2.shape[1]))

for i in range(m1.shape[0]):

for j in range(m2.shape[1]):

r1[i,j] = (m1[i]*m2.transpose()[j]).sum()

t2 = time.time()

print("Native implementation: ",r1)

print("Time: ",t2-t1)

t1 = time.time()

r2 = m1.dot(m2)

t2 = time.time()

print("\nEfficient implementation: ",r2)

print("Time: ",t2-t1)

else:

print("Wrong dimensions!")

Explanation:

We define a function (matrixMul) that receive two arrays representing the two matrices to be multiplied, then we verify is the dimensions are appropriated for matrix multiplication if so we proceed with the native implementation consisting of two for-loops and prints the result of the operation and the execution time, then we proceed with the efficient implementation using .dot method then we return the result with the operation time. As you can see from the image the execution time is appreciable just for large matrices, in such a case the execution time of the efficient implementation can be 1000 times faster than the native implementation.

7 0

3 years ago

A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.

denpristay [2]

Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is $8.621 \times 10^{-4} in/in$

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

$\frac{M}{I}=\frac{F}{y}$

Here

M is the moment which is to be calculated
I is the moment of inertia given as

$I=\frac{bd^3}{12}$

Here

b is the breath given as 0.75"
d is the depth which is given as 8"

$I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4$

y is given as

$y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\$

Force is 50 ksi

$\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in$

The yield moment is 400 k.in.

Part b:

The strain is given as

$Strain=\frac{Stress}{Elastic Modulus}$

The stress at the station 2" down from the top is estimated by ratio of triangles as

$F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi$

Now the steel has the elastic modulus of E=29000 ksi

$Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in$

So the strain is $8.621 \times 10^{-4} in/in$

Part c:

For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

$shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi$

The plastic moment is 600 ksi.

3 0

3 years ago