Answer:
a) P* = 0.5283 MPa , T* = 624.75 K , ρ* = 2.945 kg/m^3 , V* = 501.023 m/s
b) Pe = 0.1278 MPa , Te = 416.7 K , ρe = 1.069 kg/m^3 , Ve = 818.36 m/s, Ae = 33.75 cm^2
c) m' = 2.915 kg/s
Explanation:
Given:-
- The inlet pressure, Pi = 1.0 MPa
- The inlet temperature, Ti = 750 K
- Inlet velocity is negligible
- Steady, Isentropic Flow
- The specific heat ratio of air, k = 1.4
- Exit Mach number, Mae = 2
- The throat area, Ath = 20 cm^2
- Gas constant of air, R = 0.287 KJ / kg.K
Find:-
(a) the throat conditions (static pressure, temperature, density, and mach number)
b) the exit plane conditions
c) the mass flow rate
Solution:-
- For this problem we will assume air to behave like an ideal gas with constant specific heat at RTP. Also the flow of air through the nozzle is assumed to be steady, one dimensional, and Isentropic with constant specific heat ratio ( k ).
- First we will scrutinize on the exit conditions. We have a Mach number of 2 at the exit. The flow at the exit of converging-diverging nozzle is in super-sonic region this is only possible only if sonic ( Ma = 1 ) conditions are achieved by the flow at the throat area ( minimum cross-sectional area ).
- Moreover, the flow is almost still at the inlet. Hence, we can assume that the flow has negligible velocity ( vi = 0 m/s ) at the inlet and the reservoir temperature and pressure can be assumed to be stagnation temperature and pressures as follows:
[tex]P_o = 1.0 MPa\\\\T_o = 750 K[/tex]
- Using the ideal gas law we can determine the stagnation density ( ρo ) as follows:
[tex]p_o = \frac{P_o}{RT_o} = \frac{1000}{0.287*750} = 4.64576\frac{kg}{m^3}[/tex]
- We will use the already developed results for flow which has reached sonic velocity ( Ma = 1 ) at the throat region. Use Table A - 13, to determine the critical static values at the throat region:
[tex]\frac{P^*}{P_o} = 0.5283\\\\P^* = 0.5283*1 = 0.5283 MPa\\\\\frac{T^*}{T_o} = 0.8333\\\\T^* = 0.8333*750 = 624.75 K\\\\ \frac{p^*}{p_o} = 0.6339\\\\p^* = 0.6339*4.64576 = 2.945 \frac{kg}{m^3} \\\\[/tex]
[tex]V^* = \sqrt{kRT^*} =\sqrt{1.4*287*624.75} = 501.023 \frac{m}{s}[/tex]
- Similarly, we will again employ the table A - 13 to determine the exit plane conditions for ( Ma = 2 ) as follows:
[tex]\frac{P_e}{P_o} = 0.1278 \\\\P_e = 0.1278*1.0 = 0.1278 MPa\\\\\frac{T_e}{T_o} = 0.5556 \\\\T_e = 0.5556*750 = 416.7 K\\\\\frac{p_e}{p_o} = 0.23 \\\\p_e = 0.23*4.64576 = 1.069 \frac{kg}{m^3} \\\\\frac{A_e}{A_t_h} = 1.6875 \\\\A_e =1.6875*20 = 33.75 cm^2\\[/tex]
- The velocity at the exit plane ( Ve ) can be determined from the exit conditions as follows:
[tex]V_e = Ma_e*\sqrt{kRT_e} = 2*\sqrt{1.4*287*416.7} = 818.36 \frac{m}{s}[/tex]
- For steady flows the mass flow rate ( m' ) is constant at any section of the nozzle. We will use the properties at the throat section to determine the mass flow rate as follows:
[tex]m' = p^* A_t_h V^*\\\\m' = 2.945*20*10^-^4*501.023\\\\m' = 2.951 \frac{kg}{s}[/tex]
If gear X turns clockwise at constant speed of 20 rpm. How does gear y turns?
Answer:
Gear Y would turn Counter-Clockwise do to the opposite force created from gear X.
Hope this helped! Have a great day!
Determine the largest load P that can be a applied to the frame without causing either the average normal stress or the average shear stress at section a-a to exceed σ = 160 MPa and τ = 60 MPa , respectively. Member CB has a square cross section of 26 mm on each side.
Answer:
(The diagram of the question is given in Attachment 1)
The largest load which can be applied is:
P=67.62 kN
Explanation:
Make a Free body Diagram:All the forces are shown in the diagram in Attachment 2.
