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trapecia [35]
3 years ago
15

The costs of mining and transporting coal are roughly independent of the heating value of the coal. Consider:

Engineering
1 answer:
Paul [167]3 years ago
5 0

Answer:

2

Explanation:

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What is kirchoff's current law?​
Anastaziya [24]

Answer:

The sum of all currents entering a junction is 0

Explanation:

Current cannot disappear, so the currents leaving a junction have to equal the currents entering a junction. If you would give the currents leaving the junction an opposite sign (e.g., negative) it implies that the sum of all these currents is exactly 0.

4 0
3 years ago
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In the construction of a large reactor pressure vessel, a new steel alloy with a plane strain fracture toughness of 55 MPa-m1/2
sp2606 [1]

Answer:

l=24mm

Explanation:

From the question we are told that:

Plane strain fracture toughness of T=55 MPa-m1/2

Y value Y=1.0

Stress level of\sigma =200 MPa

Generally the equation for length of a surface crack is mathematically given by

l=\frac{1}{\pi}(\frac{T}{Y*\sigma})^2

l=\frac{1}{3.142}(\frac{55}{1*200})^2

l=0.024m

Therefore

in mm

l=24mm

6 0
2 years ago
10. Which of these requires a wheel alignment after replacement?
Elden [556K]
C. Both; require a wheel alignment after replacement
5 0
2 years ago
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A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125
Gekata [30.6K]

Answer:

a) \dot m_{3} = 135\,\frac{lbm}{s}, b) h_{3}=168.965\,\frac{BTU}{lbm}, c) T = 200.829\,^{\textdegree}F

Explanation:

a) The tank can be modelled by the Principle of Mass Conservation:

\dot m_{1} + \dot m_{2} - \dot m_{3} = 0

The mass flow rate exiting the tank is:

\dot m_{3} = \dot m_{1} + \dot m_{2}

\dot m_{3} = 125\,\frac{lbm}{s} + 10\,\frac{lbm}{s}

\dot m_{3} = 135\,\frac{lbm}{s}

b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:

\dot m_{1}\cdot h_{1} + \dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0

h_{3} = \frac{\dot m_{1}\cdot h_{1}+\dot m_{2}\cdot h_{2}}{\dot m_{3}}

Properties of water are obtained from tables:

h_{1}=180.16\,\frac{BTU}{lbm}

h_{2}=28.08\,\frac{BTU}{lbm} + \left(0.01604\,\frac{ft^{3}}{lbm}\right)\cdot (14.7\,psia-0.25638\,psia)

h_{2}=29.032\,\frac{BTU}{lbm}

The specific enthalpy at outlet is:

h_{3}=\frac{(125\,\frac{lbm}{s} )\cdot (180.16\,\frac{BTU}{lbm} )+(10\,\frac{lbm}{s} )\cdot (29.032\,\frac{BTU}{lbm} )}{135\,\frac{lbm}{s} }

h_{3}=168.965\,\frac{BTU}{lbm}

c) After a quick interpolation from data availables on water tables, the final temperature is:

T = 200.829\,^{\textdegree}F

8 0
3 years ago
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What is an example of a traditional career?
vladimir2022 [97]

Answer:

C, Teacher

Explanation:

7 0
2 years ago
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