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3 years ago

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An air-standard Otto cycle has a compression ratio of 6 and the temperature and pressure at the beginning of the compression pro

cess are 520 deg R and 14.2 lbf/in^2, respectively. The heat addition per unit mass of air is 600 Btu/lb. Determine (a) the maximum temperature, in deg R.

1 answer:

tangare [24]3 years ago

3 0

Answer:

The maximum temperature of the cycle is 1065⁰R

Explanation:

The maximum temperature in degree Rankin can be obtained using the formula below;

$\frac{T_2}{T_1} =[\frac{V_1}{V_2}]^{1.4-1}$

Where;

T₂ is the maximum temperature of the cycle

T₁ is the initial temperature of the cycle = 520 deg R = 520 ⁰R

V₁/V₂ is the compression ratio = 6

$T_2 = T_1(\frac{V_1}{V_2})^{0.4}$

$T_2=T_1(6)^{0.4}$

$T_2=520^0R(6)^{0.4}$

T₂ = 1064.96 ⁰R

Therefore, the maximum temperature of the cycle is 1065⁰R

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3 years ago

It is known that the connecting rod AB exerts on the crank BCa 2.5-kN force directed down andto the left along the centerline of

monitta

Answer:

M_c = 61.6 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two casesshown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

M_c = F_a*( 42 / 150 ) *88

M_c = 2.5*( 42 / 150 ) *88

M_c = 61.6 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

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3 years ago

Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series

love history [14]

Answer:

The required pumping head is 1344.55 m and the pumping power is 236.96 kW

Explanation:

The energy equation is equal to:

$\frac{P_{1} }{\gamma } +\frac{V_{1}^{2} }{2g} +z_{1} =\frac{P_{2} }{\gamma } +\frac{V_{2}^{2} }{2g} +z_{2}+h_{i} -h_{pump} , if V_{1} =0,z_{2} =0\\h_{pump} =\frac{V_{2}^{2}}{2} +h_{i}-z_{1}$

For the pipe 1, the flow velocity is:

$V_{1} =\frac{Q}{\frac{\pi D^{2} }{4} }$

Q = 18 L/s = 0.018 m³/s

D = 6 cm = 0.06 m

$V_{1} =\frac{0.018}{\frac{\pi *0.06^{2} }{4} } =6.366m/s$

The Reynold´s number is:

$Re=\frac{\rho *V*D}{u} =\frac{999.1*6.366*0.06}{1.138x10^{-3} } =335339.4$

$\frac{\epsilon }{D} =\frac{0.00026}{0.06} =0.0043$

Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941

The head of pipe 1 is:

$h_{1} =\frac{V_{1}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{6.366^{2} }{2*9.8} *(0.5+\frac{0.0294*20}{0.06} )=21.3m$

For the pipe 2, the flow velocity is:

$V_{2} =\frac{0.018}{\frac{\pi *0.03^{2} }{4} } =25.46m/s$

The Reynold´s number is:

$Re=\frac{\rho *V*D}{u} =\frac{999.1*25.46*0.03}{1.138x10^{-3} } =670573.4$

$\frac{\epsilon }{D} =\frac{0.00026}{0.03} =0.0087$

The head of pipe 1 is:

$h_{2} =\frac{V_{2}^{2} }{2g} (k_{L}+\frac{fL}{D} )=\frac{25.46^{2} }{2*9.8} *(0.5+\frac{0.033*36}{0.03} )=1326.18m$

The total head is:

hi = 1326.18 + 21.3 = 1347.48 m

The required pump head is:

$h_{pump} =\frac{25.46^{2} }{2*9.8} +1347.48-36=1344.55m$

The required pumping power is:

$P=Q\rho *g*h_{pump} =0.018*999.1*9.8*1344.55=236965.16W=236.96kW$

8 0

3 years ago

Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes

loris [4]

Answer:

a). 8.67 x $10^{-3}$ m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

We know density of air is ρ = 1.21 kg / $m^{3}$

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

= 8.67 x $10^{-3}$ m

a). Boundary layer thickness at x = 1 is δ = 8.67 X $10^{-3}$ m

b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

Re = $\frac{\rho .U.x}{\mu }$

100000 = $\frac{1.21\times5\times x}{1.822 \times10^{-5}}$

x = 0.3011 m

c). Given x = 3 m from the leading edge

The Reyonld no at x = 3 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

= 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

Re = $\frac{\rho \times U\times x}{\mu }$

5 X $10^{5}$ = $\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}$

x = 1.505 m

We know that the force acting on the plate is

$F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

= 0.2137 N

e). The flow is turbulent at the end of the plate.

Re = $\frac{\rho \times U\times x}{\mu }$

= $\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }$

= 1992316

Therefore $C_{D}$ = $\frac{0.072}{Re^{\frac{1}{5}}}$

= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

3 0

3 years ago

Small droplets of carbon tetrachloride at 68 °F are formed with a spray nozzle. If the average diameter of the droplets is 200 u

Licemer1 [7]

Answer:

the difference in pressure between the inside and outside of the droplets is 538 Pa

Explanation:

given data

temperature = 68 °F

average diameter = 200 µm

to find out

what is the difference in pressure between the inside and outside of the droplets

solution

we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is

σ = 2.69 × $10^{-2}$ N/m

so average radius = $\frac{diameter}{2}$ = 100 µm = 100 × $10^{-6}$ m

now here we know relation between pressure difference and surface tension

so we can derive difference pressure as

2π×σ×r = Δp×π×r² .....................1

here r is radius and Δp pressure difference and σ surface tension

Δp = $\frac{2 \sigma }{r}$

put here value

Δp = $\frac{2*2.69*10^{-2}}{100*10^{-6}}$

Δp = 538

so the difference in pressure between the inside and outside of the droplets is 538 Pa

7 0

3 years ago