Analyze the equilibrium of Joint C in Figure (a):∑ F(y)= 0 (Upwards is positive)
[tex]F_{BC}sin\theta-P=0\\\frac{4}{5}F_{BC} - P=0\\F_{BC}=\frac{5}{4}P\\\\F_{BC}=1.25P[/tex]
Substitute F(BC) in Figure (b):
∑ F(x)= 0 (Towards Right is positive)
[tex]N_{a-a} - F_{BC}cos\theta=0\\N_{a-a}-1.25P(\frac{3}{5})=0\\N_{a-a}=0.75P[/tex]
∑ F(y)= 0 (Upwards is positive)
[tex]F_{BC}sin\theta- V_{a-a}= 0\\(\frac{4}{5})1.25P-V_{a-a}=0\\V_{a-a}=P[/tex]
Find Cross Sectional Area:
The cross sectional area of a-a:
[tex]A_{a-a}= \frac{(0.026)(0.026)}{3/5}\\A_{a-a}= 1.127\cdot10^{-3}[/tex]
Find P from Normal Stress Equation:σ = N(a-a)/A(a-a)
Substitute values:
[tex]160\cdot10^6=\frac{0.75P}{1.127\cdot10^{-3}}\\P=240.42\cdot10^3 N\\P=240.42 kN[/tex]
Find P from Shear Stress Equation:Т= V(a-a)/A(a-a)
Substitute values:
[tex]60\cdot10^6=\frac{P}{1.127\cdot10^{-3}}\\P=67.62\cdot10^{3}N\\P=67.62kN[/tex]
Results:To satisfy both the condition, we have to choose the lower value of P.
P=67.62 kN
Initially, a mixing vessel contains 300 kg of orange juice containing 40% solids. Orange concentrate and water are continuously fed into the vessel as illustrated in Figure 2.1. If it’s desired that the orange concentrate be diluted, how long should stirring occur to reduce the solids to 35%. [25] Fig 2.1: Mixing vessel
Answer:
forever (no solution)
Explanation:
If the figure you're working with is the one shown below, no amount of mixing will reduce the solid content to 35%.
The vessel contains 40% solids.
The incoming feed is ...
(100 kg/h)×(60% solids) = 60 kg solids/h
out of a total influx of material of ...
(100 kg/h +70 kg/h) = 170 kg/h
That means the solids content of the inflow is ...
(60 kg)/(170 kg) = 0.352941 ≈ 35.3%
The solids content cannot ever be less than 35.3%. The problem has no solution.
_____
We suggest you discuss this question with your teacher to see how they would solve it.
A particle is emitted from a smoke stack with diameter of 0.05 mm. In order to determine how far downstream it travels it is important to find its terminal downward velocity. If it has a density of 1200 kg/m3, its terminal downward velocity (cm) is: (assume the drag coefficient is 24/Re and the volume of a sphere is 4/3 pi R3)
Answer: downward velocity = 6.9×10^-4 cm/s
Explanation: Given that the
Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m
Where radius r = 2.5 × 10^-5 m
Density = 1200 kg/m^3
Area of a sphere = 4πr^2
A = 4 × π× (2.5 × 10^-5)^2
A = 7.8 × 10^-9 m^2
Volume V = 4/3πr^3
V = 4/3 × π × (2.5 × 10^-5)^3
V = 6.5 × 10^-14 m^3
Since density = mass/ volume
Make mass the subject of formula
Mass = density × volume
Mass = 1200 × 6.5 × 10^-14
Mass M = 7.9 × 10^-11 kg
Using the formula
V = sqrt( 2Mg/ pCA)
Where
g = 9.81 m/s^2
M = mass = 7.9 × 10^-11 kg
p = density = 1200 kg/m3
C = drag coefficient = 24
A = area = 7.8 × 10^-9m^2
V = terminal velocity
Substitute all the parameters into the formula
V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]
V = sqrt[ 1.54 × 10^-9/2.25×10-4]
V = 6.9×10^-6 m/s
V = 6.9 × 10^-4 cm/s
A railroad runs form city A to city B, a distance of 800km, through mountainous terrain. The present one-way travel time (including time at intermediate yards) is 20 hours, and the rail freight rate is $20 per ton. There is a truck service that competes with the railroad, running over a roughly parallel road for approximately the same distance, at an average speed of 48km per hour and a rate of $30 per ton. A new highway is planned to replace the existing roads; it is expected that most of the traffic will be trucks (auto usage is expected to be negligible). The performance function of the new facility is t_T = t_0 + bV_T, where V_T is the flow in trucks per hour, t_0 = 10 hours, b = 0.08 hour per truck per hour. The railroad's estimate of the demand function is: V_T/V_R = a_0(t_r/t_R)^a_1 (c_T/c_R)^a_2 where t_T and t_R are the trip times (in hours) by truck and rail, respectively, c_T and c_R are the corresponding rates, V_T and V_R are the corresponding flows and a_0, a_1 and a_2 are parameters. The total demand is likely to remain constant at V_TOT = 200 tons per hour. The rail system is utilized at only a fraction of capacity, so its performance function is flat (travel time is constant, independent of volume). If a_0 = 1, a_1 = -1 and a_2 = -2, find the present flows of freight by truck and rail. Make an estimate of the equilibrium flows if the new highway were built. With the new highway built, what would the equilibrium flow be in each of the following two cases: if the railroad dropped its rate to $15 per ton? if truckers were taxed $5 per ton to help pay for the new highway?
