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Alex_Xolod [135]
2 years ago
6

A plant has ten machines and currently operates two 8-hr shift per day, 5 days per week, 50 weeks per year. the ten machines pro

duce the same part each at a rate of 30 pc/hr. (a) determine annual production capacity of this plant. (b) if the plant were to operate three 8-hr shifts per day, 6 days per week, 51 weeks per year, determine the annual percentage increase in plant capacity
Engineering
1 answer:
Zarrin [17]2 years ago
3 0

Answer:

  83.6%

Explanation:

<h3>(a)</h3>

On its current schedule, the plant can theoretically produce ...

  (30 pc/h/machine)(10 machine)(8 h/shift)(2 shift/day)(5 day/wk)(50 wk/yr)

  = (30)(10)(8)(2)(5)(50) pc/yr = 1,200,000 pc/yr

__

<h3>(b)</h3>

On the proposed schedule, this production becomes ...

  (30 pc/mach)(10 mach)(8 h/sh)(3 sh/da)(6 da/wk)(51 wk/yr)

  = (30)(10)(8)(3)(6)(51) pc/yr = 2,203,200 pc/yr

The increase in capacity is ...

  (2,203,200/1,200,000 -1) × 100% = 83.6% . . . capacity increase

_____

<em>Additional comment</em>

The number of parts per shift did not change. Only the number of shifts per year changed. It went up by a factor of (3/2)(6/5)(51/50) = 1.836. Hence the 83.6% increase in capacity.

We have to assume that maintenance and repair are done as effectvely as before in the reduced down time that each machine has.

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Air at 26 kPa, 230 K, and 220 rn/s enters a turbojet engine in flight. The air mass flow rate is 25 kg/s. The compressor pressur
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Explanation:

Answer:

Explanation:

Answer:  

Explanation:  

This is a little lengthy and tricky, but nevertheless i would give a step by step analysis to make this as simple as possible.  

(a). here we are asked to determine the Temperature and Pressure.  

Given that the properties of Air;  

ha = 230.02 KJ/Kg  

Ta = 230 K  

Pra = 0.5477  

From the energy balance equation for a diffuser;  

ha + Va²/2 = h₁ + V₁²/2  

h₁ = ha + Va²/2 (where V₁²/2 = 0)  

h₁ = 230.02 + 220²/2 ˣ 1/10³  

h₁ = 254.22 KJ/Kg  

⇒ now we obtain the properties of air at h₁ = 254.22 KJ/Kg  

from this we have;  

Pr₁ = 0.7329 + (0.8405 - 0.7329)[(254.22 - 250.05) / (260.09 - 250.05)]  

Pr₁ = 0.77759  

therefore T₁ = 254.15K  

P₁ = (Pr₁/Pra)Pa  

= 0.77759/0.5477 ˣ 26  

P₁ = 36.91 kPa  

now we calculate Pr₂  

Pr₂ = Pr₁ (P₂/P₁) = 0.77759 ˣ 11 = 8.55349  

⇒ now we obtain properties of air at  

Pr₂ = 8.55349 and h₂ = 505.387 KJ/Kg  

calculating the enthalpy of air at state 2  

ηc = h₁ - h₂ / h₁ - h₂  

0.85 = 254.22 - 505.387 / 254.22 - h₂  

h₂ = 549.71 KJ/Kg  

to obtain the properties of air at h₂ = 549.71 KJ/Kg  

T₂ = 545.15 K

⇒ to calculate the pressure of air at state 2

P₂/P₁ = 11

P₂ = 11 ˣ 36.913  

p₂ = 406.043 kPa

but pressure of air at state 3 is the same,

i.e. P₂ = P₃ = 406.043 kPa

P₃ = 406.043 kPa

To obtain the properties of air at  

T₃ = 1400 K, h₃ = 1515.42 kJ/Kg and Pr = 450.5

for cases of turbojet engine,

we have that work output from turbine = work input to the compressor

Wt = Wr

(h₃ - h₄) = (h₂ - h₁)

h₄ = h₃ - h₂ + h₁  

= 1515.42 - 549.71 + 254.22

h₄ = 1219.93 kJ/Kg

properties of air at h₄ = 1219.93 kJ/Kg

T₄ = 1140 + (1160 - 1140) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

T₄ = 1150.58 K

Pr₄ = 193.1 + (207.2 - 193.1) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

