What are some things that AREN’T manufactured

A pump is used to deliver water from a lake to an elevated storage tank. The pipe network consists of 1,800 ft (equivalent lengt

Nataly_w [17]

Answer:

h_f = 15 ft, so option A is correct

Explanation:

The formula for head loss is given by;

h_f = [10.44•L•Q^(1.85)]/(C^(1.85))•D^(4.8655))

Where;

h_f is head loss due to friction in ft

L is length of pipe in ft

Q is flow rate of water in gpm

C is hazen Williams constant

D is diameter of pipe in inches

We are given;

L = 1,800 ft

Q = 600 gpm

C = 120

D = 8 inches

So, plugging in these values into the equation, we have;

h_f = [10.44*1800*600^(1.85)]/(120^(1.85))*8^(4.8655))

h_f = 14.896 ft.

So, h_f is approximately 15 ft

7 0

3 years ago

Ethylene (C2H4) gas enters a well-insulated reactor and reacts completely with 400% of theoretical air, each at 25°C, 2 atm. The

GuDViN [60]

Answer:see explanation.

Explanation:

(a).The equation of reaction is;

C2H4 + 3(O2 + 3.76N2) -----> 2CO2 + 2H2O + 11.28 N2.

Theoretical air is the amount of air that will allow the complete combustion of the fuel. Air contains 21% of oxygen,79% of Nitrogen. Thus, the molecular mass of air = 29 [Kg/Kmol] . The number of oxygen needed to oxidize the hydrocarbon(C2H4) is 79/21= 3.76 moles of Nitrogen.

(b). N(total) =2+2+11.28 =15.28 kmoles of product

15.28[h(T) - h°] of air

= 15.28(Mass of air) (Cp,1000k)

T(adiabatic) = 802,310 kJ/k. Mole.

When mass of air= 29 Kg/ k.mol

Cp,1000k = 1.142 kJ.k which is the specific heat capacity of air.

T(adiabatic) = 1275k -----> assuming all the products are air.

(C). 2{∆h} co2 + 2{∆h} H20 +11.28 {∆h} N2

After series of calculations and checking of tables:

S(o2) = 320.173 - 8.314 ln 0.12= 337.46 kJ/ K.mol.k

S(H2O)= 273.986- 8.314 ln 0.1406 = 290.30 KJ/ kmol. K

S(N2) = 258.503- 8.314 ln 0.7344= 261.07 kJ/kmol.k

= 2(337.46) + 2(290.30)+11.28(261.07)-360.79-3(218.01)-11.28(193.46)

=1003.3378 kJ/kmol. K

5 0

3 years ago

1. How many pieces of 12-1/2" long wood can be cut from a piece of

Pani-rosa [81]

B. If you do 39-3/4 divided by 12-1/2 you get 3.18

8 0

3 years ago

A parallel circuit with two branches and an 18 volt battery. Resistor #1 on the first branch has a value of 220 ohms and resisto

likoan [24]

Answer:

2.455 W

Explanation:

The power dissipated in each branch is ...

P = V^2/R

So, the branch powers are ...

branch 1: 18^2/220 ≈ 1.473 W

branch 2: 18^2/330 ≈ 0.982 W

Total power is ...

1.473 W + 0.982 W = 2.455 W

8 0

3 years ago

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 75

Inessa [10]

Answer:

(a). The value of temperature at the end of heat addition process $T_{3}$ = 2042.56 K

(b). The value of pressure at the end of heat addition process $P_{3}$ = 1555.46 k pa

(c). The thermal efficiency of an Otto cycle $E_{otto}$ = 0.4478

(d). The value of mean effective pressure of the cycle $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

Explanation:

Compression ratio $r_{p}$ = 8

Initial pressure $P_{1}$ = 95 k pa

Initial temperature $T_{1}$ = 278 °c = 551 K

Final pressure $P_{2}$ = 8 × $P_{1}$ = 8 × 95 = 760 k pa

Final temperature $T_{2}$ = $T_{1}$ × $r_{p} ^{\frac{\gamma - 1}{\gamma} }$

Final temperature $T_{2}$ = 551 × $8 ^{\frac{1.4 - 1}{1.4} }$

Final temperature $T_{2}$ = 998 K

Heat transferred at constant volume Q = 750 $\frac{KJ}{kg}$

(a). We know that Heat transferred at constant volume $Q_{S}$ = $m C_{v} ( T_{3} - T_{2} )$

⇒ 1 × 0.718 × ( $T_{3}$ - 998 ) = 750

⇒ $T_{3}$ = 2042.56 K

This is the value of temperature at the end of heat addition process.

Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.

⇒ P ∝ T

⇒ $\frac{P_{3} }{P_{2} }$ = $\frac{T_{3} }{T_{2} }$

⇒ $P_{3}$ = $\frac{2042.56}{998}$ × 760

⇒ $P_{3}$ = 1555.46 k pa

This is the value of pressure at the end of heat addition process.

(b). Heat rejected from the cycle $Q_{R} = m C_{v} ( T_{4} - T_{1} )$

For the compression and expansion process,

⇒ $\frac{T_{3} }{T_{2} } = \frac{T_{4} }{T_{1} }$

⇒ $\frac{2042.56}{998} = \frac{T_{4} }{551}$

⇒ $T_{4}$ = 1127.7 K

Heat rejected $Q_{R}$ = 1 × 0.718 × ( 1127.7 - 551)

⇒ $Q_{R}$ = 414.07 $\frac{KJ}{kg}$

Net heat interaction from the cycle $Q_{net}$ = $Q_{S}$ - $Q_{R}$

Put the values of $Q_{S}$ & $Q_{R}$ we get,

⇒ $Q_{net}$ = 750 - 414.07

⇒ $Q_{net}$ = 335.93 $\frac{KJ}{kg}$

We know that for a cyclic process net heat interaction is equal to net work transfer.

⇒ $Q_{net}$ = $W_{net}$

⇒ $W_{net}$ = 335.93 $\frac{KJ}{kg}$

This is the net work output from the cycle.

(c). Thermal efficiency of an Otto cycle is given by

$E_{otto} = 1- \frac{T_{1} }{T_{2} }$

Put the values of $T_{1}$ & $T_{2}$ in the above formula we get,

$E_{otto} = 1- \frac{551 }{998 }$

⇒ $E_{otto}$ = 0.4478

This is the thermal efficiency of an Otto cycle.

(d). Mean effective pressure $P_{m}$ :-

We know that mean effective pressure of the Otto cycle is given by

$P_{m}$ = $\frac{W_{net} }{V_{s} }$ ---------- (1)

where $V_{s}$ is the swept volume.

$V_{s}$ = $V_{1} - V_{2}$ ---------- ( 2 )

From ideal gas equation $P_{1}$ $V_{1}$ = m × R × $T_{1}$

Put all the values in above formula we get,

⇒ 95 × $V_{1}$ = 1 × 0.287 × 551

⇒ $V_{1}$ = 0.6 $m^{3}$

From the same ideal gas equation

$P_{2}$ $V_{2}$ = m × R × $T_{2}$

⇒ 760 × $V_{2}$ = 1 × 0.287 × 998

⇒ $V_{2}$ = 0.377 $m^{3}$

Thus swept volume $V_{s}$ = 0.6 - 0.377

⇒ $V_{s}$ = 0.223 $m^{3}$

Thus from equation 1 the mean effective pressure

⇒ $P_{m}$ = $\frac{335.93}{0.223}$

⇒ $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

This is the value of mean effective pressure of the cycle.

4 0

3 years ago