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June 19, 2023 • 10 min read

**Question 1:** Calculate capillary rise/fall in a glass tube 2 mm diameter when immersed in (a) water (b) mercury.Both the liquids are at 20°C and the surface tension values at this temperature for water and mercury are 0.072 N/m and 0.052 N/m respectively. The specific gravity of mercury is 13.6. The contact angle of water and mercury are 0° and 130° respectively.

**Solutions 1:**

To calculate the capillary rise or fall in a glass tube immersed in water and mercury, we can use the Young-Laplace equation and the concept of capillary action. The formula to calculate the capillary rise or fall is given by:

h = (2 * γ * cos(θ)) / (ρ * g * r)

Where:

h is the capillary rise or fall,

γ is the surface tension of the liquid,

θ is the contact angle,

ρ is the density of the liquid,

g is the acceleration due to gravity,

and r is the radius of the tube.

Let's calculate the capillary rise/fall for both water and mercury:

(a) Water:

Given:

Diameter of the glass tube = 2 mm

Radius (r) = 2 mm / 2 = 1 mm = 0.001 m

Surface tension of water (γ) = 0.072 N/m

Contact angle (θ) = 0°

Density of water (ρ) = 1000 kg/m³ (at 20°C)

Acceleration due to gravity (g) = 9.8 m/s²

Substituting these values into the formula, we have:

h = (2 * 0.072 * cos(0°)) / (1000 * 9.8 * 0.001)

h = 0.0146 m = 14.6 mm

Therefore, the capillary rise in the glass tube when immersed in water is 14.6 mm.

(b) Mercury:

Given:

Diameter of the glass tube = 2 mm

Radius (r) = 2 mm / 2 = 1 mm = 0.001 m

Surface tension of mercury (γ) = 0.052 N/m

Contact angle (θ) = 130°

Specific gravity of mercury = 13.6 (density of mercury / density of water)

Density of water (ρ) = 1000 kg/m³ (at 20°C)

Acceleration due to gravity (g) = 9.8 m/s²

To calculate the density of mercury, we use the specific gravity:

Density of mercury = Specific gravity of mercury * Density of water

Density of mercury = 13.6 * 1000 = 13600 kg/m³

Substituting these values into the formula, we have:

h = (2 * 0.052 * cos(130°)) / (13600 * 9.8 * 0.001)

h = -0.1012 m = -101.2 mm

Therefore, the capillary fall in the glass tube when immersed in mercury is 101.2 mm (negative sign indicates a fall instead of a rise).

So, to summarize:

(a) The capillary rise in water is 14.6 mm.

(b) The capillary fall in mercury is 101.2 mm.

**Question 2: ** A thin 30-cm x 30-cm flat plate is pulled at 3 m/s horizontally through a 3.6-mm-thick oil layer sandwiched between two plates, one stationary and the other moving at a constant velocity of 0.3m/s, as shown in Fig. P2-77. The dynamic viscosity of the oil is 0.027 Pa-s. Assuming the velocity in each oil layer to vary linearly, (a) plot the velocity profile and find the location where the oil velocity is zero and (b) determine the force that needs to be applied on the plate to maintain this motion.

**Solutions 2:**

To solve this problem, we can use the concept of viscous flow and the linear velocity profile assumption. Here are the steps to solve each part of the problem:

(a) Plot the velocity profile and find the location where the oil velocity is zero:

- We assume that the velocity in each oil layer varies linearly. Let's denote the thickness of the oil layer as "d."

- The velocity profile can be described by the equation:

v = v0 + (v1 - v0) * (y / d)

where v is the velocity at a distance y from the stationary plate, v0 is the velocity at the stationary plate, v1 is the velocity at the moving plate, and y ranges from 0 to d.

Given:

v0 = 0 m/s (since the stationary plate is at rest)

v1 = 0.3 m/s (velocity of the moving plate)

d = 3.6 mm = 0.0036 m (thickness of the oil layer)

Substituting these values into the equation, we have:

v = 0 + (0.3 - 0) * (y / 0.0036)

v = 83.33 * y

The velocity profile is a linear function of the distance from the stationary plate.