Answer:
i) VT = 52.16
VR = 147.85
ii) VT = 61
VR = 138.99
Explanation:
The step by step solution is been done, please check the attached file below to see it
Sea B = 5.00 m a 60.0°. Sea C que tiene la misma magnitud que A y un ángulo de dirección mayor que el de A en 25.0°. Sea A ⦁ B = 30.0 m2 y B ⦁ C = 35.0 m2 . Encuentre A.
Answer:
[tex]\| \vec A \| = 6.163\,m[/tex]
Explanation:
Sean A, B y C vectores coplanares tal que:
[tex]\vec A = (\| \vec A \|\cdot \cos \theta_{A},\| \vec A \|\cdot \sin \theta_{A})[/tex], [tex]\vec B = (\| \vec B \|\cdot \cos \theta_{B},\| \vec B \|\cdot \sin \theta_{B})[/tex] y [tex]\vec C = (\| \vec C \|\cdot \cos \theta_{C},\| \vec C \|\cdot \sin \theta_{C})[/tex]
Donde [tex]\| \vec A \|[/tex], [tex]\| \vec B \|[/tex] y [tex]\| \vec C \|[/tex] son las normas o magnitudes respectivas de los vectores A, B y C, mientras que [tex]\theta_{A}[/tex], [tex]\theta_{B}[/tex] y [tex]\theta_{C}[/tex] son las direcciones respectivas de aquellos vectores, medidas en grados sexagesimales.
Por definición de producto escalar, se encuentra que:
[tex]\vec A \,\bullet\, \vec B = \|\vec A \| \| \vec B \| \cos \theta_{B}\cdot \cos \theta_{A} + \|\vec A \| \| \vec B \| \sin \theta_{B}\cdot \sin \theta_{A}[/tex]
[tex]\vec B \,\bullet\, \vec C = \|\vec B \| \| \vec C \| \cos \theta_{B}\cdot \cos \theta_{C} + \|\vec B \| \| \vec C \| \sin \theta_{B}\cdot \sin \theta_{C}[/tex]
Asimismo, se sabe que [tex]\| \vec B \| = 5\,m[/tex], [tex]\theta_{B} = 60^{\circ}[/tex], [tex]\vec A \,\bullet \,\vec B = 30\,m^{2}[/tex], [tex]\vec B\, \bullet\, \vec C = 35\,m^{2}[/tex], [tex]\|\vec A \| = \| \vec C \|[/tex] y [tex]\theta_{C} = \theta_{A} + 25^{\circ}[/tex]. Entonces, las ecuaciones quedan simplificadas como siguen:
[tex]30\,m^{2} = 5\|\vec A \| \cdot (\cos 60^{\circ}\cdot \cos \theta_{A} + \sin 60^{\circ}\cdot \sin \theta_{A})[/tex]
[tex]35\,m^{2} = 5\|\vec A \| \cdot [\cos 60^{\circ}\cdot \cos (\theta_{A}+25^{\circ}) + \sin 60^{\circ}\cdot \sin (\theta_{A}+25^{\circ})][/tex]
Es decir,
[tex]30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})[/tex]
[tex]35\,m^{2} = \| \vec A \| \cdot [2.5\cdot \cos (\theta_{A}+25^{\circ})+4.330\cdot \sin (\theta_{A}+25^{\circ}})][/tex]
Luego, se aplica las siguientes identidades trigonométricas para sumas de ángulos:
[tex]\cos (\theta_{A}+25^{\circ}) = \cos \theta_{A}\cdot \cos 25^{\circ} - \sin \theta_{A}\cdot \sin 25^{\circ}[/tex]
[tex]\sin (\theta_{A}+25^{\circ}) = \sin \theta_{A}\cdot \cos 25^{\circ} + \cos \theta_{A} \cdot \sin 25^{\circ}[/tex]
Es decir,
[tex]\cos (\theta_{A}+25^{\circ}) = 0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A}[/tex]
[tex]\sin (\theta_{A}+25^{\circ}) = 0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A}[/tex]
Las nuevas expresiones son las siguientes:
[tex]30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})[/tex]
[tex]35\,m^{2} = \| \vec A \| \cdot [2.5\cdot (0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A})+4.330\cdot (0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A})][/tex]
Ahora se simplifican las expresiones, se elimina la norma de [tex]\vec A[/tex] y se desarrolla y simplifica la ecuación resultante:
[tex]30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})[/tex]
[tex]35\,m^{2} = \| \vec A \| \cdot (4.097\cdot \cos \theta_{A} +2.865\cdot \sin \theta_{A})[/tex]
[tex]\frac{30\,m^{2}}{2.5\cdot \cos \theta_{A}+ 4.330\cdot \sin \theta_{A}} = \frac{35\,m^{2}}{4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}}[/tex]
[tex]30\cdot (4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}) = 35\cdot (2.5\cdot \cos \theta_{A}+4.330\cdot \sin \theta_{A})[/tex]
