What is the design wind speed for average-risk building:____

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

shusha [124]

Answer:

A)W'/m = 311 KJ/kg

B)σ'_gen/m = 0.9113 KJ/kg.k

Explanation:

a).The energy rate balance equation in the control volume is given by the formula;

Q' - W' + m(h1 - h2) = 0

Dividing through by m, we have;

(Q'/m) - (W'/m) + (h1 - h2) = 0

Rearranging, we have;

W'/m = (Q'/m) + (h1 - h2)

Normally, this transforms to another equation;

W'/m = (Q'/m) + c_p(T1 - T2)

Where;

W'/m is the rate at which power is developed

Q'/m is the rate at which heat is flowing

c_p is specific heat at constant pressure which from tables at a temperature of 980k = 1.1 KJ/kg.k

T1 is initial temperature

T2 is exit temperature

We are given;

Q'/m = -30 kj/kg (negative because it leaves the turbine)

T1 = 980 k

T2 = 670 k

Plugging in the relevant values;

W'/m = -30 + 1.1(980 - 670)

W'/m = 311 KJ/kg

B) The Entropy produced from the entropy balance equation in a control volume is given by the formula;

(Q'/T_boundary) + m(s1 - s2) + σ'_gen = 0

Dividing through by m gives;

((Q'/m)/T_boundary) + (s1 - s2) + σ'_gen/m = 0

Rearranging, we have;

σ'_gen/m = -((Q'/m)/T_boundary) + (s2 - s1)

Under the conditions given in the question, this transforms normally to;

σ'_gen/m = -((Q'/m)/T_boundary) - c_p•In(T2/T1) - R•In(p2/p1)

σ'_gen/m is the rate of entropy production in kj/kg

We are given;

p2 = 100 kpa

p1 = 400 kpa

T_boundary = 315 K

For an ideal gas, R = 0.287 KJ/kg.K

Plugging in the relevant values including the ones initially written in answer a above, we have;

σ'_gen/m = -(-30/315) - 1.1(In(670/980)) - 0.287(In(100/400))

σ'_gen/m = 0.0952 + 0.4183 + 0.3979

σ'_gen/m = 0.9113 KJ/kg.k

6 0

2 years ago

To revise a monthly budget, changes in which categories might need to be addressed? Check all that apply.

GREYUIT [131]

Income

amount budgeted

expenses

5 0

2 years ago

A 1-w, 350-ω resistor is connected to 24 v. Is this resistor operating within its power rating?

seraphim [82]

Answer:

No.

Explanation:

$P_r$ = Power rating = 1 W

R = Resistance = $350\ \Omega$

V = Voltage = $24\ \text{V}$

Power is given by

$P=\dfrac{V^2}{R}\\\Rightarrow P=\dfrac{24^2}{350}\\\Rightarrow P=1.65\ \text{W}$

$1.65\ \text{W}>1\ \text{W}$

So

$P>P_r$

Hence, the resistor is not operating within its power rating.

8 0

2 years ago

Dimension these please.

Svetach [21]

Answer:

86

Explanation:

3 0

2 years ago

The cross section of a heat exchanger consists of three circular pipes inside a larger pipe. The internal diameter of the three

omeli [17]

Answer:

0.0432 m^3/s

Explanation:

Internal diameter of smaller pipes = 2.5 cm = 0.025 m

pipa wall thickness = 3 mm = 0.003 m

internal diameter of larger pipes = 8 cm = 0.8 m

velocity of region between smaller and larger pipe = 10 m/s

Calculate discharge in m^3/s

First we calculate the area of the smaller pipe

A = $\pi Dt$ = $\pi ( 0.025 ) ( 0.003 )$ = 0.00023571 m^2

next we calculate area of fluid between the smaller pipes and larger pipe

A = $[\frac{\pi }{4} D^{2} _{L} ] - 3(A_{s})$

= $[ \frac{\pi }{4} (0.08 )^2 - 3 ( 0.00023571 )]$

= [ 0.00502857 - 0.00070713 ]

= 0.00432144 m^2

hence the discharge in m^3/s

Q = AV

= 0.00432144 * 10

= 0.0432 m^3/s

3 0

2 years ago

What is the design wind speed for average-risk building:_____.