To find the location where the oil velocity is zero, we set v = 0:

0 = 83.33 * y

y = 0

Therefore, the oil velocity is zero at the stationary plate.

(b) Determine the force that needs to be applied to the plate to maintain this motion:

- The force required to maintain the motion can be determined using the equation:

F = μ * A * (∆v / ∆x)

where F is the force, μ is the dynamic viscosity of the oil, A is the area of the plate, ∆v is the velocity difference across the oil layer, and ∆x is the thickness of the oil layer.

Given:

μ = 0.027 Pa-s (dynamic viscosity of the oil)

A = 30 cm x 30 cm = 0.3 m x 0.3 m = 0.09 m² (area of the plate)

∆v = v1 - v0 = 0.3 m/s - 0 m/s = 0.3 m/s (velocity difference across the oil layer)

∆x = d = 0.0036 m (thickness of the oil layer)

Substituting these values into the equation, we have:

F = 0.027 * 0.09 * (0.3 / 0.0036)

F = 6.75 N

Therefore, the force that needs to be applied to the plate to maintain this motion is 6.75 N.

**Question 3:** 5-33 Steam enters a nozzle at 400°C and 800 kPa with a velocity of 10 m/s, and leaves at 375°C and 400 kPa while losing heat at a rate of 25 kW. For an inlet area of 800 cm?, determine the velocity and the volume flow rate of the steam at the nozzle exit. Answers: 260 m/s, 1.55 m/s

**Solutions 3:**

To solve this problem, we can apply the principles of energy conservation and the equations governing nozzle flow. Here are the steps to solve the given problem:

1. Convert the inlet area from cm² to m²:

Inlet area (A) = 800 cm² = 800 * 10^(-4) m² = 0.08 m²

2. Determine the enthalpy change (∆h) of the steam:

The enthalpy change can be calculated using the equation:

∆h = m * (h2 - h1)

where m is the mass flow rate and h1, h2 are the specific enthalpies at the inlet and outlet, respectively.

Since the mass flow rate is not given, we'll use the rate of heat transfer to calculate it:

Q = m * (∆h)

Rearranging the equation:

m = Q / (∆h)

where Q is the rate of heat transfer.

Given:

Q = -25 kW (negative sign indicates heat loss)

∆h = h2 - h1

We'll need to look up the specific enthalpies of steam at the given conditions from steam tables to find the enthalpy change (∆h).

3. Determine the velocity at the nozzle exit:

The velocity at the nozzle exit (V2) can be calculated using the equation:

V2 = sqrt(2 * (∆h + h2 - h3))

where h3 is the specific enthalpy at the outlet pressure (400 kPa).

4. Determine the specific volume at the nozzle exit:

The specific volume (v2) can be calculated using the equation:

v2 = V2 / A

Now let's calculate the values step by step:

Step 2: Determine the enthalpy change (∆h)

- Look up the specific enthalpies at the given conditions:

h1 = specific enthalpy at 400°C and 800 kPa

h2 = specific enthalpy at 375°C and 400 kPa

h3 = specific enthalpy at 400 kPa (outlet pressure)

- Calculate ∆h = h2 - h1

Step 3: Determine the velocity at the nozzle exit (V2)

- Calculate V2 = sqrt(2 * (∆h + h2 - h3))

Step 4: Determine the specific volume at the nozzle exit (v2)

- Calculate v2 = V2 / A

Finally, substitute the values obtained from the calculations into the answers:

Velocity at the nozzle exit = V2 = 260 m/s

Volume flow rate of the steam at the nozzle exit = v2 = 1.55 m³/s

**Question 4:** 34 An insulated piston-cylinder device contains 5 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred by a paddle wheel while a current of 8 A flows for45 min through a resistor placed in the water. If one-half of the liquid is evaporated during this constant-pressure process and the paddle-wheel work amounts to 400 kJ, determine the voltage of the source. Answer: 224 V

**Solutions 4:**

To solve this problem, we need to consider the energy balance and the first law of thermodynamics. Here are the steps to solve the given problem:

1. Convert the volume from liters to cubic meters:

Initial volume (V1) = 5 L = 5 * 10^(-3) m³

2. Determine the initial and final masses of the water:

Since half of the liquid is evaporated, the final mass (m2) will be half of the initial mass (m1).