[tex]122.91\cdot \cos \theta_{A} + 85.95\cdot \sin \theta_{A} = 87.5\cdot \cos \theta_{A} + 151.55\cdot \sin \theta_{A}[/tex]
[tex]35.41\cdot \cos \theta_{A} = 65.6\cdot \sin \theta_{A}[/tex]
[tex]\tan \theta_{A} = \frac{35.41}{65.6}[/tex]
[tex]\tan \theta_{A} = 0.540[/tex]
Ahora se determina el ángulo de [tex]\vec A[/tex]:
[tex]\theta_{A} = \tan^{-1} \left(0.540\right)[/tex]
La función tangente es positiva en el primer y tercer cuadrantes y tiene un periodicidad de 180 grados, entonces existen al menos dos soluciones del ángulo citado:
[tex]\theta_{A, 1} \approx 28.369^{\circ}[/tex] y [tex]\theta_{A, 2} \approx 208.369^{\circ}[/tex]
Ahora, la magnitud de [tex]\vec A[/tex] es:
[tex]\| \vec A \| = \frac{35\,m^{2}}{4.097\cdot \cos 28.369^{\circ} + 2.865\cdot \sin 28.369^{\circ}}[/tex]
[tex]\| \vec A \| = 6.163\,m[/tex]
The strength and stability of tires may be enhanced by heating both sides of the rubber ( 0.14 W/m·K, 6.35 × 10^-8m^2/s) in a steam chamber for which T[infinity]= 200°C. In the heating process, a 20-mm-thick rubber wall (assumed to be untreaded) is taken from an initial temperature of 35°C to a midplane temperature of 170°C. If steam flow over the tire surfaces maintains a convection coefficient of 200 W/m^2·K. How long will it take to achieve the desired midplane temperature?
Answer:
[tex]\mathbf{t_f = 1436.96 \ sec }[/tex]
Explanation:
Given that :
The strength and stability of tires may be enhanced by heating both sides of the rubber ( 0.14 W/m·K, 6.35 × 10^-8m^2/s)
i.e
k = 0.14 W/mK
∝ = 6.35 × 10⁻⁸ m²/s
L = 0.01 m
[tex]B_1 = \dfrac{hL}{k} \\ \\ B_1 = \dfrac{200*0.01}{0.14} \\ \\ B_1 = 14.2857[/tex]
We cannot use the model of Lumped Capacitance; SO Let assume that Fourier Number [tex]F_o > 0.2[/tex]
⇒ [tex]\dfrac{T_o - T_ \infty }{T_i - T_ \infty} = C_1 exp (- \zeta_i^2 *F_o)[/tex]
From Table 5.1 ; at [tex]B_1[/tex] = 14.2857
[tex]C_1 = 1.265 \\ \\ \zeta_1 = 1.458 \ rad[/tex]
[tex]\dfrac{170-200}{35-200} = 1.265 exp [ - (1.458)^2* \dfrac{ \alpha t_f}{L^2}][/tex]
[tex]In ( \dfrac{0.1818}{1.265}) = \dfrac{-1.458^2*6.35*10^{-8}*t_f}{0.01^2}[/tex]
[tex]-1.9399=-0.001350 *t_f[/tex]
[tex]t_f = \dfrac{-1.9399}{-0.001350}[/tex]
[tex]\mathbf{t_f = 1436.96 \ sec }[/tex]
A multi-plate clutch is to transmit 12 kW at 1500 rev/min. The inner and outer radii for the plates are to be 50 mm and 100 mm respectively. The maximum axial spring force is restricted to lkN. Calculate the necessary number of pairs of surfaces if ll = 0-35 assuming constant ‘vyear. What will be the necessary axial force?
Answer:
The uniform pressure for the necessary axial force is W = 945 N
The uniform wear for the necessary axial force is W = 970.15 N
Explanation:
Solution
Given that:
r₁ = 0.1 m
r₂ = 0.05m
μ = 0.35
p = 12 N or kW
N = 1500 rpm
W = 1000 N
The angular velocity is denoted as ω= 2πN/60
Here,
ω = 2π *1500/60 = 157.07 rad/s
Now, the power transferred becomes
P = Tω this is the equation (1)
Thus
12kW = T * 157.07 rad/s
T = 76.4 N.m
Now, when we look at the uniform condition, we have what is called the torque that is frictional which acts at the frictional surface of the clutch dented as :
T = nμW R this is the equation (2)
The frictional surface of the mean radius is denoted by
R =2/3 [(r₁)³ - (r₂)³/(r₁)² - (r₂)²]
=[(0.1)³ - (0.05)³/[(0.1)² - (0.05)²]
R is =0.077 m
Now, we replace this values and put them into the equation (2)
It gives us this, 76. 4 N.m = n * 0.35* 1000 N * 0.077 m
n = 2.809 = 3
The number of pair surfaces is = 3
Secondly, we determine the uniform wear.
