Solid rockets can experience significant 2 phase flow. a) True b) False

You may wonder who the rest goes

Deffense [45]

Answer:

no not really

Explanation:

From your friendly neighborhood cereal killer,

Sir. BLOODPR1NCE

6 0

2 years ago

(1) Estimate the specific volume in cm3 /g for carbon dioxide at 310 K and (a) 8 bar (b) 75 bar by the virial equation and compa

ludmilkaskok [199]

Answer:

70.66 cm^3

The specific volume for P = 70.66 is within 1% of the experimental value while the viral equation will be inaccurate when the second viral coefficient is used )

Explanation:

Viral equation : Z = 1 + Bp + Cp^2 + Dp^3 + -----

Viral equation can also be rewritten as :

Z = 1 + B ( P/RT )

B ( function of time )

Temperature = 310 K

P1 = 8 bar

P2 = 75 bar

Determine the specific volume in cm^3

V = 70.66 cm^3

b) comparing the specific volumes to the experimental values

70.58 and 3.90

The specific volume for P = 70.66 is within 1% of the experimental value while the viral equation will be inaccurate when the second viral coefficient is used )

attached below is the detailed solution

3 0

2 years ago

A wastewater treatment plant discharges 1.0 m3/s of effluent having an ultimate BOD of 40.0 mg/ L into a stream flowingat 10.0 m

kondaur [170]

Answer:

a) 6.4 mg/l

b) 5.6 mg/l

Explanation:

Given data:

effluent Discharge Q_w = 1.0 m^3.s

Ultimate BOD L_w = 40 mg/l

Discharge of stream Q_r = 10 m^3.s

Stream ultimate BOD L_r = 3 mg/l

a) Ultimate BOD of mixture $= \frac{Q_w l_w + Q_r L_r}{Q_w + Q_r}$

$= \frac{1*40 + 10*3}{10 +1} = 6.4 mg/l$

b) utlimate BOD at 10,000 m downstream

$t =\frac{distance}{speed} = \frac{10,000}{\frac{Q_r +Q+w}{55}} \times \frac{hr}{3600} \times \frac{day}{24 hr}$

putting $Q_r + Q_w = 1+ 10 = 11 m^3/s$

t = 0.578 days

we know

$L_t = L_o e^{-kt}$

$L_t = 6.4 \times e^{-0.22 \times 0.578}$

$L_t = 5.6 mg/l$

7 0

2 years ago

Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cool

MariettaO [177]

Answer: hello your question is incomplete below is the missing part

question :Determine the temperature of the cooled water exiting the cooling tower

answer : T = 43.477° C

Explanation:

Temp of water at exit = 45°C

mass flow rate of cooling tower = 15,000 kg/s

Temp of makeup water = 20°C

Assuming an atmospheric pressure of = 101.3 kPa

Determine temperature of the cooled water exiting the cooling tower

Water entering cooling tower at 45°C

Given that Latent heat of water at 45°C = 43.13 KJ/mol

Cp(wet air) = 1.005+ 1.884(y1)

where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273

Hence : Cp(wet air) = 29.145 + (0.01273) (33.94) = 29.577 KJ/kmol°C

First step : calculate the value of Q

Q = m*Cp*(ΔT) + W(latent heat)

Q = 321.6968 (29.577) (40-30) + 43.13 (18.26089)

Q = 95935.8547 KJ/s

Given that mass rate of water = 15000 kg/s

Hence the temperature of the cooled water can be calculated using the equation below

Q = m*Cp*∆T

Cp(water) = 4.2 KJ/Kg°C

95935.8547 = (15000)*(4.2)*(45 - T)

( 45 - T ) = 95935.8547/ 63000. ∴ T = 43.477° C

5 0

2 years ago

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t

Strike441 [17]

Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / $0.02^{0.22$

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

$l_i$ = $l_0e^{ET$

given that $l_0$ = 480 mm

we substitute

$l_i$ = $480mm$ × $e^{0.04481$

$l_i$ = 501.998 mm

Now we find the elongation;

Elongation = $l_i - l_0$

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm

6 0

2 years ago