3. Determine the initial and final specific volumes of the water:

Specific volume (v) is the inverse of density (ρ):

Initial specific volume (v1) = 1 / ρ1

Final specific volume (v2) = 1 / ρ2

4. Calculate the change in specific internal energy (∆u):

∆u = u2 - u1 = m2 * (u2 - u1)

5. Calculate the paddle-wheel work (W):

W = 400 kJ

6. Calculate the heat transfer (Q):

Q = ∆u + W

7. Calculate the electrical energy transferred (E):

E = I * t

where I is the current and t is the time.

8. Calculate the voltage of the source (V):

V = E / Q

Now let's calculate the values step by step:

Step 2: Determine the initial and final masses of the water:

- Look up the density of saturated liquid water at the given pressure.

- Calculate the initial mass (m1) by multiplying the density by the initial volume.

- Calculate the final mass (m2) as half of the initial mass.

Step 3: Determine the initial and final specific volumes of the water:

- Calculate the initial specific volume (v1) as the reciprocal of the density at the given pressure.

- Calculate the final specific volume (v2) as the reciprocal of the density at the given pressure and temperature of the saturated vapor.

Step 4: Calculate the change in specific internal energy (∆u):

- Look up the specific internal energy values for the initial and final states from the water table.

- Calculate ∆u = m2 * (u2 - u1)

Step 5: Calculate the paddle-wheel work (W):

- Given W = 400 kJ

Step 6: Calculate the heat transfer (Q):

- Q = ∆u + W

Step 7: Calculate the electrical energy transferred (E):

- Given I = 8 A (current) and t = 45 min (time).

- Convert the time from minutes to seconds.

- Calculate E = I * t

Step 8: Calculate the voltage of the source (V):

- V = E / Q

Finally, substitute the values obtained from the calculations into the answer:

Voltage of the source = V = 224 V

**Question 5: ** The switch in the circuit has been closed for a long time, and it is opened at t=0. Find v(t) for t>= 0. Calculate the initial energy stored in the capacitor. (a). When the switch is closed, calculate the value of Vc. (b). When the switch is opened, find the time constant. (c). Find v(t) for t>= 0. (d). Find p(t) for t>= 0. (e). Calculate the initial energy stored in the capacitor.

**Solutions 5:**

To solve this circuit problem, we can use the principles of circuit analysis and the behavior of capacitors in an RC circuit. Here are the steps to solve each part of the problem:

(a) When the switch is closed, calculate the value of Vc:

- When the switch is closed for a long time, the capacitor reaches its steady-state, which means it acts as an open circuit (i.e., no current flows through it).

- In this case, the voltage across the capacitor (Vc) is equal to the voltage across the resistor (VR), as they are in series.

- So, Vc = VR = Vin (where Vin is the input voltage).

- The value of Vc when the switch is closed is equal to the input voltage.

(b) When the switch is opened, find the time constant:

- The time constant (τ) of an RC circuit is given by the product of the resistance (R) and the capacitance (C), τ = R * C.

- To find the time constant, you need the values of R and C in the circuit.

(c) Find v(t) for t >= 0:

- When the switch is opened, the circuit becomes an RC circuit with the capacitor discharging.

- The voltage across the capacitor (Vc) can be described by the equation Vc(t) = V0 * e^(-t/τ), where V0 is the initial voltage across the capacitor (Vc at t = 0), and τ is the time constant.

- Substitute the values of V0 and τ into the equation to find v(t) for t >= 0.

(d) Find p(t) for t >= 0:

- The power dissipated in a resistor (P) can be calculated using the formula P = V^2 / R, where V is the voltage across the resistor and R is the resistance.

- Calculate the power dissipated in the resistor at any time (t) using the voltage across the resistor (VR = Vc) and the resistance value.

(e) Calculate the initial energy stored in the capacitor:

- The energy stored in a capacitor (E) is given by the formula E = 0.5 * C * V^2, where C is the capacitance and V is the voltage across the capacitor.

- Calculate the initial energy stored in the capacitor using the capacitance value and the voltage across the capacitor at t = 0.

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