So, the mean radius is denoted as follows:
R = r₁ + r₂/ 2
=0.1 + 0.05/2
=0.075 m
Now, we replace the values and put it into the equation (2) formula
76. 4 N.m = n *0.35* 1000 N * 0.075 m
n= 2.91 = 3
Again, the number of pair surfaces = 3
However, for the uniform pressure with regards to the number of clutch plates is 3 we can derive the necessary axial force from the equation (2)
76. 4 N.m = 3 * 0.35 * W *0.077 m
W = 945 N
Also, for the uniform wear with regards to the number of clutch plates is 3 we can derive the necessary axial force from the equation (2)
76. 4 N.m = 3 * 0.35 * W *0.075 m
W = 970. 15 N
A sandy soil has a natural water content of 12% and bulk unit weight of 18.8 kN/m3 . The void ratios corresponding to the densest state (emin) and loosest state (emax) of this soil are 0.48 and 0.88, respectively. Find the relative density and degree of saturation for this soil.
Answer:
The value of relative density is 75 % while that of degree of saturation is 54.82%.
Explanation:
Given Data:
Bulk density of Sandy Soil=[tex]\gamma_b=18.8\ kN/m^3[/tex]
Void Ratio in Densest state=[tex]e_{max}=0.88[/tex]
Void Ratio in Loosest state=[tex]e_{min}=0.48[/tex]
Water content=[tex]w=12\%[/tex]
To Find:
Relative Density=[tex]D_R=\dfrac{e_{max}-e}{e_{max}-e_{min}} \times 100 \%[/tex]
Degree of Saturation=[tex]S=\dfrac{w\times G_s}{e}[/tex]
Now all the other values are given except e. e is calculated as follows
e is termed as In situ void ratio and is given as
[tex]e=\dfrac{\gamma_w \times G_s-\gamma_d}{\gamma_d}[/tex]
Here
γ_w is the density of water whose value is 1
G_s is the constant whose value is 2.65
γ_d is the dry density of the sandy soil which is calculated as follows:
[tex]\gamma_d=\dfrac{\gamma_b}{1+\dfrac{w}{100}}[/tex]
Putting values
[tex]\gamma_d=\dfrac{18.8}{1+\dfrac{12}{100}}\\\gamma_d=16.78\ kN/m^3=1.678 g/cc \\[/tex]
Putting this value in the equation of e gives
[tex]e=\dfrac{1 \times 2.65-1.678}{1.678}\\e=0.579=0.58[/tex]
So the value of Relative density is given as
[tex]D_R=\dfrac{e_{max}-e}{e_{max}-e_{min}} \times 100 \%\\D_R=\dfrac{0.88-0.58}{0.88-0.48} \times 100 \%\\D_R=75 \%[/tex]
So the value of relative density is 75 %
Now the value of degree of saturation is given as
[tex]S=\dfrac{w\times G_s}{e}\\S=\dfrac{12\times 2.65}{0.58}\\S=54.82 \%[/tex]
The value of degree of saturation is 54.82%.
Answer:
The relative density = 0.83 which is equivalent to 83%
The degree of saturation, S = 0.58 which gives 58% saturation
Explanation:
The parameters given are;
Water content W% = 12%
Bulk unit weight, γ = 18.8 kN/m³
Void ratio of [tex]e_{min}[/tex] = 0.48
Void ratio of [tex]e_{max}[/tex] = 0.88
[tex]G_S[/tex] = Constant (As learnt from an answer to the question on the current page) = 2.65 for Sandy soil
[tex]\gamma =\dfrac{W}{V} = \dfrac{W_{w}+W_{s}}{V}[/tex]
Where, V = 1 m³
W = 18.8 KN
Bulk unit weight, γ = [tex]\gamma_d[/tex] × (1 + W)
∴ 18.8 = [tex]\gamma_d[/tex] × (1 + 0.12)
[tex]\gamma_d[/tex] = 18.8/ (1.12) = 16.79 kN/m³
[tex]\gamma_d =\dfrac{W_s}{V} = \dfrac{W_{s}}{1} = 16.79 \, kN/m^3[/tex]
[tex]W_s[/tex] = 16.79 kN
∴ [tex]W_w[/tex] = 18.8 kN - 16.79 kN = 2.01 kN
[tex]m_w = 2.01/9.81 = 0.205 \, kg[/tex]
Volume of water = 0.205 m³
[tex]\gamma = \dfrac{GS \times \gamma _{w}\times \left (1+w \right )}{1 + e} = \dfrac{GS \times 9.81\times \left (1+0.12 \right )}{1 + e} =18.8[/tex]
e + 1 = 0.58×GS = 0.58×2.65 =
e = 1.54 - 1 = 0.55
The relative density is given by the relation;
[tex]Relative \ density, Dr=\dfrac{e_{max} - e}{e_{max} - e_{min}}[/tex]
[tex]Relative \ density, Dr=\dfrac{0.88 - e}{0.88 - 0.48} = \dfrac{0.88 - 0.55}{0.4} = 0.83[/tex]
The relative density = 0.83
The relative density in percentage = 0.83×100 = 83%
S·e = GS×w = 0.12·2.65
S×0.55 = 0.318
The degree of saturation, S = 0.58
The degree of saturation, S in percentage = 58%.