Pr₄ = 200.5636

Calculating the ideal enthalpy of the air at state 4;

Лr = h₃ - h₄ / h₃ - h₄*

0.9 = 1515.42 - 1219.93 / 1515.42 - h₄  

h₄* = 1187.09 kJ/Kg

now to obtain the properties of air at h₄⁻ = 1187.09 kJ/Kg

P₄* = 179.7 + (193.1 - 179.7) [(1187.09 -1184.28) / (1207.57 - 1184.28)]

P₄* = 181.316

P₄ = (Pr₄/Pr₃)P₃       i.e. 3-4 isentropic process

P₄ = 181.316/450.5 * 406.043

P₄ = 163.42 kPa

For the 4-5 process;

Pr₅ = (P₅/P₄)Pr₄

Pr₅ = 26/163.42 * 200.56 = 31.9095

to obtain the properties of air at Pr₅ = 31.9095

h₅= 724.04 + (734.82 - 724.04) [(31.9095 - 3038) / (32.02 - 30.38)]

h₅ = 734.09 KJ/Kg

T₅ = 710 + (720 - 710) [(31.9095 - 3038) / (32.02 - 30.38)]

T₅ = 719.32 K

(b) Now we are asked to calculate the rate of heat addition to the air passing through the combustor;

QH = m(h₃-h₂)

QH = 25(1515.42 - 549.71)

QH = 24142.75 kW

(c). To calculate the velocity at the nozzle exit;

we apply steady energy equation of a flow to nozzle

h₄ + V₄²/2 = h₅ + V₅²/2

h₄  + 0  = h₅₅ + V₅²/2

1219.9 ˣ 10³ = 734.09 ˣ 10³ + V₅²/2

therefore, V₅ = 985.74 m/s

cheers i hope this helps

6 0
3 years ago
A 0.19-m3 rigid tank equipped with a pressure regulator contains steam at 2 MPa and 300°C. The steam in the tank is now heated.
AVprozaik [17]

Answer:

576.21kJ

Explanation:

#We know that:

The balance mass m_{in}+m_{out}=\bigtriangleup m_{system}

so, m_e=m_1-m_2

Energy \ Balance\\E_{in}-E_{out}=\bigtriangleup E_{system}\\\\\therefore Q_i_n+m_eh_e=m_2u_2-m_1u_1

#Also, given the properties of water as;

(P_1=2Mpa,T_1=300\textdegree C)->v_1=0.12551m^3/kg,u_1=2773.2kJ/kg->h_1=3024.2kJ/kg\\\\(P_2=2Mpa,T_1=500\textdegree C)->v_2=0.17568m^3/kg,u_1=3116.9kJ/kg->h_1=3468.3kJ/kg

#We assume constant properties for the steam at average temperatures:h_e=\approx(h_1+h_2)/2

#Replace known values in the equation above;h_e=(3024.2+3468.3)/2=3246.25kJ/kg\\\\m_1=V_1/v_1=0.19m^3/(0.12551m^3/kg)=1.5138kg\\\\m_2=V_2/v_2=0.19m^3/(0.17568m^3/kg)=1.0815kg

#Using the mass and energy balance relations;

m_e=m_1-m_2\\\\m_e=1.5138-1.0815\\\\m_e=0.4323kg

#We have Q_i_n+m_eh_e=m_2u_2-m_1u_1: we replace the known values in the equation as;

Q_i_n+m_eh_e=m_2u_2-m_1u_1\\\\Q_i_n=0.4323kg\times3246.2kJ/kg+1.0815kg\times3116.9-1.5138kg\times2773.2kJ/kg\\\\Q_i_n=573.21kJ

#Hence,the amount of heat transferred when the steam temperature reaches 500°C is 576.21kJ

5 0
3 years ago
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