Convert 25 inches / min to mm/hour

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

Assoli18 [71]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

8 0

4 years ago

Basic concepts surrounding electrical circuitry?

Elanso [62]

Hopefully that helps you out and is this for history or science?

3 0

3 years ago

What is a smooth flow of air over a surface is called?

pochemuha

Answer:

laminar flow

Explanation:

In fluid dynamics, laminar flow is characterized by fluid particles following smooth paths in layers, with each layer moving smoothly past the adjacent layers with little or no mixing.

4 0

3 years ago

A sinusoidal wave of frequency 420 Hz has a speed of 310 m/s. (a) How far apart are two points that differ in phase by π/8 rad?

Olin [163]

Answer:

a) Two points that differ in phase by π/8 rad are 0.0461 m apart.

b) The phase difference between two displacements at a certain point at times 1.6 ms apart is 4π/3.

Explanation:

f = 420 Hz, v = 310 m/s, λ = wavelength = ?

v = fλ

λ = v/f = 310/420 = 0.738 m

T = periodic time of the wave = 1/420 = 0.00238 s = 0.0024 s = 2.4 ms

a) Two points that differ in phase by π/8 rad

In terms of the wavelength of the wave, this is equivalent to [(π/8)/2π] fraction of a wavelength,

[(π/8)/2π] = 1/16 of a wavelength = (1/16) × 0.738 = 0.0461 m

b) two displacements at times 1.6 ms apart.

In terms of periodic time, 1.6ms is (1.6/2.4) fraction of the periodic time.

1.6/2.4 = 2/3.

This means those two points are 2/3 fraction of a periodic time away from each other.

1 complete wave = 2π rad

Points 2/3 fraction of a wave from each other will have a phase difference of 2/3 × 2π = 4π/3.

8 0

4 years ago

A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the othe

vodomira [7]

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V $_f$ = V $_g$ = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v $_f$ = 0.001057 m³/kg

v $_g$ = 1.0037 m³/kg

u $_f$ = 486.82 kJ/kg

u $_g$ 2524.5 kJ/kg

h $_g$ = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V $_f$ /v $_f$ + V $_g$ /v $_g$

we substitute

m₁ = 0.002/0.001057 + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v $_g$

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m $_{out$ = m₁ - m₂

m $_{out$ = 1.89414 - 0.003985

m $_{out$ = 1.890155 kg

so, Initial internal energy will be;

U₁ = m $_f$ u $_f$ + m $_g$ u $_g$

U₁ = (V $_f$ /v $_f$ )u $_f$ + (V $_g$ /v $_g$ )u $_g$

we substitute

U₁ = (0.002/0.001057)(486.82) + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E $_{in$ - E $_{out$ = ΔE $_{sys$

QΔt - m $_{out$ h $_{out$ = m₂u₂ - U₁

QΔt - m $_{out$ h $_g$ = m₂u $_g$ - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

5 0

3 years ago

Convert 25 inches / min to mm/hour​

Convert 25 inches / min to mm/hour