Answer every question of this quiz
Please note: you can answer each question only once.
Which number shows the intake valve?
OK
I'd say number 4, number 3 looks like an exhaust valve
The thrust F of a screw propeller is known to depend upon the diameter d,speed of advance \nu ,fluid density p, revolution per second N, and the coefficient of viscosity μ of the fluid. Determine the dimensions of each of the variables in terms of L,M,T,and find an expression for F in terms of these quantities
Answer:
screw thrust = ML[tex]T^{-2}[/tex]
Explanation:
thrust of a screw propeller is given by the equation = p[tex]V^{2}[/tex][tex]D^{2}[/tex] x [tex]\frac{ND}{V}[/tex]Re
where,
D is diameter
V is the fluid velocity
p is the fluid density
N is the angular speed of the screw in revolution per second
Re is the Reynolds number which is equal to puD/μ
where p is the fluid density
u is the fluid velocity, and
μ is the fluid viscosity = kg/m.s = M[tex]L^{-1}[/tex][tex]T^{-1}[/tex]
Reynolds number is dimensionless so it cancels out
The dimensions of the variables are shown below in MLT
diameter is m = L
speed is in m/s = L[tex]T^{-1}[/tex]
fluid density is in kg/[tex]m^{3}[/tex] = M[tex]L^{-3}[/tex]
N is in rad/s = L[tex]L^{-1}[/tex][tex]T^{-1}[/tex] =
If we substitute these dimensions in their respective places in the equation, we get
thrust = M[tex]L^{-3}[/tex][tex](LT^{-1}) ^{2}[/tex][tex]L^{2}[/tex][tex]\frac{T^{-1} L}{LT^{-1} }[/tex]
= M[tex]L^{-3}[/tex][tex]L^{2}[/tex][tex]T^{-2}[/tex]
screw thrust = ML[tex]T^{-2}[/tex]
This is the dimension for a force which indicates that thrust is a type of force
A flashed steam geothermal power plant is located where underground hot water is available as saturated liquid at 700 kPa. The well head pressure is 600 kPa. The
flashed steam enters a turbine at 500 kPa and expands to 15 kPa, when it is condensed. The flow rate from the well is 29.6 kg/s. determine the power produced in
kW.
Answer:
The power produced by the turbine is 74655.936 kW.
Explanation:
A turbine is a device that operates at steady-state. Let suppose that turbine does not have heat interactions with surroundings, as well as changes in potential and kinetic energies are neglictible. Power output can be determined by First Law of Thermodynamics:
[tex]-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0[/tex]
[tex]\dot W_{out} = \dot m\cdot (h_{in}-h_{out})[/tex]
Let suppose that water enters as saturated vapor and exits as saturated liquid. Specific enthalpies are, respectively:
[tex]h_{in} = 2748.1\,\frac{kJ}{kg}[/tex]
[tex]h_{out} = 225.94\,\frac{kJ}{kg}[/tex]
The power produce by the turbine is:
[tex]\dot W_{in} = \left(29.6\,\frac{kg}{s} \right)\cdot \left(2748.1\,\frac{kJ}{kg} - 225.94\,\frac{kJ}{kg} \right)[/tex]
[tex]\dot W_{in} = 74655.936\,kW[/tex]
To find the reactance XLXLX_L of an inductor, imagine that a current I(t)=I0sin(ωt)I(t)=I0sin(ωt) , is flowing through the inductor. What is the voltage V(t)V(t)V(t) across this inductor?
