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Xelga [282]
3 years ago
9

Three piezometers located 1000 m apart bottom in the same horizontal aquifer. Piezometer A is due south of piezometer B and piez

ometer C is to the east of the line AB. The surface elevations of A, B, and C are 95, 110, and 135 m, respectively. The depth to water in A is 5 m, in B is 30 m, and in C is 35 m. Determine the direction of ground-water flow through the triangle ABC and calculate the hydraulic gradient.

Engineering
1 answer:
leva [86]3 years ago
3 0

Answer:

Hydraulic gradient=0.0173

Explanation:

The hydraulic head at the points A, B and C will be as follows

A=95-5=90 m

B=110-30=80 m

C=135-35= 100 m

By drawing the equipotential lines, the direction of ground water can be seen as in the triangle

The length of flow will be

L=\sqrt {1000^{2}-500^{2}}\approx 866.025 m

The hydraulic gradient will be given by

i=\frac {\triangle h}{L}=\frac {95-80}{866.025}\approx 0.0173

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A 1 turn coil carries has a radius of 9.8 cm and a magnetic moment of 6.2 X 10 -2 Am 2. What is the current through the coil?
Alexus [3.1K]

Answer:

The current through the coil is 2.05 A

Explanation:

Given;

number of turns of the coil, N = 1

radius of the coil, r = 9.8 cm = 0.098 m

magnetic moment of the coil, P = 6.2 x 10⁻² A m²

The magnetic moment is given by;

P = IA

Where;

I is the current through the coil

A is area of the coil = πr² = π(0.098)² = 0.03018 m²

The current through the coil is given by;

I = P / A

I = (6.2 x 10⁻² ) / (0.03018)

I = 2.05 A

Therefore, the current through the coil is 2.05 A

6 0
4 years ago
A material has the following properties: Sut = 275 MPa and n = 0.40. Calculate its strength coefficient, K.
Tems11 [23]

Answer:

The strength coefficient is K = 591.87 MPa

Explanation:

We can calculate the strength coefficient using the equation that relates the tensile strength with the strain hardening index given by

S_{ut}=K \left(\cfrac ne \right)^n

where Sut is the tensile strength, K is the strength coefficient we need to find and n is the strain hardening index.

Solving for strength coefficient

From the strain hardening equation we can solve for K

K = \cfrac{S_{ut}}{\left(\cfrac ne \right)^n}

And we can replace values

K = \cfrac{275}{\left(\cfrac {0.4}e \right)^{0.4}}\\K=591.87

Thus we get that the strength coefficient is K = 591.87 MPa

6 0
4 years ago
Does jupiter have a permanent storm thats looks like a large red spot?
alekssr [168]
Yes it does have a large spot
8 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:

  1. Constant properties
  2. Steady state conditions
  3. Negligible effect of radiation
  4. Negligible constant resistance between tube and insulation
  5. 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
4. If a hot wire is shorted to ground, what will usually happen?
barxatty [35]

Answer:

nothing will happen

Explanation:

except for whatever that wire feeds will lose power

help support our cause by spreading this flag all over brainly

5 0
2 years ago
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