Answer:
V(t) = XLI₀sin(π/2 - ωt)
Explanation:
According to Maxwell's equation which is expressed as;
V(t) = dФ/dt ........(1)
Magnetic flux Ф can also be expressed as;
Ф = LI(t)
Where
L = inductance of the inductor
I = current in Ampere
We can therefore Express Maxwell equation as:
V(t) = dLI(t)/dt ....... (2)
Since the inductance is constant then voltage remains
V(t) = LdI(t)/dt
In an AC circuit, the current is time varying and it is given in the form of
I(t) = I₀sin(ωt)
Substitutes the current I(t) into equation (2)
Then the voltage across inductor will be expressed as
V(t) = Ld(I₀sin(ωt))/dt
V(t) = LI₀ωcos(ωt)
Where cos(ωt) = sin(π/2 - ωt)
Then
V(t) = ωLI₀sin(π/2 - ωt) .....(3)
Because the voltage and current are out of phase with the phase difference of π/2 or 90°
The inductive reactance XL = ωL
Substitute ωL for XL in equation (3)
Therefore, the voltage across inductor is can be expressed as;
V(t) = XLI₀sin(π/2 - ωt)
What is 90 to the power of 46
Answer:Just multiply 90 by itself 46 times
Explanation:
do it
A particular Table in a relational database contains 100,000 Data Records/rows, each of which Data Record/row requires 200 bytes. A select statement returns all Data Records/rows in the Table that satisfy an equality search on an attribute. Estimate the time in milliseconds to complete the query when each of the following Indexes on that attribute is used.
A. No Index (Heap File of Data Records)
B. A Static Hash Index (with no overflow buckets/Pages). Assume the cost of applying the hash function is H, negligible.
The correct question is;
A particular table in a relational database contains 100,000 rows, each of which requires
200 bytes of memory. Estimate the time in milliseconds to to insert a new row into the
table when each of the following indices on the related attribute is used. Assume a page
size of 4K bytes and a page access time of 20 ms.
a. No index (heap file)
b. A clustered, non-integrated B+ tree index, with no node splitting
required. Assume that each index entry occupies 100 bytes. Assume that the
index is 75% occupied and the actual data pages are 100% occupied. Assume
that all matching entries are in a single page.
Answer:
A) 20 ms
B) 120 ms
Explanation:
A) Append (at the end of file). Just one IO, i.e., 20 ms
B) Now, when we assume that each entry in the index occupies 100 bytes, then an index page can thus hold 40 entries. Due to the fact that the data file occupies 5000 pages, the leaf level of the tree must contain at least 5000/40 pages which is 125 pages.
So, the number of levels in the tree (assuming page 75% occupancy in the
index) is (log_30 (125)) + 1 = 3. Now, if we assume that the index is clustered and not integrated with the data file and all matching entries are in a single
page, then 4 I/O operations and 80ms are required to retrieve all matching
records. Two additional I/O operations are required to update the leaf page
of the index and the data page. Hence, the time to do the insertion is
120ms.
A 1000 mm wide steel sheet made of C35 is normalized by cold rolling 10 mm thick
deformed to 5 mm. The rollers, 600 mm in diameter, run at a peripheral speed of 0.12 m/s.
The deformation efficiency is 55%.
Find out:
a) the roller force
b) the roller torque
c) the performance on the pair of rollers.
Answer:
a. 20.265 MN
b. 0.555 MNm
c. 403.44 KW
Explanation:
Given:-
- The width ( w ) = 1000 mm
- Original thickness ( to ) = 10 mm
- Final thickness ( t ) = 5 mm
- The radius of the rollers ( R ) = 600 mm
- The peripheral speed of the roller ( v ) = 0.12
- Deformation efficiency ( ε ) = 55%
Find:-
a) the roller force ( F )
b) the roller torque ( T )
c) the performance on the pair of rollers. ( P )
Solution:-
- The process of flat rolling entails a pair of compressive forces ( F ) exerted by the rollers on the steel sheet that permanently deforms.
- The permanent deformation of sheet metal is seen as reduced thickness.
- We will assume that the compressive force ( F ) acts normal to the point of contact between rollers and metal sheet.
- The roll force ( F ) is defined as:
[tex]F =L*w*Y_a_v_g[/tex]
Where,
L: The projected length of strip under compression
Y_avg: The yielding stress of the material = 370 MPa
- The projected length of strip under compression is approximated by the following relation:
[tex]L = \sqrt{R*( t_o - t_f )} \\\\L = \sqrt{0.6*( 0.01 - 0.005 )} \\\\L = 0.05477 m[/tex]
- The Roll force ( F ) can be determined as follows:
[tex]F = (0.05477)*(1 )*(370*10^6 )\\\\F = 20.265 MN[/tex]
- The roll torque ( T ) is given by the following relation as follows:
[tex]T = \frac{L}{2} * F\\\\T = \frac{0.05477}{2} * 20.265\\\\T = 0.555 MNm[/tex]
- The rotational speed of the rollers ( N ) is determined by the following procedure:
[tex]f = \frac{v}{2\pi* R} = \frac{0.12}{2*\pi 0.6} = 0.03181818 \frac{rev}{s} \\\\N = f*60 = 1.9090 rpm[/tex]
- The power consumed by the pair of rollers ( P ) is given by:
[tex]P = \frac{2\pi * F * L * N}{e*60,000} KW \\\\P = \frac{2\pi * ( 20.265*10^6) * (0.05477) * (1.90909 ) }{60,000*0.55} KW\\\\P = 403.44 KW[/tex]
Steam flows steadily through a turbine at a rate of 45,000 lbm/h, entering at 1000 psia and 9008F and leaving at 5 psia as saturated vapor. If the power generated by the turbine is 4 MW, determine the rate of heat loss from the steam.
A piston–cylinder device contains 0.78 kg of nitrogen gas at 140 kPa and 37°C. The gas is now compressed slowly in a polytropic process during which PV1.3 = constant. The process ends when the volume is reduced by one-half. Determine the entropy change of nitrogen during this process. The gas constant of nitrogen is R = 0.2968 kJ/kg·K. The constant volume specific heat of nitrogen at room temperature is cv = 0.743 kJ/kg·K. (Round the final answer to five decimal places.)
Answer:
The entropy change of nitrogen during this process. is - 0.32628 kJ/K.
Explanation:
Solution
Given that:
A piston cylinder device contains =0.78 kg of nitrogen gas
Temperature = 37°C
The nitrogen gas constant of R = 0.2968 kJ/kg.K
At room temperature cv = 0.743 kJ/kg.K
Now,
We assume that at specific condition the nitrogen can be treated as an ideal gas
Nitrogen has a constant volume specific heat at room temperature.
Thus,
From the polytropic relation, we have the following below:
T₂/T₁ =(V₁/V₂)^ n-1 which is,
T₂ = T₁ ((V₁/V₂)^ n-1
= (310 K) (2)^1.3-1 = 381.7 K
So,
The entropy change of nitrogen is computed as follows:
ΔSN₂ = m (cv₁ avg ln T₂/T₁ + ln V₂/V₁)
= (0.78) ((0.743 kJ/kg .K) ln 381.7 K/310K + (0.2968 kJ/kg. K) ln (0.5))
= 0.57954 * 0.2080 + (-0.2057)
= 0.12058 + (-0.2057) = -0.32628
Therefore the entropy change of nitrogen during this process. is - 0.32628 kJ/K.
Q2. Air at 400°C and 1.7 bar flows through a horizontal 8.2-cm D pipe at a velocity of 50.0 m/s.
I. Calculate Éx (W)? Assuming ideal gas behavior and Mol.Wt (Air)=29 g/mol. [2 Marks]
II. If the air is cooled to 250°C at constant pressure, what is AĖ,? [2 Marks]
Why would it be incorrect to say that the rate of transfer of heat to the gas must equal the rate of
change of kinetic energy? [1 Mark]
(Felder and Rousseau, 2005)
Answer:
The correct answer is (I) 290.81 W (II) 83.413 W (III) It is not correct to say that the rate of heat transfer must be the same to the heat of change in kinetic energy
Explanation:
Solution
Recall that:
Pressure (p) = 1.7 bar
Temperature (T₁) = 400°C which is = 673k
The velocity (v) = 50.0 m/s.
The pipe diameter (D)= 8.2 cm approximately 8.2 * 10 ^⁻2 m
The molecular air weight (M)= 29 g/mol
Suppose Air is seen as an ideal gas
where pv = mrT
p =(m/v) r T = p = ρrT
So,
r = the characteristics of gas constant (R/m)
p = pressure
R =The universal gas constant
T = temperature
ρ = density which is (kg/m³)
R is 8.314 J/mole -k
Then
1.7 * 10 ^5 = ρ * (8.314 /29) * 673
The density ρ = 881.09 g/ m³
(I) The mass flow rate = ρAV
thus,
m = 881.09 *π/4 ( 8.2 * 10^⁻2)² * 50
Therefore m = 232.65 g/s
We already know that
k= 1/2 mv²
k =1/2 *232.65/1000 * (50)²
so at 400°C k = 290.81 W
(II) Now in solving the process of the constant pressure we recall that
P = ρrT
Air is cooled to 250°C
p/r = ρT this is constant
So,
ρ₁T₂ = ρ₂T₂
881.09 * 673 = ρ₂ * 523
ρ₂ = 1133.79 g/m³
Thus,
m = ρ₂AV = 1133.79 * π /4 (8.2 * 10^ ⁻2) * 50
Hence m = 299.37 g/s
now,
k =1.2 mv² = 1.2 *(299.37)/1000 * (50)²
At 250°C, k = 374.22 W
Thus,
Δk = k ( 250°c) - k ( 400°c)
Δk = 374.22 - 290.81
Therefore,
Δk=83.413 W
(III) The steady state formula is given below
Q = W + ΔkE +ΔPE + ΔH
Now,
W = work (shaft)
Q =The rate of transfer of heat
ΔkE = The change in kinetic energy
ΔPE= The change in potential energy
ΔH =Change in enthalphy
For no shaft work, W =0
The horizontal pipe ΔPE = 0
Therefore,
The rate of heat transfer is explained as follows:
Q =ΔkE + ΔH
Because of the enthalphy, Q is not equal to ΔkE
Finally, it is not correct to say that the rate of heat transfer must be the same to the heat of change in kinetic energy.
